West Bengal Board Class 10 Physical Science Book Solution in English WBBSE

WBBSE Class 10 Physical Science Question Answer West Bengal Board

WBBSE Class 10 Physical Science Book Solutions West Bengal Board in English Medium

WBBSE Class 10 Physical Science Book Solutions West Bengal Board in Hindi Medium

WBBSE Class 10 Physical Science Syllabus West Bengal Board 2024

Chapter 1 Concerns About Our Environment

Chapter 1.1 Physical Science and Environment:
How can we interpret our environment in terms of physical science? – Structure of the atmosphere, The ozone layer, Greenhouse effect, and global warming.

Chapter 1.1 Rational Use of Energy:
Should we not conserve energy for future generations? – Harnessing energy resources for sustainable development.

Chapter 2 Behaviour of Gases

Chapter 2.1 Behaviour in the Bulk:
How does a gas behave under different conditions? – Pressure exerted by a gas and its volume, Boyle’s law, Charles’ law, Absolute temperature scale, Corrjbining Boyle’s and Charles’ laws, Ideal gas, Avogadro’s law, Ideal gas equation.

Chapter 2.2 Behaviour at the Molecular Level:
How can we describe behaviour of gases at the molecular level? – Molecular picture of an ideal gas, Deviation from ideal behaviour.

Chapter 3 Chemical Calculations

Chapter 3.1 Stoichiometric Equations:
How to answer ‘how much?’ – Conservation of mass in chemical reactions, Weight versus weight calculations.

Chapter 4 Thermal Phenomena

Chapter 4.1 Thermal Expansion:
How do different materials expand on heating? – Coefficient of expansion for solids, liquids, and gases.

Chapter 4.2 Thermal Conduction:
Does heat flow equally well through all solids? – Thermal conductivity.

Chapter 5 Light

Chapter 5.1 Reflection at Spherical Mirror:
What happens when the reflecting surface of a mirror is convex or concave? – Reflection of light at the spherical surface, Geometry of a spherical mirror, Reflection in the spherical mirror.

Chapter 5.2 Refraction of Light:
What happens when a light ray travels from one medium to another? – Laws of refraction, Familiarity with the structure of a glass slab and a prism, Refraction through a glass slab and a prism. Deviation of light rays due to refraction.

Chapter 5.3 Lenses:
How does a lens differ from a glass slab? – Familiarity with the structure of spherical lenses (concave and convex), Refraction through a lens (convex and concave), Image formation by thin lenses, Simple camera, and the Human eye as application image formation by a lens.

Chapter 5.4 Dispersion of Light:
What happens when white light passes through a prism?
Dispersion of white light as a result of refraction through a glass slab and a prism the idea of monochromatic and polychromatic and polychromatic light.

Chapter 5.5 Light Wave:
What kind of wave is a light wave, and how many different kinds of light waves are there? – Frequency, wavelength, and velocity of a wave, Uses and adverse effects of UV, X-ray, and gamma-ray, and Scattering of light.

Chapter 6 Current Electricity

Chapter 6.1 Electric Current, Potential Difference & EMF:
How does the charge start moving? – Electric charge, Electric potential difference, EMF, and electrical cell as a source of EMF, Electric current.

Chapter 6.2 Ohm’s Law:
What is resistance? – Relation between potential difference and current in a wire: Concept of resistance from Ohm’s law, EMF and internal resistance of a cell, Resistivity and conductivity, Series and parallel combination of resistances, Domestic circuits.

Chapter 6.3 Heating Effect of Electric Current:
How do we measure the consumption of electrical energy? – Joule’s law on the heating effect of current: Concept of electrical energy, Domestic uses of the heating effect of current.

Chapter 6.4 Electrical Power:
Electrical Power, the concept of a kilowatt-hour, B.O.T.

Chapter 6.5 Electromagnetism:
How does a current-carrying wire interact with a magnet? – Action of electric current on magnet: Ampere’s swimming rule, right-hand grasp rule, Action of magnet on current carrying wire: Fleming’s left-hand rule, Working principle of the motor.

Chapter 6.6 Electromagnetic Induction:
How can we get electricity from motion? – Concept of induced EMF and induced current, Basic idea of direct current and alternating current.

Chapter 6.7 Electric Generator:
How does a generator work? – Working principle and functioning of an electric generator.

Chapter 6.8 Domestic Electrical Circuit:
How are electrical wirings done in households? – Components used in domestic electrical circuit, Schematic representation of domestic electrical circuit in simplest form.

Chapter 7 Atomic Nucleus

Chapter 7.1 Radioactivity:
What is radioactivity and how is it related to the atomic nucleus? – Nature of α, β and γ rays, and their origin.

Chapter 7.2 Nuclear energy:
How can we get energy from the nucleus? – Concept of mass defect binding energy and fission, Fusion.

Chapter 8 Physical and Chemical Properties of Matter

Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements:
What is meant by periodicity? – Brief history of the periodic table, Modern periodic table, Periodicity of properties of elements.

Chapter 8.2 Ionic and Covalent Bonding:
What holds the ions in a solid or atoms in a molecule? – Properties of ionic compounds, Ionic bonding, Properties of covalent compounds, Covalent bonding.

Chapter 8.3 Electricity and Chemical Reactions:
How are electricity and chemical changes related? – Electrolytes, Strong and weak electrolytes, Mechanism of electrical conduction in molten/solution states, Electrolysis, Applications of electrolysis.

Chapter 8.4 Inorganic Chemistry in the Laboratory and Industry:
Just how important are the inorganic chemicals? – Laboratory preparation of ammonia, Properties of ammonia, Major industrial uses and industrial manufacture of NH3 and urea, Laboratory preparation of hydrogen sulfide, Properties of H2S, Laboratory preparation and major uses of N2, Properties of N2, Industrial manufacture of HCl, HNO3, and H2SO4.

Chapter 8.5 Metallurgy:
How can we get the metals from their ores and use them? – Uses of Fe, Cu, Zn, and Al and their alloys, Ores, and minerals, Brief introduction to electronic theory of redox processes, Example of Thermite reaction, Metal corrosion: (a) Rusting of iron and its prevention (b) Corrosion of other metals and its health implications.

Chapter 8.6 Organic Chemistry:
How are some carbon compounds distinctly different from inorganic compounds of carbon? – Organic compounds are compounds of carbon, Tetravalency and catenation property of carbon, Structures of C2H6, C2H4, C2H2, Functional groups, Isomerism, Homologous series, IUPAC nomenclature of simple organic compounds, Industrial source and major uses of CH4, C2H4, C2H2, LPG & CNG, Reactions of CH4, C2H4, C2H2, Some other synthetic organic polymers, Biodegradable polymers, Uses and properties of ethyl alcohol and acetic acid, Harmful effects of methanol and ethanol, Denatured spirit.

WBBSE Class 10 Physical Science Blueprint for 1st 2nd Summative Evaluation WBBSE Class 10 Physical Science Blueprint for 3rd Summative Evaluation

WBBSE Class 10 Physical Science Blueprint for 1st Summative Evaluation (Total Marks – 40)

Theme/Sub-theme MCQ (Group A) VSA (Group B) SA (Group C) LA (Group D) Total
1. Concerns About Our Environment 1 × 1 1 × 2 2 × 1 5
2. Behaviour of Gases 1 × 1 1 × 2 2 × 1 3 × 1 8
3. Light 1 × 3 1 × 3 2 × 1 3 × 2 14
4. Periodic Table and Periodicity of the Properties of Elements 1 × 1 1 × 1 2 × 1 3 × 1 7
5. Ionic and Covalent Bonding 1 × 1 1 × 1 2 × 2 6
Total 7 9 12 12 40

WBBSE Class 10 Physical Science Blueprint for 2nd Summative Evaluation (Total Marks – 40)

Theme/Sub-theme MCQ (Group A) VSA (Group B) SA (Group C) LA (Group D) Total
1. Chemical Calculations 1 × 1 1 × 1 3 × 1 5
2. Thermal Phenomena 1 × 1 1 × 2 3 × 1 6
3. Current Electricity 1 × 2 1 × 2 2 × 1 3 × 2 12
4. Electricity and Chemical Reactions 1 × 1 1 × 2 3 × 1 6
5. Inorganic Chemistry in the Laboratory and in Industry 1 × 1 1 × 1 2 × 1 3 × 1 7
6. Metallurgy 1 × 1 1 × 1 2 × 1 4
Total 7 9 6 18 40

WBBSE Class 10 Physical Science Blueprint for 3rd Summative Evaluation/Selection Test (Total Marks – 90)

Theme/Sub-theme MCQ (Group A) VSA (Group B) SA (Group C) LA (Group D) Total
1. Concerns about Our Environment 1 × 1 1 × 2 2 × 1 5
2. Behaviour of Gases 1 × 1 1 × 2 2 × 1 3 × 1 8
3. Chemical Calculations 1 × 1 3 × 1 4
4. Thermal Phenomena 1 × 1 1 × 1 3 × 1 5
5. Light 1 × 2 1 × 2 2 × 1 3 × 2 12
6. Current Electricity 1 × 2 1 × 2 2 × 1 3 × 2 12
7. Atomic Nucleus 1 × 1 1 × 1 3 × 1 5
8. Periodic Table and Periodicity of the Properties of Elements 1 × 1 1 × 2 3 × 1 6
9. Ionic and Covalent Bonding 1 × 1 1 × 1 2 × 2 6
10. Electricity and Chemical Reactions 1 × 1 1 × 2 3 × 1 6
11. Inorganic Chemistry in the Laboratory and Industry 1 × 1 1 × 2 2 × 1 3 × 1 8
12. Metallurgy 1 × 1 1 × 2 2 × 1 5
13. Organic Chemistry 1 × 1 1 × 2 2 × 1 3 × 1 8
Total 15 21 18 36 90

WBBSE Class 10 Solutions

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 5 Light offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 5 Question Answer – Light

Very Short Answer Type Questions :

Question 1
On normal incidence of a ray of light on a plane mirror, what are the angle of incidence and angle of reflection?
Answer:
∠ i = ∠ r = 0°

Question 2.
What is the linear magnification of image in a plane mirror ?
Answer:
Magnification, m = \(\frac{I}{O}\) = pm 1 as size of image, I = size of object, O

Question 3.
Name the type of mirror which has focal length equal to infinity.
Answer:
A plane mirror.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 4.
What is leteral inversion of image?
Answer:
The side of an object appears as left in the image of the object and vice versa in an image formed in a plane mirror. This is known as lateral inversion of image.

Question 5.
What is the radius of curvature and focal length of a plane mirror?
Answer:
In a plane mirror both radius of curvature and focal length of a plane mirror are infinity.

Question 6.
A man approaches a vertical plane mirror at a speed u. At what speed does he approach his image?
Answer:
The image is formed at the same distance behind the plane mirror as the object is infront of it. Thus, the man will approach it image with a speed = u-(-u) = 2 u.

Question 7.
Define principal axis of a spherical mirror.
Answer:
The straight line passing through the pole and the centre of curvature is called principal axis of the mirror.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 8.
Why is a convex mirror used as driver’s mirror?
Answer:
As a convex mirror always produces an erect image of diminished size providing a wide field of view.

Question 9.
Which mirror is convergent, concave or convex?
Answer:
Concave mirror.

Question 10.
Can we cast the image formed by a convex mirror on a screen?
Answer:
The image of real object cannot be cast, but the image of a virtual object can be cast.

Question 11.
Can a convex mirror form a magnified image?
Answer:
No.

Question 12.
When does a concave mirror form a virtual image?
Answer:
When the object lies between the pole and focus of the mirror.

Question 13.
What is the relation between f and r of a spherical mirror?
Answer:
f = \(\frac{r}{2}\)

Question 14.
Can a virtual image be photographed?
Answer:
Yes, it can be photographed with the help of the lens of the camera.

Question 15.
What is the number of images of an object held between two parallel plane mirror?
Answer:
n = \(\left(\frac{360}{\theta}-1\right)\) = α when θ = 0°

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 16.
What is the effect of the size of a mirror on the nature of the image?
Answer:
Nature of the image does not depend on the size of the mirror.

Question 17.
Does the focal length of a concave mirror change if it be immersed in water?
Answer:
No.

Question 18.
Which of the following does not change when light goes from one medium to other frequency, wavelength speed and intensity?
Answer:
Frequency

Question 19.
What is the absolute refractive index of vacuum?
Answer: 1.

Question 20.
For which medium is refractive index (i) minimum (ii) maximum?
Answer:
(i) μ min for vacuum (=1) and (ii) μ max for diamond (= 2.42)

Question 21.
Can total internal reflection occur when light travels from a rarer to a denser medium?
Answer:
No.

Question 22.
Does critical angle depend on the colour of light?
Answer:
Yes, as μ is different for different colour of light.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 23.
What is the cause of refraction of light?
Answer:
Different velocity of light is different media.

Question 24.
Give the relation between critical angle and refractive index.
Answer:
θc = sin-1 \(\left(\frac{1}{\mu}\right)\)

Question 25.
Can light travelling from air to glass suffer total internal reflection?
Answer:
No.

Question 26.
What is the principal of optical fibre?
Answer:
Total internal reflection of light.

Question 27.
What do you mean by refraction of light ?
Answer:
The phenomenon of bending of light rays from its original path while passing from one medium to another is called refraction of light.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 28.
Define refractive index of the material.
Answer:
For a given pair of media and a colour of light, refractive index,
\(\mu=\frac{\sin i}{\sin r}=\frac{C}{C^{\prime}}\)
where i and r and angles of incidence and refraction respectively and C and C’ are velocities of light in vacuum and in the medium respectively.

Question 29.
What is the basic cause of refraction of light?
Answer:
The speed of light is different in different media.

Question 30.
Can absolute refractive index of light of any material be loss than one? Why?
Answer:
No, because μ = \(\frac{c}{v}\) and c is always greater than v.

Question 31.
What is the main use of optical fibres?
Answer:
It can transmit light from one place to another without any loss of intensity of light.

Question 32.
What will be the focal length and power of a plane glass plate?
Answer:
f = α ; power = 0

Question 33.
An object is placed at the focus of concave lens where will be the image?
Answer:
At the midpoint of focus and optical centre of the concave lens.

Question 34.
Define power of a lens.
Answer:
The reciprocal of the focal length of the lens is called its power.

Question 35.
How can a convex lens behave like a diverging lens ?
Answer:
It can behave a diverging lens, if it placed in a medium of refractive index greater than the refractive index of the material of the lens.

Question 36.
An object is placed at the focus of a concave lens, where will be its image formed?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 5 Light 1

Question 37.
How does the power of a concave lens vary, if the incident red light is replaced by violet light?
Answer:
As the focal length is minimum for violet light, so power increases.

Question 38.
What is deviation produced in a prism?
Answer:
The angle between the emergent ray and the direction of incident ray is called angle of deviation or the deviation produced in a prism.

Question 39.
How is angle of incidence at the first face of a prism related to the angle of emergence at the other face at minimum deviation position?
Answer:
i1 = i2

Question 40.
What is meant by dispersion of light?
Answer:
The phenomenon of splitting of white light into its constituent colours is called dispersion of light.

Question 41.
For which colour, m of the material of the prism is (i) minimum (ii) maximum?
Answer:
(i) red
(ii) violet.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 42.
Which colour of light deviate (i) most (ii) least on passing through a prism?
Answer:
(i) violet
(ii) red

Question 43.
Is dispersion of white light in vacuum possible?
Answer:
No.

Question 44.
How is deviation produced in a prism related to the angle of the prism and angles of incident and emergency?
Answer:
δ = i1 + i2A

Question 45.
Why is a angled prism a better reflector than a plane mirror?
Answer:
As in the prism no part of the light is refracted.

Question 46.
Does a beam of light give spectrum on passing through a hollow prism containing air, explain?
Answer:
No, because all colours of light passes through vacuum or air same speed and their is no dispersion.

Question 47.
What is principal section of prism ?
Answer:
Any section of the prism by a plane at angle to the refracting edge is called a principal section of a prism.

Question 48.
What is the edge of a prism ?
Answer:
The intersection of the refracting surfaces is called the edge of the prism.

Question 49.
Which of the colours has (i) maximum velocity and (ii) minimum velocity in a glass prism ?
Answer:
(i) red (ii) violet.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 50.
What is the cause of dispersion?
Answer:
The difference in the deviations suffered by the two extreme colours (i.e. red and violet) of white light in passing through a prism is called angular dispersion.

Question 51.
What is the cause of dispersion?
Answer:
The deviations of different colours of light through a prism are different.

Question 52.
What is the unit of angular dispersion?
Answer:
Degrees.

Question 53.
What is the dispersive power of a prism?
Answer:
The dispersive power of a prism is the ratio of angular dispersion to the mean deviation produced by the prism.

Question 54.
Does dispersive power of a prism depend on the angle of the prism?
Answer:
No, it depends on the nature of the material of the prism.

Question 55.
What is the essential condition for observing rainbow?
Answer:
The sun should be at the back of the observer after rainfall.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 56.
Can a short sighted person read books without spectacles?
Answer:
Yes, A short sighted person can see objects at short distance.

Question 57.
What kind of lens one should use to correct myopic eye?
Answer:
Concave lens.

Question 58.
What kind of lens one should use to correct astigmatic eye?
Answer:
Cylindrical lens.

Question 59.
What is the least distance of distinct vision?
Answer:
25 cm.

Question 60.
What is the far point of a normal eye?
Answer:
Infinity.

Question 61.
Which colour of light travels faster in vacuum?
Answer:
Light rays of all colour travel with same speed in vacuum.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 62.
How many colours of light mix to produce white light?
Answer:
Seven colours of light mix to produce white light.

Question 63.
Will focal length of a convex lens vary for refraction of light of different colours through it?
Answer:
For different colours of light the convex lens will have different focal length.

Question 64.
What will be the position and nature of the image formed with a convex lens when the object is situated at a distance twice the focal length of the lens?
Answer:
The image will be at a distance twice the focal length on the other side of the lens and the nature of the image will be real, inverted and same size as that of the object.

Question 65.
What is a convex lens?
Answer:
If the lens is thicker at its centre than at its edges, it is a convex or converging lens.

Question 66.
Where would you place an object before a convex lens to get a virtual and magnified image?
Answer:
If the object is at a distance less than the focal length of the lens, the image becomes virtual, erect and magnified.

Question 67.
Who discovered first the dispersion of light?
Answer:
Dispersion of light was first discovered by Sir Issac Newton in 1666.

Question 68.
What is dispersion of light?
Answer:
The splitting up of white light into seven colours on passing through a refracting medium like prism is known as dispersion of light.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 69.
What is a spectrum?
Answer:
The band of different colours obtained due to dispersion of white light is known as a spectrum.

Question 70.
What is a lens?
Answer:
A lens is a piece of transparent refracting substance bounded by two spherical or one spherical and another plane surfaces.

Question 71.
What is a concave lens?
Answer:
The lens which is thinner of the centre and wider at the two edges is called a concave lens.

Question 72.
What is prism?
Answer:
A homogenous transparent refracting medium enclosed by three rectangular and two triangular surfaces is known as a prism.

Question 73.
What is the principle section of the prism?
Answer:
The surface which intersects the refracting surface of the prism perpendicularly is known as the principle section of the prism.

Question 74.
What is the refracting angle of the prism?
Answer:
The angle made by the two refracting surface of the prism is known as the refracting angle of the prism.

Question 75.
What is polychromatic light?
Answer:
The light which is composed of more than one colour is known as polychromatic light.

Question 76.
What is monochromatic light?
Answer:
The light which consists of only one colour is known as monochromatic light.

Question 77.
What are primary colours ?
Answer:
Red, green and blue are called primary colc ars.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 78.
Give the name of a secondary colour.
Answer:
Yellow is called a secondary colour.

Question 79.
What is linear magnification ?
Answer:
The ratio of the length of the image and that of the object is called the linear magnification.

Question 80.
What is focal plane of the lens?
Answer:
The plane passing through principal focus of a lens and perpendicular to the principal axis of the lens is known as the focal plane of the lens.

Question 81.
How does a green cloth appear in red light?
Answer:
The green cloth appears black in red light.

Question 82.
What is a double convex lens ?
Answer:
If both the surfaces of convex lens are convex then the lens is called double convex lens. It is used in camera, telescope etc.

Question 83.
What is a plano convex lens ?
Answer:
If one surface of convex lens is plane and the other is convex then it is called plano convex lens. e.g. eye piece of telescope.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 84.
What is concavo convex lens?
Answer:
If one surface of convex lens is convex and the other is concave then it is called concavo convex lens. It is used in spectacles.

Question 85.
What colour of light is used for danger signal ?
Answer:
Red.

Question 86.
Name a phenomenon occuring in nature due to dispersion of light.
Answer:
Rainbow.

Question 87.
What type of reflection is used in torch?
Answer:
A torch uses concave reflector with the bulb placed very near to the focus of the reflector.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 88.
When a point object is placed in front of a concave mirror in between centre of curvature and focus, where will the image formed?
Answer:
The image will be formed at a distance greater than the radius of curvature.

Question 89.
What type of mirror is represented by the front side of a shining stainless spoon?
Answer:
The front side of a shining stainless steel spoon acts like a concave mirror.

Question 90.
What type of mirror is represented by the backside of a shining stainless steel spoon?
Answer:
The backside of a shining stainless steel spoon acts like a convex mirror.

Question 91.
Refractive Index is 2.41. What is meant by that?
Answer:
Refractive Index is 2.41 – it means that light travels 2.41 times faster in air (vacuum) than in diamond.

Question 92.
What do you mean by refracting faces or refracting surfaces of a prism?
Answer:
The two rectangular surfaces through which light can enter the prism or come out from the prism are called refracting surfaces of a prism.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 93.
What are the factors on which the lateral shift depends?
Answer:
Lateral shift depends on
(a) thickness of the glass slab
(b) angle of incidence
(c) refractive index of the slab.

Question 94.
What shall be the angle of refraction when angle of incidence is 90° ?
Answer:
There is no refraction taking place in this case and the ray of light passes along the line of separation of the two media.

Question 95.
If the speed of light in two different transparent media is same, will a ray of light suffer refraction?
Answer:
If the speed of light in both the media is same, the ray of light will not be refracted.

Question 96.
Why is a normal eye unable to clearly see objects placed closer than 25 cm ?
Answer:
The least distance of distinct vision for a normal eye is 25 cm. So, a normal eye is not able to clearly see objects, placed closer than 25 cm.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 97.
Your friend can read a book properly but cannot see the writings on the blackboard from the last bench. What type of defect of viston he/she has?
Answer:
He/She is suffering from myopia or short-sightedness.

Question 98.
Your friend has difficulty in reading a book properly but can see the writings on the blackboard from the last bench clearly. What type of defect of vision he/she has?
Answer:
He/She is suffering from hypermetropia.

Question 99.
What is the focal length of a parallel sided glass slab ?
Answer:
A parallel sides glass slab has infinite focal length and radius of curvature.

Question 100.
What is the nature of image formed by eye lens-real or virtual?
Answer:
The convex eye lens forms and inverted, real image of the object on the retina.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 101.
Is dispersion possible without refraction?
Answer:
No, dispersion is not possible without refraction. In an optical medium rays of different colour travel with different speeds and so, they deviate as per their infrangibility. Hence a polychromatic light is dispersed only after refraction.

Question 102.
What is monochromatic light?
Answer:
Rays of light which do not undergo splitting into different colours due to dispersion are known as monochromatic light. Each colour of the spectrum is monochromatic.

Question 103.
Does dispersion of white light take place in vacuum?
Answer:
In vacuum, rays of different colours travel with same speed. So, travelling through vacuum, they will not be separated from each other and hence no dispersion will take place.

Question 104.
Hod does the refractive index of glass with respect to air change with change in wavelength of light?
Answer:
The refractive index of glass with respect to air increases with the decrease in the wavelength of light.

Question 105.
What do you mean by vistble spectrum?
Answer:
The part of the spectrum which is visible to human eye is known as visible spectrum. The wave length of visible spectrum ranges from 400 nm to 700 nm.

Question 106.
Whick ray has the highest frequency?
Answer:
Gamma-ray has the highest frequency.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 107.
What would have been the colour of the sky if there were no atmosphere?
Answer:
If the earth did not have any atmosphere, there would have been no particles to scatter the sunlight and the sky would have appeared black.

Question 108.
What is the nature of light wave?
Answer:
Light wave is transverse electromagnetic waves.

Short answer type questions :

Question 1.
What is centre of curvature of a lens ?
Answer:
Centre of curvature: The spherical surface of a lens is a part of sphere. The centre of this sphere is known as centre of curvature of the surface of lens.

Question 2.
What do you mean by the principal axis of a convex lens?
Answer:
Principal axis of a convex lens : The line joing the centres of curvature of the two spherical surfaces of a convex lens is called its principal axis.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 2

Question 3.
What is radius of curvature ?
Answer:
Radius of Curvature : Radius of curvature of a lens is the radius of the glass sphere from which the surfaces of the lens are cut.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 4.
What is focus of a convex lens ?
Answer:
Focus of a convex lens : If a beam of parallel rays, travelling parallel to the principal axis of a cc the lens, the rays become converging and intersect each other at a particular point on the axis. The point is known as the focus of the convex lens.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 3

Question 5.
What is optical centre ?
Answer:
Optical Centre : If a ray of light strikes on surface of a lens in such a way that the emergent ray from the other surface is parallel to it then the corresponding refracted ray passes through a definite point on the principal axis. This point (A) is the optical centre of the lens.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 4

Question 6.
What is focal length ?
Answer:
Focal length : The distance between the optical centre and the focus is known as focal length.

Question 7.
What is Focal plane ?
Answer:
Focal plane : The plane passing through principal focus of a lens and perpendicular to the principal axis of the lens is known as the focal plane of the lens.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 8.
What is optical plane?
Answer:
Optical plane : It is an imaginary vertical plane that cuts the principal axis perpendicularly and passes through the optical centre of a lens.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 5

Question 9.
What is linear magnification of lens ?
Answer:
Linear magnification of lens : The ratio of the length of the image and that of the object is called the linear magnification.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 6

Question 10.
What does the term ‘thin’ signify when it is related to a convex lens? Do all rays of light suffer deviation while crossing through a thin convex lens?
Answer:
A thin convex lens is one, the thickness at the middle of which is extremely small compared to the radii of curvature of its surfaces. Rays directed towards the optical centre do not suffer deviation while crossing through a convex lens.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 11.
Which rays should be considered in drawing ray diagrams for images formed by a convex lens ?
Answer:
The following points should be considered :

  1. A beam of rays parallel to the principal axis after refraction through the lens pass through the principal focus.
  2. A beam of rays passing through principal focus emerge parallel to the principal axis, after refraction through the lens.
  3. A beam of rays through the optical centre pass out undeviated.

Question 12.
Draw a neat diagram of the formation of image of an object at infinity by a convex lens.
Answer:

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 7
Formation of image of an object at infinity by a convex lens : Distance of object : at infinity Image formed: On the focal plane. Nature of image : real and inverted. Size of image : very much smaller than the object.

Question 13.
Draw a neat diagram of the formation of image of an object which is placed in between infinity and 2f from the lens.
Answer:
Object is between infinity and 2 f from the lens:

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 8

  • Distance of object : between infinity and 2 f from the lens.
  • Image formed: On the opposite side of the object and is situated between ‘f and 2f,
  • Nature of image : real and inverted.
  • Size of image : smaller than the object in size.

Question 14.
Draw a neat diagram of the formation of image of an object at 2 f from the lens.
Answer:
Object is at 2f from the lens :

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 9

  • Distance of object : at 2f from the lens.
  • Image formed : at a distance 2 f from the lens.
  • Nature of image : real and inverted.
  • Size image : equal to the size of the object.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 15.
Draw a neat diagram of the formation of image of an object which is placed in between ‘f’ and 2f
Answer:
Object is between ‘f’ and 2 f from the lens :

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 10

  • Distance of object: between f and 2 f from the lens.
  • Image formed: Or, the opposite side of the object at a distance greater than 2f. Nature of image : real and inverted.
  • Size of image: Greater in size than the object.

Question 16.
Draw a neat diagram of the formation of image of a object which is at the focus of the lens.
Answer:
Object is at the focus of the lens :

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 11

  • Distance of object : at the focus.
  • Image formed : at infinity.
  • Nature of image : real and inverted.
  • Size of image : Highly magnified.

Question 17.
Draw a neat diagram of the formation
Fig. for Question 16 of image of a object which is situated between the lens and the focus.
Answer:
Object is within the focal length of the lens :

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 12

  • Distance of object : Within the focal length.
  • Image formed : On the same side of the object.
  • Nature of image : virtual and erect.
  • Size of image : Magnified.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 18.
(i) Draw a neat diagram to explain the principle of a magnifying glass
(ii) To read the tiny scripts of a book with ease, where should be book be kept in front of a magnifying glass ?
Answer:

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 13
(i) Nature and size of image : An erect, virtual and magnified image is formed on the same side of the object. This property of a convex lens is working principle of magnifying glass.
(ii) To read tiny scripts of a book, the book must be placed within the focal length of the lens.

Question 19.
What is a spectrum ?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 5 Light 14
Spectrum : The band of different colours obtained due to dispersion of white light is known as a spectrum.

Question 20.
What is a pure specturm ?
Answer:
Pure spectrum : The spectrum in which the constituent colours do not overlap on each other and are separated distinctly into elementary colours is known as a pure spectrum.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 21.
What is an impure spectrum ?
Answer:
Impure spectrum : The spectrum in which the constituent colours partially superpose on each other and are not separated distinctly into elementary colours is known as an impure spectrum.

Question 22.
What are the differences between Pure and Impure spectrum ?
Answer:
Distinction between Pure and impure spectruin :

Pure spectrum Impure spectrum
(i) In a pure spectrum all the seven colours occupy distinct and different positions on a screen. (i) In an impure spectrum all the seven colours do not occupy different and distinct positions on a screen.
(ii) In a pure spectrum, monochromatic colours presents in white light, are placed in order of refrangibility. (ii) In an impure spectrum, monochromatic colours are not placed in order of refrangibility.

Question 23.
Is dispersion possible without refraction?
Answer:
Explanation Dispersion without refraction is not possible. In an optical medium different colours travel with different speed and so deviate with different magnitudes. Thus, the constituent colours of a polychromatic light deviate by different amounts i.e. they are dispersed after refraction.

Question 24.
Why blue is used after washing white shirts ?
Answer:
Explanation : Blue whitens yellow or orange colour. So, white shirts after wash may have some yellow or orange stain. To convert such stains to white, blue is used.

Question 25.
How would you determine whether a mirror is plane, concave or convex ?
Answer:
A plane mirror produces an erect image of the same size as that of the object. A concave mirror produces an erect and magnified image of an object when placed within its focus.
A convex mirror produces always an erect and diminished image of an object.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 26.
What type of mirror would you use as shaving mirror?
Answer:
A concave mirror of larger focal length is used for shaving purpose. Because it produces a magnified erect image of the face when placed within the focus of the mirror.

Question 27.
A concave mirror of small aperture forms a sharper image why ?
Answer:
It is because such a mirror is free from the defect due to spherical aberration.

Question 28.
The refractive index of diamond is much greater than the ordinary glass. Is this fact of some use to a diamond-cutter ?
Answer:
An critical angle [sin-1(1/μ)] of diamond is much smaller and total internal reflection takes place easily.

Question 29.
Why does a diamand sparkle with great brilliance?
Answer:
Multiple total internal reflections of light occur within the diamond. Diamond cutter uses this fact.

Question 30.
Why does a convex lens of glass μ = 1.5 behave as a diverging lens when immersed in a liquid of μ = 1.65 ?
Answer:
The refractive index of glass with respect to the liquid \(\left(\frac{1.5}{1.65}\right)\) is less than I.

Question 31.
On what factors does the angular dispersion depend?
Answer:
The angular dispersion depends on

  • refracting angle of the prism
  • nature of the material of the prism
  • wavelength of the incident light.

Question 32.
How does the ray passes through the prism in the minimum deviation position?
Answer:
In the minimum deviation position, the ray passes symmetrically through the prism i.e. the incident ray and emergent ray are equally inclined to the respective faces of prism.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 33.
What is myopia?
Answer:
Myopia: The defect where far point is less than infinity is known as myopia. The defect is also called short sightedness.

Broad Answer Type Questions :

Question 1.
Draw a labelled diagram to show refraction of a light ray through a prism.
Answer:
Path of a ray of light through a prism : let A B C be the principal section of a prism A B and A C are two refracting surfaces. B C is the base. PQRS is the path of a ray. It is to be noticed that in a prism, the refracted ray and the emergent ray bend towards the base of the prism, when the surrounding medium is optically rarer than the material of the prisms. But, if the medium outside the prism be optically denser than the prism material, the refracted and emergent rays bend oppositely i.e. they bend towards the angle of the prism.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 15

Question 2.
What are the conditions of formation of pure spectrum ?
Answer:
Conditions of formation of pure spectrum :
(i) Slit should be very narrow to allow rays of light to pass as minimum as possible.
(ii) The slit should be placed at the focus so that the emergent light rays may be parallel.
(iii) Prism should be placed at the minimum deviation position of spectrum i.e. at the mean deviation of yellow spectrum.
WBBSE Class 10 Physical Science Solutions Chapter 5 Light 16
(iv) The second lens (L2 L2‘) should be placed in such a position so that the screen may be in the focal plane of the lens.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Question 3.
How will you explain twinkling of stars ?
Answer:
The light coming from stars is refracted continuously by different layers of the atmosphere before coming to us. Due to these repeated refractions, the apparent position of a star is different from its actual position. The temperature and density of air in the atmosphere continuously changing and so apperent position of the star also changes continuously. Due to this continuous change in the apparent position of a star, it appears to be twinkling.

Question 4.
A ray of light shows no dispersion on emerging from a glass slab. Explain why?
Answer:
A rectangular glass slab cut diagonally into two pieces behaves like a pair of prisms placed side by side in reverse order. So the light dispersed by the first prism, recombines by the second prism. Thus there is no dispersion of light in a rectangular glass slab, but it is only laterally displaced.

Numerical Problems :

Example 1:
How far from a concave mirror of radius 2 m} would you place the object to get an image magnified 3 times?
Answer:
Given m = ± 3 = – \(\frac{v}{u}\) i.e. v = mp 3 u
For real image v = -3 u, r = -2 m and u is negative.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 17

Example 2:
A concave mirror forms an image of 20 cm high object on a screen placed 5 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between ma.c and the object.
Answer:
Given size of object =20 cm
size of the image =50 cm
v = -5 m

WBBSE Class 10 Physical Science Solutions Chapter 5 Light 18

And distance between the mirror and the object = 2 m.

Example 3:
Light falls from glass of refractive index 1.5 to air. Find the angle of incidence for which the angle of deviation is 90°.
Answer:
Refractive index of glass is 1.5 . So, critical angle of glass to air medium θ is given by
WBBSE Class 10 Physical Science Solutions Chapter 5 Light 19
So, reflection in the glass medium takes only when the angle of incidence exceeds 41.8°. Now the angle of deviation would be 90° when the angle between the incident ray and the reflected ray becomes 90° i.e. the angle of incidence be equal to 45°.

WBBSE Class 10 Physical Science Solutions Chapter 5 Light

Example 4:
Where should an object be placed from a converging lens of focal length 20 cm, so as to obtain a real image of magnification 2 ?
Answer:
Given m = \(\frac{v}{u}\) = 2, or, v = +2 u, f = +20 cm
WBBSE Class 10 Physical Science Solutions Chapter 5 Light 20
i.e. object to be placed 10 cm} in front of the lens.

Example 5:
Two thin lens of focal lengths 15 cm and 30 cm respectively are kept in contact with each other. What is the power of the combined system?
Answer:
Given f1 = 15 cm, f2 =30 cm}
Then focal length F of the equivalent lens
WBBSE Class 10 Physical Science Solutions Chapter 5 Light 21

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 7 Atomic Nucleus offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 7 Question Answer – Atomic Nucleus

Very Short Answer Type Questions :

Question 1.
Out of α, β and γ rays which one is called hand rays?
Answer:
γ-rays

Question 2.
Out of α, β and γ-rays which one causes maximum damage to the body tissues ?
Answer:
α-rays.

Question 3.
What is the density of nucleus?
Answer:
1014 / cm3

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 4.
What is the binding energy for a mass defect of 1 amu?
Answer:
931.5 Mev.

Question 5.
What is the n / p ratio of lighter nuclides ( Z up to 20 )?
Answer:
1.

Question 6.
What is the n / p ratio of \({ }_1^1 \mathrm{H}\) ?
Answer:
0 .

Question 7.
What is the n / p ratio of \({ }_{83}^{209} \mathrm{Bi}\) ?
Answer:
1.52 .

Question 8.
Which emission increases n/p ratio?
Answer:
positron.

Question 9.
Which emission decreases n/p ratio?
Answer:
β-rays

Question 10.
What is the value of packing fraction for stable nuclides?
Answer:
Zero or negative.

Question 11.
What is the value of packing fraction for unstable nuclides?
Answer:
positive.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 12.
What are the magic numbers?
Answer:
Magic numbers are 2, 8, 20, 50, 82 and 126.

Question 13.
Do nuclear forces obey inverse square law?
Answer:
Nuclear forces are not governed by inverse square law.

Question 14.
What is the cause of nuclear isomerism?
Answer:
Nuclear isomerism arises because of the difference in arrangement of nucleons and difference in energy of nucleons.

Question 15.
Is rate of nuclear decay dependent of concentration, pressure and temperature?
Answer:
Rate of nuclear decay is independent of concentration, pressure and temperature.

Question 16.
What is the life span of a radioactive element?
Answer:
The life span of a radioactive element is infinite.

Question 17.
Is the half-life of a radioactive element dependent of its physical and chemical state?
Answer:
The half-life of a radioactive element is independent of its physical and chemical state.

Question 18.
What is the end product of (4n + 1) series?
Answer:
In (4n+1) series, the end product is bismuth.

Question 19. What is the end product of 4 n series?
Answer:
The end product of 4 n series is lead.

Question 20.
Why (4n+1) series is called neptunium series though this series starts from plutonium?
Answer:
The (4n+1) series actually starts from plutonium but is called neptunium series because neptunium is the longest lived isotope of the series.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 21.
What is the starting element of (4n + 3) series?
Answer:
The (4n+3) series starts from \({ }_{92}^{235} \mathrm{U}\).

Question 22.
Is there any change is group number during a and b-emissions in case of lanthanides and actinids?
Answer:
There is no change in group number during α and β-emissions in case of lanthanides and actinides.

Question 23.
How many a and b particles are emitted from thorium series?
Answer:
The thorium series emits 6 α and 4β particles.

Question 24.
How many a and b particles are emitted from uranium series?
Answer:
The uranium series emits 8 α and 6β particles.

Question 25.
Name the natural radioaclive series.
Answer:
The 4n, (4n+2),(4n+3) are natural radioactive series.

Question 26.
What is artificial series?
Answer:
(4n+1) is artificial series.

Question 27.
What is the cause of b-emission?
Answer:
β-particle emission is due to breaking down of a neutron into a proton and an electron.

Question 28.
What is the relation of nuclear forces with electrostatic forces?
Answer:
The nuclear forces are about 1021 times as strong as the electrostatic forces.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 29.
What is the (n/p) ratio of \({ }_1^3 \mathrm{H}\) ?
Answer:
The (n/p) ratio of \({ }_1^3 \mathrm{H}\) is 2.0.

Question 30.
What amount of energy is released during the fission of an atom of \({ }^{235} \mathrm{U}\) ?
Answer:
The fission of an atom \({ }^{235} \mathrm{U}\) releases 211.5MeV of energy.

Question 31.
What are the uses of P-32, Co-60, Na-24, I-131?
Answer:

  • P-32 : Leukemia (Blood cancer)
  • Co-60 : Treatment of tumar and cancer
  • Na-24 : Circulation of blood.
  • I-131 : Treatment of gioter (Hyper thyroidism)

Question 32.
Why Nuclear fusion reactions are called thermonuclear reactions?
Answer:
Nuclear fusion reactions require very high temperature (≈106 K) these are therefore, known as thermonuclear reactions.

Question 33.
What are the types of meson particles?
Answer:
The meson particles are of three types

  1. Positive meson (π+)
  2. Negative meson (π)
  3. Neutral meson (π0)

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 34.
What are the factors on which stability of nucleus depends?
Answer:
The stability of the nucleus is governed mainly by mass defects, binding energy, magnetic numbers and n / p ratio.

Question 35.
How mass defect can be converted into energy?
Answer:
The mass defect can be converted into energy by using Einstein equation ΔE = m c2.

Question 36.
What is called binding energy?
Answer:
Binding energy is the energy which binds the nucleons together and binding energy divided by the number of nucleons gives binding energy per nucleon, used to compare the stability of various nuclei.

Question 37.
Who discovered natural radioactivity?
Answer:
Natural radioactivity was discovered by Henry Becquerel.

Question 38.
What is the effect of mass number and atomic number during emission of a-particle?
Answer:
The emission of α-particle decreases the mass number by 4 unit and decreases atomic number by two unit.

Question 39.
State the different unit of rate of decays.
Answer:
nit of rate of decays are :

  • 1 Curie (c) = 3.7 × 1010 dps
  • 1 Rutherford (Rd) = 106 dps
  • 1 Becquerel (Bq) = 1 dps

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 40.
What is the use of carbon dating?
Answer:
Carbon dating is used to determine the age of wood, animal fossils etc having carbon.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 1

Question 41.
What are the parts of nuclear reactor?
Answer:
Nuclear reactor consists of –

  • A fissionable material → Pu or \({ }^{235} \mathrm{U}\)
  • A moderator → Graphite or D2O to slow down of neutrons.
  • Control rods → B, steel or Cd to capture some neutrons.
  • Coolent → Heavy water or sodium or an alloy of Na and K (used to carry away the heat)

Question 42.
What happens when Δ-ray is emitted from an atom?
Answer:
When Δ-ray is emitted from an atom, no new element is produced. It results in a change in the energy state of the nucleus.

Question 43.
What is β-ray?
Answer:
β-ray is nothing but a stream of fast moving electrons with a negative charge of 1.6 × 10-19 coulomb.

Question 44.
Which element is formed when an α-particle takes up two electrons?
Answer:
When an α-particle (He2+) takes up two electrons, the element Helium (He) is formed.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 45.
Name a radioactive inert element.
Answer:
The inert gas Radon (Rn) with atomic number, Z = 86, is a radioactive inert element.

Question 46.
Name an instrument to record radioactive emissions.
Answer:
Geiger Counter is an instrument used to record the rate of emission from radioactive substance.

Question 47.
Which nuclear reaction takes place inside the sun?
Answer:
The enormous amount of heat energy produced in the sun is due to the thermonuclear fusion reactions taking place in the core of the sun.

Question 48.
What is the role of a neutron in nuclear fission?
Answer:
When a neutron (nuclear projectile) is absorbed by a nucleus, it provides energy that destabilizes the nucleus leading to fission.

Question 49.
Is the mass per nucleon of a nucleus greater than, less than or the same as the mass of a nucleon outside the nucleus?
Answer:
The mass per nucleon is greater outside the nucleus.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 50.
Why is neutron a better nuclear projectile than a proton or an electron?
Answer:
Neutron has no charge, so it dges not experience electrostatic forces from surrounding nucleons. It can thus penetrate further before colliding with other particles.

Question 51.
Mention the function of heavy water in a nuclear reaction.
Answer:
Heavy water (D2O) is used as moderator and coolant in a nuclear reactor. It is the best moderator avalable.

Short Answer Type Questions :

Question 1.
What are nuclear reactions?
Answer:
Nuclear reactions: Nuclear reactions are the reactions in which nucleons of an atom undergo a change.

Question 2.
What do you mean n y ‘isotopes’?
Answer:
Isotopes: The atoms of an element having same atomic number but different mass number are called isotopes.
Example :
\({ }_1^1 \mathrm{H}, \quad{ }_1^2 \mathrm{H}, \quad{ }_1^3 \mathrm{H}\)

Question 3.
What are nuclear isomers?
Answer:
Nuclear isomers: The nuclear species having same atomic number and same mass number but different radioactive properties are called nuclear isomers.
Examples : \({ }^{69} \mathrm{Zn}\)(t1/2 = 13.8 hour) nd \({ }^{69} \mathrm{Zn}\) (t1/2 = 57 minute)

Question 4.
What is isoster?
Answer:
Isoster: Molecules or ions with same number of atoms and also the same number of electrons are said to form isosteric group or more simply isosters.
Examples : N2O and CO2

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 5.
What do you mean by nuclear forces?
Answer:
Noclear forces: The forces which held the nucleons together within the small nucleus are called nuclear forces.
These force exist among p-p, p-n and n-n.

Question 6.
What is packing fraction ?
Answer:
Packing fraction: Packing fraction was proposed by Aston and defined as the difference of actual isotopic mass and the mass number.

Question 7.
What are magic numbers ?
Answer:
Magic numbers: Magic numbers are the numbers 2, 8, 20, 50, 82 and 126 . Nuclides having magic number of either proton or neutrons or both are more stable.

Question 8.
What is Radioactivity ?
Answer:
Radioactivity : It is a process in which nucleus of certain elements undergo spontaneous disintegration without excitation by any external means.

Question 9.
What do you mean by ‘Group Displacement Law’?
Answer:
Group Displacement Law : We know that an α-emission decreases the atomic number of the parent by 2 and β-emission increases the atomic number by 1 .
Thus ‘In an α-emission, the parent element will be displaced to a group two places to the left and in β-emission, it will be displaced to a group one place to the right’.

Question 10.
What are disintegration series ?
Answer:
Disintegration series: The whole series of elements starting with the parent radioactive element to the stable end product is called a radioative disintegration series. These series are-
4n(4n + 1),(4n + 2) and (4n+3)

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 11.
What is half-life period ?
Answer:
Half-life period: Half-life period of a radioactive isotope is the time required for one-half of the isotope ot decay.
t1/2 = \(\frac{0.693}{\lambda}\) (λ = disintegration constant)

Question 12.
What do you mean by ‘Rutherford’?
Answer:
Rutherford : If a radioactive substance has 106 discitegrations per second, it is said to have an activity of one Rutherford.

Question 13.
What is Radioactive equilibrium ?
Answer:
Radioactive equilibrium: Radioactive change being an irreversible process shows equilibrium when a daughter element disintegrates at the same rate at which it is formed from parent element.

Question 14.
What is ‘Articicial radioactivity’ ?
Answer:
Artificial radioactivity : It is the phenomenon in which artificial transmutation of a stable nucleus leads to the formation of a radioactive nuclide.

Question 15.
What do you mean by ‘Alchemy’?
Answer:
Alchemy: The process of transforming one element into other is known as alchemy and the person involved in such experiments is called alchemist.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 16.
What is spallation reaction ?
Answer:
Spallation reaction: High speed projectiles with energies approximately 40 \mathrm{MeV} may chip fragments from a heavy nucleus, leaving a small nucleus. This type of reaction is called spallation
Examples :
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 2

Question 17.
What is nuclear fusion?
Answer:
Nuclear fusion : A nuclear reaction in which two lighter nuclei are fused together to form a heavier nuclei is called nuclear fusion.
Examples :
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 3

Question 18.
What is Nuclear fission?
Answer:
Nuclear fission : The process of artificial transmutation in which heavy nucleus is broken down into two lighter nucli of nearly comparable masses with release of large amount of energy is termd as nuclear fission.
Examples :
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 4

Question 19.
What is meant by ‘projectile capture reactions’?
Answer:
Projectile capture reaction: The bombarding particle is absorbed with or without the emission of γ-radiations.
Examples :
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 5

Question 20.
What is Natural transmutation?
Answer:
Natural transmutation: It is a process in which elements such as radium undergoes transmutation on their own.

Broad Answer Type Questions :

Question 1.
What are the symbol, mass and charges of electron,proton and neutron?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 6

Question 2.
What do you mean by Isobars and Isotones ?
Answer:
Isobars : The nuclides of different chemical elements having same mass number but different atomic numbers are called isobars.
Examples : \({ }_{18}^{40} \mathrm{Ar},{ }_{19}^{40} \mathrm{~K},{ }_{20}^{40} \mathrm{Ca}\)
Isotnes: The nuclides of different chemical elements having the same number of neutrons but different atomic numbers are called isotones.
Examples : \({ }_1^3 \mathrm{H},{ }_2^4 \mathrm{He}\)

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 3.
What are the nature, penetrating power and ionising power of a, b, g ravs.
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 7

Question 4.
What do you mean by ‘Binding energy’?
Answer:
Binding energy-Atomic nucleus is made of protons and neutrons closely is a small volume. Although there exist intensive repulsive forces between the component protons, the nucleus is not slit apart. This is so because the nucleons are bound to one another by very powerful forces. The energy that binds the nucleus together in the nucleus is called the nuclear binding energy.

Question 5.
What are the parent and end product of different disintegration series?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 8

Question 6.
What do you mean by ‘Curie’ and ‘Becquerel’?
Answer:
Curie : If a radioative substance disintegrates at the rate of 37 × 1010 disintegrations per second, its activity is said to be 1 curie.
Becquerel (SI unit) : If a radioactive substance has 1 disintegration per second, it is said to have an activity of one Becquerel.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Question 7.
Give differences between Nuclear Reactions and Chemical Reactions.
Answer:
Difference between Nuclear Reactions and Chemical Reactions.

Nuclear Reactions Chemical reactions
i. Proceed by distribution of nuclear particles i. Proceed by the rearrangement of extranuclear electrons.
ii. One element may be converted into another. ii. No new element can be produced.
iii. Rate of reaction is unaffected by external factors such as concentration temperature, pressure and catalyst. iii. Rate of reaction is influenced by external factor.

Question 8.
What are the differences between nuclear fission and nuclear fusion.
Answer:
Difference between nuclear rission and nuclear fusion.

Nuclear fission Nuclear fusion
i. A bigger nucleus splits into smaller nuclei. i. Larger nuclei fuse together to form the heavier nucleus.
ii. It does not require high temperature ii. Extremely high temperature is required
iii. A chain reaction sets in. iiii. It is not a chain reaction.

Numerical Problems :

Example 1:
What is the packing fraction of \({ }_{26}^{56} \mathrm{Fe}\) (isotope mass =55.92066 )?
Answer:
Packing fraction = \(\frac{\text { (isotopic mass – mass number) }}{\text { mass number }}\) × 104
= \(\frac{(55.92066-56)}{560}\) × 104 = -14.167
So, the packing fraction of \({ }_{26}^{56} \mathrm{Fe}\) is -14.167.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Example 2:
How many α and β-particles wil be emitted by \({ }_{84}^{218} \mathrm{Ra}\) in changing to \({ }_{82}^{206} \mathrm{~Pb}\) ?
Answer:
Let x and y be the number of α and β-particles involved in bringing about the change

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 9

Comapring the mass number = 218 = 206 + 4x + 0y
4 x = 12
x = 3
or,
x = 3
comparing the atomic numbers
84 = 82 + 2x – y
2x – y = 2
y = 4
The number of α nd β-particles emitted in the given nuclear reaction is 3 and 4.

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus

Example 3.
Calculate the number of neutrons in the remaining atom after emission of an α particle from \({ }^{238}{ }_{92} \mathrm{U}\) atom.
Answer:
On emission of an α-particle atomic mass
of daughter element = 238 – 4 = 234
atomic number of daughter element = 92 – 2 = 90
Number of neutron = Atomic mass – atomic number = 234 – 90 = 144

Example 4:
A radioactive substance decays at such a rate that after 46 days only 0.25 of its original amount is left. Calculate its disintegration constant.
Answer:
Let, the original amount be NO, then
Amount after 46 days, Nt =0.25 × N0
For a nuclear decay
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 10

Example 5 :
A certain nuclide has a helf-life of 60 min. If a sample containing 600 atoms is allowed to decay for 90 min, what will be the remaining atoms?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 11

Example 6 :
The activity of a sample of radioactive element A100 is 6.02 curie. Its decay constant is 3.7 × 104 S-1. Calculate the initial mass of the sample.
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 12

Example 7:
At radioactive equilibrium, the ratio between the atoms at two radioactive elements A and B was found to be 3.1 × 109: 1 respectively. If t1/2 of element A is 2 × 1010 year then what will be the t1/2 of the element B ?
Answer:
At radioactive equilibrium

WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 13

Example 8:
Half-life period of \({ }^{14} \mathrm{C}\) is 5760 years. An old piece of wood have a disintegration rate which is 25% of the disintegration rate of an equal weight of a new piece of wood?
Answer:
Let rate of disintegration of new piece =100
The rate of disintegration of old piece =25
WBBSE Class 10 Physical Science Solutions Chapter 7 Atomic Nucleus 14

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 2 Behaviour of Gases offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 2 Question Answer – Behaviour of Gases

Very Short Answer Type Questions :

Question 1.
Why are gaseous molecules always in incessant motion?
Answer:
Because gaseous molecules have no mutual force of attraction.

Question 2.
What is the cause of pressure of a gas taken in a vessel?
Answer:
Gaseous molecules continuously strike against the walls of the container.

Question 3.
In which direction pressure exerted by a gas in vessel does vary, if at all?
Answer:
Equal in all direction.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 4.
What is the effect of supplying heat to a gas at constant volume ?
Answer:
Pressure of the gas increases.

Question 5.
What is the cause of temperature of gas ?
Answer:
Kinetic energy possessed by gaseous molecules.

Question 6.
What is the effect of withdrawal of heat from a gas at constant volume ?
Answer:
Temperature of the gas falls.

Question 7.
What are the two constant quantities in Boyle’s law ?
Answer:
Mass of the gas and temperature.

Question 8.
What is the temperature inCelsius scale corresponding to absolute zero?
Answer:
-273°C

Question 9.
What Is meant by ideal gases?
Answer:
Gases which obey Boyle’s law and Charle’s law at all pressure and temperature.

Question 10.
How does the volume of a gas change for each degree celsius rise of temperature at constant pressure?
Answer:
\(\frac{1}{273}\) of volume of the gas at 0°C increases.

Question 11.
What will happen when a gas-filled balloon heated?
Answer:
The balloon will burst as the supplied heat increases the volume of the gas inside the balloon.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 12.
What will be the volume of an ideal gas at absolute zero?
Answer:
The volume will be zero.

Question 13.
At what temperature speed of gas molecules becomes zero?
Answer:
At -273°C or 0 K speed of gas molecule becomes zero.

Question 14.
What are the constant quantities inCharle’s law?
Answer:
In Charle’s law, the constant quantities are mass of the gas and pressure.

Question 15.
What are the variable quantities inCharle’s law?
Answer:
In Charle’s law, the variable quantities are temperature and volume.

Question 16.
What are the variable quantities in Boyle’s law?
Answer:
In Boyle’s law the variable quantities are pressure and volume.

Question 17.
What is the freezing temperature of water in absolute scale?
Answer:
273 K

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 18.
Give a mathematical expression of Boyle’s law.
Answer:
V × P= constant

Question 19.
Give a mathematical expression ofCharle’s law.
Answer:
[\(\frac{V}{T}\) = constant]

Question 20.
What is the SI unit of pressure?
Answer:
SI unit of pressure is Pascal.

Question 21.
How is atmospheric pressure determined ?
Answer:
Atmospheric pressure is determined by Barometer invented by Torricelli.

Question 22.
How is pressure of a gas determined ?
Answer:
The pressure of a gas is determined by Manometer.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 23.
What is absolute zero?
Answer:
The temperature at which the volume of a gas reduces to zero is called absolute zero.

Question 24.
IsCharle’s law applicable for liquids ?
Answer:
Charle’s law is not applicable for liquids.

Question 25.
Write the ideal gas equation for one gram-mole gas.
Answer:
PV = RT

Question 26.
What is the value of absolute temperature in Fahrenheit scale ?
Answer:
(-) 4594° F

Question 27.
What is the relation between absolute scale and celsius scale of temperature?
Answer:
If any temperature is t°C in celsius scale and if is TK in Kelvin scale then
T = t + 273

Question 28.
What is Torr?
Answer:
Torr is the unit of pressure used at high vacuum. 1 Torr = 1 mm of Hg

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 29.
What is Loschmidt number?
Answer:
The numher of molecules present in 1 ml of gas or vapour at STP is known as Loschmidt number.

Question 30.
What is Boyle’s temperature?
Answer:
The temperature above which a real gas behaves like an ideal gas is called Boyle’s temperature.

Question 31.
Give a main property of a gas.
Answer:
Diffusion is a main property of a gas.

Question 32.
How does rate of diffusion of a gas vary with temperature ?
Answer:
Rate of diffusion of a gas increases with increase in temperature.

Question 33.
State the names of the scientists related to kinetic theory of gases.
Answer:
Kinetic theory of gases was proposed by Bernouli and extended by Clausius, Maxwell, Kronig, Boltzmann etc.

Question 34.
What is collision frequency?
Answer:
The number of collisions made by all gas molecules in a unit volume per second is known as collision frequency.

Question 35.
Define torr ?
Answer:
Pressure exerted by exactly 1 mm of mercury column at 0°C and at standard gravity is termed as 1 torr.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 36.
Which state of matter has no definite volume and no definite shape?
Answer:
gaseous state.

Question 37.
What is the value of gas constant in SI unit ?
Answer:
8.314 JK-1 mol-1

Question 38. How is the unit ‘L’ related to dm3 ?
Answer:
IL=1 dm3

Question 39.
Name the SI unit of pressure ?
Answer:
Pascal

Question 40.
Under what conditions is Boyle’s law is applicable?
Answer:
At constant mass and at constant temperature.

Question 41.
What is the effect of increase of temperature on vapour of a liquid at constant pressure?
Answer:
With rise of temperature vapour of a liquid increases at constant pressure.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 42.
Write down the value of R in L atm mol -1 k-1.
Answer:
0.0821 L atm mol-1 k-1

Short Answer Type Questions :

Question 1.
State one point of difference between expansion of gas and expansion of solid.
Answer:
For any range of temperature rise all gases expand equally, but different solids expand by different amount for same range of temperature rise.

Question 2.
Why does not the volume of a real gas become zero at absolute zero?
Answer:
Explanation : Any gas or vapour liquefies much before reaching this low temperature -273°C. Again,Charle’s law is valid for gases, not for liquids. Hence, this situation of zero volume of gas is never attained practically.

Question 3.
Why tyre of an automobile is inflated to a lesser pressure in summer than in winter?
Answer:
Explanation : In summer, the room temperature is much higher as compared to winter.Consequently, the same volume of air will exert greater pressure in summer. Therefore, it is always advisable to inflate the tyre to lesser extent in summer as compared to winter to avoid the bursting of tyres.

Question 4.
Is it possible to cool a gas below absolute zero ?
Answer:
Explanation : No, it is not possible to cool a gas below absolute zero. At absolute zero, the kinetic energy of the gas molecules becomes zero. The molecules come to rest at the bottom of the container.
Therefore, pressure of the gas is also zero. It is therefore, impossible to cool a gas below absolute zero, since there is no heat left to be removed from the gas.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 5.
What is effusion of a gas?
Answer:
Effusion: It is the passage of a gas under pressure through a tiny hole in a container. Process of effusion is used to separate the isotopes of an element. During effusion, the lighter isotopes escape first from the pores of the container leaving behind the heavier isotopes.

Question 6.
How does an ideal gas differ from a real gas?
Answer:
Explanation : As ideal gas obeys gas laws or the gas equation P V=n R T all temperatures and pressures whereas a real gas obeys the gas laws or the gas equation only at high temperatures and low pressures. There is no cooling or heating effect observed when an ideal gas expands in vacuum whereas a real gas shows either cooling or heating effect.

Question 7.
Is it possible to keep liquid nitrogen in a sealed cylinder at ordinary temperature like ammonia?
Answer:
Reason : No, it is not possible to keep the liquid nitrogen in a sealed container at ordinary temperature like ammonia because critical temperature of nitrogen is very low whereas the critical temperature of ammonia is lighter than that of ordinary temperature.

Question 8.
Why aerated bottles are kept under water during summer ?
Answer:
Explanation : Areated water bottles contain carbon dioxide gas dissolved in aqueous solution under pressure. In summer, the solubility of the gas in solution decreases with rise in temperature.Consequently, the amount of free gas in bottle increases. This leads to increase in pressure. Thus, the bottle may explode. To avoid explosion, bottles are kept under cold water in summer.

Question 9.
State two characteristics of gases.
Answer:
Characteristics of gases :
(i) A gas has no definite shape or volume. It takes the shape and attains the volume in which it is kept. This tendency of scattering of the gas molecules is called expansion of gas.
(ii) If a few gases which do not react with each other, are kept in a container, these gases mix with each other and a homogeneous mixture is prepared. This property is known as diffusion.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 10.
What is normal temperature and pressure (NTP) ?
Answer:
NTP or STP (Standard Temperature and Pressure): It is defined as the pressure exerted by a column of mercury 76 cm in height at 0°C, at 45° latitude and at mean sea-level.

Question 11.
Calculate the value of normal preseure inCGS and SI system.
Answer:
Normal pressure inCGS system : 76 × 13.6 × 980 = 1.013 × 106 dyne / cm2 Normal pressure in SI system : 1.013 × 105 N / m2=1.013 × 105 pascal

Question 12.
State Boyle’s law.
Answer:
Boyle’s law (1662) : At constant temperature, the volume of a definite mass of any gas’varies inversely as the pressure of the gas.

Question 13.
Deduce the mathematical expression of Boyle’s law.
Answer:
Mathematical expression of Boyle’s law : If P be the pressure and V the volume of a given mass of a gas, then according to Boyle’s law,
V ∝ \(\frac{1}{P}\) (the mass and the temperature remaining constant)
or, V = \(\frac{K}{P}\) ( K is a proportionality constant)
or, PV = K (constant)
This is the mathematical experession of Boyle’s law.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 14.
What will be the nature of graph if P is plotted against V for a given mass of gas at constant temperature?
Answer:
Explanation of graph : On plotting P against volume V fat a given temperature, the graph will be a rectangular hyperbola as shown below.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 1

Question 15.
When a balloon infiates, It seems to violate Boyle’s law. Explain.
Answer:
Explanation : When a rubber balloon inflated with the end of a pump, the pressure inside increases but the volume instead of suffering a decrease in accordance with Boyle’s law increases. This happens because with the introduction of air, the mass of the gas increases resulting an increase in volume. Thus, inflation of a balloon by mouth or pump appears to violate Boyle’s law but it is not so in the true sense.

Question 16.
StateCharle’s law.
Answer:
Charle’s law (1787): At constant pressure, the volume of a given mass of a gas increases or decreases by \(\frac{1}{273}\) of its volume at 0°C for each one degree rise or fall in temperature.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 17.
Deduce the mathematical expression ofCharle’s law.
Answer:
Mathematical expression of Charle’s law :
Let V0 be the volume of adiefnite mass of a gas at 0°C, then the volume of 1°C rise in temperature will be =(V0+V0 × \(\frac{1}{273}\))
∴ The volume of t°C rise in temperature will be =V0 + V0 × \(\frac{1}{273}\) If Vt represents this volume at t°C then
Vt + V0+V0 × \(\frac{t}{273}\) = V0(1 + \(\frac{t}{273}\))
Similarly, when temperature at the same mass of gas be gradually made to decrease, its volume at -t°C is given by
Vt + V0 + V0 × \(\frac{t}{273}\) = V0(1-\(\frac{t}{273}\))

Question 18.
What will be the nature of graph if Vt is plotted against t for a given mass of gas at constant pressure?
Answer:
Explanation of graph : On plotting Vt against t, at a given pressure the graph will be a straight line. If this straight line is extraploted then the straight line meets at ‘ X ‘-axis which represents the absolute temperature (-273°C).

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 2

Question 19.
What is absolute zero? Define absolute scale of temperature.
Answer:
Absolute zero: The lowest possible temperature at which a given mass of gas does not occupy any volume or does not exert any pressure is known as absolute zero.
Absolute scale of temperature : The temperature scale which has been devised by taking -273°C as zero and using the magnitude of each degree as the celsius degree is known as absolute or Kelvin scale of temperature.

Question 20.
What is the alternative form ofCharle’s law?
Answer:
Alternative form ofCharle’s law : Suppose at constant pressure, the volume of a certain mass of gas at 0°C = V0 and at t1°C and t2°C temperatures, the volumes are V1 and V2 respectively.

According toCharle’s law,

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 3

Alternative statement ofCharle’s law : The volume of a given mass of a gas kept at constant pressure is directly proportional to its absolute (Kelvin) temperature.

Question 21.
State Pressure-Temperature relationship or Gay-Lussac’s law.
Answer:
Pressure-Temperature relationship or Gay-Lussac’s law : At constant volume, the pressure of a given mass of a gas increases by \(\frac{1}{273}\) of its original value at 0°C for every 1°C rise in temperature.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 22.
State the mathematical expression of Gay-Lussac’s law.
Answer:
Mathematical expression of Gay-Lussac’s law :
WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 4

Therefore, Gay. Lussac’s law can also be stated. At constant volume, the pressure of a given mass of a gas is directly proportional to its absolute temperature.

Question 23.
Establish the combined law of Boyle’s andCharle’s laws.
Answer:
Combination of Boyle’s law andCharle’s law :
Let V be the volume of a given mass of gas at pressure P and temperature T (absolute).
From Boyle’s law, V ∝ \(\frac{I}{P}\)
[When temperature and mass of the gas are constant]
FromCharle’s law, V ∝ T
[ When pressure and mass of the gas are constant]
∴ V∝ \(\frac{T}{P}\) when both T and P vary.
or, V = \(\frac{KT}{P}\) or, \(\frac{PV}{T}\) = K [K= constant]
Now, if V1 is the volume of a gas at pressure P1 and temperature T1 and V2 is the volume of the same amount of gas at pressure P2 and temperature T2 then,
\(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 24.
What Is equation of state? What is universal gas constant ?
Answer:
Equation of state: From the combination of Boyle’s andCharle’s laws we get, \(\frac{PV}{T}\) = K
The numerical value of the constant of proportionality (K) depends upon quantity of gas and the units in which volume and pressure are expressed but is totally independent of the nature of the gas.
For 1 mole of gas, the constant is termed as universal gas constant.
∴ \(\frac{PV}{T}\) = R
or, PV = RT
This is known as general gas equation.
For ‘ n ‘ moles of the gas therefore, the gas equation may be written as:
PV = nRT

Question 25.
What is the relation between density and pressure of an ideal gas.
Answer:
According to ideal gas equation PV = nRT = \(\frac{m}{MRT}\) where ‘m’ is the mass of gas and ‘m’ its molecular mass,
P = \(\frac{m}{V}\) \(\frac{RT}{M}\)
or, P = d \(\frac{RT}{M}\) where d = \(\frac{m}{V}\) = density of the gas.
∴ d α P constant temperature.

Question 26.
What is real gas? What is ideal gas ?
Answer:
Real gas : It is a gas which does not obey general gas equation and all other gas laws strictly but tends towards ideality at low pressure and high temperature.
Ideal gas : It is a gas obeys the general gas equation and other gas laws under all conditions of temperature and pressure.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 27.
State the values of universal gas constant R in different units.
Answer:
Values of universal gas constant R in different units ;

  1. 0.0821 lit atm mol-1 K-1
  2. 8.314 × 107 erg mol-1 K-1
  3. 8.314 J mol K-1
  4. 1.987 cal mol-1 K-1

Broad Answer Type Questions :

Question 1.
What are the characteristic properties of gases ?
Answer:
Characteristic properties of gases :

  1. Shape and Volume: Since particles of gases are not held in fixed positions and move freely, gases neither have definite shapes nor definite volumes.
  2. Homogeneous nature : All the parts of a gas or a gaseous mixture have similar composition throughout.
  3. Density : Due to large separation of molecules, gases have large volume and thus low density.
  4. Compressibility : On increasing pressure gases can be readily compressed due to the presence of large empty spaces.
  5. Random motion : The molecules or atoms of a gas are in a state constinuous zig-zag motion in all directions.
  6. Pressure : Due to their random motion, the molecules of the gas collide on the walls of the container and thus exert certain force on walls of the container. The force per unit area is called the gas pressure.
  7. Diffusion : Gases mix (diffuse) with each other freely due to the free movement of their molecules.
  8. Liquefaction : On cooling and applying pressure, gases can be liquefied. However, a gas has to be cooled below a certain characteristic temperature called critical temperature before it can be liquefied by the application of pressure alone.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 2.
What are the postulates of Kinetic theory of gases?
Answer:
Postulates of Kinetic theory of gases :

  1. All gases consist of very large number of tiny particles called molecules that are in constant rapid motion.
  2. The gas molecules are perfectly round, very hard and separated by large distances. Their actual volume is thus, negligible as compared to the total volume of the gas.
  3. The collisions between the gas molecules are perfectly elastic, i.e. there is no loss of energy during these collisions.
  4. The distance between gas molecules being very large, there is no effective force of attraction or repulsion between them.
  5. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas.
  6. The gas molecules collide with one another and with the walls of the container. The pressure exerted by a gas is due to the bombardment of its molecules on the walls of the vessel.
  7. The gas molecules move freely in all directions. Their speed and direction change continously due to collisions among them. As a result, their motion becomes zig-zag or random.

Brain Storming Questions :
Very Short Answer Type Questions :

Question 1.
What is the significance of S.T.P. or N.T.P?
Answer:
Temperature and pressure affects the volume of gas. In order to compare the volumes of any two gases, they are to be brought under some standard condition of temperature and pressure. For this reason, S.T.P. or N.T.P. are introduced during study of gas.

Question 2.
Why do hot air balloons rise higher than cold air balloons?
Answer:
At higher temperature, a given mass of gas occupies more volume as per charles’ law and is thus dense.Cold air being denser than hot air, raises the balloon to a lesser height.

Question 3.
Why do gas balloons brust at high altitudes?
Answer:
We know that, at high altitudes, the atmospheric pressure on the balloon decreases, resulting in increase in volume as per Boyles’law. So the balloon brusts at high altitudes.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 4.
Under what conditions the value of \(\frac{P}{T}\) will be constant irrespective of the value of T ?
Answer:
We know that, for a certain mass of gas at constant volume, \(\frac{P}{T}\) = K (constant). So, the value of \(\frac{P}{T}\) is constant for a fixed mass of gas at constant volume no matter what the value of T is. This is obtained from Gay Lussac’s law.

Question 5.
Carbondioxide is heavler than air, yet it does not form the lower layer of air-why?
Answer:
From the definition of diffusion we know that gases in the air get intermixed. Although CO2 is heavier than air yet it does not form the lower layer.

Question 6.
Wet air is more penetrating than dry air-why?
Answer:
As wet air being lighter than dry air, so wet air diffuses more readily than dry air.

Question 7.
Name an instrument to measure the pressure of a gas.
Answer:
Manometer is used to measure the pressure of a gas.

Question 8.
What do you mean by STP?
Answer:
STP means standard temperature and pressure. Standard temperature is 0°C. 76 cm mercury pressure or 1 atm. pressure is taken as standard.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 9.
Can temperature in Keloin scale be negative?
Answer:
No. starting with the zero reading at absolute zero temperature, temperature in kelvin scale is always positive.

Question 10.
What is the dimension of R ?
Answer:
We know that, universal gas constant, R

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 5

Question 11.
On which two factors does the molar volume of a gas depend?
Answer:
Molar volume of a gas depends on pressure and temperature.

Question 12.
What is the Boyle temperature of hydrogen gas?
Answer:
Boyle temperature of hydrogen is -156°C.

Question 13.
Mention an example of diffusion applicable to daily life.
Answer:
Detection of fragrance of perfumes occurs due to gaseous diffusion.

Question 14.
What is the nature of collisions between the molecules of a gas?
Answer:
The collisions between the molecules are perfectly elastic i.e.there is no loss of energy during these collisions.

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Question 15.
What is the value of kinetic energy of an Ideal gas at absolute zero temperature ?
Answer:
At absolute zero temperature all the molecular motions cease and hence the kinetic energy of an ideal gas becomes zero.

Question 16.
What is molar volume?
Answer:
The volume occupied by one mole of substance at a given temperature and pressure is called the molar volume of the substance.

Numerical Problems :

(i) Tt=t_c+273
(ii) P1 V1=P2 V2 (at constant temperature and constant mass of gas)
(iii) V1=V_{\circ(1+\frac{t{273)
(iv) \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) (at constant pressure and constant mass of gas)
(v) \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\) (at constant volume and constant mass of gas)
(vi) PV = nRT = \(\frac{m}{MRT}\)

Erample 1:
What will be the reading of 30°C in absolute scale of temperature?
Answer:
The reading of 30°C in absolute scale of temperature
= (273 + 30) K = 303K

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 2:
What is the value of absolute zero in the celsius scale?Calculate its value in Farhenheit scale.
Answer:
The value of absolute zero in celsius scale = -273°C.
Let the value of the above temperature in Farhenheit scale = X° F.
We know, \(\frac{C}{5}\) = \(\frac{F-32}{9}\)
or, \(\frac{-273}{5}\) = \(\frac{X-32}{9}\)
or, 5 x-160 = -2457 or, 5 x = -2297
or, x = -459.4° F
∴ The temperature in Farhenheit scale = -459.4° F

Example 3 :
27°C or 300 K, which temperautre is higher?
Answer:
27°C = (27 + 273)K = 300 K
So, the two temperatures are equal.

Example 4 :
30°C or 300 K, which temperature is higher ?
Answer:
30°C=(30+273) K=303 K
So, 30°C is higher than 300 K

Example 5:
A weather balloon has a volume of 175 L when filled with hydrogen at a pressure of 1.00 atm.Calculate the volume of the balloon when it rises to a height of 2000 m, where the atmospheric pressure is 0.80 atm. Assume that the temperature is constant.
Answer:
P1 = 1.00 atm
V1 = 175 L
P2 = 0.80 atm
V2 = ?

According to Boyle’s law
P1 V1=P2 V2
or, V2=\(\frac{P_1 V_1}{P_2}\) = \(\frac{1 \times 175}{0.80}\)

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 6:
A sample of oxygen has a volume of 880 mL and a pressure of 740 torr. What additional pressure is required to reduce the volume to 440 mL.
Answer:
P1 = 740 torr
V1 = 880 mL
P2 = ?
V2 = 440 mL

According to Boyle’s law
P1V1 = P2 V2
or, P2 = \(\frac{P_1 V_1}{V_2}=\frac{740 \times 880}{440}\)
= 218.75 L
∴ Additional pressure required
= (1480-740) torr
= 740 torr

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 7:
A bottle of volume 9 litres contains a gas at 10 atmospheric pressure. How many bottles each of 1 litre capacity can be filled with the gas at 2 atmospheric pressure, at constant ternerature.
Answer:
P1 = 10 atm
V1 = 9 lit
P2 = 2 atm
V2 = ?

According to Boyle’s law
P1 V1=P2 V2
or, V2=\frac{P1 V1{P2=\frac{10 × 9{2=45 lit
∴ 45 bottles, each of 1 litre volume can be filled.

Example 8 :
A sample of helium has a volume of 520 cm3 at 373 K.Calculate the temperature at which the volume will become 260 cm3. Assume that the pressure is constant.
Answer:
According toCharle’s law
V1 = 520 cm3
T1 = 373 K
V2 = 260 cm3
T2 = ?
or, = 186.5 K

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 9:
Find the volume of N2 at 27°C at a pressure of 760 torr, if it occupies 40 mL at STP.
Answer:
V1 = 40 mL
T1 = 373 K
T2 = (27+373) K=300 K
V2 = ?

According toCharle’s law

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
or, V2= \(\frac{V_1 \times T_2}{T_1}\) = \( \frac{40 \times 300}{273}\) = 44 mL

Example 10:
The volume of a certain mass of gas at of 400 K and pressure 202600 Pascal is 2 cubic metres. What would be the volume of the gas at 327°C and 152 cm of mercury pressure?
[given 1.013 × 105 pascal =1 atmospheric pressure]
Answer:
1.013 × 105 Pascal = 76 cm of Hg
202600 Pascal = \(\frac{76 \times 202600}{1.013 \times 10^5}\)
= 125 cm of Hg
T1 = 327°C = (327 + 273) K = 600K
V1 = 2 m3
V2 = ?
T1 = 400 K

According toCharle’s law

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
or, V2 = \(\frac{V_1 \times T_2}{T_2}\)
or, V2 = \(\frac{2 \times 600}{400}\)
∴ V2 = 3 m3

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 11:
A steel tank contains carbon dioxide at 27°C and a pressure of 10.0 atm.Calculate the internal gas pressure when the tank and the gas are heated to 100°C.
Answer:
P1 = 10.0 atm
T1 = (27 + 273) K = 300 K
P2 = ?
T2 = (100 + 273) K = 373 K

According to Gay-Lussac’s law.

\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
or, P2 = \(\frac{P_1 \times T_2}{T_1}\)
or, P2 = \(\frac{10 \times 373}{300}\) = 12.43 atm.

Example 12 :
An iron cylinder contains helium at a pressure at 250 K Pa at 300 K. The cylinder can withstand a pressure of 1 × 106 Pa. The room in which cylinder is placed catches fire. Predict whether the cylinder will blow up before it melts or not.
(M.P. of the cylinder =1800 K )
Answer:
P1 = 250 KPa = 250 × 103 Pa
V1 = 300 K
P2 = ?
T2 = 1800 K

According to Gay-Lussac’s law

\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
or, P2 = \(\frac{P_1 \times T_2}{T_{21}}\)
or, P2 = \(\frac{250 \times 10^3 \times 1800}{300}\) = 1.5 × 106 Pa

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 13:
A gas having temperature 0°C is heated so that its pressure and volume are doubled. What will be the final temperature of the gas?
Answer:
Let, P1 = P (say)
by condition, P2 = 2 P
again V1 = V (say)
So, V1 = 2 V
T1 = (0 + 273) K = 273 K
T2 = ?

By combining Boyle’s law andCharle’s law we get

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 6

Example 14:
A certain quantity of a gas occupies a volume of 1000 cm3 at 760 mm and 27°C. Find the volume of the gas if the pressure and temperature are 1520 mm and 327°C.
Answer:
P1 = 760 mm
V1 = 1000 cm3
T2 = (273 + 27) K = 300 K
P2 = 1520 mm
T2 = (273 + 327) K = 600 K
V2 = ?

By combining Boyle’s law andCharle’s law we get,

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases 7

Example 15:
7.0 g of a gas at 300 K and 1 atmospheric pressure occupies a volume of 4.1 littres. What is the molecular mass of gas?
Answer:
w = 7.0 g
R = 0.0821 lit atm mol-1 K-1
T = 300 K
M = ?
P = 1 atm
V = 4.1 lit

Applying ideal gas equation,

PV = nRT
PV = \(\frac{w}{M}\)
or, M = \(\frac{wRT}{PV}\)
or, M = \(\frac{7 \times 0.0821 \times 300}{1 \times 4.1}\)
∴ M = 42 g-mol-1

WBBSE Class 10 Physical Science Solutions Chapter 2 Behaviour of Gases

Example 16:
10 g of oxygen are introduced in a vessel of 5 lit capacity at 27°C.Calculate the pressure of the gas in atmospheres in the container.
Answer:
w = 10 g
M = 32
R = (27+273) K=3000 K
V = 5 lit
R = 0.0821 lit atm mol-1 K-1
P = ?

Applying ideal gas equation,

P V =n R T
∴ P V = \(\frac{w}{M}\)
or, P = \(\frac{w R T}{V \times M}\)
∴ P = 1.539 atm

Example 17:
The volume of 1 gram mole of oxygen gas is 22400 mL at 760 mm.Calculate the temperature of the gas (Given R=0.082 L-atm mol{ -1 K-1 )
Answer:
P = 670 mm = 1 atm
V = 22400 mL = 22.4 L
R = 0.082 L – atm mol-1 K-1

We know from ideal gas equation for 1 mole of an ideal gas

PV = RT
or, T = \(\frac{P V}{R}\) or, T = \(\frac{1 \times 22.4}{0.082}\)
or, T = 273.17 K
Temperature in celsius scale = 273.17 – 273 = 0.17°C

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 4 Thermal Phenomena offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 4 Question Answer – Thermal Phenomena

Very Short Answer Type Questions :

Question 1.
Is it possible for a body to have negative temperature in kelvin scale?
Answer:
No, because below 0K temperature, volume of a gas would be negative which is impossible.

Question 2.
What is the SI unit of heat ?
Answer:
SI unit of heat is Joule.

Question 3.
Mention the CGS unit of heat?
Answer:
CGS unit of heat is calorie.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 4.
Is volume coefficient of a gas different from its pressure coefficient?
Answer:
No, they are equal.

Question 5.
What is the value of specific heat of water in SI unit?
Answer:
4200 J kg-1 K-1.

Question 6.
Is it possible that heat is supplied to a body without changing the temperature?
Answer:
Yes, during change of state there is no change in temperature although heat is supplied.

Question 7.
What is the value of latent heat of fusion of ice in CGS system?
Answer:
80 cal g-1.

Question 8.
Do water and ice have the same specific heat?
Answer:
No, specific heat of water is 1 cal g-1 °C-1 and that of ice is 0.5 cal g-10 C-1.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 9.
What is difference between 1° C and 1 C° ?
Answer:
1° C is a particular temperature, while 1 C° is the interval of one degree Celsius of temperature.

Question 10.
What is the lower fixed point of a celsius scale?
Answer:
0° C

Question 11.
What is the upper fixed point of a Fahrenheit Scale?
Answer:
212° F

Question 12.
Is there any upper fixed point in Kelvin Scale?
Answer:
273.13 K, which is the triple point of water.

Question 13.
Which is more expansible-solid, liquid or gas ?
Answer:
Gas.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 14.
A metal scale does not give correct value at all temperature-explain.
Answer:
At different temperature distance between different marks of the metal scale is different due to thermal expansion of the metal.

Question 15.
The Coefficient of linear expansion of a metal is 19 × 10-6° C-1. What is its value per degree Fahrenheit ?
Answer:
∝ = \(\frac{5}{9}\) × 19 ×10-6 = 10.6 × 10-6 °F.

Question 16.
Is coefficient of linear expansion possible in the case of a liquid?
Answer:
No, liquid has no definite length.

Question 17.
There is co-efficient of apparent expansion in a liquid but not in a gas-why ?
Answer:
For same rise of temperature, liquid expands only about 10 times than solid, while gas expands about 100 times than solid. So expansion of vessel can be neglected in case of gases.

Question 18.
What is anomalous expansion of water?
Answer:
In the range 0°C to 4°C, volume of water decrease instead of increasing, like other liquids. This is known as anomalous expansion of water.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 19.
Why are there two types of coefficient of expansion of a gas?
Answer:
The change of temperature is very much affected by both its pressure and volume, so gas has two types of coefficient of expansion.

Question 20.
Is a temperature lower than absolute zero of temperature possible?
Answer:
No.

Question 21.
What is thermal conductivity of a perfect heat conductor?
Answer:
Infinite.

Question 22.
What is thermal conductivity of a perfect heat insulator?
Answer:
Zero.

Question 23.
Can heat flow by canvection take place in a straight line?
Answer:
In convection heat travels in straight or curved path in upward direction only.

Question 24.
Can convection take place in solid ?
Answer:
No, for convection a fluid medium is required.

Question 25.
What is the absorbing power of a perfectly black body ?
Answer:
One

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 26.
What is the reflecting power of a perfectly black body ?
Answer:
Zero.

Question 27.
What is a perfectly black body ?
Answer:
A body which absorbs all radiations incident it.

Question 28.
By which mode the heat reaches to the earth from the Sun?
Answer:
Radiation.

Question 29.
Explain for cooking, a black vessel is more convenient than a polished one.
Answer:
Black surface is better absorber of heat.

Question 30.
Heat is generated continuously in an electric heater but its temperature becomes constant offers some time. Why?
Answer:
When rate at which heat is generated by electric current becomes equal to the rate at which heat is lost by radiation, thermal equilibrium is reached and hence temperature becomes constant.

Question 31.
Why ventilator are placed near the roof at a room?
Answer:
Warm air of the room becomes lighter and moves upward. So for better convection current ventilators are placed near the roof of the room.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 32.
Name three modes of transfer of heat.
Answer:
Conduction, convection and radiation.

Question 33.
Name the element which is the best conductor of heat ?
Answer:
Silver.

Question 34.
What is the unit of coefficient of thermal conductivity in SI ?
Answer:
Wm-1 K-1.

Question 35.
What is thermal expansion?
Answer:
Dimensions of all substances generally increases with increase in temperature. This phenomenon is known as thermal expansion.

Question 36.
Name the three types of expansion in solid?
Answer:
The types of expansion in solid –

  1. Linear expansions
  2. Superficial expansions
  3. Volume expansions.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 37.
Define the term ‘coefficient’ of linear expansion of solld.
Answer:
Coefficient of linear expansion of solids : The coefficient of linear expansion of a solid is the increase in its length per unit length per degree rise of temperature.

Question 38.
Give the SI unit of coefficlent of linear expansion.
Answer:
K-1.

Question 39.
Write the relationship among α, β and γ.
Answer:
α = \(\frac{β}{2}\) = \(\frac{γ}{3}\)

Question 40.
Does unit of coefficient of linear expansion depend on unit of length?
Answer:
No.

Question 41.
What is thermal conduction ?
Answer:
Conduction is the process of transfer of heat through a substance without any motion of the particles of the substance.

Question 42.
Name two substances which are good conductors of heat.
Answer:
Silver, Copper.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 43.
Name two bad conductors of heat.
Answer:
Glass and wooden rod.

Short answer type questions :

Question 1.
What do you mean by heat ?
Answer:
Heat : It is a form of energy which produces sensation of warmth.

Question 2.
What are the different types of motion?
Answer:
Different types of motions are

  1. Translational motion
  2. Vibrational motion
  3. Rotational motion

Question 3.
What is definition of heat with respect to motion?
Answer:
Definition of heat with respect to motion: Heat possessed by a body is the total thermal energy of the body and is the sum of kinetic energies of all the individual molecules forming the body due to translational, vibrational and rotational motions of the molecules.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 4.
What do you mean by calorie?
Answer:
Calorie: It is the quantity of heat required to raise the temperature of 1 g} water through 1° \mathrm{C}.

Question 5.
What is temperature?
Answer:
Temperature : It is the thermal condition of a body which would determine the direction of flow of heat, when the body is placed in thermal contact with another body.

Question 6.
What is upper fixed point of a thermometer ?
Answer:
Upper fixed point : It is the temperature of steam from water boiling under a pressure of 76 cm} of mercury at sea level and 45° latitude.

Question 7.
What is lower fixed point of a thermometer ?
Answer:
Lower fixed point: It is the temperature of melting ice under a pressure of 76 cm} of mercury at sea level and 45° latitude.

Question 8.
What is fundamental interval ?
Answer:
Fundamental interval. The difference between the fixed points of a scale is called fundamental interval.

Question 9.
What do you mean by thermal expansion ?
Answer:
Thermal Expansion: Dimension of all substances generally increases with increase in temperature. This phenomenon is known as thermal expansion.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 10.
What are the types of expansions in solids ?
Answer:
Types of expansions in solids:

  1. Linear expansions
  2. Superficial expansions
  3. Volume expansions

Question 11.
What is apparent expansion of the liquid ?
Answer:
Apparent expansion of the liquid: If the expansion of the liquid is measured ignoring the expansion is called the apparent expansion of the liquid.

Question 12.
What is real expansion of the liquid ?
Answer:
Real expansion of the liquid: When the actual expansion of the liquid is measured by considering the expansion of the containing vessel, it is called real expansion of the liquid.

Question 13.
What is the conclusion of Hope’s experiment ?
Answer:
Conclusion of Hope’s experiment: From Hope’s experiment it is proved that water at bottom which is densest at 4°C. After sufficiently long time the temperature of the lower thermometer falls slightly due to loss of heat by conduction to the upper regions.

Question 14.
What is molar specific heat at constant volume (Cv) ?
Answer:
Molar specific heat at constant volume (Cv) : The molar specific heat of a gas at constant volume (Cv) is the amount of heat required to raise the temperature of one mole of the gas through one degree keeping the volume constant throughout.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 15.
What is molar specific heat at constant pressure (Cp) ?
Answer:
Molar specific heat at constant pressure (Cp) : The molar specefic heat of a gas at constant pressure (Cp) is the amount of heat required to raise the temperature of one mole of the gas through one degree keeping the pressure constant throughout.

Question 16.
What is water equivalent of a body ?
Answer:
Water equivalent: Water equivalent of a body is the mass of water which will be heated through one degree by the amount of heat that raises the temperature of the body through one degree.

Question 17.
What is Regelation ?
Answer:
Regelation: The phenomenon of melting of ice under pressure and freezing again on releasing the pressure is called regelation.

Question 18.
What do you mean by vapour pressure of a liquid ?
Answer:
Vapour pressure: Whenever a liquid evaporates at any temperature, the vapour exerts a definite pressure on everything in contact. This pressure is called vapour pressure of the liquid at that temperature.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 19.
What is conduction of heat ?
Answer:
Conduction: It is the process of the transfer of heat through a substance without any detectable motion of the particles of the substance.

Question 20.
What is convection of heat ?
Answer:
Convection of heat: It is the process by which heat is transmitted through a liquid or gas from a hotter point to a colder point due to the bodily motion of the heated particles of the substances.

Question 21.
What is Radiation of heat ?
Answer:
Radiation: It is the transmission of heat from a hot body to a cold body without the help of any medium and without appreciable heating at the intervening medium if any.

Question 22.
What is perfectly black body ?
Answer:
Perfectly black body: A perfectly black body is that which absorbs completely the radiations of all the wavelengths on it.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 23.
What is green house effect ?
Answer:
Green house effect: Green house effect is an example of selective absorption of heat by glass. The amount of heat transmitted through a substance depends on temperature of the source of heat.

Question 24.
What do you mean by global warming ?
Answer:
Global warming: Carbon dioxide, water vapour, methane, nitrus oxide, tropospheric ozone, chlorofluoro carbon compounds, halogens campounds etc. on increasing the amount of greenhouse gases in the atmosphere, the temperature of earth increases. It is known as global warming.

Broad answer type questipns :

Question 1.
State the units and dimensions of different coefficient.
Answer:
The unirs and dimensions of different coefficient.

Question 2.
What is anomalous expansion of water ?
Answer:
Aromalous expancion of water: Usually liquids expand on heating. But in the case of water we find deviation from this general behaviour of the liquids within a certain range of temperature. The volume of water is minimum at 4°C and hence its density is maximum at 4°C. This phenomenon is called the anomalous expansion of water.

Question 3.
What is the fundamental principle of calorimetry ? What are the conditions ?
Answer:
Fundamental principle of calorimetry:
Heat lost by hot body = Heat gained by cold body.
Cotditicns: (i) during the process of heat transfer, there is no heat exchange with the surrounding.
(ii) no chemical reaction takes place between the bodies.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 4.
What is thermometric conductivity ?
Answer:
Thermometric conductivity: The rate of rise of temperature during the variable state is proportional to K/ps (where K = thermal conductivity of the material, Ps = specific heat). This ratio is known as thermometric conductivity or the thermal conductivity per thermal capacity per unit volume.

Question 5.
Give few applications of conductivity ?
Answer:
Applications of conductivity:
(i) Ice is packed in saw dust, because air which is bad conductors of heat being trapped in the saw dust prevents transfer of heat from the surrounding to the ice. So ice does not melt.
(ii) Cooking utensils are provided with wooden or plastic handles, as wood or plastic are bad conductors of heat, so we can hold the hot utensiles with the help of these handles.
(iii) In winter, birds often swell their feathers, ther by enclosing air between their bodies and feathers and thus does not allow flow of heat from the bodies of the birds and keeps them warm.

Frain Storing Questions :
Very Short Answer Type Questions :

Question 1.
The value of α for iron is 1.2 × 10-5. What do you mean by that?
Answer:
This means 1 cm of an iron rod (at 0°c) expands linearly by 1.2 × 10-5 cm for 1°C rise in temperature.

Question 2.
Name two devices which use bimetallic strip.
Answer:
Automatic fire alarm and thermostat.

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 3.
The diameter of an iron tyre is kept slightly smaller than the wooden wheel on which this tyre is to be mounted. why?
Answer:
The diameter of an iron tyre is kept slightly smaller than the wooden wheels because of heating, the iron tyre expands and perfectly fits over the wooden wheels.

Question 4.
Between apparent and real expansion coefficients of a liquid which one is its own characteristic?
Answer:
Apparent expansion of liquid happens simply because the container expands. Therefore, real expansion coefficient of a liquid is its own characteristic.

Question 5.
How δr is related to δa & δg ?
Answer:
The relation is :
δr = δa + δgr = Real expansion of liquid
δa = Apparent expansion of the liquid
δg = Volume expansion of the container)

Question 6.
What is the conductivity for vacuum ?
Answer:
Conduction does not take place through vacuum. So, pure vacuum has zero conductivity.

Question 7.
What is thermal resistivity?
Answer:
Reciprocal of coefficient of thermal conduction is called thermal resistivity (\(\frac{1}{K}\)).

Question 8.
What is the unit of thermal resistivity?
Answer:
The unit of thermal resistivity is j-1 m.k.s

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Question 9.
What should be the value of thermal conductivity for an ideal conductor ?
Answer:
The value of thermal conductivity for an ideal conductor is infinite.

Question 10.
Write the dimension of thermal conductivity (K)
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena 2

Numerical Problems

Example 1:
On a certain day the maximum temperature was found to be 120°F. What is it in celsius scale?
Answer:
We know that
\(\frac{C}{5}\) = \(\frac{F-32}{9}\)
Let 120°F = x°C
or, \(\frac{x}{5}\) = \(\frac{120-32}{9}\)
or, x = 48.9°C

Example 2:
A thermometer reads 5° in melting ice and 99° in dry steam at normal pressure. Find the temperature on the Fahrenheit scale when the thermometer reads 52°.
Answer:
Let the temperature in the Fahrenheit scale be x°F corresponding to 52° in the faulty thermometer whose fundamental interval in 99° – 5° = 94°. Thus,
WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena 3
or,
x = 90 + 32 = 122°F

Example 3:
Find the temperature at which the reading in Fahrenheit scale is double of the reading in the celsius scale.
Answer:
Let x°C = 2 x°F
We know \(\frac{C}{5}\) = \(\frac{F-32}{9}\)
or, \(\frac{x}{5}\) = \(\frac{2 x-32}{9}\)
or, x = 160°C and 2 x = 320°F

Example 4:
What is the reading of a Fahrenheit thermometer when a centigrade thermometer reads 25°C ? The temperature of a body is raised by 25°C. What is the value of this rise of temperature in Fahrenheit scale?
Answer:
We know, \(\frac{C}{5}\) = \(\frac{F-32}{9}\)
We have, \(\frac{25}{5}\) = \(\frac{F-32}{9}\)
∴ F = 45 + 32 = 77°F
We know, 1 division on centigrade scale = \(\frac{9}{5}\) division on the Fahrenheit scale
∴ 25°C = \(\frac{9}{5}\) × 25 = 45°F

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Example 5:
Find the increase in energy per atom of aluminium when temperature of a piece of aluminium increases by 1° C .27 g of aluminium contains 6 ×1023 atoms and its specific heat is 0.2 .
Answer:
Let us take 27 g of aluminium heat absorbed by it in raising its temperature by 1°C = 27 × 1 × 0.2 = 5.4 cal
= 5.4 × 4.2 J
Now, 27 g of aluminium contains 6 ×1023 atoms
Thus, energy per atom = \(\frac{5.4 ×4.2}{6 ×1023}\) = 3.78 × 10-23 J

Example 6:
Two rods, one of glass and the other of steel, are of equal length of 0°C. At 100°C they differ in length by 1 mm. Find their length at 0°C. Given α for glass = 8 × 10-6 C and α for steel = 12 ×10-6 C-1.
Answer:
Let length of each rod l0 m at 0° C and at 100° C. Let the lengths of glass and steel rod be l g and l s metres respectively.
This, ls = l (1 + 8 × 10-6: 100) and ls = l0(1 + 12 × 10-6 × 100)
∴ ls – lg + l0 × 1 v-4(12-8) = 4 × 1 J-4 l0
Now, ls – ls = 1 mm = 10-3 m
So, 10-3 = 4 × 10-4 l0
∴ l0 = 25 m

WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena

Example 7:
Two vessels connected by a thin pipe with a sliding plug contain a liquid at temperature 20° C and 80° C. What is the ratio of the heights of liquid columns in the vessel if the coefficient of volume expansion of the liquid be 10-3 °C-1 ?
Answer:
Let h1 and h2 be the heights of the liquid columns at 20°C and 80°C and corresponding densities be ρ1 and ρ2.
Now at equilibrium condition,

h1 ρ1 g=h2 ρ2 g
or,
WBBSE Class 10 Physical Science Solutions Chapter 4 Thermal Phenomena 4

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 8.1 Question Answer – Periodic Table and Periodicity of the Properties of Elements

Very Short Answer Type Questions :

Question 1.
Periodic changes of what occur in a periodic table?
Answer:
Physical and chemical properties of sets of elements arranged in some manner change periodically.

Question 2.
How does atomic size of elements change in a group?
Answer:
Atomic size gradually increases downward in a yroup.

Question 3.
How does electronegativity of elements change along a period of periodic table?
Answer:
Electronegativity of elements gradually increases along a period from left to right of periodic table.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 4.
How many groups are there in the long form periodic table ?
Answer:
There are 18 groups in the long form periodic table.

Question 5.
How does metallic character of elements change in a group in periodic table?
Answer:
Metallic character of elements gradually increases downward in a group.

Question 6.
Which is more fundamental-atomic number or atomic weight of elements ?
Answer:
Atomic number is more fundamental than atomic weight of elements.

Question 7.
How are elements arranged in Mendeleev’s modern periodic table?
Answer:
Elements are urianged in Mendeleev’s modern periodic table in ascending order of their atomic number.

Question 8.
What is the position of chlorine in the Mendeleev’s periodic table?
Answer:
The position of chlorine in the Mendeleev’s periodic table:
Period : 3
Group : VIIB

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 9.
In which group of Mendeleev’s periodic table are kept the alkali metals ?
Answer:
Alkali metals are kept in group IA in Mendeleev’s periodic table.

Question 10.
What is the name ‘Oxygen’ group ?
Answer:
Chalcogens, group.

Question 11.
What are ‘d’-block elements called?
Answer:
‘d’-block elements are called transition elements.

Question 12.
What are ‘f’-block elements called ?
Answer:
‘f’-block elements are called inner transition elements.

Question 13.
How many elements are there in first period of Mendeleev’s periodic table?
Answer:
The first period contains two elements. (Hydrogen and Helium)

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 14.
What is the name of first group elements of periodic table ?
Answer:
The first group elements of periodic table are called alkali metals.

Question 15.
What is the decreasing order of electron affinity of F, Cl, Br ?
Answer:
The decreasing order of electron affinity of F, Cl, Br is Cl>Br>F.

Question 16.
Is Cobalt a transition element ?
Answer:
Yes, Cobalt is a transition element.

Question 17.
Is the second I.P. is always greater than the first I.P. of a given species?
Answer:
The second I.P. is always greater than the first I.P. of a given species.

Question 18.
Is radioactivity a periodic property ?
Answer:
Radioactivity is not a periodic property.

Question 19.
Between K and C a atom which one has larger in size ?
Answer:
K atom has larger in size than Ca atom.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 20.
What is the symbol of the element having atomic number 109 ?
Answer:
The symbol of the element having atomic number 109 is Mt.

Question 21.
Who proposed modern periodic law ?
Answer:
Modern periodic law was proposed by Mosely.

Question 22.
What are representative elements ?
Answer:
Elements of s-and p-blocks are collectively known as normal or representative elements.

Question 23.
What are the elements of second period called ?
Answer:
Elements of second period are called bridge elements.

Question 24.
What are the elements of third period called ?
Answer:
Elements of third period are called typical elements.

Question 25.
Between Na and \(\mathbf{N a} a^{\oplus}\) which one has greater is size ?
Answer:
Na has greater in size than \(\mathbf{N a} a^{\oplus}\).

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 26.
Between Cl and \(\mathrm{Cl}^{\Theta}\) which one has smaller in size ?
Answer:
Cl has smaller in size than \(\mathrm{Cl}^{\Theta}\).

Question 27.
Between \(\mathbf{N a} a^{\oplus}\) and Mg2+ which one has smaller in size ?
Answer:
Mg2+ has smaller in size than \(\mathbf{N a} a^{\oplus}\).

Question 28.
Are Van der Waal’s radii larger than covalent radii ?
Answer:
Van der Waal’s radii are larger than covalent radii.

Question 29.
Atomic radii decrease across a period. However, at the end of each period there is increase in atomic radii. Why?
Answer:
Because atomic radii in case of noble gases are Van der Waal’s radii.

Question 30.
Have ionisation energies a positive value?
Answer:
lonisation energies have a positive value.

Question 31.
Why halogens have high values of electron affinity ?
Answer:
Halogens have high values of electron affinity due to their ns2 np5 configuration.

Question 32.
What is the classification of elements ?
Answer:
The method of arranging similar elements together and separating them from dissimilar elements is called classification of elements.

Question 33.
What is the law of triads?
Answer:
According to law of triads, in a triad of elements, having similar properties, the atomic weight of the central atom was merely the arithmetic mean of the atomic weights of the other two elements.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 34.
Between Br and I which one has metallic character ?
Answer:
Between Br and I, iodine has metallic character.

Question 35.
What is the valency of an alkali metal ?
Answer:
The valency of an alkali metal is 1.

Question 36.
What is the relation between N3- and O2- ?
Answer:
N3- and O2- are iso-electronic ions.

Question 37.
Which halogen is radioactive element?
Answer:
Astatine (At) is the only radioactive halogen.

Question 38.
Which group of elements was missing from Mendeleev’s original periodic table?
Answer:
Mendeleev’s original periodic table did not contain noble or inert gases belonging to Group 18 of the modern periodic table.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 39.
In. which period of the periodic table are first transitonal elements found?
Answer:
The first series of transitional elements belong to the 4th period.

Question 40.
X, Y and Z are the elements of a Dobereiner’s triad. If the atomic mass of X be 20 and that of Z be 40, What should be the atomic mass of Y ?
Answer:
Atomic mass of Y = Arithmetic mean of atomic masses of X and Z = \(\frac{20+40}{2}\) = 30

Question 41.
Mention the position of the radioactive alkall metal.
Answer:
Francium (Fr), the radioactive alkali metal, has the position in 1st group of 7 th period.

Question 42.
Name the lanthanide having the lowest atomic number.
Answer:
Cerium \({ }_{58} \mathrm{Ce}\) is the lanthanide having lowest atomic number.

Question 43.
Which halogen has the lowest electronegativity?
Answer:
Electronegativity of an atom decreases from top to bottom in a group. There fore, Astatine (At) has the lowest electronegativity of 2.2 (Pauling scale).

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 44.
The atomic number of an element is 9. In which group should tt be placed in the modern periodic table?
Answer:
The electronic configuration of the element is 2,7. Since it has 7 electrons in its outermost shell, it should be placed in Group 17.

Question 45.
Name one element which was not known during Mendeleev’s time, but a place was left vacant in Mendeleev’s periodic table.
Answer:
The element is Germenium (Ge) which was not known during Mendeleev’s time, but its properties were predicted by Mendeleev and accordingly a place was kept vacant for it in his periodic table.

Short Answer Type Questions :

Question 1.
What is the necessity of arrangingrelements in the periodic table?
Answer:
Necessity of arranging elements in the periodic table It was observed from the early days that there were few groups of elements each of which had almost identical chemical and physical properties. So, a systematic arrangement of elements is essential since it is difficult to remember the individual properties of elements.

Question 2.
What is the ‘law of triads’?
Answer:
Law of triads (Dobereiner, 1817) : In a triad of elements having similar properties, the atomic weight of the central element was merely the arithmetic mean of the atomic weights of the other two elements.

Question 3.
What is the ‘law of octaves’ ?
Answer:
Law of octaves (Newlands, 1864) : If the elements are arranged in the increasing order of their atomic weights, the eighth succeeding element was the repetition of the first one like the 8 th note of musical scale i.e. first and similar characteristics.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 4.
What is Mendeleev’s periodic law ?
Answer:
Mendeleev’s periodic law (1869) : The physical and chemical properties of elements are the periodic function of their atomic weights.

Question 5.
What is modern periodic law ?
Answer:
Modern periodic law (Moseley) : It states that the physical and chemical properties of elements are the periodic function of their atomic numbers.

Question 6.
What is periodicity ?
Answer:
Periodicity : It is the recurrence of elements with similar properties after certain regular intervals when these are arranged in the increasing order of their atomic numbers.

Question 7.
Define group and period of modern periodic table.
Answer:
Period: The horizontal rows along which elements occur in the periocic table in ascending order of atomic number having dissimilar chemical properties are called periods.
Group: The vertical columns in which elements of similar physical and chemical properties occur in the periodic table are called groups.

Question 8.
How many periods and groups are present in Mendeleev’s periodic table and in the long form periodic table?
Answer:
Mendeleev’s periodic table contains seven periods and nine groups. Long form periodic table contains seven periods and eighteen groups.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 9.
What are the structural features of long or extended form of the periodic table (Bohr’s table)?
Answer:
Bohr’s table has the following features :
(i) All the elements have been arranged in the increasing order of atomic members.
(ii) Elements with similar electronic configurations have similar properties and hence have been placed together at one place.
The periodic table consists of :
(a) Seven horizontal rows called periods
(b) Eighteen vertical columns called groups or families.
(c) Four blocks (s, p, d, f)

Question 10.
What are the defects of long form of the periodic table ?
Answer:
Defects of long form of the periodic table :
(a) Position of hydrogen. It does not resemble fully with alkali metals but has been placed along with them. It also resembles halogens.
(b) Position of lanthanides and actinides are not proper in the periodic table.
(c) According to electronic configuration, helium should be placed in s-block where as it is placed in p-block.

Question 11.
What is atomic volume?
Answer:
Atomic volume : Atomic volume of an element is the volume in cm3 occupied by one gram atom of the element in the solid state and hence it is also called gram-atomic volume.

Question 12.
Explain how the atomic size of the elements change in periodic table?
Answer:
Atomic size : The distance of the outermost orbit from the nucleus of spherically shaped atom is called the atomic size.
The atomic size gradually decreases from left to right of the period upto the group VIIB, again increases in the end element.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 13.
What is Covalent radius ?
Answer:
Covalent radius : It is defined as one half of the distance between the centres of nuclei of two similar atoms held together by a purely covalent single bond.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements 1

Question 14.
What is Van der Waal’s radius?
Answer:
Van der Waal’s radius : It is defined as one half of the internuclear distance between two similar, adjacent atoms belonging to two neighbouring molecules of the same substance in the solid state.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements 2

Question 15.
What is metallic radius ?
Answer:
Metallic radius : It is defined as one half of inter-nuclear distance between two nearest metal atoms in a metallic lattice is called metallic radius.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements 3

Question 16.
What is ionic radius ?
Answer:
Ionic radius : It is defined as the effective distance from the nucleus of an ion upto which it has an influence in the ionic bond.

Question 17.
What is lonisation energy (IE) or lonisation potential (IP) ?
Answer:
lonisation energy or lonisation potential : First ionisation energy or simply ionisation energy (IE) is defined as the amount of energy required to remove one valence electron from an isolated neutral gaseous atom resulting in the formation of a monovalent positive ion.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 18.
What is electron affinity ?
Answer:
Electron affinity (EA) : First electron affinity or simply electron affinity is the amount of energy released when one electron is added to a neutral gaseous atom to form a monovalent negative ion.

Question 19.
What is the definition of electronegativity ?
Answer:
Elecironegativity (Pauling’s definition): It is the attractive force which an atom, bonded by a co-valent bond exerts on the bond pair of electrons responsible for the co-valent bond.
The electro-negativity of the elements increase from left to right of the periods in a periodic table.

Question 20.
What are s-block elements ?
Answer:
s-block elements : These are the elements of IA or 1 group (alkali metals; configuration n s1 ) and IIA or 2 group (alkaline earth metals ; configuration n s2 ). These are so named because the last electron in them enters s-oribitals.

Question 21.
What are p-block elements ?
Answer:
p-block elements These are the elements in which the last electron enters p-orbital of valence shell. The elements with configurations n s2 n p1 to n s2 n p6 constitute this block. Thus p-block consists of elements of group IIIA(13), IVA(14), VA(15), VIA (16), VIIA (17) and zero group (18).

Question 22.
What are d-block elements ?
Answer:
d-block elements: These are the elements in which the last electron enters a (n-1) d orbital. These have configuration. (n-1) d1-10 n s0-2. This block is situated in between s and p-block elements. d-block elements are also called transition elements.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 23.
What are f-block elements ?
Answer:
f-block elements: These are the elements in which the last electron enters a ( n-2) f-orbital. Lanthanoids and actinoids constitute the f-block and are also shown separately at the bottom of the periodic table. They have (n-2)f1-14 (n-1) d1-10 n s2 configuration. Elements of f-blocks are also called inner-transition elements.

Question 24.
What are inert gases or noble gases ?
Answer:
Inert gases or noble gases: The elements which belong to the group 18 of the periodic table are called inert gases or noble gases. They have 8 electrons (except He ) in their outermost shell. Therefore, their combining capacity or valency is zero. Hence, they are inert in nature.

Question 25.
What are representative elements ?
Answer:
Representative elements: The elements in which only the outer s or p-electron sub shell is incompletely filled are called representative or normal elements. Elements of s or p-blocks excluding the inert gases are included in this category. These elements have the outer electronic configuration from ns1 to np5.

Question 26.
What are transition elements ?
Answer:
Transition elements: These are the elements in which the last shell is not completely filled. These elements have the configuration (n-1) d1-9 ns1-2 i.e., these elements contain 1-9 electrons in the penultimate shell and 1-2 electrons in the valence shell. The elements having unfilled or partially filled d-orbitals are named transition elements because these present a transition (change) from the most electro-positive elements to the most electro-negative elements.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 27.
What are inner transition elements ?
Answer:
Inner transition elements : Lanthanides \(\left({ }_{53} \mathrm{Ce}-{ }_{71} \mathrm{Lu}\right)\) and actinides \(\left({ }_{90} \mathrm{Th}-{ }_{103} \mathrm{Lw}\right)\) tare collectively known as inner transition elements. In these elements, three outermost shells are not completely filled. The last electron in them enters in the (n-2) subshell.

Question 28.
What is diagonal relationship?
Answer:
Diagonal relationship : Elements of second and third row which are present diagonally have similar properties because on moving along the diagonal, the decrease and increase of size, electropositive character and polarizing power partly cancel each other.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements 4

Question 29.
What is the IUPAC Nomenclature of elements (z >100) ?
Answer:
IUPAC Nomenclature of elements (z >100) : Elements with atomic number beyond 100(z >100) have been named according to IUPAC system, which is connected with the elements atomic number. The first two letters tell us how many hundred (e.g. Un = one) the next three how many tens (e.g. nil = 0 ) the rest units (e.g. quad = 4, pent = 5, hex = 6, hept = 7 and so on).

Question 30.
Ca2+, K+, Cl have 18 extra nuclear electrons. How will they be arranged in the increasing order of size?
Answer:
Explanation : Size of iso-electronic ions decreases with increase in atomic number. Therefore increasing order of size is :
Ca2+< K+< Cl< S2-

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 31.
Explain the reason why the first ionization energy of Na is smaller than that of Mg and second ionisation energy of Mg is smaller than that of Na.
Answer:
Explanation : Electronic configuration of Na and Mg are 1s2 2s2 2p6 3s1 respectively. Therefore magnesium has higher first ionization potential than Na due to its higher value of nuclear charge and more stable configuration (fully filled s-orbital).
After removal of one electron, Na+ and Mg+ ions have configuration 1s2 2s2 2p6 and 1s2 2s2 2s2 p6 3s1 respectively and now it will be easier to remove 3s1 electron rather than electron is lower than that of Na.

Question 32.
Differentiate between covalent radius and Van der Wal’s radius.
Answer:
Difference between Covalent radius and Van der Waal’s radius : Covalent radius is one half of the distance between the centres of nuclei of two similar atoms held together by a purely covalent bond but van der waal’s radius is one half of the inter-nuclear distance between two similar adjacent atoms belonging to two neighbouring molecules of the same substance in the solid state.

Broad Answer Type Questions :

Question 1.
State the different periods with the elements.
Answer:
The different period swith the elements :
WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements 5

Question 2.
What are group and sub-groups?
Answer:
Groups : Mendeleev’s periodic taule shows that the elements with related chemical properties fall in c ne below the other forming vertical columns called groups. There are nine groups from group I to group VIII and 0 (zero). Sub-groups : Each of the groups from I to VII has been divided into two sub-groups A and B.

Question 3.
What are called alkali metals ? In which group do they belong ? State the similarity of their chemical properties.
Answer:
Alkali metals : Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb) and Caesium (Cs). These five metals are generally known as alkali metals. Position in periodic table: The alkali metals are included in the A subgroup of the first group (IA).
Similarity : (i) Alkali metals are strong electro positive.
(ii) The oxides of these metals are strongly basic.
(iii) Each has valency 1.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Question 4.
Where are the inert gas elements placed and why ?
Answer:
Position of inert gas elements in periodic table : He, Ne, Ar, Kr, Xe, and Rn. These are inert elements. They are each gaseous and monoatomic. These are chemically non-reactive. As these elements do not react, these have zero valency.
So they are included in 0 (zero) group of the periodic table.
Besides Helium, all the inert gases have eight (8) electrons in their outermost orbit. There are only 2 electrons in the outermost orbit of Helium.

Question 5.
What are halogen elements ? In which do they belong ?
Answer:
Halogen elements : The four strongly electro-negative non-metals fluorine (F), chlorine (Cl), bromine (Br) and iodine (I) form a family of closely allied elements known as the halogens meaning literally sea salf-producer as these elements react with most metals to form compounds similar to sea-salt, sodium chloride. Position in periodic table: Halogen elements are placed in group VIIB of periodic table. The electro-negative character of the elements increases from left to right in a period. So, the strongly electronegative elements are in the almost extreme right group i.e. in group VIIB of the periodic table.

Question 6.
Discuss the position of hydrogen in periodic table.
Answer:
Position of hydrogen in periodic table :
Reasons for placing in group IA :

  1. Like the alkali metals, hydrogen is electro-positive.
  2. Valency of hydrogen is I, alkali metals are also mon-valent.
  3. During electrolysis of a fused chloride of an alkali metal, the alkali metal deposits at the cathode; in the electrolysis of dilute aqueous solution of hydrogen chloride, hydrogen collects at the cathode.
  4. Like the alkali metals, hydrogen has reducing property and forms stable oxide, like water (H2O).
  5. Like hydrogen, each of the alkali metals, Li, Na, K, Rb, Cs, Fr possesses one valence-electron.

WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements

Reasons for placing in Group VIIB :

(i) Like the halogens, hydrogen is a non-metallic element.
(ii) Valency of hydrogen and that of each halogen is 1.
(iii) Halogen molecules are diatomic, a hydrogen molecule is also diatomic.
(iv) The halogens from metallic halides like, NaCl, KBr, Kl etc. Hydrogen also produces salt-like hydrides like LiH, NaH, CaH9 etc with the strongly electropositive elements and hydrogen here acts as an electro-negative element.
(v) Atoms of halogen elements may transform to negative halide ions by taking one electron each. (e.g. X + e → X ; X = F, Cl, Br, I) and hydrogen atom also may transform to a negative hydride ion by capturing oen electron : H + e → H.
(vi) Like the hologens, hydrogen also can produce covalent comj unds reacting with electro-negative elements like S, O etc.
Due to such varied tendency for occupying a seat in the periodic table, hydrogen is often called a naughty element or a rogue elements.

Question 7.
Draw a diagram by which change of periodic properties are shown.
Answer:
Diagram.
WBBSE Class 10 Physical Science Solutions Chapter 8.1 Periodic Table and Periodicity of the Properties of Elements 6

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 8.6 Organic Chemistry offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 8.6 Question Answer – Organic Chemistry

Very Short Answer Type Questions :

Question 1.
What is the common element of organic compounds ?
Answer:
Carbon is the common element of organic compounds.

Question 2.
What is Catenation ?
Answer:
Catenation is the property of organic compounds where a large number of carbon atom link together.

Question 3.
What is monomer ?
Answer:
Monomer is a small organic molecule that joins with other similar molecules to form a polymer.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 4.
What is mustard gas ?
Answer:
It is a gas produced from ethylene.

Question 5.
What is a primary alcohol ?
Answer:
It is a type of alcohol in which the carbon atom bearing the OH group con­tains at least 2 hydrogen atoms.

Question 6.
What is an unsaturated hydrocarbon ?
Answer:
It is an organic compound in which at least two carbon atoms join by covalent double or triple bond.

Question 7.
What is a saturated hydrocarbon ?
Answer:
It is hydrocarbon in which carbon atoms are linked with single covalent bond.

Question 8.
What is a functional group ?
Answer:
Functional group is a group of atoms present in organic compounds of a class determining almost the common properties of the compounds.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 9.
What type of bonding exists in organic compounds ?
Answer:
Covalent type of bonding exists in organic compounds.

Question 10.
What is DNA ?
Answer:
DNA stands for deoxyribonucleic acid.

Question 11.
What are biomolecules ?
Answer:
These are some organic molecules involved in metabolic process.

Question 12.
What is isomerism ?
Answer:
Isomerism is the property due to which a group of organic compounds of same molecular formula but differ from each other in their structural formulae and properties.

Question 13.
State one use of methane.
Answer:
Use of methane : As a domestic and industrial fuel.

Question 14.
State one use of ethylene.
Answer:
Use of ethylene : To prepare polythene.

Question 15.
State one use of acetylene ?
Answer:
Use of acetylene : To produce benzene, artificial rubber.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 16.
What are the essential two components of amino acids ?
Answer:
The essential two components of amino acids are carboxyl (-COOH) and an amino (-NH2) group attached to the same carbon atom.

Question 17.
Name and write down the formula of a ketone.
Answer:
The name of a ketone is acetone (CH3COCH3).

Question 18.
Which gas burns in a carbide gas lamp ?
Answer:
Acetylene gas burns in a carbide gas lamp.

Question 19.
What change of litmus will occur in alcohol ?
Answer:
Alcohol is a neutral compound, so no change of colour of a litmus paper dipped in it occurs.

Question 20.
What is the functional group of alcohol ?
Answer:
Hydroxyl group (-OH) is the functional group of alcohol.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 21.
Which gas is known as marsh gas ? Why is it called so ?
Answer:
Methane gas is known as marsh gas. It is present in marshy (i.e. always wa­tery) lands, so it is called marsh gas.

Question 22.
Name three hydrocarbons available from coal gas.
Answer:
Methane, ethylene, acetylene are three hydrocarbons available from coal gas.

Question 23.
Give two examples of alkyne with corresponding formulae.
Answer:
Two alkyne :

  • Acetylene (HC ≡ CH)
  • Propyne (CH3– C = C – H)

Question 24.
What is alkane ? Write down its general formula.
Answer:
An alkane is an open-chained saturated hydrocarbon.
The general formula of alkane : CnH2n + n(n – integer)

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 25.
Name and draw structural formula of a saturated organic compound.
Answer:
Saturated organic compound :
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 1

Question 26.
Name and draw structural formula of an unsaturated organic compound.
Answer:
Unsaturated organic compound : Ethylene
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 4

Question 27.
Name and draw structural formula of an aromatic compound.
Answer:
Aromatic compound Benzene
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 3

Question 28.
Name three organic compounds used in daily life.
Answer:
Three organic compounds used in dialy life : Carbohydrates, proteins, glucose.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 29.
Name two plymers.
Answer:
Two polymers are :

  • Polythene
  • Teflon

Question 30.
What is hydrocarbon ? Name the simplest hydrocarbon.
Answer:
Hydrocarbon : A binary compound formed of carbon and hydrogen by covalency is called hydrocarbon.
Simplest hydrocarbon : Methane (CH4).

Question 31.
What is PVC ? State its use.
Answer:
PVC : It is polyvinyl chloride, its monomer is vinyl chloride.
Uses of PVC : To prepare water and rain pipes, toys etc.

Question 32.
Why butane is used in LPG ?
Answer:
Butane is the most convenient fuel for domestic use because both isomers of butane (n-butane and isobutane) are easily liquefied and can be transported in steel cylinders easily.

Question 33.
How does the boiling point of alkane changes ?
Answer:
The boiling point generally increase with increase in the number of carbon atoms.

Question 34.
What is mineral oil ?
Answer:
Oil obtained from petroleum is called mineral oil.

Question 35.
What is fire damp ?
Answer:
The gas responsible for explosions in coal mines is methane and is also known as fire damp.

Question 36.
Why are the alkenes more reactive than alkanes ?
Answer:
The chemical reactivity of alkenes are mote than alkanes due to presence of π-electron in alkenes.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 37.
How many s and p bonds are present in ethylene ?
Answer:
Ethylene has one π-bond and five σ-bonds.

Question 38.
Why the hydrogens of alkynes are acidic in nature ?
Answer:
Hydrogen has appreciable acidic character when bond to a carbon atom of a(C = C) triple bond (in case of alkyne).

Question 39.
What are the end products of combustion of hydrocarbons ?
Answer:
End products of combustion of all hydrocarbons are carbon dioxide and water.

Question 40.
What is the most oxidised form of hydrocarbon ?
Answer:
RCOOH (R = alkyl group) represents the most oxidised form of hydrocarbons.

Question 41.
What is sabatier and senderens reaction ?
Answer:
Hydrogenation of unsaturated aliphatic hydrocarbons in presence of nickel catalyst is called sabatier and senderens reaction.

Question 42.
What is the condition for halogenation of alkanes ?
Answer:
Halogenation of alkanes does not occur in absence of light or heat.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 43.
What is aldose ?
Answer:
Carbohydrate containing aldehyde WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 5 group is called aldose.

Question 44.
What is ketose ?
Answer:
Carbohydrate containing keto WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 6 group is called ketose.

Question 45.
Give an example of a monosaccharide.
Answer:
Glucose (C6H12O6)

Question 46.
Give an example of polysaccharide.
Answer:
Starch.

Question 47.
What is pyrofax ?
Answer:
Propane is used as a fuel under the name pyrofax.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 48.
What is an artificial camphor ?
Answer:
Hexachloroethane is called artificial camphor.

Question 49.
What is TEL ?
Answer:
TEL is commonly used antiknock compound tetraethyl lead, [(C2H5)4 Pb].

Question 50.
What is AK – 33X ?
Answer:
To avoid lead pollution, a new compound called AK – 33X (cyclopentadienyl manganese carbonyl) is used as an antiknock these days.

Question 51.
What is dextrose ?
Answer:
Glucose is called grape sugar or dextrose.

Question 52.
What is fruit sugar ?
Answer:
Fructose is called laevolase or fruit sugar and is the sweetest sugar.

Question 53.
What is milk sugar ?
Answer:
Lactose is known as milk sugar.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 54.
What is amylum ?
Answer:
Starch is called amylum.

Question 55.
State an example of globular protein.
Answer:
Insulin is an example of globular protein.

Question 56.
What is nucleotide ?
Answer:
Both DNA and RNA are polymers of a basic repeating unit, called a nucle­otide.

Question 57.
What is the sugar present In RNA ?
Answer:
Sugar present in RNA is ribose.

Question 58.
What is the sugar present in DNA ?
Answer:
Sugar present in DNA is 2-deoxyribose.

Question 59.
What are the main functions of nucleic acids ?
Answer:
Main functions of nucleic acids are the direct synthesis of protein in living cells and transference of genetic information.

Question 60.
Who discovered the double helix structure of DNA ?
Answer:
The double helix structure of DNA was proposed by Watson and Crick (1953).

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 61.
What is called gene ?
Answer:
DNA sequence that acts as a code for a specific protein or a polypeptide is called gene.

Question 62.
What is called an anabolism ?
Answer:
Anabolism is the process in which macro molecules are synthesised by the cell.

Question 63.
What is called catabolism ?
Answer:
Catabolism is the process in which macro molecules break into smaller ones.

Question 64.
What is metabolism ?
Answer:
Metabolism is the combination of anabolism and catabolism.

Question 65.
What is proteases ?
Answer:
Proteases is the breakdown of large protein molecules into peptides.

Question 66.
Who converts peptides into amino acids ?
Answer:
Peptidases convert peptides into amino acids.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 67.
What is soda lime ?
Answer:
Soda lime is a mixture of sodium hydroxide and calcium oxide (NaOH + CaO).

Question 68.
What is the source of acetylene ?
Answer:
The source of acetylene is calcium carbide (CaC2) from which it is obtained by the action of water at ordinary temperature.
CaC2 + 2H2O → C2H2 + Ca(OH)2.

Question 69.
Name one biodegradable polymer.
Answer:
One biodegradable polymer is cellulose.

Question 70.
Name the hydrocarbon formed when ethanol is heated with concentrated H2SO4
Answer:
When ethanol is heated with concentrated H2SO4, ethene (or ethylene) is formed.

Question 71.
What is the common name of ethyne?
Answer:
The common name for ethyne is acetylene.

Question 72.
What is the term used for the compounds which have same molecular formula but different structures?
Answer:
They are called isomers.

Question 73.
What is next homologue of C2H5OH?
Answer:
The next homologue of C2H5OH (Called ethanol or ethyl alcohol) is C2H7OH (propanol).

Question 74.
How is biopolymer protein biodegraded in nature?
Answer:
Proteins are degraded by some specific enzymes produced by some particular Phy. Sc.bacteria (e.g. Bacillus subtilis, Bacillus mycoides, Bacillus vulgaris etc) through ammonification.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 75.
How many hydrogen atoms are there in alkyne containing three carbon atoms?
Answer:
The general formula of alkyne is CnH2n-2
Here, n = 3
∴ Number of hydrogen atoms = 2n-2 = (2×3)-2=4

Question 76.
What is power alcohol?
Answer:
In countries having shortage of petrol, a mixture of petrol (80%), ethanol (20%) and benzene is used as fuel. This petrol mixed with alcohol is known as power alcohol.

Question 77.
Name two functional groups one containing carbon atom and the other devoid of carbon.
Answer:
Carbon containing functional group : carboxyl
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 26
Functional group devoid of carbon : Amino
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 27

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 78.
Give the structural formula of propanal.
Answer:
CH3 – CH2 -CHO

Short Answer Type Questions :

Question 1.
What is organic chemistry ?
Answer:
Organic chemistry : All carbon containing compounds except oxides of carbon, metal carbonate, bicarbonate, hydrogen cyanide and metallic cyanides are organic compounds and the chemistry of organic compounds is called organic chemistry.

Question 2
What is Biochemstry ?
Answer:
Biochemistry : It is the branch of chemistry which deals with the com¬position and behaviour of living system.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 3.
What are Biomolecules ?
Answer:
Biomolecules : These are the macro-molecules like carbohydrates, amino acids, proteins, vitamins, fats, RNA, DNA etc., which are essential for the continuation of life process of biological species like plants and animals (including human beings) are called biomolecules.

Question 4.
What are Carbohydrates ?
Answer:
Carbohydrates These are the polyhydroxy aldehydes or ketones or these are the compounds which on acidic hydrolysis give polyhydroxy aldehydes or ketones. Their molecules are composed of carbon, hydrogen and oxygen. Carbohydrates are mainly used as food which give us energy to work.

Question 5.
What are Monosaccharides ?
Answer:
Monosaccharides : These are the compounds such as glucose and fructose which do not break into more simpler compounds on hydrolysis.

Question 6.
What are Disaccharides ?
Answer:
Disaccharides : These are the compounds which on hydrolysis give two monosaccharide units.
[Sucrose, Moltose and lactose (milk sugar)]
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 7

Question 7.
What are Polysacharides ?
Answer:
Polysaccharides : These are the compounds which on hydrolysis produce very large number of monosaccharide molecules. Cellulose and starch are two important examples of polysaccharides.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 8

Question 8.
What is Moltose ?
Answer:
Moltose : It is obtained by the partial hydrolysis of starch by enzyme diastase present in malt.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 9.
What is Lactose ?
Answer:
Lactose : It occurs in the milk of all animals cow’s milk contains about 5% Lactose while human milk contains about 7% lactose.

Question 10.
What is Cellulose ?
Answer:
Cellulose It is the main structural material of wood and other plants. Cotton is about 80% cellulose.

Question 11.
What is Glycogen ?
Answer:
Glycogen : It is a higher sugar stored in liver. In case of emergency, this gets hydrolysed to glucose which then gets oxidised to produce energy.

Question 12.
What are Amino acids ?
Answer:
Amino acids : These are organic compounds containing both amino and carboxylic group in their molecules. They are represented by the general formula.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 28

Question 13.
What are essential amino acids ?
Answer:
Essential amino acids : Human body can synthesis 10 out of 20 a-amino acids found in proteins. The remaining 10 must be present in our diet and are called essential amino acids.

Question 14.
What is Zwitter ion ?
Answer:
Zwitter ion In aqueous solution, the acidic carboxyl group donates a proton to the basic amino group to form an internal salt called a dipolar ion or zwitter ion. Although it is neutral overall, it contains both a positive and a negative charge.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 10

Question 15.
What are Peptides ?
Answer:
Peptides These are condensation products of self-amides formed by the reaction of two or more amino acid molecules.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 16.
What is peptide linkage ?
Answer:
Peptide linkage : The bond
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 11
between the carboxyl group of an acid molecule and nitrogen of the other amino acid molecule is known as the peptide bond or peptide linkage.

Question 17.
What are Proteins ?
Answer:
Proteins These are complex nitrogeneous substances present in all forms of living matter. These are obtained by the condensation polymerisation of a-amino acids through the formation of peptide bonds. Actually proteins are polypeptides with very high molecular weights (more than 10,000). These are formed by the combination of more than 100 molecules of amino acids.

Question 18.
What is Denaturation ?
Answer:
Denaturation : After coagulation, proteins lose their physiological activity and certain other properties. This phenomenon is known as denaturation.

Question 19.
What are Enzymes ?
Answer:
Enzymes : These are also proteins which acts as catalyst in many bio-chemical reactions. Enzymes are specific in action due to their specific structural arrangement.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 20.
What is the biuret test ?
Answer:
Biuret test : To alkaline solution of proteins add a dilute solution of CuS04. Formation of violet colour confirms proteins.

Question 21.
What are Vitamins ?
Answer:
Vitamins : These are the biomolecules needed in small quantities, that regulate many biochemical function and prevent the development of many deficiency diseases.

Question 22.
What are Fats ?
Answer:
Fats : These are organic compounds composed of carbon, hydrogen and oxygen. They are made of glycerol and fatty (organic) acids. Fats may be of animal or vegetables origin.

Question 23.
What are the types of simple proteins ?
Answer:
Types of simple proteins :

  • Albumins — e.g. milk, serum etc.
  • Globumins — e.g. egg yolk, tissues etc.
  • Glutemins — e.g. wheat, rice etc.
  • Prolamins — e.g. barli, wheat etc.
  • Scleroproteins — e.g. keratin, fibroin etc.

Question 24.
State examples of conjugated proteins and derived proteins.
Answer:
Conjugated proteins :
(a) Phosphoproteins
(b) Glycoproteins
Derived proteins : This type of protein is obtained after the partial hydrolysis of protein of very high molecular weight by acid, base or enzyme to simpler proteins.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 25.
What is DNA ?
Answer:
DNA : It is deoxyribonucleic acid. DNA molecule contains three different chemical constituents :
(a) Phosphoric acid
(b) Deoxyribose
(c) Pyridine like base adenine and guanine, pyrimidine like base thymine and cytosine. DNA consists of two strands of polynucleotides coiled around each other by hydrogen bond in the form of a double helix.

Question 26.
What is RNA ?
Answer:
RNA : It is consisted by ribose along with phosphoric acid and bases. In pyrimidine base uracil is present instead of thymine in RNA. RNA carries the message of DNA and acts accordingly. Mainly there are three types of RNA :

  • m-RNA
  • t-RNA
  • r-RNA

t-RNA and m-RNA have taken an important role in the protein synthesis.

Question 27.
What is Catenation ?
Answer:
Catenation : The property by virtue of which carbon forms covalent linkage chains is called catenation.

Question 28.
What is Functional group ?
Answer:
Functional group An atom or a group of atoms which being present within the molecule of an organic compound, causes the compound to function or behave chemically in a particular way is called a functional group, e.g. alcohol (-OH), aldehyde (-CHO)

Question 29.
What are Hydrocarbons ?
Answer:
Hydrocarbons : These are organic compounds containing only carbon and hydrogen atoms in their molecules. They are divided into two classes.

  • Saturated hydrocarbons
  • Unsaturated hydrocarbons.

Question 30.
What are saturated hydrocarbons ?
Answer:
Saturated hydrocarbons : Saturated hydrocarbons are those in whose molecules the carbon atoms are joined to each other by single bond and the remaining valencies of the carbon atoms are satisfied by hydrogen atoms. e.g. Methane (CH4), Ethane (C2H6).

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 31.
What are unsaturated hydrocarbons ?
Answer:
Unsaturated hydrocarbons :
The compounds in which carbon atoms are connected among themselves by double bond or triple bond are called unsaturated hydrorcarbons.

Question 32.
What are Alkenes ?
Answer:
Alkenes Those unsaturated hydrocarbons containing double bond (=) between two adjacent carbon atoms in their molecules are called as alkenes.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 12

Question 33.
What are Alkynes ?
Answer:
Alkynes : Those unsaturated hydrocarbons containing triple bonds (s) be­tween two adjacent carbon atoms in their molecules are called as alkynes.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 13

Question 34.
What are Cyclic compounds ?
Answer:
Cyclic compounds : Organic compounds containing closed of atoms are called cyclic compounds. Cyclic compounds are of two types,

  • Homocyclic compounds
  • Heterocyclic compounds.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 35.
What are Homocyclic compounds ?
Answer:
Homocyclic compounds These are the compounds where all the atoms present in the cyclic compounds in the series are carbon atoms. These are of two types,

  • Aromatic compounds
  • Alicyclic compounds.

Question 36.
What are Aromatic compounds ?
Answer:
Aromatic compounds Aromatic compounds are those which have a hexagonal ring structure formed by six carbon atoms bonded each other by a single or double bonds.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 14

Question 37.
What are Alicyclic compounds ?
Answer:
Alicyclic compounds : These compounds are the cyclic or ring compounds without having any benzene ring.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 15

Question 38.
What are Heterocyclic compounds ?
Answer:
Heterocyclic compounds : Sometimes N, O, S atoms are linked along with carbon atoms in series during the formation of closed ring structure. These compounds are called heterocyclic compounds.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 16

Question 39.
What are Homologous series ?
Answer:
Homologous series A homologous series may be defined as a group of compounds in which various members

  • have similar chemical properties
  • can be represented by the same general formula
  • possess the same functional group the different members of homologous series are called homologues and the phenomenon is called homology.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 40.
What is Isomerism ?
Answer:
Isomerism In organic chemistry when the same molecular formula rep­resents two or more compounds which differ in their physical and chemical properties, then such compounds are called isomers and the phenomenon is called isomerism.

Question 41.
What are Structural isomers ?
Answer:
Structural isomers : Isomers that have same molecular formula but differ in the arrangement of atoms within the molecule are known as structural isomers.

Question 42.
What are Stereoisomers ?
Answer:
Stereoisomers The isomers that have the same constitution but different in the spatial arrangement of their atoms are called stereoisomers.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 43.
What are the types of structural isomers ?
Answer:
(a) Chain isomerism :
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 17
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 18

Question 44.
What is the “will-o-the wisp”?
Answer:
In marshy lands and in sewage sludge, methane gas is set free by the bacterial decomposition of vegetable matters under water. Methane itself is highly inflammable. Moreover, a little phosphine (PH3) and diphosphine (P2H4) accompany the gas and make it spontaneously inflammable. Thus the ‘will-o-the-wisp’ results.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 45.
Draw position isomers of butene(C4H8)
Answer:
Position isomers of butene (C4H8)are shown below :
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 19

Question 46.
Name an organic compound which is
(a) used to illuminate country houses
(b) used to make polythene
(c) poisonous containing -OH group
(d) consumed as a drink
Answer:
(a) Acetylene
(b) Ethene
(c) Methanol
(d) Ethanol.

Question 47.
Which gas is produced when sodium carbonate reacts with acetic acid ?
Answer:
Acetic acid decomposes sodium carbonate to liberate CO2
2CH3COOH + Na2O2CO3→ 2CH3COONa + CO2↑+ H2O

Question 48.
How can traces of ethylene mixed with methane be identified?
Answer:
When the gas mixture is passed through reddish-brown solution of bromine in CCl4the solution becomes colourless due to the formation of colourless 1, 2-dibromoethane if the gas mixture contains ehhylene.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 20
This is a test for detecting ethylene and ethylenic unsaturation in the organic compounds.

Short Answer Type Questions 

Question 1.
State the difference between organic and inorganic compounds.
Answer:
Difference between organic and inorganic compounds :

Organic Compounds

  • The catenation property of carbon atoms among themselves and other atoms give rise to limidess number of organic compounds.
  • They form covalent bonds.
  • They have low melting and low boiling point.
  • These are not ionised at all.
  • They show isomerism.
  • They are soluble in organic solvents like alcohol, ether etc.

Inorganic compounds

  • Due to absence of catenation property of its atoms the number of inorganic compounds formed by the rest of the elements is not so large .
  • Inorganic compounds may be covalent or electrovalent.
  • They have high melting and boiling point.
  • Most of the inorganic compounds are electrolytes.
  • They have high rate of reaction.
  • Very few Inorganic compounds show isomerism.
  • They are soluble in ionising solvent like water, but are insoluble in organic solvent.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 2.
State the following facts of Methane.
(i) Source/preparation
(ii) Important reaction
(iii) Uses
(iv) Structure
Answer:
The following facts of methane :
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 22

Question 3.
State the following facts of Ethylene.
(i) Source/preparation
(ii) Important reaction
(iii) Uses
(iv) Structure
Answer:
The following are of ethylene.
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 23

Question 4.
What are Polymers and what are Monomers. Give few examples.
Answer:
Polymer : Under suitable conditions, many molecules of alkene and alkyne combine together forming a very high molecular weight compound known as polymer. The small moleculers are known as monomers and this process is known as polymerisation.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Some common polymers
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 24

Question 5.
What are Biodegradable and Non-biodegradable materials.
Answer:
Biodegradable materials : Materials of vegetable and animal origin invariably decay and decompose into CO
2, H2O, N2 or NH3 by the com­bined action of sometimes of natural agencies like air, water, subshine etc. These are called biodegradable materials.

Non-biodegradable materials : Synthetic materials like plastics, polythene, teflon, PVC etc, are not decomposed by these natural agencies even on for a long time. These are called non-biodegradable materials.

Question 6.
State the following facts of acetylene.
(i) Source/preparation
(ii) Important reaction
(iii) Uses
(iv) Structure
Answer:
The following facts of acetylene
WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry 25

Question 7.
Discuss about the hazards of using Polyethylene, Teflon and PVC.
Answer:
Explanation : Polyethylene, teflon and PVC are non-biodegradable material. These substances do not decompose or decay naturally on long standing. These articles are usually left in soil or water. This practice causes pollution of soil, water and air.

(a) Solid wastes containing PVC when incinerated (consumed by fire) produce the highly toxic substance dioxin. Soil contains two main bios, detrivore and decomposer which analyse or detoxify many bio-degradable substances.

(b) But the polyethylene, teflon, PVC etc. are not bio-digradable substances, they rather spoil detrivore and decomposer of soil. In presence of air and water these synthetic substances excreate some poisonous chemicals in top soil spoiling the minerals necessary for plants.

WBBSE Class 10 Physical Science Solutions Chapter 8.6 Organic Chemistry

Question 8.
State possible alternatives to avoid the hazards due to polythene materials.
Answer:
Possible alternatives : The possible alternative that can eliminate the hazards of using polythene is to restrict their uses to a minimum extent. In places of polythene bags for carrying purchases, bags made of cloth, paper or jute should be used. Paper or dry leaf packets should be used instead of polythene packets to carry sweets and other edible solid materials.

The use of polythene glass, tea-cups etc. should be discontinued, and those made of hard paper or earthenwares should be used instead. Polythene sheets for covering should be replaced by thin terpaulin, waxed paper or tar-laminated papers, jute cloth etc.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 6 Current Electricity offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 6 Question Answer – Current Electricity

Very Short Answer Type Questions :

Question 1.
What is electric current?
Answer:
Electric current is general means a continuous flow of electrons, ions or any electrically charged particles through a medium.

Question 2.
What are conductors ?
Answer:
The substances through which electric charge flows easily are known as conductors.

Question 3.
What are non-conductors ?
Answer:
The substances which do not allow electric charge to pass through them are called non-conductors or insulators.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 4.
How can we define electric current in a quantitative way ?
Answer:
In a quantitative way we can say that the rate of flow of electric charge through any cross section of a conductor per unit time is called the electric current.

Question 5.
What do you mean by D.C. ?
Answer:
If the electric current always flow in the same direction then the current is called D.C. (Direct Current.)

Question 6.
What do you mean by A.C.?
Answer:
If the direction of current alternates i.e. changes periodically from one direction to opposite direction then the current is called A.C. (Alternating current).

Question 7.
What is electric cell?
Answer:
A device in which electrical energy is obtained from the chemical energy is known as electric cell.

Question 8.
What is negative electrode?
Answer:
The metal rod in which there is excess of negative charge is called negative electrode of cell.

Question 9.
What is positive electrode?
Answer:
The metal rod in which there is excess of positive charge is called positive electrode.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 10.
What is open circuit?
Answer:
When the electrodes of a cell are not connected by a conductor, externally, the cell is said to be in open circuit.

Question 11.
What is close circuit?
Answer:
When the electrodes are connected internally with a conductor, the cell is said to be in closed circuit.

Question 12.
What is emf of a cell?
Answer:
The potential difference between the electrodes of a cell in open circuit is called emf (electromotive force).

Question 13.
What is the direction of current in the external circuit of a cell?
Answer:
Current flows from positive electrode to negative electrode in the external circuit of a cell.

Question 14.
What is the emf of a simple voltaic cell?
Answer:
The emf of a simple voltaic cell is 1.08 volt.

Question 15.
What is the physical nature of electromotive force?
Answer:
Electromotive force is some energy.

Question 16.
What is the unit of charge in SI system?
Answer:
The unit of charge in the SI system is coulomb.

Question 17.
What is the SI unit of current strength?
Answer:
The SI unit of current strength is ampere.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 18.
What are the units of electromotive force and potential difference in SI system?
Answer:
The SI unit of both electromotive force and potential difference is volt.

Question 19.
What is the SI unit of resistance?
Answer:
The SI unit of resistance of Ohm.

Question 20.
State the cases in which Ohm’s law is not valid.
Answer:
Ohm’s law is not valid for current flowing through gases under low pressure, electrolytes and semi-conductors.

Question 21.
How does the resistance of a conductor depend on the cross section of the conductor?
Answer:
Resistance of the conductor decreases with increase of cross section.

Question 22.
In a system of resistors connected in parallel, how is the magnitude of the equivalent resistance related to the magnitude of the resistor of least value?
Answer:
Equivalent resistance is smaller than the resistance of smallest resistor.

Question 23.
How is the heat generated due to electric current through a resistor related to the strength of current?
Answer:
Heat is directly proportional to square of current.

Question 24.
How is the heat generated due to electric current in a resistor related to resistance of the resistor.
Answer:
Heat is directly proportional to resistance.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 25.
What is 1 B.O.T.?
Answer:
The total electric energy expended in 1 hour at the rate of 1 kilowatt is known as 1 B.O.T. (Board of Trade unit).

Question 26.
Which energy is converted to which other in an electric motor?
Answer:
In an electric motor, electric energy is converted to mechanical energy.

Question 27.
What is earthing?
Answer:
Earthing : It means to connect the metal case of an electrical appliance to the earth with the help of a metal wire.

Question 28.
What is the usual colour of a live wire?
Answer:
The usual colour of a live wire is red.

Question 29.
What is the usual colour of a neutral wire?
Answer:
The usual colour of a neutral wire is black.

Question 30.
What is the usual colour of the earthing wire?
Answer:
The usual colour of the neutral wire is green.

Question 31.
Which effect of electric current is demonstrated in an electromagnet?
Answer:
Magnetic effect.

Question 32.
How is current related to potential difference?
Answer:
Current is directly proportional to potential difference.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 33.
What is the unit of resistivity in CGS system?
Answer:
The unit of resistivity in CGS system is Ohm-Cm.

Question 34.
Current flows through a conductor from east to west, in which direction electrons flow through it?
Answer:
Electrons flow from west to east.

Question 35.
What is an electromagnet?
Answer:
An electromagnet is a temporary magnet produced by passing electtric current through an insulated copper wire coiled around a soft iron body.

Question 36.
What is the core of an electromagnet?
Answer:
The soft iron body around which an insulated copper wire is coiled is called the core of the electromagnet.

Question 37.
Mention two uses of electromagnet.
Answer:
Electromagnet is used in electric calling bell, electric motor.

Question 38.
What happens if the soft iron used in the electromagnet of an electric bell be replaced with steel?
Answer:
The bell will not work, since the steel body turns to a permanent magnet.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 39.
How does the strength of an electromagnet depend on the nubmer of turns of a solenoid?
Answer:
Number of turns of insulated wire per unit length of a solenoid increases strength of electromagnet.

Question 40.
What is the function of an electric switch?
Answer:
Electric switch is a very common and simple device used to stop or allow flow of electric current to an electric current to an electric appliance as and when needed.

Question 41.
What is voltmeier?
Answer:
A voltmeter is used to measure the potential difference between two points in a section of an electrical circuit.

Question 42.
What is an ammeter?
Answer:
An ammeter is used to measure the current flowing through an electric circuit.

Question 43.
What should be the resistance of an ideal ammeter?
Answer:
Ideally the resistance of ammeter should be zero.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 44.
What should be the reisstance of an ideal volmeter?
Answer:
Ideally the resistance of voltmeter should be infinity.

Question 45.
What are the metals used to prepare fuse wire?
Answer:
Usually fuses consist of fine wires made of an alloy of lead (75%) and a small amount of tin (25%).

Question 46.
What is the rating of a fuse wire?
Answer:
The maximum current that may be allowed to flow through fuse wire before it melts is called the rating of fuse wire.

Question 47.
What is the nature of electric charge on an ebonite rod, when it is rubbed with cut skin?
Answer:
Negative charge.

Question 48.
A glass rod is rubbed with silk. What type of charges do they acquire?
Answer:
Glass is positively charged and the silk is negatively charged.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 49.
What is the SI unit of electric charge ?
Answer:
Coulomb (C)

Question 50.
What is the least value of electric charge available?
Answer:
1.6 × 10-19C

Question 51.
How many electrons will have a total charge of one coulomb?
Answer:
Number of electrons = \(\frac{q}{e}\) = \(\frac{1}{1.6 \times 10^{-19}}\) = 6.25 × 1018.

Question 52.
Name two basic properties of electric charge.
Answer:
(i) The electtric charge of an isolated system is conserved and (ii) the electric charge is quantised.

Question 53.
What is meant by conservation of charge?
Answer:
It means that the total electric charge in an isolated system remains constant.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 54.
How is the mass of a body affected on charging?
Answer:
The charging is due to transfer of electrons from one body to another. So, mass increases in the case of negatively charged body and decreases in the case of positively charged body.

Question 55.
In coulomb’s law, on what factors does the value of electrostatic force constant K depend?
Answer:
It depends upon the system of units used and on the interverning medium.

Question 56.
Electrostatic force between two charges is called central force. why?
Answer:
It acts along the line joining the centres of the charges, and hence, it is a central force.

Question 57.
Does a dipole experience a force, when placed in the non-uniform plectric field?
Answer:
Yes, it experiences a force.

Question 58.
What is electric flux ? Write its SI unit?
Answer:
The electric flux linked with a surface is the total number of lines passing through the surface. Its SI unit is Nm2 C-1.

Question 59.
What is electric potential?
Answer:
The electric potential at a point is the work done in bringing a unit positive charge from infinity to that point in the electric field.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 60.
What is electric potential energy?
Answer:
The electric potential energy is the work done in bringing charges from infinity to their respective positions to form the system.

Question 61.
Define electric potential difference between two points in an electric field.
Answer:
The electric potential difference between two points in an electric field is the work done in bringing a unit positive charge from one point to the other point.

Question 62.
Illustrate a condition in which electric field is zero but potential is not zero.
Answer:
Electric field inside a hollow charged sphere is zero but potential is not zero.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 63.
Illustrate a condition in which electric field is not zero but potential is zero.
Answer:
The electric field on the equatorial line of an electrical dipole is not zero but the potential is zero.

Question 64.
Define the unit of electric unit.
Answer:
The unit of current is ampere. It is the one coulomb of charge flowing through a conductor in one second.

Question 65.
What is the direction of conventional current?
Answer:
It is the direction opposite to the flow of electrons in a conductor under the influence of electric field.

Question 66.
What is current density?
Answer:
It is the current flowing per unit area of a conductor.

Question 67.
What is the direction of current density?
Answer:
Its direction is same as that of applied electric field.

Question 68.
What is meant by steady current?
Answer:
A current whose magnitude does not change with time.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 69.
What is meant by varying current?
Answer:
A current whose magnitude changes with time.

Question 70.
What is drift velocity?
Answer:
It is the average velocity with which a free-electron gets drifted in an conductor under the influence of an external electric field applied to the conductor.

Question 71.
How does the drift velocity of electrons in a metallic conductor vary with temperature?
Answer:
Drift velocity of a metal decreases with increase in temperature.

Question 72.
What do you mean by relaxation time or mean free time?
Answer:
It is average time interval between the two successive collisions between electron and ion in a conductor.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 73.
On what factors does the resistance of the conductor depend?
Answer:
The resistance varies with length of the conductor and inversely proportional to area of cross section of the conductor and increases with temperature of the conductor.

Question 74.
What is the difference between Ohmic and Non-Ohmic conductors?
Answer:
Ohmic conductors strictly obey Ohm’s law while Non-Ohmic conductors do not obey Ohm’s law.

Question 75.
What do you mean by conductivity of a material? Give its SI unit.
Answer:
It is the reciprocal of resistivity. Its SI unit is mho m-1.

Question 76.
What is CFL ?
Answer:
CFL is compact fluorescent lamp.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 77.
What is LED ?
Answer:
LED is light emitting diode.

Question 78.
What is the value of EMF of a voltaic cell and that of a Leclanche cell ?
Answer:
The value of EMF of a voltaic cell is 1.08 volt and that of Leclanche cell is 1.5 volt.

Question 79.
What is the value of permitivity of vacuum?
Answer:
Its value is, ε0 = 8.854 × 10-12 C2 N-1 m-2

Question 80.
What do you mean by EMF of a cell?
Answer:
EMF of a cell may be defined as the energy spent (or work done) per unit charge in taking one unit positive charge around the complete circuit.

Question 81.
What do you mean by internal resistance of a cell?
Answer:
The resistance offered by a cell to the flow of current when it passes through it, is called the internal resistance of that cell. It is generally represented by r.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 82.
Name of metal which behaves as a superconductor?
Answer:
Mercury below 4.2 K behaves as a superconductor.

Question 83.
Mention one use of superconductivity?
Answer:
Superconductons are used to construct very strong magnets.

Question 84.
Which effect of current is utilized in an electric bulb?
Answer:
The heating effect of current is utilized in an electric bulb.

Question 85.
Is Joule heating a reversible effect?
Answer:
The phenomenon of joule heating is an irreversible one-change of the direction of current does not reverse the heating effect.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 86.
Why are coils of electric toasters and irons made of an alloy rather than a pure metal ?
Answer:
The resistivity of an alloy is higher than the pure metal. Moreover, at higher temperature, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and irons are made of an alloy rather than a pure metal.

Question 87.
When an electric bulb is connected to a 12 V battery, it draws a current of 1A. What is the power of the bulb?
Answer:
Power of the bulb, P = 12 V × 1 A = 12 W

Question 88.
What is the full form of EER?
Answer:
The full form of EER is Energy Efficiency Ratio.

Question 89.
How can you define EER, say, for an air conditioner?
Answer:
Energy Efficiency Ratio (EER) for an air conditioner can be defined as follows: EER of AC = cooling capacity (in watt)/Power Consumption (in watt)

Question 90.
Show that, potential difference × electric current = electric power
Answer:
Electric power is the electric work done per unit time.
So, power P = Work done/time or, P = W/t
Putting W = V.Q = Vlt, we have, P = \(\frac{\mathrm{VIt}}{\mathrm{t}}\) = VI
∴ Electric Power (P) = Potential difference (V) × Electric current (I).

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 91.
Which effect of current can be utilized in detecting a current carrying wire concealed in a wall?
Answer:
The electromagnetic effect can be utilized for the purpose.

Question 92.
Which rule is known as Motor rule?
Answer:
Fleming’s left hand rule is known as Motor rule.

Question 93.
Who first showed that a momentary electric current is set up in closed coll of wire by moving it near a magnet or in any other magnetic field?
Answer:
Michael Faraday (in 1831)

Question 94.
What type of energy transfer takes place in a dynamo?
Answer:
A dynamo converts mechanical energy into electrical energy and generates DC.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 95.
Mention a way to increase the magnitude of e.m.f generated in a generator.
Answer:
The magnitude of e.m.f. generated can be increased by increasing the strength of the magnetic field in the generator.

Question 96.
For a tube light, which circuit should be used?
Answer:
Since a tube light consumer less power, so it should be connected to the lighting circuit (5A).

Question 97.
For an air conditioner, which circuit should be used?
Answer:
Since an air conditioner consumes more power, so it must be connected to the 5 A power circuit.

Short answer type questions :

Question 1.
What do you mean by potential difference?
Answer:
Potential difference It is the electrical condition of a point in an electric field or on a current carrying conductor that indicates whether electrons will flow from it or it from another connected point.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 2.
What is the definition of coulomb ?
Answer:
Coulomb : It is the quantity of electric charge that passing through silver nitrate solution deposits 0.001118 g silver at the cathode.

Question 3.
What is the definition of ampere?
Answer:
Ampere : It is the current that flowing through silver nitrate solution for one second during electrolysis deposits 0.001118 g silver at the cathode.

Question 4.
What is the definition of ohm ?
Answer:
Ohm : If one ampere current flowing through a conductor establishes 1 volt potential difference between the two ends of the conductor, the resistance of the conductor is one ohm.

Question 5.
State Ohm’s law.
Answer:
Ohm’s law (1826) : The temperature and other physical conditions remaining constant the current flowing between any two points of a conductor is proportional to the potential difference between them.

Question 6.
Deduce the mathematical form of Ohm’s law.
Answer:
Mathematical form of Ohm’s law :
Let VA and VB are the potentials at the ends A and B of the conductor A B respectively.
So, the potential difference between the points is VA – VB=V (say).

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 1

Now, if current I flows through the conductor, then following ohm’s law, V α I
or, V = R I(R= constant, the resistance of conductor )
or, \(\frac{V}{I}\) = R

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 7.
What is the definition of resistance from Ohm’s law ?
Answer:
Definition of resistance from Ohm’s law : The resistance of a conductor is a ratio of the potential difference between its ends to the current flowing through it.

Question 8.
On what factors does the resistance of a conductor depend ?
Answer:
Factors upon which the resistance of a conductor depends :
(i) Effect of length : Temperature, material and area of cross section remainning constant, the resistance (R) of a conductor is proportional to its length (1)
∴ R ∝ l (when temperature, material and cross-section are constant)
(ii) Effect of cross-section : Temperature, material and length remaining constant the resistance (R) of a conductor is inversely proportional to its area of cross-section (A)
∴ R ∝ \(\frac{1}{A}\) (When temperature, material and length are constant)

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 9.
What is specific resistance ?
Answer:
Specific resistance : We know, R ∝ l when A is constant
R ∝ \(\frac{1}{A} \) when l is constant
When both length and the area of cross-section of a conductor vary, then from the law of joint variation, we can write,
R ∝ \(\frac{1}{A}\)
or, R = ρ \(\frac{1}{A}\) [ρ (rho) is the constant of proportionality and is known as the specific resistance of resistivity]
Now, if l = 1 and A = 1, then R = ρ

Definition: The specific resistance or resistivity of a material is numerically equal to the resistance of a conductor of the material of length 1 metre and area of cross-section 1 m2.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 10.
Specific resistance of copper 20°C is 1.6 × 10-6 ohm-cm, what do you mean by it ?
Answer:
Specific resistance of copper 20°C is 1.6 × 10-6 ohm-cm}, it means: Resistance across the opposite faces of a copper cube of 1 cm side, at 20° C is 1.6 × 10-6 ohm.

Question 11.
What do you mean by combination of resistance ?
Answer:
Combination of resistance : In different electrical circuit more than are resistance are connected together. This is known as combination of resistances. Usually two types of combination are used-
(i) Series combination
(ii) Parallel combination

Question 12.
Define equivalent resistance.
Answer:
Equivalent resistance : The single resistance, instead of multiple resistance in a circuit, keeps the voltage and current unchanged, is called equivalent resistance of those resistances.

Question 13.
What do you mean by series combination?
Answer:
Series combination: In this combination resistances are so connected that extreme end of one resistance is joined to the begining end of next resistance and so on. In this connection same current flows through all the resistances. If three resistors r1 r2 r3 connected in series, the same current I passes through each then their equivalent resistance R will be.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 2

Question 14.
What do you mean by parallel combination?
Answer:
Parallel combination : A number of resistors are said to be connected in parallel when they are placed side by side and their corresponding ends joined together so that the main current is distributed among them.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 3

If the individual resistances is parallel combination are r1 r2 r3 then their equivalent resistance R is given by
\(\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\)
So, to create a low resistance out of few relatively high resistances, those needed to be connected to parallel combination. Equivalent resistance in parallel combination is lesser than the lowest of the individual resistances.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 15.
What is internal resistance of a cell ?
Answer:
Internal resistance: The small resistance offered by the electrolyte of a cell to the electric charges flowing through it from the negative to the positive plate is known as the internal resistance of the cell.

Question 16.
What is watt-hours ? Whose unit is it ?
Answer:
Watt-hour : If an elecrtrical machine of power one watt operates for one hour then one watt hour amount of energy is said to be spent.
1 watt-hour = 1 watt × 1 hour
= 1 watt × 3600 sec = 3600 J
Unit of electrical energy is watt-hour.

Question 17.
Define 1 kilowatt-hour ?
Answer:
If a machine of 1 kilo-hour operates for one hour the amount of energy spent is known as kilo-watt hour. This amount of energy is also known as Board of Trade Unit (B. O. T, units).

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 4

Question 18.
Why is nichrome wire used in electrically heated appliances ?
Answer:
(i) Explanation : Nichrome, an alloy of iron (24%), nickel (60%) and chromium (16%) was much larger resistance than copper and it is about 80 times. Thus, large quantity of heat produces when electricity flows through it. It neither melts nor gets oxidized even when it is red hot. For these reasons nichrome wire is used in most of electrically heated appliances.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 19.
What is short circuiting and what is overloading.
Answer:
Short circuiting : It means direct connection between the two terminals of a source of electricity or through a metallic wire of feeble resistance. As a result, large current flows through the connecting wire which melts due to overheating and therefore further flow of current stops.
Overloading : It is another dangerous effect which is caused by drawing excessive current when a number of high-powered electrical appliances are simultaneously switched on.

Question 20.
What is a fuse ? What is fuse rating ?
Answer:
Fuse : For the safety of the electrical gadgets one thin wire made of an alloy of lead (75 %) and tin (25%) and which has high resistance and low melting is kept in a insulator box of china clay. The thin coil is called fuse wire.
The wire is kept in series with the main circuit of the household electrical appliances.
If for some reason or other there is a surge of the current the household electical appliances may get burn. Under such condition the fuse wire melts and there by cut of the entire circuit and saves the domestic electrical equipments. Rating of fuse wire: The maximum current that may be allowed to flow through a fuse wire before it melts is called the rating of fuse wire.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 21.
What is earthing? How is a person saved from electric shock even if the live wire accidentally touches the metal body of an electric iron the person working with.
Answer:
Earthing : It means to connect a body or the metal case of an electrical appliance to the earth with the help of a metal wire, called, earth wire.
This wire protects a person using the appliance from electric shocks.
If the live wire by chance touches the metal part of an electrical appliance which has been earthed, the current passes directly to the earth through the earth wire.

Question 22.
What is Oersted’s experiment ?
Answer:
Oersted’s experiment : In this experiment, a conductor is held above and parallel to a freely rotatable pivoted magnetic needle. It is found that the magnetic needle deflects due to current flow in the conductor. The deflection increases with increase of current and it reverses when current is reversed.

Question 23.
State Ampere’s swimming rule.
Answer:
Ampere’s swimming rule : If a man be imagined to be swimming along a current carrying wire in the direction of the current (south to north) with his face turned towards a freely rotating magnetic needle, then the north pole of the needle will be deflected towards his left hand.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 24.
State Fleming’s left hand rule.
Answer:
Fleming’s left hand rule: If the thumb, the first finger and the middle finger of the left hand be held mutually perpendicular to each other in such a way that the first finger points in the direction of the magnetic field and the second finger to that of the current, then the thumb will indicate the direction at motion of the conductor.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 5

Question 25.
Show how is Fleming’s left hand rule verified by Barlow’s wheel.
Answer:
Explanation : Barlow’s wheel is an arrangement to demonstrate action of magnet on current that verifies Fleming’s left hand rule.
WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 6
According to the diagram, the current carrying conductor, the toothed wheel is situated in between the magnetic poles. So, when current is allowed to flow, the wheel rotates following Fleming’s left hand rule. On reversing the current the wheel rotates in oppsoite direction following the same rule.

Question 26.
On what factors does the relocing of rotation of Barlow’s wheel depend?
Answer:
Factors responsible for the speed of rotation :

  1. Rotational speed increases with current and vice versa.
  2. Rotational speed increases with intensity of magnetic field and vice versa.
  3. If alternating current is used instead of Direct current, the wheel will try to reverse its direction with the change in direction of current resulting no rotation.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 27.
How can the direction of rotation of Barlow’s wheel be changed?
Answer:
Factors governing the direction of rotation of Barlow’s wheel :

  1. The direction of rotation will reverse if direction of current is reversed keeping direction of magnetic field unchanged.
  2. The direction of rotation will reverse if direction of magnetic field is reversed keeping direction of current unchanged.
  3. The direction of rotation will remain unchanged in both direction of magnetic field and direction of current are reversed.

Question 28.
A bulb is marked ‘230 V-60 W’ what does it indicate?
Answer:
Explanation: When the bulb is used in an electrical line of potential difference 230 volts, 60 joules electrical Work per second is performed due to flow of electric current through the filament of the bulb.

Question 29.
What do you mean by the statement. ‘Potential difference between two points in an electric field is 5 volts’?
Answer:
Explanation: ‘Potential difference between two points in an electric field is 5 volts’- this statement means an external agent has to do 5 joules work to carry 1 coulomb positive charge from a point at lower potential to a point at higher potential in the electric field.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 30.
Why resistivity is also called specific resistance ?
Answer:
Explanation : Resistivity is also called specific resistance as resistivity of a material is the resistance offered by the material of specified dimensions unit length unit cross sectional area.

Question 31.
What is electric motor ?
Answer:
Electric Motor: The device or machine which converts the electrical energy into mechanical energy is known as Electric Motor.
An insulated copper coil, wound over a suitable frame rotates in a magnetic field when electric current passes through the coil.

Question 32.
How can the strength of the motor be increased ?
Answer:
The strength of an electric motor can be increased by :

  1. increasing the current in the armature.
  2. increasing the strength of magnetic field.
  3. increasing the number of turns in the armature.

Question 33.
What is solenoid ?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 7
Solenoid: If many turns an insulated wire wound around a cylinder the resulting coil is known as solenoid. The soft iron body around which an insulated copper wire is coiled is known as the core of the electromagnet.

Question 34.
What is electromagnet ?
Answer:
Electromagnet: If soft iron is kept in a current carrying solenoid, then that soft iron behaves like a magnet so long as the current passes. This magnet is known as electromagnet.

Question 35.
How can the strength of electromagnet be increased ?
Answer:
The strength of electromagnet be increased by :

  1. increasing the number of turns of solenoid.
  2. increasing the current through the solenoid.
  3. changing the core of the electromagnet.
  4. if the core is ‘U’ shaped, then the distance between the poles is minimum.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 36.
State a few applications of eleciromagnet in daily life.
Answer:
Uses of electromagnets :

  1. They are widely used in cranes for lifting and removing heavy beams etc. made of iron or steel in factories.
  2. Physicians use special types of electromagent to remove small pieces of iron from the eye of a patient.
  3. They are used in loudspeakers, receivers of telephone etc.

Question 37.
What is voltmeter ?
Answer:
Voltmeter : A voltmeter is used to measure the potential difference between two points in a section of an electrical circuit ; the voltmeter is connected parallel to the section.

Question 38.
What is ammeter ?
Answer:
Ammeter : It is used to measure the current flowing through an electric circuit where the ammeter is connected in series.

Question 39.
Why is e.m.f. (electromotive forces) of an electric cell is greater than p.d. (potential difference) ?
Answer:
Explanation : The difference of potential between two poles of a cell in open circuit is e.m.f. of the cell, the potential difference between the poles in closed circuit is p.d. (potential difference) of the cell. Now, in a closed circuit, a part of the e.m.f. is expended to overcome the internal resistance of the ce: is offered by the electrolyte of the cell. So, p.d. is less than the e.m.t. cell or e.m.f. is greater than p.d.

Broad Answer Type Questions

Question 1.
What is the difference between e.m.f. and p.d. ?
Answer:
Difference between e.m.f and p.d. :

e.m.f (Electromotive Foce) p.d (Potential Difference)
i. It causes conversion of some other form of energy into electrical energy. i. It causes conservation of electrical energy into some other form of energy.
ii. It is the potential difference across two terminals of an open circuit cell. ii. It is the potential difference between two terminals of a closed circuit cell.
iii. The magnitude of e.m.f. is greater than p.d. in a cell. iii. The magnitude of p.d. is lesser than e.m.f. in a cell.
iv. It is considered as the cause of p.d. iv. Potential difference is the result/effect of e.m.f.
v. It does not depend on the resistance of the circuit. v. Potential difference across any two points of a circuit depends on the resistance of that portion of the circuit.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 2.
State Joule’s laws of heating effect of current.
Answer:
Joule’s Laws (1841) :
(i) First law : The amount of heat produced in a conductor in a given interval of time is proportional to the square of the current passed. Thus if H be the amount of heat generated in a conductor having resistance R when current I passes through it in time t, then
H ∝ P2 (when R and t are kept constant).

(ii) Second Law : The amount of heat produced by a given current in a given time is proportional of the resistance of the conductor.
H ∝ R (When I and t are kept constant)

(iii) Third Law : The amount of heat produced in a given conductor by a given current is proportional to the time for which the current passes.
H ∝ t (When I and R are kept constant)
Combining the three laws, we have
H ∝ I2 R T (when I, R and t vary)
or, H = \(\frac{I^2 R T}{\mathrm{~J}}\)
(J = mechanical equivalent of heat = 4.2 joule/calorie)
If I be in ampere, R in ohm, t in second and H in calorie, then
H = \(\frac{I^2 R T}{4 \cdot 2}\)
∴ H = 0.24 F 2 RT calorie

Question 3.
What is the difference between electromagnet and a natural or permanent magnet ?
Answer:
Difference between electromagnet and natural magnet :

Electromagnet Natural magnet
i. The magnetism of this type of magnet almost disappears as soon as the current is stopped i.e. it is a temporary one. i. If properly maintained magnetism of this type of magnet is almost permanent.
ii. It can be made very strong. ii. It is usually not very strong.
iii. Strength of the magnet by changing the strength of the current. iii. Strength of the magnet cannot be changed.
iv. By changing the number of turns per unit length of the coil, the strength of the magnet can be changed. iv. Strength of the magnet cannot be changed.
v. By changing the direction of current polarity of the magnet can be easily reversed. v. In this case polarity of the magnet cannot be reversed easily.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Question 4.
An electric heater works both in A.C. and D. C. lines but Barlow’s wheel works only in D. C.- Why ?
Answer:
Explanation Generation of heat in a resistor due to a current depends on magnitude of the current not on its direction. Moreover, the produced heat H ∝ I2, so even if the oppositely directed current of A.C. be taken with negative direction, on being squared, it becomes positive. So, a heater can work both in A.C. and D.C.
the rotation of Barlow’s wheel in a given magnetic field depends upon the direction of current. The direction of rotation will reverse if direction of current is reversed keeping direction of magnetic field infact.
Now, in A.C., the direction of current reverses alternately and so frequently that, as soon as Barlow’s wheel is set to rotate in one direction, the reverse current flows into it tending to rotate it oppositely. This happens continuously and very frequently, so the wheel remains static.

Numerical Problems :

Example 1:
The resistance of a conductor is 100 hm; a current of 5 ampere is flowing through it. What will be the p.d. at the two ends of the conductor?
Answer:
R = 10 ohm
I = 5 ampere
V = p.d. = ?
We know,
V = RI
or, V = 10 × 5 = 50 volts

Example 2:
A current of 2A flows through a conductor. The p.d. across the conductor is 50 volts. What is the resistance?
Answer:
We know,
V = R I
or, R = \(\frac{V}{I}\) = \(\frac{50}{2}\)
I = 2 volts
R = 25 ohm

Example 3:
10 ampere current is allowed to a conductor for 1 minute. Find the amount of charge flown.
Answer:
I = 10 ampere
t = 1 minute = 60 sec
Q = ?
We know,
Q = I × t
or, Q = 10 × 60 = 600
∴ Q = 600 coulomb

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Example 4:
Potential difference across two points of a conductor is 60 volts. Resistance of the conductor is 10 ohm. Find the’current through the conductor.
Answer:
V = 60 volts
R = 10 ohm
I = ?
We know,
V = R I
or, I = \(\frac{V}{R}\) = \(\frac{60}{10}\) = 6
or, I = 6 ampere

Example 5:
A current of 2 ampere is passed through a conductor of resistance 4 \mathrm{ohm}. Find the potential difference across the conductor.
Answer:
I = 2 unit
R = 2 unit
V = ?
We know,
V = I × R
or, V = 2 × 4 = 8
∴ V = 8 volts

Example 6:
Resistance of one conductor is double that of the other. What will be the ratio of current through these two conductor if they are subjected to same potential difference.
Answer:
Let the resistance of the conductors are R and 2 R. Let both of them are subjected to same potential difference of V. So, naturally currents in the conductors will be as follows :

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 8

Example 7:
2 units of current is flowing through a conauctor or of 2 units. What is the potential difference at the ends of the conductor?
Answer:
I = 2 ampere
R = 4 ohm
V = ?
We know,
V = R . I
or, V = 2 × 2 = 4 unit

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Example 8:
The resistance per metre of a wire is 1.2 ohm and its specific resistance is 60.28 × 10-8 ohm-m. Find the radius of the wire.
Answer:
R = 1.2 ohm
I = 1 meter
ρ = 60.28 × 10-18ohm-m
A = ?
We know,
R = ρ \(\frac{1}{A}\) or, A= \(\frac{\rho l}{R}\)
or, A = \(\frac{60 \cdot 28 \times 10^{-8} \times 1}{1 \cdot 2}\)
∴ A 50.2 × 10-8 m2
Again, if r be the radius of the wire,
πr2 = A = 50.2 × 10-8 or, r2=\(\frac{50 \cdot 2 \times 10^{-8}}{3 \cdot 14}\) 16× 10-8 m2
∴ r = 4 × 10-14 m = 0.04 cm

Example 9:
There are two copper wires of equal length. The radius of one is twice the other. Find the ratio of their resistances.
Answer:
r = radius of the wire
We know,
WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 9

∴ The thinner wire has a resistance four times the resistance of the thicker wire.

Example 10:
The resistivity of a substance is 9 \times 10-8 \mathrm{ohm} . \mathrm{m}. What length of the wire of that material, having diameter 0.3 cm} will give a resistance of 100 \mathrm{ohm} ?
Answer:
We know,

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 10

Example 11:
The resistance of wire of cross-section area 0.01 cm2 is 10 ohm. What is the length of the wire? The specific resistance of the wire is 50 × 10-6 ohm-cm.
Answer:
We know,
R = 10 ohm
A = 0.01 cm2
ρ = 50 × 10-6 ohm-cm
l = ?
R = ρ \(\frac{1}{A}\) or, l = \(\frac{RA}{\rho}\)
or, l = \(\frac{10 \times 0.01}{50 \times 10^{-6}}\)
∴ l = 2000 cm

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Example 12 :
The length and area of cross-section of a wire are double of these of another wire of same material. Find the ratio of their resistances.
Answer:
Let the length of wires be 2 l and l and their areas of cross. Section be 2 A and A respectively. If \rho be the resistivity of the material of the wires, then their resistance R1 and R2 will be given by
R1= \(\frac{\rho 2 l}{2 A}\) and R2 = \(\frac{\rho l}{A}\)
∴ \(\frac{R_1}{R_2}\) = \(\frac{1}{1}\) or, R1 : R2 = 1 : 1

Example 13 :
Find the specific resistance of the material of a wire of length 100 cm, area of cross-section 0.2 cm2 and resistance 2 ohm.
Answer:
R = 2 ohm
l = 100 cm
A = 0.2 cm2
ρ = ?
R = ρ × \(\frac{1}{A}\)
∴ ρ = \(\frac{R A}{l}\) = \(\frac{2 \times 0.2}{100}\) =40 ×s 10-4 ohm-cm

Example 14 :
Three resistors 1 ohm, 2 ohm and 3 ohm are connected in parallel. What will be their equivalent resistance?
Answer:
We know, for parallel combination,

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 11

Example 15 :
Three wires of resistance 2 ohm, 3 ohm and 5 ohm are connected in series with a cell of e.m.f. 5 volt. What will be the current through each wire?
Answer:
We know, for series combination the equivalent resistance
R = r1 + r2 + r3 = 2 + 3 + 5 = 10 ohm
We also know,
I = \(\frac{V}{R}\) = \(\frac{5}{10}\) = 0.5 A

Example 16 :
Find effective resistance of the resistors 2 ohm, 4 ohm, 5 ohm connected in (i) series and (ii) parallel.
Answer:
We know, for series combination equivalent resistance
R = r1 + r2 + r3 = 2 + 4 + 5 = 11 ohm
We also know, for parallel combination equivalent resistance

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity 12

Example 17 :
In a circuit 2 ohm and 3 ohm resistors are connected in series to the parallel arrangement of 2 ohm and 3 ohm resistors. Calculate the resistance of the circuit.
Answer:
Let R be the equivalent reistance of 2 ohm and 3 ohm arranged in parallel.
So, \(\frac{1}{R}\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\)
∴ \(\frac{1}{R}\) = \(\frac{5}{6}\) or, R = \(\frac{6}{5}\) = 1.2 ohm
Now, 2 ohm and 3 ohm resistors are connected in series with 1.2 ohm. So, the resistance of the circuit = (2 + 3 + 1.2) ohm = 6.2 ohm.

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Example 18 :
A current of 0.5 ampere passes through a wire of resistance 2.5 ohm for 1 hour. Find the heat produced.
Answer:
I = 0.5 ampere
R = 2.5 ohm
t = 1 hr = 3600 sec
H = ?
We know, H = \(\frac{I^2 R T}{J}=\frac{I^2 R T}{4 \cdot 2}\)
∴ H=\(\frac{(0.5)^2 \times 2.5 \times 3600}{4.2}\) = 535.7 calorie

Example 19 :
Two electric bulbs marked 220V-60W are used daily for 5 hours for 10 days. If the price of each unit of electrical energy be 65 paise. What will be the cost for it ?
Answer:
From the problem we get, Total power of the two bulbs = 2 × 60 watt =120 watt
Electrical energy consumed per day = 120 watt × 5 hour = 600 watt-hour
∴ Total energy consumed in 10 days = 600 × 10 = 6000 watt hour
= \(\frac{6000}{1000}\) = 6 kilo watt-hour = 6 B.O.T. unit
∴ cost of electrical energy at 65 paise per unit = 6 × 65 paise = Rs. 3.90

WBBSE Class 10 Physical Science Solutions Chapter 6 Current Electricity

Example 20 :
The same current passes for the same time through two wires the resistance of one is double that of the other. What is the ratio of quantities of heat developed in the two wries?
Answer:
Let the resistances of the wires be 2 R and R and the quantities of heat produced in them be H1 and H2respectively. Then from Joule’s second law we get,
\(\frac{H_1}{H_2}=\frac{2 R}{R}=\frac{2}{1}\) or, H1: H2 = 2: 1

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 8.4 Question Answer – Inorganic Chemistry in the Laboratory and in Industry

A. Ammonia

Very Short Answer Type Questions

Question 1.
Name the chemicals required for the preparation of ammonia in laboratory
Answer:
Chemicals required :

  • Ammonium chloride (NH4Cl)
  • Quick lime (CaO) or dry slaked lime [Ca(OH)2]

Question 2.
What is the characteristic smell of ammonia ?
Answer:
Characteristic smell of ammonia Pungent

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 3.
What is the nature of aqueous solution of ammonia ?
Answer:
Nature of aqueous solution of ammonia : alkaline

Question 4.
State one method for identification of ammonia.
Answer:
Ammonia has a typical pungent smell.

Question 5.
Which particular substance is used to dry ammonia gas ?
Answer:
Ammonia is dried with calcium oxide (CaO).

Question 6.
What colour is obtained when ammonia burns in presence of oxygen?
Answer:
Ammonia burns in presence of oxygen with a greenish yellow flame forming nitrogen and water.
4NH3 + 3O2 = 2N2 + 6H2O

Question 7.
What is Nessler’s reagent ?
Answer:
Nessler’s reagent Nessler’s reagent is an alkaline solution (KOH) of po­tassium mercuric iodide [K2Hgl4]

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 8.
What happens when red litmus paper is placed in aqueous solution of ammonia ?
Answer:
The aqueous solution of ammonia is basic in nature so it turns red litmus to blue.

Question 9.
State one use of ammonium sulphate.
Answer:
Use of ammonium sulphate : as fertilizer.

Question 10.
What is the amount of ammonia present in liquor ammonia ?
Answer:
35% (by weight)

Question 11.
Is ammonia heavier or lighter than air ?
Answer:
Ammonia is lighter than air.
[Vapour density of air = 144 ; vapour density of ammonia = 8-5 ]

Question 12.
State one use of liquid ammonia.
Answer:
Use of liquid ammonia : as a refrigerant in ice making.

Question 13.
Give an example of formation of solid by the combination of two gases.
Answer:
Ammonia gas reacts with hydrogen chloride gas forming solid ammonium chloride.
NH3(gas) + HCl (gas) = NH4Cl (solid)

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 14.
How is ammonia collected in laboratory ?
Answer:
Collection of ammonia Ammonia is lighter than air, it may be collected by the downward displacement of air.

Question 15.
What colour is obtained when ammonia reacts with Nessler’s reagent?
Answer:
Ammonia produces brown colouration in Nessler’s reagent.

Question 16.
Which compound of nitrogen has fractional oxidation number ?
Answer:
Hydrazoic acid (HN3)
Nitrogen has \(\left(-\frac{1}{3}\right)\) oxidation number in hydrazoic acid.

Question 17.
Which hydride of nitrogen has no oxidising property ?
Answer:
Ammonia has no oxidising property.

Question 18.
Which gas is obtained when a mixture of ammonium sulphate and sodium hydroxide is heated ?
Answer:
Ammonia (NH3) gas.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 19.
How ammonium sulphate is obtained without using sulphuric acid ?
Answer:
CaSO4 + 2NH3 + CO2 + H2O = CaCO3 + (NH4)2SO4

Question 20.
Two gaseous compounds of nitrogen react with each other to form nitrogen. What are the two gaseous compounds of nitrogen ?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 1

Short answer type questions

Question 1.
Why is conc. H2SO4 not used for drying ammonia gas ?
Answer:
Ammonia is a basic compound. So it forms salt with cone. H2SO4.
For this reason cone. H2SO4 is not used for drying ammonia gas.
2NH3 + H2SO4 = (NH4)2SO4

Question 2.
Why is anhydrous CaCl2 not used for drying ammonia gas ?
Answer
Ammonia forms an additive compound with anhydrous CaCl2.
CaCl2 + 8NH3 = CaCl2, 8NH3 (additive compound)

Question 3.
Why is P2Os not used for drying ammonia gas ?
Answer:
Ammonia is a basic compound. P2Os is an acidic oxide. So it forms salt with P2O5 SO P2O5 is not used for drying ammonia gas.
6NH3 + P2Os + 3H2O = 2(NH4)3PO4

Question 4.
Why CaO is used for drying agent of ammonia ?
Answer:
CaO is used for drying agent of ammonia : Ammonia is a basic com­pound. It is dried of using a basic drying agent i.e. Quick lime (CaO) which has no action on ammonia.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 5.
What is liquor ammonia ?
Answer:
Liquor ammonia A saturated (35% by weight) aqueous solution of am­monia (sp. gr. = 088) is called liquor ammonia.

Question 6.
Prove that ammonia has reducing property.
Answer:
Reducing property of ammonia : When ammonia is passed over strongly heated black cupric oxide which is reduced to red metallic copper and am­monia is oxidised to nitrogen.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 2

Question 7.
State about solubility of ammonia in water.
Answer
Solubility of ammonia is water : Ammonia is highly soluble in water. One volume of water can dissolve about 1299 volumes ammonia at N.T.P A concentrated aqueous solution of ammonia is called ammonia.

Question 8.
What are the differences between liquid ammonia and liquor ammonia ?
Answer
Difference between liquid ammonia and liquor ammonia :
There is a sharp difference between liquid ammonia and liquor ammonia. The first one is the liquified ammonia gas which has formula NH3, while the second one is the concentrated solution of ammonia which has formula NH4OH.

Question 9.
How will you prove that ammonia contains nitrogen ?
Answer
Ammonia contains nitrogen : Ammonia gas when passed over heated cupric oxide produces colourless gas having no smell.
3CuO + 2NH3 = 3Cu + 3H2O + N2
When the colourless gas is passed over heated magnesium, magnesium nitride (Mg3N2), the white powder produces.
3Mg + N2 = Mg3N2
When the white powder is boiled with water it will produce pungent smell gas NH3
Mg3N2+ 6H2O = 3Mg(OH)2 + 2NH3
This reaction proves that the colourless gas is nitrogen. This reaction also proves that one of the constituents of ammonia is nitrogen.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 10.
What precautions should be taken to combat the effect of ammonia leaked from industries and ammonia tanks ?
Answer
Precautions for ammonia If accidentally, ammonia gas leaks from indus­tries and ammonia tanks, heavy shower of water should be applied in the atmosphere that is filled with the leaked gas. If at any time, tanks filled with ammonia gas are required to be shifted somewhere, the carrier of tanks must also carry plenty of water that may be used to spray in the air filled with accidental leakage of ammonia gas.

Question 11.
How does ammonia react with sodium ?
Answer
Reaction of ammonia with sodium Heated sodium metal at 400°C reacts with ammonia producing sodamide and hydrogen gas.
2NH3 + 2Na = 2NaNH2 + H2
This reaction proves that ammonia contains hydrogen.

Question 12.
How does ammonia react with chlorine ?
Answer
Reaction of ammonia with chlorine : Ammonia reacts with chlorine in two ways.
(i) When ammonia is excess : Excess ammonia is oxidised by chlorine forming nitrogen and is reduced to hydrochloric acid. Hydrochloric acid thus formed combines with ammonia producing ammonium chloride.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 3
(ii) When chlorine is excess : When excess chlorine reacts ammonia forming nascent nitrogen which again combines with chlorine producing nitrogen trichloride, an oily yellow explosive compound.
NH3 + 3Cl2 = 3HCl + NCl3

Question 13.
Show with an equation what happens when ammonia is burnt in oxygen.
Answer:
Ammonia is burnt in oxygen Ammonia is neither combustible nor a supporter of combustion. But ammonia burns in presence of oxygen with a greenish yellow flame forming nitrogen and water.
4NH3 + 3O2 = 2N2↑ + 6H2O

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 14.
How does aqueous ammonia react with aluminium chloride and ferric chloride solutions ?
Answer:
(i) Reaction of aqueous ammonia with aluminium chloride solution: Aqueous solution of ammonia (NH4OH) reacts with aluminium chloride forming white gelatinus precipitate of aluminium hydroxide.
AlCl3 + 3NH4OH = Al(OH)3↓(white) + 3NH4Cl

(ii) Reaction of aqueous ammonia with ferric chloride solution :
Aqueous solution of ammonia (NH4OH) reacts with ferric chloride solution forming brown precipitate of ferric hydroxide.
FeCl3 + 3NH4OH = Fe(OH)3↓(brown) + 3NH4Cl

Question 15.
State with reasons two precautions taken during preparation of ammonia in laboratory.
Answer:
Precautions taken during preparation of ammonia in laboratory :

  • The components ammonium chloride (NH4Cl) and calcium hydroxide [Ca(OH)2] should be dry to avoid evolution of ammonia while mixing the two.
  • The mixture of the components should fill about half the flask to leave enough space for easy passage of the evolved gas.

Question 16.
State some uses of ammonia.
Answer:
Uses of ammonia :

  • It is used for the industrial preparation of nitric acid and sodium carbonate.
  • Large quantities of ammonia are used in the manufacture of fertilizers, such as urea, ammonium phosphate etc.
  • Ammonia is used as solvent.
  • Liquid ammonia is used as a refrigerant in ice making.
  • Ammonia is used as laboratory reagent.
  • It is also used in pharmaceutical industries and also in the preparation of smelling salt.

Question 17.
Give an example of formation of a solid by combination of two gases.
Answer:
Formation of a solid by combination of two gases : Ammonia gas reacts with hydrogen chloride gas forming a solid ammonium chloride. This is an example of formation of a solid by the combination of two gases.
NH3 (gas) + HCl (gas) = NH4Cl (solid)

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 18.
State how does ammonia react with HCl, H2SO4, HNO, acids.
Answer:
Reaction of ammonia with acids : Ammonia is a base, it readily reacts with an acid to form salt.
NH3 + HCl = NH4Cl
2NH3 + H2SO4 = (NH4)2SO4
NH3 + HNO3 = NH4NO3
Aqueous solution of ammonia reacts with acids to form salt and water.
NH4OH + HCl = NH4Cl(salt) + H2O3
2NH4OH + H2SO4 = (NH4)2SO4(salt) + 2H2O
NH4OH + HNO3 = NH4NO3(salt) + H2O

Question 19.
How is urea obtained from ammonia?
Ans. Preparation of urea from ammonia: Ammonia reacts with carbon dioxide at 200°C and 150 atm pressure to form urea.
CO2 + 2NH3 = CO(NH2)2 (urea) + H2O

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 20.
How does aqueous ammonia react with copper sulphate solution?
Answer:
Reaction of aqueous ammonia with copper sulphate solution When aqueous ammonia i.e. ammonium hydroxide is slowly added to a copper sul­phate solution, a pale blue precipitate of basic copper sulphate is first formed. On adding excess of ammonium hydroxide, the precipitate dissolves forming a deep solution of the complex compound called copper tetramine sulphate.
2CuSO4+ 2NH4OH = CuSO4. Cu(OH)2 + (NH4)2SO4
CuSO4 .Cu(OH)2 + 6NH4OH + (NH4)2SO4 = 2[Cu(NH3)SO4+ 8H2O

Broad Answer Type Questions

Question 1.
How ammonia is prepared in laboratory ?
Answer:
Laboratory preparation of Ammonia :
(a) Chemicals required : Ammonium chloride (NH4Cl) and quick lime (CaO) or dry slaked lime [Ca(OH)2

(b) Condition Normally ammonia gas is obtained in the laboratory by heating a mixture of ammonium chloride with slaked lime. Instead of using slaked lime quick lime may also be used. NH4>Cl, Ca(OH)2 or CaO should be in powdered form and dry. The round bottomed flask should be half filled with the mixture of the two for easy escape of ammonia gas formed.

(c) Collection Ammonia is lighter than air, it may be collected by the downward displacement of air. Ammonia is not collected through the downward displacement of water because, it is highly soluble in water.

(d) Equations of the reaction :
2NH4Cl + Ca(OH)2 = 2NH3↑+ CaCl2 + 2H2O
2NH4Cl + CaO = 2NH3↑ + CaCl2 + H2O

(e) Precautions : The ingradients, the test-tube, the delivery pipes and the gas-jar should be absolutely dry. All the connections in the apparatus should be leak-proof.

(f) Drying of ammonia : As ammonia is a basic substance, it cannot be dried by acidic drying agents like cone. H2SO4 or P2Os. The gas is absorbed by fused CaCl2 with the formation of an addition compound CaCl2.8NH3. So, fused CaCl2 cannot be used to dry ammonia. It is best dried with the basic drying agent, quick lime (CaO).

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 2.
Prove by experiment that ammonia is dissolved in water and produces alkaline solution.
Answer:
Fountain experiment :
Arrangement of the experiment : A flask is filled with dry ammonia gas and its mouth is corked. The flask is kept in inverted position and is damped to a stand. Through a hole in the cork one end of a glass tube is introduced inside the flask. This end of the tube inside the flask is shaped into a jet. The other end of the tube dips in some red litmus solution taken in a beaker.

Operation : Now ice or ether is poured upon the flask.

Observation : A blue fountain produces inside the flask.

Explination Due to evaporation of ether the flask is cooled. Ammonia gas inside the flask contracts, as a result there is a partial vacuum inside the flask. At this stage if stop cock is opened the red litmus solution rushes inside the flask and ammonia is dissolved in it. Due to vaccum created inside there is a formation of fountain and the solution becomes blue.

Conclusion : This experiment proves that ammonia is highly soluble in water and the aqueous solution is alkaline.

Question 3.
How is ammonia identified ?
Answer:
Identification of ammonia :
(a) Ammonia is easily detected by its pungent smell.

(b) A glass rod dipped in into hydrochloric acid is then held into ammonia a white fumes is observed due to formation of ammonium chloride.
NH3 + HCl = NH4Cl (white fumes)

(c) Ammonia turns moist red litmus paper into blue.

(d) Ammonia produces brown colouration or precipitate in Nessler’s reagent.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 4

(e) A strip of filter paper, soaked in mercurous nitrate solution, when exposed to ammonia gas, turns black.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 5

B. Sulphuretted Hydrogen (H2S)

Very Short answer type questions 

Question 1.
Which acid is used to prepare H2S gas in laboratory ?
Answer:
Dilute sulphuric acid.

Question 2.
During preparation of H2S gas in laboratory, how Is the gas collected?
Answer:
By upward displacement of air.

Question 3.
What happens when H2S gas is passed through acidified yellow ferric chloride solution ?
Answer:
Yellow ferric chloride solution reduces to colourless ferrous chloride solution.

Question 4.
How is H2S gas dried In laboratory preparation ?
Answer:
By passing the gas through phosphorus pentoxide (P2O5).

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 5.
How does H7S gas smell ?
Answer:
It has a smell of rotten egg.

Question 6.
How does H2S gas react with blue litmus ?
Answer:
It turns blue litmus red.

Question 7.
What gas is used in detecting metallic radical in laboratory ?
Answer:
Sulphuretted hydrogen (H2S)

Question 8.
Is H2S gas heavier or lighter than air ?
Answer:
H2S gas is heavier than air.

Question 9.
Is H7S combustible or a supporter of combustion ?
Answer:
H2S is a combustible gas.

Question 10.
Why is an old silver made substance turned to black ?
Answer:
Due to formation of silver sulphide (Ag2S).

Question 11.
What is the common use of hydrogen sulphide in laboratory ?
Answer:
For the identification of basic radical (metal ion).

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 12.
Name the chemicals required for preparation of H2S in laboratory.
Answer:
Chemicals required for preparation of H2S :

  • Ferrous sulphide (FeS)
  • Dilute sulphuric acid (H2SO4)

Question 13.
Is HCl acid suitable for preparation of H2S ?
Answer:
No. As HCl is very volatile.

Question 14.
Mention one identifying test for H2S.
Answer:
Lead acetate paper turns black when it is held in H2S gas.

Question 15.
What is the density of H2S at NTP ?
Answer:
Density of H2S gas is 1-53 g/L at NTP

Question 16.
Give the name of a gas which is soluble in cold water but insoluble in hot water.
Answer:
Hydrogen sulphide (H2S)

Question 17.
Which gas is used to identify basic radical ?
Answer:
Hydrogen sulphide (H2S)

Question 18.
What happens when H2S gas is passed through a blue coloured solution of CuSO2 ?
Answer:
Black precipitate of CuS is formed.

Question 19.
Is H2S acidic or basic gas ?
Answer:
H2S is acidic gas.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 20.
How many types of salts are formed by H2S ?
Answer:
Since H2S is a bibasic acid, it gives two types of salt, one is acid salt and other is normal salt.

Question 21.
Give the name of a poisonous gas.
Answer:
Hydrogen sulphide (H2S).

Short Answer Type Questions

Question 1.
Name the chemicals required for preparation of H2S in laboratory. Write down the equation of reaction.
Answer:
Chemicals required for preparation of H2S in laboratory :
Ferrous sulphide (FeS) and dilute sulphuric acid (H2SO4)
Equation : FeS + H2SO4 = FeSO4 + H2S↑

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 2.
Is HCl acid suitable for preparation of H2S ? Give reason.
Answer:
HCl acid is not suitable for preparation of H2S.
Cause : HCl acid is not chosen due to volatile nature of HCl. During the preparation, HCl vapour that may be formed will accompany the evolved H2S and will make it impure.

Question 3.
Is HNO3 acid suitable for preparation of H2S? Give reason.
Answer:
HNO3 acid is not suitable for preparation of H2S.
Nitric acid cannot be used in the preparation, because it is an oxidising agent and H2S is a reducing agent. Nitric acid will oxidise H2S to sulphur and will thus hamper the reaction.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 4.
Mention one identifying test for H2S.
Answer:
Identification of H2S: Hydrogen sulphide is passed through a solution of sodium hydroxide solution and then sodium nitroprusside solution is added to it. The solution turns violet.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 7

Question 5.
How does H2S react with cone. HNOa?
Answer:
Reaction of H2S with cone HNO3 :
H2S-reduces cone. HNO3 to brown nitrogen dioxide (NO2) and is itself oxidised to sulphur.
Equation : H2S + 2HNO3 = 2NO2↑ + S↓+ 2H2O

Question 6.
How does H2S react with cone. H2SQ4 ?
Answer:
Reaction of H S with cone. H2SO4 : H2S) reduces cone. H2SO4 to SO2 and is itself oxidised to sulphur.
Equation : H2S + H2SO4 = SO2↑ + S↓ + 2H2O

Question 7.
Write down the reaction of H2S with NaOH.
Answer:
Reaction of H2S with NaOH : H2S reacting with NaOH produces acid salt in the first step and normal salt in the second step along with water in both the steps.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 6

Question 8.
Which substance is used for drying H2S and why ?
Answer:
Drying of H2S gas : The most suitable drying agent for H2S is the acidic oxide P2O5, that does not react with H2S which is also of acidic nature.

Question 9.
What is the density of H2S and state about its solubility in water at ordinary temperature and in hot water.
Answer:
Density of H2S : Density of the gas is T53 g/L at NTR
Solubility of H2S in water : It is moderately soluble in water at ordinary temperature but insoluble in hot water.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 10.
What happens when H2S is burnt in small and excess supply of oxygen ?
Answer:
H2S is burnt in small supply of oxygen : During burning of H2S if the supply of oxygen is low, the gas burns with a blue flame and deposits sulphur. 2H2S + O2 = 2S↓ + 2H2O
H2S is burnt in excess supply of oxygen : In an excessive supply of oxygen the gas burns with a blue flame and produces sulphur dioxide and water.

Question 11.
Show that H2S is a reducing agent.
Answer:
Reducing property of H2S : Hydrogen sulphide is a very good reducing agent. H2S reduces acidified yellow solution of ferric chloride to colourless ferrous chloride.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 8

Question 12.
What happens when H2S is passed through acidified solution of FeCla?
Answer:
Yellow ferric chloride solution is reduced by H2S giving rise colourless ferrous chloride and H2S itself on oxidation gives a precipitate of sulphur.
Equation : 2FeCl3 + H2S = 2FeCl2 + 2HCl + S4

Question 13.
What happens when H2S is passed through acidified solution of CuSO4?
Answer:
If H2S is passed through acidified blue coloured solution of CuSO4 then black precipitate of cupric sulphide (CuS) is formed.
Example:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 9

Question 14.
What happens when H2S is passed through acidified solution of Pb(NO3)2
Answer:
If H2S is passed through acidified colourless solution of Pb(NO3)2 then black precipitate of lead sulphide (PbS) is formed.
Example:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 10

Question 15.
What happens when H2S comes in contact with a paper soaked in lead acetate ?
Answer:
A paper soaked in lead acetate [Pb(CH3COO)2] turns black with H2S gas due to formation of black lead sulphide (PbS).
Equation:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 11

Question 16.
Mention the uses of hydrogen sulphide.
Answer:
Uses of hydrogen sulphide :

  • H2S is used as a reagent in the separation of metal ions in group analysis.
  • H2S is sometimes used as a reducing agent.

Broad answer type questions

Question 1.
Describe laboratory method of preparation of H2S with the points:
(i) chemicals required
(ii) condition
(iii) equation of reaction
(iv) drying agent
(v) collection
Answer:
(i) Chemicals required Ferrous sulphide (FeS) and dilute sulphuric acid (H2SO4)
(ii) Condition Hydrogen sulphide is prepared in the laboratory by the action of dil. H2SO4 on pieces of ferrous sulphide (FeS) at ordinary temperature.
(iii) Equation of reaction : FeS + H2SO4 = FeSO4 + H2S↑
(iv) Drying agent The gas is passed through P2O5 to dry it.
(v) Collection : Because the gas is heavier than air it is collected over the upward displacement of air.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 2.
What precautions should be taken during preparation and handling of H2S ?
Answer:
Precautions taken during preparation of H2S : Hydrogen sulphide is a poisonous gas and is harmful at concentrations above 0-1% by volume in air. So, due precautions should be taken not to inhale the gas not to allow its prolonged contact with skin during handling.

During preparation of the gas caution must be taken so that excessive gas does not spread in the atmosphere of laboratory. To do this, sulphuric acid should be added in the wolfe’s bottle in small quantities and in steps.

C. Nitrogen

Very short answer type questions

Question 1.
Name the chemicals used In laboratory method of preparation of nitrogen.
Answer:
Chemicals required : Ammonium chloride (NH4Cl) and sodium nitrite (NaNO2).

Question 2.
State what you know about solubility of nitrogen in water.
Answer:
Solubility of nitrogen in water is about 23-5 ml/L at NTR It is almost half that of oxygen.

Question 3.
What is the density of nitrogen?
Answer:
Density of nitrogen 125 g/L at NTR

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 4.
How is nitrogen prepared in laboratory dried?
Answer:
Nitrogen has is dried by passing the gas through concentrated sulphuric acid (H2SO4).

Question 5.
What is the percentage of nitrogen in air by volume?
Answer:
Percentage of nitrogen in air by volume : 77 17%

Question 6.
What would happen if there were no nitrogen in the atmosphere?
Answer:
In absence of nitrogen burning in atmosphere would be vigorous and uncontrollable.

Question 7.
What is the use of liquid nitrogen?
Answer:
Liquid nitrogen (b.p.-195’8°C) is used to produce low temperature in scientific research.

Question 8.
Is nitrogen heavier or lighter than air?
Answer:
Nitrogen is slightly lighter than air.

Question 9.
Give a name of nitrogen containing inorganic fertilizer.
Answer:
Nitrogen containing inorganic fertilizer : Ammonium nitrate (NH4NO3).

Question 10.
Which gas is used in gas-thermometer?
Answer:
Nitrogen gas is used in gas-thermometer.

Question 11.
Which gas is used in electric bulb?
Answer:
Nitrogen gas is used in electric bulb.

Question 12.
What is nitrogen trihydride?
Answer:
Nitrogen trihydride is ammonia

Question 13.
Which gas is presentjarge amount in atmosphere?
Answer:
Nitrogen gas is present large amount in atmosphere.

Question 14.
What is nitrolim?
Answer:
Nitrolim : Mixture of calcium cyanamide and carbon is known as nitrolim.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 15.
Which gas is used to create inert atmosphere during^hemical reactions.
Answer:
Nitrogen gas is used to create inert atmosphere during chemical reactions.

Question 16.
What are the two essential gaseous components of the atmospheric air ?
Answer:
Two essential gaseous components are :
(i) Oxygen
(ii) Nitrogen

Question 17.
Which plants absorb nitrogen directly from air ?
Answer:
A few leguminous plants such as pea, bean, clover etc. can absorb nitrogen directly from air.

Question 18.
What are the ways of fixation of nitrogen?
Answer:
There are two ways of fixation of nitrogen:
By electric discharge
Bio-chemical reaction through bacteria.

Short Answer Type Questions

Question 1.
How does nitrogen react with oxygen?
Answer:
Reaction of nitrogen with oxygen : Nitrogen reacts with oxygen forming nitric oxide under the influence of electric arc at a temperature of 3000°C.
\(\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \text { – heat (Endothermic reaction) }\)

Question 2.
Why Is the solution of ammonium chloride and sodium nitrite is heated to produce nitrogen instead of heating ammonium nitrite directly?
Answer:
Nitrogen gas is not prepared by heating ammonium nitrite (NH4NO2) directly, for in that case rapid explosive dissociation of ammonium nitrite takes place.

Question 3.
How does nitrogen react with magnesium ?
Answer:
Reaction with magnesium : When turnings of magnesium metal are strongly heated in nitrogen gas, the metal directly combines with nitrogen to form magnesium nitride
(Mg3N2).
3Mg + N2 = Mg3N2

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 4.
What happens if magnesium nitride is boiled with water ?
Answer:
Magnesium nitride is boiled with water to form magnesium hydroxide and ammonia.
Mg3N2 + 6H2O = 3Mg(OH)2 + 2NH

Question 5.
What is nitrolim? How it is formed? Mention its important use.
Answer:
Nitrolim: When small lumps of white calcium carbide (CaC2) are heated in nitrogen gas at about 1100°C, calcium cyanamide (CaNCN) is formed, along with the separation of carbon particles (black). The blackish brown mixture of the products (CaNCN + C) is commercially known as nitrolim. Use of nitrolim : It is largely used as a fertilizer.

Question 6.
State the uses of nitrogen.
Answer:
Uses of nitrogen :

  • It is used in the industrial preparation of ammonia and nitric acid.
  • It is used for the production of fertilizer, such as ammonium sulphate, am­monium nitrate.
  • Liquid nitrogen (b.p. -195’8°C) is used as condenser.
  • In many chemical reactions, nitrogen is used as inert medium.
  • It is used in the preparation of electric bulb and gas thermometer.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 7.
Why gaseous nitrogen is chemically quite inactive at ordinary temperatures, but it is active at higher temperatures ?
Answer:
Chemical nature of gaseous nitrogen : The unreactivity of nitrogen at ordinary temperatures, is due to the structure of nitrogen molecule (N2), in which two nitrogen atoms are bonded together by three covalent bonds (N ≡ N). The force of attraction of two nitrogen atoms so bonded is very high. So large amount of energy (generally heat energy) is required to break up this bond to set free the two nitrogen atoms so that they can enter into chemical combination.

Question 8.
State the physical properties of nitrogen.
Answer:
Physical properties of nitrogen :

  • It is colourless gas without smell or taste,
  • Density of the gas is 1.25 g/L at NTR
  • It is slightly soluble in water.
    Solubility of nitrogen in water is about 23-5 ml/L at NTR

Broad Answer Type Questions 

Question 1.
Describe the laboratory method of preparation of nitrogen. State the following matters :
(a) chemicals required
(b) condition
(c) drying
(d) chemical equation
(e) collection of gas.
Answer:
Laboratory method of preparation of nitrogen :
(a) Chemicals required :

  • Sodium nitrite (NaNO2)
  • Ammonium chloride (NH4Cl)

(b) Condition : The chemicals ammonium chloride and sodium nitrite must not be in solid state but as solution. The mixture is heated gently to evolve nitrogen gas.

(c) drying : The nitrogen gas is dried by passing through a U-tube containing cone., sulphuric acid.

Chemical equation:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 12

(e) Collection of gas : Though nitrogen gas is slightly soluble in water, it is collected by the downward displacement of water.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 2.
State briefly the significance of the presence of nitrogen in air.
Answer:
Significance of the presence of nitrogen in air : Nitrogen in air dilutes oxygen present therein. Oxygen is a very active oxidant. The mixture of nitrogen with oxygen in air moderates the intense activity of pure oxygen in the processes of respiration, combustion etc, which would have otherwise occurred very rapidly.

The intake of only oxygen during respiration would have caused increased rate of respiration, causing excessive burning of body tissues. This would have caused high and unchecked rise in the temperature of body, which results death. Without nitrogen in air, the combustion of earthly materials would have been explosively vigorous.

The huge amount (about 78% by volume) of nitrogen in air can be ab­sorbed by certain symbiotic bacteria, such as Rhizobium leguminouserum, R. japonicum etc. Nitrogen is converted to nitrogenous salts in nature by many chemical and biochemical process. Proteins, the complex nitrogenous compounds, which are absolutely essential for living beings are also indirectly formed from the nitrogen of air.

Question 3.
State the two possible ways of reaction of nitrogen with hydrogen, also write down the connected equations of reaction.
Answer:
First type Nitrogen combines with hydrogen under the influence of electric spark producing ammonia gas.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 13

Second type : Nitrogen also combines with hydrogen at 550°C under 200 atmospheric pressure in presence of iron catalyst to form nitrogen gas.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 14

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 4.
State two important ways of fixation of nitrogen.
Answer:
Two ways of fixation of nitrogen :
By electric discharge : During electric discharge in the atmosphere nitrogen and oxygen present in air combine to produce nitric oxide. The produced nitric oxide is then oxidised by atmospheric oxygen to form nitrogen dioxide. Later this oxide upon missing with water vapour or rain water product nitric acid which falls upon our earth.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 15

Nitric acid is then reacted with the bases present in the soil forming nitrate salts.

iii. Fixation of nitrogen due to bacteria Some micro organism and blue green algae convert nitrogen present in air to ammoiate salt by bio-chemical process. When the animal and plant bodies undergo decay most of the nitrogen (proteins) is liberated as ammonia which is oxidised in the soil by the combined action of nitrosifying and nitrifying becteria into nitrate to be again assimilated by plants. A portion of the fixed nitrogen is set free by the action of denitrifying bacteria of the soil.

(d) Hydrogen Chloride (Hydrochloric acid),
Nitric acid and Sulphuric acid

Very Short Answer Type Questions )

Question 1.
Why is concentrated sulphuric acid used to prepare HCl in laboratory ?
Answer:
Concentrated sulphuric acid is used to prepare HCl in laboratory because concentrated sulphuric acid is less volatile.

Question 2.
What is muriatic acid ?
Answer:
Muriatic acid is the commercial name of hydrochloric acid (HCl).

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 3.
Why is not nitric acid used to prepare HCl ?
Answer:
Nitric acid is not used to prepare HCl as nitric acid is highly volatile and a strong oxidant.

Question 4.
State about solubility of HCl gas in water ?
Answer:
HCl gas is highly soluble in water.

Question 5.
Why is not hydrochloric acid used to prepare nitric acid ?
Answer:
Hydrochloric acid is not used to prepare nitric acid as hydrochloric acid is more volatile than nitric acid.

Question 6.
What is fuming nitric acid ?
Answer:
Fuming nitric acid is a product produced by dissolving excess nitrogen di­oxide in nitric acid.

Question 7.
What is acid rain ?
Answer:
Dissolved NO2and SO2 of atmosphere produce acids which come down along with rain. This is called acid rain.

Question 8.
What is oleum ?
Answer:
Oleum is sulphuric acid in which excess sulphur trioxide is dissolved.

Question 9.
What is stone cancer ?
Answer:
Corrosion of marble or stone due to acid rain is called stone cancer.

Question 10.
What is the commercial name of nitric acid ?
Answer:
The commercial name of nitric acid is aqua fortis.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 11.
How does copper react with sulphuric acid ?
Answer:
Copper reacts with hot and concentrated sulphuric acid.

Question 12.
Why does sugar of white colour turns to a black mass in contact with concentrated sulphuric acid ?
Answer:
Concentrated sulphuric acid absorbs the water molecules present in sugar molecules leaving behind black carbon.

Question 13.
What is passive iron ?
Answer:
Chemically inactive iron is called passive iron.

Question 14.
What is the commercial name of sulphuric acid ?
Answer:
The commercial name of sulphuric acid is oil of vitriol.

Question 15.
Which acid produces hydrogen reacting with metals Mg and Mn only?
Answer:
Dilute and cold nitric acid (HNO3) produces hydrogen reacting with Mg and Mn.

Question 16.
Name an acid which has great affinity to water.
Answer:
Concentrated sulphuric acid (H2SO4) has great affinity to water.

Question 17.
Which acid is identified by ring test ?
Answer:
Nitric acid is identified by ring test.

Question 18.
Which acid turns blue copper sulphate to white anhydrous salt ?
Answer:
Concentrated sulphuric acid (H2SO4) turns blue copper sulphate to white anhydrous salt.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 19.
Among HCl, H2SO4 and HNO3 acid which acid has the lowest boiling point ?
Answer:
HCl acid has the lowest boiling point.

Question 20.
What is aqua regia ?
Answer:
Aqua regia is a mixture of 3 volumes concentrated hydrochloric acid and 1 volume concentrated nitric acid.

Question 21.
Which acid is detected by AgNO3 solution ?
Answer:
Hydrochloric acid is detected by AgNO3 solution.

Question 22.
What is royal water ?
Answer:
Aqua regia is also known as royal water.

Question 23.
What is fuming sulphuric acid ?
Answer:
Fuming sulphuric acid or oleum is obtained when sulphur trioxide is passed over 98% sulphuric acid.

Question 24.
How is hydrogen chloride gas collected in laboratory ?
Answer:
Hydrogen chloride gas is heavier than air so it is collected by upward dis­placement of air.

Question 25.
Name one air pollutant gas.
Answer:
Air polutant gas is nitrogen dioxide (NO2)

Question 26.
Which gas pollutes air form Goldsmith’s workshop ?
Answer:
Nitrogen dioxide (NO2)

Question 27.
Which mineral acid has no reducing property ?
Answer:
Nitric acid (HNO3)

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 28.
What is the commercial name of nitric acid ?
Answer:
The commercial name of nitric acid is aquafortis.

Question 29.
State one use of aqua regia.
Answer:
Noble metals like gold, platinum are soluble in aqua regia.

Question 30.
State one use of sulphuric acid.
Answer:
Sulphuric acid is used to prepare fertilizer like ammonium sulphate [NH2SO4]

Question 31.
State one use of nitric acid.
Answer:
Nitric acid is used for the manufacture of high explosive like nitroglycerine.

Question 32.
Name a reducing acid.
Answer:
Hydrochloric acid (HCl) is an example of reducing acid.

Question 33.
What is the catalyst used during preparation of nitric acid by the Ostwald process ?
Answer:
Platinum gauze catalyst is used during preparation of nitric acid by the Ostwald process.

Question 34.
Which catalyst is used during preparation of sulphuric acid by contact process ?
Answer:
Platinum asbestos or vanadium pentoxide catalyst is used during preparation of sulphuric acid by contact process.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 35.
Which acid is detected by BaCl, solution ?
Answer:
Sulphuric acid (H2SO4) is detected by BaCl solution.

Question 36.
What is the basicity of sulphuric acid ?
Answer:
The basicity of sulphuric acid is 2.

Question 37.
Which acid is used as a cleaning agent in bathroom ?
Answer:
Muriatic acid (HCl) is used as a cleaning agent in bathroom.

Question 38.
Among HCl, H2SO4 and HNO which acid has the highest point?
Answer:
Sulphuric acid has the highest boiling point [338°C] among HCl, H2SO4 and HNO3

Question 39.
Which acid is called ‘king of chemicals’ ?
Answer:
Sulphuric acid (H2SO4) is called ‘king of chemicals’.

Question 40.
Which acid is gas at ordinary temperature ?
Answer:
Hydrochloric acid.

Question 41.
Dense white fumes are formed when a glass-rod, moistened with strong ammonia solution, is held in a gas. Identify the gas.
Answer:
The gas is ammonia.

Question 42.
What is laughing gas ?
Answer:
Nitrous oxide (N2O) is called laughing gas.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 43.
Can dry hydrogen chloride gas turn blue litmus paperred ?
Answer:
No. Dry hydrogen chloride gas turns a moist blue litmus paper red.

Question 44.
What is the molecular weight of nitric acid ?
Answer:
The molecular weight of nitric acid is 63.

Question 45.
What is the specific gravity of sulphuric acid ?
Answer:
The specific gravity of sulphuric acid is 1.84.

Question 46.
Name one metal ion that can be identified by using H2S.
Answer:
Cu2+

Question 47.
What is commercial name of pyrosulphuric acid?
Answer:
Oleum (H2S2O7)

Question 48.
What is the catalyst used for preparation of HNO3 by ‘ostwald process’
Answer:
Platinum wie gauze is used as the catalyst for the purpose.

Question 49.
What is liquor ammonia?
Answer:
Liquor ammonia is a saturated solution of ammonia in water.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 50.
What is Nessler’s reagent?
Answer:
Alkaline solution of potassium tetra iodo mercurate, K2[Hgl4] is known as Nessler’s neagent.

Question 51.
What is aqua fortis?
Answer:
Nitric acid (HNO3) was known to the alchemists as aqua fortis.

Question 52.
Why do old oil paintings turn black on exposure to H2S?
Answer:
The paints used in oil paintings contain lead that react with weak dibasic H2S to form black lead sulphide (PbS).

Question 53.
Name a gas which can be prepared in kipp’s apparatus.
Answer:
H2S can be readily and intermittently prepared in kipp’s apparatus.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 54.
What happens when calcium cyanamide is heated with super heated steam?
Answer:
When calcium cyanamide is heated with super heated steam (especially in an autoclave), it is hydrolysed to form ammonia.

Question 55.
What happens when ammonia is burnt in air?
Answer:
When ammonia is burnt in air, it burns with a greenish-yellow flame forming nitrogen and steam.

Question 56.
H2SO4 is industrially produced by which process?
Answer:
H2SO4 is industrially produced by contact process.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 57.
Why H2S gas is collected by upward displacement of air?
Answer:
H2S is soluble in cold water and is heavier than air. So it is not collected over water and is collected by upward displacement of air.

Short Answer Type Questions

Question 1.
Describe the preparation of hydrogen chloride in laboratory with mention of
(i) chemical equation
(ii) collection
Answer:
Preparation of hydrogen chloride in laboratory : Chemical equation:
(i) If the mixture of sodium chloride and concentrated sulphuric acid is heated at 150°C – 200°C then sodium bisulphate and hydrogen chloride gas are produced.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 16
(ii) If it is heated at 500°C then sodium sulphate and hydrogen chloride gas are evolved.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 17
Collection : Since dry hydrogen chloride gas is heavier than air it is col­lected by upward displacement of air.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 2.
Why the reaction is kept at lower temperature during the preparation of hydrogen chloride in laboratory from sodium chloride and cone, sulphuric acid ?
Answer:
Reason In the laboratory the reaction is kept at lower temperature because:

  • at high temperature the flask used in the preparation may be cracked.
  • Sodium sulphate at higher temperature forms a hard crust and sticks to the glass. Its removal is then difficult.

Question 3.
Why hydrogen chloride is not collected over displacement of water?
Answer:
Reason : Hydrogen chloride is highly soluble in water. It has been found that at 0°C, 450 ml of hydrogen chloride is dissolved in 1 ml of water. So, it is not collected over displacement of water.

Question 4.
What is the compound necessary for drying of hydrogen chloride gas?
Answer:
Drying of hydrogen chloride : To remove water vapour, hydrogen chloride gas is passed over cone. H2SO4.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 18
Hydrogen chloride is acidic nature. Hence, it cannot be dried over quick lime (CaO), caustic soda (NaOH) or caustic potash (KOH).

Question 5.
How is hydrochloric acid prepared ?
Answer:
Preparation of hydrochloric acid : Hydrochloric acid is a solution of hydrogen chloride gas in water. Cone, hydrochloric acid, as used in the laboratory is not 100% pure hydrochloric acid. It is about 38% solution of hydrogen chloride gas in water.

Hydrogen chloride gas is passed into a beaker containing cold water till the water is saturated with the gas. The contents of the beaker is a concentrated solution of hydrochloric acid.

Question 6.
Hydrochloric acid is not prepared by dissolving hydrogen chloride gas in water directly—why ?
Answer:
Reason As hydrogen chloride gas is highly soluble in water and the rate of dissolving of HCl gas in water is much higher than the rate of formation of HCl gas as a result there is a vacuum in the flask. To fill up the vacu­um water in the beaker enters into the flask and creates explosion when it contacts with sulphuric acid present in the flask.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 7.
Nitric acid is not used in the preparation of hydrochloric acid—why?
Answer:
Explanation :
(a) Nitric acid is not used because during the preparation of hydrogen chloride gas, nitric acid oxidises the evolved hydrogen chloride gas into chlorine gas.
Reaction HNO3 + ;3HCl = NOCl + 2[Cl] + 2H2O

(b) Hydrochloric acid (b. p. 86°C) and hydrogen chloride are volatile in nature. So they will be collected jointly in a receiver where nitric acid may oxidise hydrogen chloride into chlorine. On the other hand H2SO4(b.p. 338°C) is non-volatile; so, this is not happened.

Question 8.
State two physical properties of HC:
Answer:
Physical properties of HCl:

  • It is a colourless gas with a choking smell and strongly fuming in moist air, hydrogen chloride is 127 times as heavy as air ; it neither burns nor supports burning.
  • It is highly soluble in water. At 0°C, 450 ml of hydrogen chloride gas is dissolved in 1 ml of water.

Question 9.
How is pure hydrogen chloride prepared ?
Answer:
Preparation of pure hydrogen chloride Pure hydrogen chloride is obtained by the action of water upon silicon tetrachloride. Equation : SiCl4 + 4H2O = Si(OH)4 + 4HCl

Question 10.
State the reaction of HCl with Na.
Answer:
Hydrogen chloride neither burns in air nor supports combustion. However burning sodium continues to burn in the gas with a bright yellow flame producing hydrogen and anhydrous sodium chloride.
Equation : 2Na + 2HCl = 2NaCl + H2

Question 11.
Show the acidic property of hydrochloric acid.
Answer:
The aqueous solution of hydrochloric acid is a strong acid :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 19
dissolves base metals, liberating hydrogen and forming metallic chlorides,
e.g. Zn + 2HCl = ZnCl2 + H2
The order of activity of the metals for liberating hydrogen may be stated as:
Ca > Mg > Al > Zn > Fe > Sn.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 12.
How copper and silver react with hydrochloric acid ?
Answer:
Generally, copper, silver are not attacked by hydrochloric acid. In presence of air, copper and silver react with very slowly producing corresponding chlorides and water.
Equations :

  • 2Cu + 4HCl + O2 = 2CuCl22 + 2H2O
  • 4Ag + 4HCl + O2 = 4AgCl + 2H2O

Question 13.
How does ammonia react with hydrogen chloride
Answer:
Reaction of ammonia with hydrogen chloride : Ammonia in contact with hydrogen chloride gives dense white fumes of solid ammonium chlo­ride. This reaction is an example of the formation of a solid product by the interaction of two gases.
Equation : NH3(g) + HCl(g) = NH4Cl(s)

Question 14.
What is reaction between Na2CO3 and HCl ?
Answer:
Reaction of Na2CO3 with HCl :
HCl reacts with Na2COs to literate carbon dioxide.
Na2CO3 + 2HCl = 2NaCl + CO2↑ + H2O

Question 15.
State the action of hydrochloric acid with oxidising agent like MnO2 and KMnO4.
Answer:
Reaction of hydrochloric acid with MnO2and KMnO4
Manganese dioxide (MnO2 oxidises hot and concentrated hydrochloric acid to greenwish yellow chloride.Potassium permanganate oxidises the hydrochloric acid at the ordinary tem­perature.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 20

Question 16.
Show the presence of chloride ion in hydrochloric acid.
Answer:
Presence of chloride ion in hydrochloric acid : When hydrochloric acid is added to a solution of silver nitrate, a white precipitate of the chloride of silver is obtained. Silver chloride is insoluble in nitric acid but soluble in ammonium hydroxide.
Equation : AgNO3 + HCl = AgCl↓ + HNO

Question 17.
How is hydrochloric acid identified?
Answer:
Identification of hydrochloric acid : When silver nitrate solution is added to hydrochloric acid taken in a test tube, curdy white precipitate of silver chloride is formed. The precipitate is insoluble in nitric acid but dissolves in ammonium hydroxide forming a clear solution of complex salt argento ammonium chloride. Equations :

  • AgNO3 + HCl = AgCl↓ + HNO3
  • AgCl + 2NH4OH = [Ag(NH3)2]Cl + 2H2O

Question 18.
State the identification of hydrogen chloride gas.
Answer:
Identification of hydrogen chloride gas : Dense white fumes are formed when a glass-rod, moistened with strong ammonia solution, is held in the hydrogen chloride gas.
Equation : NH3(g) + HCl (g) = NH4Cl (s)

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 19.
What are the uses of hydrochloric acid ?
Answer:
Uses of hydrochloric acid :

  • It is used in dyeing and calico-printing,
  • It is used as a cleaning agent in galvaning and tinplating,
  • It is used to prepare aqua-regia,
  • It is used in the manufacture of glucose, glue and many useful metal-chlorides.

Question 20.
Describe the preparation of nitric acid in laboratory with mention of
(i) chemical required
(ii) chemical reaction.
Answer:
Preparation of nitric acid in laboratory :
Chemical required :

  • Chille salt petre (NaNO3) or nitre (KNO3)
  • Cone. Sulphuric acid.

Reaction occurs in two steps :

(a) If the reaction is kept at 200°C-300°C then sodium bisulphate and nitric acid are produced.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 21

(b) At 800°C, sodium sulphate and nitric acid are formed in the reaction between sodium nitrate and concentrated sulphuric acid.
2NaNO2+ H2SO4 = Na2SO4 + 2HNO4

Question 21.
Why nitric acid is prepared at lower temperature (200°C) in laboratory?
Answer:
Nitric acid is prepared at lower temperature (200°C) in laboratory because :

  • At high temperature nitric acid decomposes.
    4HNO3 = 4NO2 + O2 + 2H2O
  • Nitric acid vapour attacks glass surface of the retort.
  • Sodium sulphate (Na2SO4) formed at the higher temperature sticks to glass and is difficult to remove from the glass retort.

Question 22.
Why is cone, sulphuric acid used instead of cone, hydrochloric acid in the preparation of nitric acid in laboratory ?
Answer:
Choice of acid : In the preparation of an acid, the reacting acid must be less volatile than the acid to be prepared.Hence, cone. H2SO4(b.p. 338°C) is chosen because it is less volatile than nitric acid (b.p. 86°C). Hydrochloric acid is more volatile than nitric acid, so hydrochloric acid is not used. Beside this hydrochloric acid reacts with produced nitric acid to form nitrosyl chloride, chlorine and water.
3HCl + HNO3 = NOCl + Cl2 + 2H2O

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 23.
What is fuming nitric acid ?
Answer:
Fuming nitric acid : It is brown coloured liquid. It is actually cone. HNO3 containing dissolved nitrogen dioxide (NO2) gas. The dissolved nitrogen di­oxide spontaneously comes out of the liquid as brown fumes. That is why, it is called fuming nitric acid.

Preparation of fuming nitric acid : It is prepared by distilling cone, nitric acid with starch or arsenious oxide (As2O3).

Property and use of fuming nitric acid : Fuming nitric acid is a stronger oxidising agent than ordinary cone, nitric acid. It is used for many oxidation purposes where ordinary nitric acid does not give satisfactory results.

Question 24.
State the preparation of nitric acid by Ostwald process.
Answer:
Preparation of nitric acid by Ostwald process (1914) :

  • In this process ammonia is oxidised in presence of heated (750°C-900°C) platinum gauze catalyst by oxygen in air to produce nitric oxide.
    4NH3 + 5O2 = 4NO↑ + 6H2O + 218,660 cal
  • The gases containing nitric oxide (NO) and steam (H2 are cooled and then mixed with air to oxidise nitric oxide to nitrogen dioxide.
    2NO + O2 = 2NO2
  • Nitrogen dioxide, is then absorbed in water to yield a 50% solution of nitric acid and nitric oxide.
    3NO2 + H2O = 2HNO3 + NO↑

Question 25.
What are the physical properties of nitric acid ?
Answer:
Physical properties of nitric acid :

  • Pure nitric acid is a colourless, fuming liquid of specific gravity 152 at 15°C,
  • It boils at 86°C and freezes at -42°C into a transparent crystalline substance.
  • It is highly soluble in water.

Question 26.
State the reaction of nitric acid with alkalis.
Answer:
Reaction of nitric acid with alkalis :
As nitric acid is a strong acid, it rapidly reacts with and neutralises alkali to form salt and water.
Equations :
(a) NaOH + HNO3 = NaNO3 + H2O
(b) NH4OH + HNO3 = NH4NO3 + H2O

Question 27.
State the reaction of nitric acid with metals.
Answer:
Reaction of nitric acid with metals :
(i) Reaction with iron : Iron is oxidised to ferric nitrate on reacting with hot concentrated nitric acid.
Equation : Fe + 6HNO3 = Fe(NO3)3 + 3NO2 + 3H2O

(ii) Reaction with magnesium :

  • Cold and moderately concentrated nitric acid on reaction with magnesium produces magnesium nitrate, nitric oxide and water.
    Equation : 3Mg + 8HNO3 = 3Mg(NO3)2 + 2NO + 4H2O
  • Magnesium liberates hydrogen from cold and very dilute nitric acid.
    Equation : Mg + 2HNO3 Mg(NO3)2 + H2
  • Reaction with copper: Hot concentrated nitric acid reacts with copper metal producing copper nitrate and nitrogen dioxide.
    Equation : Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 28.
State the oxidising action of concentrated nitric acid.
Answer:
Oxidising action of concentrated nitric acid :
Cone. HNO3 is a very good oxidising agent.
It oxidises many metals and non-metals.

(i) Cone. HNO3 oxidises copper turnings to copper nitrate and itself is reduced to brown coloured nitrogen dioxide gas.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 23

(ii) Concentrated HNO3 oxidises charcoal i.e. carbon to carbon dioxide and is itself reduced to nitrogen dioxide.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 22

Question 29.
What is passive iron?
Answer:
Passive iron: Cold and concentrated nitric acid or fuming nitric acid when comes in contact with iron produces passive iron. Passive iron is chemically inactive.

Cause if passivity : The nibic acid when comes in contact with iron, an insoluble coating of iron oxide (Fe3O4) forms on iron at initial stage. The coating makes iron chemically inactive.

Question 30.
What is aqua regia ? What is its use ?
Answer:
Aqua regia : A mixture of cone. HNO3 (1 vol.) and cone, hydrochloric acid (3 vols.) is known as aqua regia.
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 24

Use of aqua regia : It dissolves gold and platinum.

Question 31.
What is ring test experiment of nitric acid ?
Answer:
Ring test experiment of nitric acid : A little amount of dilute nitric acid is taken in a test tube. Freshly prepared ferrous sulphate is added to it. Now cone. H2SO4 is taken in another test tube. cone. H2SO4 taken in a tube is carefully poured into the above liquid mixture so as to form a heavy layer at the bottom. A brown ring is formed at the junction of two liquids.

Thus, the reaction involved in ring tests as :WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 25

Question 32.
State uses of nitric acid.
Answer:
Uses of nitric acid :

  • It is used in the production of explosives such as dynamite, gun cotton, nitro-glycerine, picric acid, and trinitro toluene (TNT).
  • It is used to prepare aqua regia in electroplating and in battery.
  • It is also used to prepare rayon, artificial silk and dyes.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 33.
What is acid rain ?
Answer:
Acid rain The oxides NO2 and SO2 reacting with moisture and oxygen of air correspondingly produce HNO3 and H2SO4. The acids dissolve in rain water. The rain coming down on earth carrying these acids is known as acid rain. Function of acid rain : The acids present in the rain water damage buildings monuments, statues by corrosion. Soil also becomes acidic which causes degradation of soil that in turn causes decline in forest area and agricultural productivity.

Question 34.
Sulphuric acid is not prepared directly by adding water to SO3 why ?
Answer:
Explanation : It is because of the fact that water is not used for absorbing SO3 as the reaction is exothermic. Result is this that large amount of heat evolved will be a cause of most of SO3 to be volatilised, the rest amount of it does not form any homogeneous mixture but a dense fog of H2SO4 particles is formed. H2SO4 is not easily available from the fog. Beside this at high temperature some parts of H2SO4 decompose to SO2 and O2. Hence water is not directly added to SO3.

Question 35.
What is fumming sulphuric acid or oleum ?
Answer:
Fumming sulphuric acid or oleum : Fumming sulphuric acid or oleum is obtained when SO3 is passed over 98% (approximately) sulphuric acid.
Equation:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 26

Question 36.
State the effect of SO2 pollution
Answer:
of SO2 pollution ;

  • SO2 creates problems in eyes and also in lungs,
  • Many diseases like asthama, bronchitis etc. effect if SO2 is inhaled.

Question 37.
Historical monuments like Tajmahal is also affected by sulphuric acid – why ?
Answer:
Reason : Historical monuments like Tajmahal are damaged due to corrosion when rain carrying H2SO4 acid pour on these. Environmentalists say the corrosion of Tajmahal is due to presence of SO2 in high concentration in the air of Agra. The SO2 gas emitted from petroleum refinery situated on the other side of the Jamuna river reacts with oxygen and moisture of air to produce H2SO4.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 38.
What are possible remedies to save our historical monument—Tajmahal?
Answer:
Possible remedy :

  • Banning of this factories in the neighbourhood region of Taj and also in Mothura is promulgated by an ordinance.
  • Excessive use of motor vehicles should be restricted. Petroleum refining plant should be kept 90 km apart.
  • Electrostatic precipitators which are used for treatment of the harmful emitted gas S02 before emission.

Question 39.
What is stone cancer ?
Answer:
Stone cancer : In atmosphere, Sulphur dioxide (SO2) gas reacts with oxygen and water vapour producing sulphuric acid. The acid being dissolved in rain water comes down on marble walls. The marble walls thus corrode, the corrosion of the marble is called stone cancer.

Question 40.
State physical properties of H2SO4.
Answer:
Physical properties of H2SO4 :

  • It is colourless, odourless heavy oily liquid.
  • Its specific gravity is 1.84 and its b.p. (boiling point) is 338°C.
  • It is soluble in water and it is a corrosive acid.

Question 41.
Water is not added on concentrated sulphuric acid to make it dilute — why ?
Answer:
Explanation : Huge amount of heat is produced if water is added to con­centrated sulphuric acid. As a result water all on a sudden being volatilised spreads all around and creates problem. So, cone. H2SO4 acid is slowly added to with constant stirring.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 42.
State the reaction of H2SO4 with alkalis.
Answer:
Reaction of H2SO4 with alkalis : As H2SO4 is a dibasic acid containing two replaceable hydrogen atoms, it reacts with bases and alkalis to produce two types of salts acid salts and normal salts and water.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 27

Question 43.
State the reaction of H2SO4 with carbonates and bicarbonates.
Answer:
Reaction of H2SO4 with carbonates and bicarbonates : At ordinary temperature it liberates carbon dioxide from carbonates and bicarbonates.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 28

Question 44.
State the reaction of H2SO4 with metals.
Answer:
Reaction of H2SO4 with metals :
Metals lying above hydrogen in the electrochemical series react with dilute sulphuric acid to form hydrogen gas.
Equation :
(a) Mg + H2SO4 = MgSO4 + H2
(b) Zn + H2SO4 = ZnSO4 + H2

Question 45.
State the oxidising property of concentrated sulphuric acid.
Answer:
Oxidising property of concentrated sulphuric acid :
(a) Reaction with metals : Hot and concentrated sulphuric acid oxidises metallic copper, silver, zinc etc. into their sulphates, itself being reduced to sulphur dioxide.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 29

(b) Reaction with non-metals Non-metals carbon and sulphur on being treated with hot concentrated sulphuric acid are oxidised to carbon dioxide, sulphur dioxide respectively.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 30

Question 46.
State the dehydration action of concentrated sulphuric acid.
Answer:
Dehydration action of concentrated sulphuric acid :
(a) Concentrated sulphuric acid chars sugar. The acid takes up all its hydrogen and oxygen in the form of water molecules from the compound separating black carbon as residue.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 34

(b) When a mixture of formic acid and concentrated sulphuric acid is heated carbon monoxide is produced by loss of water molecule from the formic acid.

Equation
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 35

Question 47.
What is the identification reaction of sulphuric acid ?
Answer:
Identification reaction of sulphuric acid : When barium chloride solution is added to dilute sulphuric acid, a white precipitate of barium sulphate insoluble in dilute or concentrated hydrochloric acid or nitric acid is obtained.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 33

Question 48.
State the uses of sulphuric acid.
Answer:
Uses of sulphuric acid :

  • It is used for the preparation of different chemical compounds like HCl, HNO3, ether, alcohol etc.
  • It is used as an important raw material for dyes, medicine, plastic industries.
  • It is used to prepare different explosives like trinitrotoluene (TNT) nitroglecer- ine etc.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 49.
Explain the following :
Concentrated nitric acid turns yellow in sun light.
Answer:
Explanation Nitric acid turns yellow because of its decomposition forming nitrogen dioxide (NO2gas.
Equation : 4HNO3= 4NO2+ 2H2O + O2

Broad Answer Type Questions

Question 1.
State the contact process for the manufacture of sulphuric acid.
Answer:
Contact process for the manufacture of sulphuric acid :
Principle : (a) Formation of SO2: SO2 is prepared by burning sulphur or iron pyrites in excess of air.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 38

(b) Formation of Sulphur trioxide (SO3) SO3 is prepared by the oxidation of sulphur dioxide with oxygen (from air) in presence of platinised asbestos or, V2O5 as catalyst at 450°C.
Equation:
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 37

(c) Formation of oleum : Now sulphur trioxide thus produced are not allowed to react with water directly. Sulphur trioxide absorbs concentrated sulphuric acid (98%) turning it as fumming sulphuric acid or oleum.
Equation :
WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 36

(d) Dilution of oleum to sulphuric acid : Pure and concentrated sulphuric acid is produced by adding slowly requisite amount of water in fumming sulphuric acid.
Equation : H2S2O7 + H2O = 2H2SO4

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Question 2.
State an experiment to illustrate the solubility of HCl in water and the acidic character of the aqueous solution.
Answer:
Fountain experiment : A round bottomed flask is fitted with a cork through which passes a long tube ending in a jet in­side the flask. The flask is filled with hydrogen chloride and clamped in an inverted position so that the other end of the tube dips in water coloured blue by litmus.

The flask is then cooled when the gas contracts inside. As a result, the blue solution rises up through the tube and enters the flask in the form of fountain. The blue litmus solution in the flask turns red. This experiment proves simultaneously the solubility of hydrogen chloride in water and the acidic character of its aqueous solution.

WBBSE Class 10 Physical Science Solutions Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 39

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 3 Chemical Calculations offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 3 Question Answer – Chemical Calculations

Very Short Answer Type Questions

Question 1.
What is the molecular weight of ammonium phosphate ?
Answer:
149

Question 2.
How much CaO is obtained from 10 kg CaCO3 ?
Answer:
5.6 kg

Question 3.
How much CaCO3 will react with dil HCl to produce 22 g CO2 ?
Answer:
50 g

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 4.
How much hydrogen is required to produce 72 g water?
Answer:
8g

Question 5.
How much potassium chlorate is to be heated to produce as much oxygen as required to burn 6 g carbon completely?
Answer:
40.83 g

Question 6.
How much quantity of silver chloride (AgCl) can be obtained from 1.0 g silver nitrate?
Answer:
0.844 g

Question 7.
How many grams limestone are needed to get 48g CO2 ?
Answer:
109.09 g

Question 8.
How much hydrogen is produced when steam is passed over 28 kg red hot iron?
Answer:
1.33 kg

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 9.
What is mole number of a substance ?
Answer:
Mole number of a substance

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 1

Question 10.
What is the percentage of water in blue vitriol ?
Answer:
36.07%.

Question 11.
2 Mg + O2 = 2 MgO; what type of chemical reaction is it?
Answer:
2 Mg + O2 = 2 MgO is direct combination type of chemical reaction.

Question 12.
HCl + NaOH = NaCl + H2 O; what type of chemical reaction is it?
Answer:
HCl + NaOH = NaCl + H2 O is the type of neutralisation reaction.

Question 13.
Zn + CuSO4 = ZnSO4 + Cu; what type of chemical reaction is it?
Answer:
Zn + CuSO4 = ZnSO4 + Cu is the type of substitution reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 14.
NH4 CNO (8) CO(NH2)2; what type of chemical reaction is it?
Answer:
NH4 CNO arrow CO(NH2)2 is the type of rearrangement reaction.

Question 15.
Give an example of double decomposition type of reaction.
Answer:
AgNO3 + NaCl = AgCl + NaNO3 is an example of double decomposition type of reaction.

Question 16.
State a limitation of a chemical equation.
Answer:
A reaction whether exothermic or endothermic is not known from a chemical equation.

Question 17.
Give an example of addition reaction.
Answer:
Carbon monoxide reacts with chlorine to form carbonyl chloride (COCl2) is an example of addition reaction.

Question 18.
Give an example of a substitution reaction.
Answer:
Chlorine reacts with methane (CH4) to form successively CH3 Cl, CH2 Cl2, CHCl3 and ultimately CCl4 in presence of sunlight is an example of a substitution reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 2

Question 19.
Give an example of a thermal decomposition type of reaction.
Answer:
Calcium carbonate on heating decomposes into calcium oxide and carbon dioxide is an example of a thermal decomposition type of reaction.

Question 20.
Give an example of redox type of reaction.
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 3
is an example of redox type of reaction.

Question 21.
Balance the following reaction: Fe + H2 O(Fe3 O4 + H2.
Answer:
Balanced equation of Fe + H2 O arrow Fe3 O4 + H2 is
3 Fe + 4 H2 O = Fe3 O4 + 4 H2

Question 22.
2 SO2 + O2 ⇌ 2 SO3; what type of reaction is it ?
Answer:
2 SO2 + O2 ⇌ 2 SO3 is an example of reversible type of reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 23.
A + B2+ = A2+ + B; what type of reaction is it ?
Answer:
A + B2+ = A2+ + B is an example of oxidation-reduction type of reaction.

Question 24.
Balance the following reaction :
MnO2 + HCI → MnCI2 + CI2 + H2O
Answer:
Balanced equation of MnO2 + HCl → MnCl2 + Cl2 + H2O is
MnO2 + 4 HCl = MnCl2 + Cl2 + 2 H2O

Question 25.
Balance the following reaction :
Ag2 SO4 + BaCI2 → BaSO4 + AgCl.
Answer:
Balanced equation of Ag2 SO4 + BaCl2 → BaSO4 + AgCl is
Ag2 SO4 + BaCl2 = BaSO4 + 2 AgCl

Question 26.
Write the equation relating energy with mass.
Answer:
E = m c2

Question 27.
Write the relation between molecular mass and vapour density.
Answer:
Molecular mass = 2 × Vapour density.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 28.
Calculate the vapour density of oxygen relative to hydrogen.
Answer:
Vapour density of oxygen
= \(\frac{Molecular mass of oxygen}{2} = [latex]\frac{32}{2}\) = 16

Question 29.
6 g carbon on complete combustion produce how much volume of CO2 at STP ?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 4
12g carbon on complete combustion produce 22.4 L CO2 at STP
6g carbon on camplete combustion produce 11.2 L CO2 at STP

Question 30.
One mole of zinc reacts with excess H2 SO, to produce how many moles of hydrogen ?
Answer:
1 mole of hydrogen.

Short Answer type Questions :

Question 1.
What is empirical formula?
Answer:
Empirical formula : It is the formula of a compound which gives the simple whole number ratio of the atoms of various elements present in one molecule of compound.

Question 2.
What is molecular formula ?
Answer:
Molecular formula : It is the formula of a compound which gives the actual number of atoms of various elements present in one molecule of the compound.

Question 3.
What are the limitations of chemical formula ?
Answer:
Limitations of chemical formula :

  1. It fails to convey whether the elements in a molecule are present in the form of atoms or ions.
  2. It does not tell anything about the binding force that holds atoms in a molecule together.
  3. It does not tell us about the arrangement of various atoms with respect to one another within the molecule.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 4.
What is structural formula ?
Answer:
Structural formula : A formula which gives the actual arrangement of the different atoms in the molecule or show how the different atoms in the molecule are linked together is called a structural or a graphic formula of the compound.

Question 5.
What is percentage of an element?
Answer:
Percentage of an element : Percentage of an element in a chemical compound is the number of parts by weight of it present in 100 parts by weight of the compound.

Question 6.
What are mass-mass relationship problems ?
Answer:
Mass-mass relation hip problems : In this type of problems, mass of one of the reactants/products is to be calculated if that of the other reactants/ products is to be calculated if that of the other reactants/ products are given.

Question 7.
What are mass-volume relationship problems ?
Answer:
Mass-volume relationship problems : In this type of problems, mass or volume of one of the reactants or products is calculated from the volume or mass of other substances.

Question 8.
What are volume-volume relationship problems ?
Answer:
Volume-volume relationship problems : In this type of problems, the volume of one of the reactants or products is to be calculated from the volume of some other reactant and product.

Question 9.
What is limiting reactant ?
Answer:
Limiting reactant : The reactant which is completely used and determines the amount of product formed is known as limiting reactant.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 10.
What is theoretical yield ?
Answer:
Theoretical yield : The theoretical yield of a product is the maximum yield obtainable as calculated on the basis of the amount of limiting reactant used.

Question 11.
What is percent yield?
Answer:
Percent yield : Percent yield which is the ratio of the actual yield to the theoretical yield multiplied by 100 .

Question 12.
What do you mean by chemical reaction ?
Answer:
Chemical reaction: Any chemical change in matter which involves the transformation of matter into a new substance or new substances is termed as chemical reaction.

Question 13.
Define reactants in a chemical reaction.
Answer:
Reactants: The substances with which a chemical reaction is started are called the reactants.

Question 14.
Define products in a chemical reaction.
Answer:
Products: The substances formed as the result of the chemical reaction are called the products.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 15.
Define direct combination reaction.
Answer:
Direct combination : The reaction in which two or more reactions combine directly forming molecules of new substance is called direct combination.
Example : Burning magnesium wire reacts directly with oxygen forming new molecules of magnesium oxide.
2 Mg + O2 = 2 MgO

Question 16.
What is meant by decomposition reaction?
Answer:
Decomposition : When molecule of a substance is broken down or decomposed in a chemical reaction to form two (or more) new substances in presence of heat or electricity. It is called decomposition.
Example : (a) Calcium carbonate on heating decomposes into calcium oxide and carbon dioxide.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 5

(b) Electricity is passed through acidulated water hydrogen and oxygen are produced due to decomposition of water.
2H2O = 2H2 + O2

Question 17.
Define double decomposition reaction.
Answer:
Double decomposition : The reaction in which the constituents of the molecules of reactants Change their position and forming molecules of new substances is called double decomposition.
Example : AgNO3 + NaCl = AgCl↓+ NaNO3

Question 18.
What is meant by acid-base reaction or neutralisation reaction ?
Answer:
Acid-base reaction or neutralisation reaction : In this reaction an acid reacts with a base forming salt and water. Equivalent amount of an acid neutralises an equivalent amount of a base. This is known as neutralisation reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 6

Question 19.
Define oxidation-reduction type reaction.
Answer:
Oxidation-reduction reaction : The chemical reaction in which a chemical species loses electron(s) is an oxidation reaction and the reaction in which a chemical species gains electron(s) is called oxidation-reduction reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 7

Question 20.
What is meant by addition reaction?
Answer:
Addition reaction: The reaction in which one reactant molecule directly combines with other molecule of reactant forming new molecules of product without leaving any part of molecules of reactants is known as addition reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 8

Question 21.
Define substitution reaction.
Answer:
Substitution reaction: It is a chemical reaction in which one atom or molecule of a compound is replaced by another atom or molecule of another substance.
Example : Zn + CuSO4 = ZnSO4 + Cu

Question 22.
What is meant by rearrangement reaction?
Answer:
Rearrangement reaction: A reaction where a compound changes by internal arrangement of its atoms into another substance with different properties but having the same composition is known as rearrangement reaction.
Example : Ammonium cyanate changes to urea on heating.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 9

Question 23.
Define chemical equation.
Answer:
Chemical equation : A brief representation of a chemical reaction by using symbols of atoms of the elements and formulae of molecules of reactants and products maintaining the balance in between is known as chemical equation. It reveals both the qualitative ancL quantitative aspects of a chemical change.

Broad Answer Type Questions :

Question 1.
What steps should be followed to solve a problem based on chemical equation by weight-weight method ?
Answer:
Following steps are followed :

  1. First write down the complete balanced chemical equation.
  2. The molecular weight of the reactants and products are to be calculated from their respective formulae by adding atomic weights of the concerned elements.
  3. The unknown weight of the substance asked for in the question is to be calculated from the calculated weight in the equation.
  4. Same units are to be used for all quantities.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 2.
What are the informations avallable from the formula CaCO3 ?
Answer:
The formula CaCO3 conveys the following informations :

  1. It stands for calcium carbonate.
  2. It tells that calcium carbonate is composed of calcium, carbon and oxygen elements.
  3. It represents that a molecule of calcium carbonate.
  4. It represents that a molecule of calcium carbonate is made up of One atom each of calcium and carbon and three atoms of oxygen.
  5. It stands for 100 parts by weight of calcium carbonate i.e. its molecular weight and signifies that the ratio of calcium, carbon and oxygen by weight in it is 40: 12: 48.

Question 3.
What is the method of writing chemical equation ?
Answer:
Method of writing chemical equation :
i. Initially the symbols of atoms of elements and formulae of molecules of reactants and products are written.

ii. Symbols and formulae of the reactants are written in the left side and these for products are in the right side. An arrow mark (→) is placed between reactants and products. For more than one reactants and products a plus sign (+) is to be given between the reactants and also between the products.

iii. Number of atoms of the reactants in the left hand side must be equal to the number of atoms of the products in the right hand side so as to maintain the law of conservation of mass. To equalise both sides, proper multiplication is required so that the number of atoms in both sides are same.

iv. Now the equation is expressed by replacing the arrow sign by a sign of equal ( = ).

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 4.
What informations are obtained from the chemical equations ?
Answer:
A chemical equation gives the following informations :
(a) Qualitative informations :

  • From chemical equation, the naming of elements and compounds taking part in the reaction and also the products are known.
  • From chemical equation, the symbols and the formulae of the reactants and products are also known.

(b) Quantitative informations :

  • Number of atoms or molecules of the reactants and products involved in the reaction are known.
  • How many parts of elements and compounds taking part in the reaction and producing are also known.
  • If the reactants and products are all gaseous then at the same temperature and pressure the ratio in volumes is known.

Question 5.
What informations are obtained from the equation :
C + O2 = CO2
Answer:
The following informations are obtained from the equation :
C + O2 = CO2

(a) Qualitative informations :
i. From the above equation it is known that carbon combines with oxygen producing carbon dioxide.
ii. Their respective symbols and formulae are also known.

(b) Quantitative informations :
i. One carbon atom combines with one oxygen molecule producing one carbon dioxide molecule.
Hence total number of atoms in the left hand side is (1 + 2) = 3 and the number of atoms in the right hand side is also 3.
ii. One gram-atom carbon reacts with one gram-molecule of oxygen giving rise one gram-molecule of carbon dioxide.
12 g carbon combines with 32 g oxygen yield 44 g carbon dioxide.
Hence, total mass in the left hand side (12 + 32) g = 44 g is equal to the total mass in the right hand side 44 g. It proves law of conservation of mass.
iii. Again 12 g carbon combines with 32 g oxygen producing 22.4 litre of carbon dioxide at STP.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 6.
What informations are obtained from the equation :
2 H2 + O2 = 2 H2O
Answer:
The following informations are obtained from the equation :
2 H2O + O2 = 2 H2O

(a) Qualitative informations : The reactants are hydrogen and oxygen, water is the product.
(b) Quantitative informations :

  1. Two molecules of hydrogen combine with one molecule oxygen to produce two molecules water.
  2. Two volumes hydrogen and one volume oxygen combine to produce two volumes water at the same temperature and pressure and their ratio is 2 : 1 : 2
  3. Two moles of hydrogen and one mole oxygen combine chemically to produce two moles water.

Question 7.
What informations are obtained from the equation :
3 H2 + N2 = 2 NH3
Answer:
The following informations are obtained from the equation :
3 H2 + N2 = 2 N H3
(a) Qualitative informations : The reactants are hydrogen and nitrogen, ammonia is the product.
(b) Quantitative informations :

  1. Three molecules of hydrogen combine with one molecule nitrogen to produce two molecules ammonia.
  2. Three volumes hydrogen and one volume nitrogen combine to produce two volumes ammonia at the same temperature and pressure and their ratio is 3 : 1 : 2.
  3. Three moles of hydrogen and one mole nitrogen combine chemically to produce two moles ammonia.

Question 8.
What are the limitations of a chemical equation ?
Answer:
Limitations of a chemical equation :

  1. Under what conditions a chemical reaction is occurred i.e. pressure, temperature, catalyst are not known.
  2. A reaction whether exothermic or endothermic is not known from chemical equation.
  3. Chemical equation cannot provide information about completion of the reaction or attainment of equilibrium.
  4. A chemical equation does not give any idea about the reversibility of the reaction.
  5. From a chemical equation, the time required for the completion of the equation is not known.
  6. A reaction whether slow or fast does not know from the equation.
  7. Equation does not provide information regarding the nature of the reactants and products viz. solid, liquid or gas.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 9.
What is meant by ‘balancing’ of a chemical equation ? Why it is necessary?
Answer:
Balancing of a chemical equation : It means making both the sides of the equation equal in respect of the kind and number of the atoms of the elements involved.
Balancing of a chemical equation is necessary : Due to the conservation of mass and indestructibility of matter, no atoms can be created or destroyed in a chemical reaction. Hence, the same kind of atoms in same numbers must be present in both the sides of the chemical equation.

Question 10.
Explain the method of balancing a chemical equation by trial and error method.
Answer:
Balancing of a chemical equation by trial and error method :
(i) In this method the proper numbers are putting before reactants and products so that the number of atoms in both sides are equal.
(ii) In doing so, we must cautious to see that all the reactants and products exist as molecules, and not as atoms, as free existence of atoms is not possible.

Question 11.
Balance the equation KClO3 ⟶ KCl + O2 by trial and error method.
Answer:
KClO3 ⟶ KCl + O2; it is seen that the number of atoms of oxyen in left hand side is 3 whereas in the right hand side it is 2. To equalize the number of atoms of oxygen in both sides it requires to multiply KClO3 by 2 and oxygen by 3.
2 KClO3 ⟶ KCl + 3 O2. So as to balance this equation if KCl is multiplied by 2 then the equation is properly balanced.
2 KClO3 = 2 KCl + 3 O2

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 12.
Write the following as balanced equations.
(i) Cu + HNO3 ⟶ Cu(NO3)2 + NO + H2 O
(ii) Pb(NO3)2 ⟶ PbO + NO2 + O2
(iii) Al + NaOH + H2 O ⟶ NaAlO2 + H2
(iv) CuO + NH3 ⟶ Cu + N2 + H2 O
Answer:
Balanced equations are :
(i) 3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2NO + 4H2 O
(ii) 2 Pb(NO3)2 = 2 PbO + 4 NO2 + O2
(iii) 2 Al + 2 NaOH + 2 H2 O = 2 NaAlO2 + 3H2
(iv) 2 NH3 + 3 CuO = 3 Cu + 2 N2 + 3H2O

Numerical problems :

Working formula :

(i) Molecular weight (M) = n × Emperical formula
(ii) Atomic weight of a gaseous element = \(\frac{\text { Actual yield }}{\text { Theoretical yield }} \times 100\)
Problems related to mass-mass relationship

Example 1:
How many grams of oxygen evolve when 122.5 g potassium chlorate is heated ? (Given, K = 39, Cl = 35.5, O = 16)
Answer:
The balanced equation is :
2 KClO = 2 KCl + 3 O2
2[39 + 35.5 + 3 × 16] g & 3[2 × 16] g
= 245 g = 96 g
By heating 245 g KClO3 96 g O2 is obtained
∴ By heating 122.5 g KClO3 \(\frac{96 \times 122.5}{245}\) = 48 g O2 is obtained

Example 2:
2.6 g zinc is treated with excess dil H2 SO4; how many gram oxygen combines with the evolved hydrogen?
Answer:
The balanced equation for production of hydrogen is :
Zn + H2 SO4 = ZnSO4 + H2
65g (2 × 1) g = 2g
So, 65 g zinc produces 2 g hydrogen
∴ 2-6 zinc produces \(\frac{2 \times 2.6}{65} \mathrm{~g}\) = 0.089 g hydrogen
Now, the reaction where hydrogen and oxygen combine is :
2 H2 + O2 = 2 H2 O
So, 4 g hydrogen combine with 32 g oxygen 0.08 g hydrogen combine with \(\frac{32 \times 0.08}{4} \mathrm{~g}\) = 0.64 g oxygen

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Example 3:
What is the observed loss in weight of 5 g calcium carbonate when it undergoes thermal decomposition?
Answer:
The balanced equation is :
CaCO3 = CaO + CO2
(40 + 12 + 16 × 3)g (12 + 16 × 2)g
= 100g = 44g
Loss in weight in the weight of CO2 that escapes
100 g CaCO3 produces 44g CO2
∴ 5 g CaCO3 produces \(\frac{44 \times 5}{1000}\) g = 2.2g CO2

Example 4:
What is the percentage of ammonia in that quantity of ammonium chloride that can produce 5g ammonia?
Answer:
The balanced equation of reaction is :
2 NH4 Cl + CaO = 2 NH3 + CaCl2 + H2 O
2[14 + 1 × 4 + 35.5] g 2[14 + l × 3] g
= 107 g = 34 g
So, 34 g ammonia is obtained form 107 g NH4 Cl

Example 5:
What weight of potassium chlorate of 96% purity will yield 4.8 g oxygen on complete thermal decomposition?
(Given, K = 39, Cl = 35.5, O = 16)
Answer:
The balanced equation is :
2 KClO3 = 2 KCl + 3 O2
245 g 96 g
∴ 96 g oxygen is obtained from 245 g KClO2 of 100% purity
∴ 4.8 g oxygen is obtained from \(\frac{245 \times 4 \cdot 8}{96}\) g
= 12.25 g KClO3 of 100% purity.
Let x gram of 96% purity contains 12.25 g KClO3 of 100% purity
∴ \(\frac{96}{100}\).x = 12.56
or, x = \(\frac{12 \cdot 25 \times 100}{96}\) = 12.76 (approx)
So, the required quantity of KClO3 = 12.76g

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Example 6:
On strong heating limestone decomposes into quicklime and carbon dioxide. How much quantity of limestone will produced on complete decomposition, 30 g of quicklime by the above reaction?
Answer:
The balanced equation is :
CaCO3 = CaO + CO2
[40 + 12 + 3 × 16] g [40 + 16] g
= 100 g = 56 g
So, 56 g CaO is obtained by the complete decomposition of 100 g CaCO3
∴ 30 g CaO is obtained by the complete decomposition of
\(\frac{100 \times 30}{56}\)g = 53.6 g CaCO3
Thus 53.6 g limestone will have to be decomposed.

Example 7:
How many grams of magnesium metal will give 1.2 g hydrogen on complete reaction with dilute H2 SO4 (Mg = 24, H = 1)
Answer:
The balanced equation is :
Mg + H2 SO4 = MgSO4 + H2
24 g (1 × 2) g = 2 g
So, 2 g H2 is obtained from 24 g Mg
∴ 1.2 g H2 is obtained from \(\frac{24 \times 1.2}{2}\)g Mg = 14.4 g Mg
So, 14.4 g Mg will be required.
Problems based on mass-volume relationship

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Example 8:
What volume of carbon dioxide measured at 300 K and 720 mm pressure be obtained on treatment of 1 g CaCO3 with dilute HCl ?
Answer:
The balanced equation is :
CaCO3 + 2 HCl = CaCl2 + H2O + CO2
1 mol(40 + 12 + 3 × 16) g        1 mol (22.4 lit at NTP)
= 100 g
So, 100g CaCO3 gives 22.4 lit CO2 at NTP
1g CaCO3 gives 0.224 lit CO2 at NTP

Reducing the volume to the given condition applying gas equation,

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 10

P1 = 760 mm
V1 = 0.224 lit
T2 = 273 K
P2 = 720 mm
V1 = ?
T2 = 300 K

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 11

= 0-2598 lit = 259-8 ml
Hence, 259.8 ml CO2 will be obtained at 300 K and 720 mm pressure on treatment of 1 g CaCO3 with HC 1.

Example 9:
How much potassium nitrate should be heated to get enough oxygen required to completely burn 56 lit of hydrogen at NTP ?
Answer:
The balanced equation is :

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 12

= 2 × 22-4 lit
= 44.8 lit (at NTP)
So, 44.8 lit H2 requires 22.4 lit O2 at NTP
56 lit H, requires \(\frac{2 \cdot 24 \times 56}{44 \cdot 8}\) at NTP
= 28 lit O2 at NTP

The other balanced equation is :
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 13
= 202 g
Now, 22.4 lit O2 at NTP are obtained by heating 202 g KNO3
∴ 28 lit O2 at NTP are obtained by heating \(\frac{202}{22.4}\) × 28 g KNO3
= 252.5 g KNO3

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Problems based on volume-volume relationship.

Example 10:
In the Ostwald process for the manufacture of nitric acid, ammonia gas is burnt in oxygen in the presence of a pt-catalyst. What volume of O2 is required and what volume of NO is formed in the combustion of 500 lit of NH3.
Answer:
The balanced equation is :
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 14

(i) 4 lit of NH3 requires 5 lit of O2 for combustion
500 lit of NH3 requires \(\frac{5 \times 500}{4}\) lit = 625 lit O2
(ii) 4 lit of NH3 produces 4 lit NO
∴ 500 lit of NH3 produces \(\frac{4}{4}\) × 500 lit NO = 500 lit NO

Example 11:
Calculate the volume of oxygen necessary to burn completely 5 lit butane gas. What is the volume of carbon dioxide formed?
Answer:
The balanced equation is :
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 15

So, 2 lit of butane at NTP requires 13 lit O2 at NTP
∴ 5 lit of butane at NTP requires \(\frac{13 \times 5}{2}\) lit = 32.5 lit of O2 at NTP.
Also 2 lit of butane at NTP produce 8 lit CO2 at NTP
∴ 5 lit of butane of NTP produce
= \(\frac{8}{2}\) × 5 lit = 20 lit CO2 at NTP