Class 10 Science

Chapter 3: Thermal Physics – Questions & Answers

Class 10 Science Chapter 3 Choose the Correct Answer

Question 1.
The value of universal gas constant:
(a) 3.81 mol^-1 K^-1
(b) 8.03 mol^-1 K^-1
(c) 1.38 mol^-1 K^-1
(d) 8.31 mol^-1 K^-1

Answer:
(d) 8.31 mol^-1 K^-1

Question 2.
If a substance is heated or cooled, the change in mass of that substance is:
(a) positive
(b) negative
(c) zero
(d) none of the above

Answer:
(b) negative

Question 3.
If a substance is heated or cooled, the linear expansion occurs along the axis of ______.
(a) X or -X
(b) Y or -Y
(c) both (a) and (b)
(d) either (a) or (b).

Answer:
(c) both (a) and (b)

Hint: When a substance is heated its expansion is positive i,e, can be taken along either +X or +Y direction. But when substance is cooled it’s either length or area or volume decreases i.e. with respect expansion, it is opposite direction i.e. either -X or -Y direction respectively.

Question 4.
Temperature is the average of the molecules of a substance.
(a) difference in K.E and P.E
(b) sum of P.E and K.E
(c) difference in T.E and P.E
(d) difference in K.E and T.E

Answer:
(b) sum of P.E and K.E

Question 5.
In the Given diagram, the possible direction of heat energy transformation is:

(a) A ← B, A ← C, B ← C
(b) A → B, A → C, B → C
(c) A → B, A ← C, B → C
(d) A ← B, A → C, B ← C

Answer:
(a) A ← B, A ← C, B ← C

Class 10 Science Chapter 3

Fill in the blanks.

1. The value of Avogadro number ………..
2. The temperature and heat are ……….. quantities.
3. One calorie is the amount of heat energy required to raise the temperature of ……….. of water through
4. According to Boyle’s law, the shape of the graph between pressure and reciprocal of volume is …………

Answer:

1. 6.023 × 10²³
2. Inter convertible
3. 1 gram, 1°C
4. A straight line

Class 10 Science Chapter 3  State whether the following statements are true or false, if false explain why?

1. For a given heat in liquid, the apparent expansion is more than that of real expansion.
2. Thermal energy always flows from a system at higher temperature to a system at lower temperature.
3. According to Charles’s law, at constant pressure, the temperature is inversely proportional to volume.

Answer:

1. True
2. True
3. False – According to Charles law, at constant pressure, the volume is directly proportional to temperature.

Class 10 Science Chapter 3  Match the items in column-I to the items in column-II

Column I	Column II
A Linear expansion	(p) change in volume
B Superficial expansion	(q) hot body to cold body
C Cubical expansion	(r) 1.381 × 10-23 JK-1
D Heat transformation	(s) change in length
E Boltzmann constant	(t) change in area

Answer:
A. (s)
B. (t)
C. (p)
D. (q)
E. (r)

Class 10 Science Chapter 3  Assertion and Reason type questions.

(a) Both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) Both the assertion and the reason are true but the reason is not the correct explanation of the assertion.
(c) The assertion is true but the reason is false.
(d) The assertion is false but the reason is true.
1. Assertion: There is no effects on other end when one end of the rod is only heated.
Reason: Heat always flows from a region of lower temperature to higher temperature of the rod.

2. Assertion: Gas is highly compressible than solid and liquid
Reason: Interatomic or intermolecular distance in the gas is comparably high.
Answer:
1. (b)
2. (a)

Class 10 Science Chapter 3  Answer in briefly.

Question 1.
Define one calorie.
Answer:
One calorie is defined as the amount of heat energy required to rise the temperature of 1 gram of water through 1°C.

Question 2.
Distinguish between linear and superficial areal expansion.
Answer:

Linear Expansion	Areal and Superficial Expansion
In this expansion, length of a body increases.	In this expansion, area of a body increases.
Coefficient of linear expansion is different for different materials.	Coefficient of areal expansion is different for different materials.

Question 3.
What is the coefficient of cubical expansion?
Answer:
The ratio of increase in the volume of the body per degree rise in temperature to its unit volume is called a coefficient of cubical expansion.

Question 4.
State Boyle’s law
Answer:
When the temperature of a gas is kept constant, the volume of a fixed mass of gas is inversely proportional to its pressure.
P ∝ 1 / V

Question 5.
State-the law of volume.
Answer:
When the pressure of a gas is kept constant, the volume of a gas is directly proportional to the temperature of the gas.
i.e., V ∝ T.
(or)
VT = constant.

Question 6.
Distinguish between ideal gas and real gas.
Answer:

Ideal gas	Real gas
In this gas, molecules or atoms of a gas interact with each other with some interatomic force.	In this gas, atoms or molecules of a gas do not interact with each other.
They obey Boyle’s law, Charles law and Avogadro’s law.	They do not obey Boyle’s law, Charles law and Avogadro’s law.

Question 7.
What is co-efficient of real expansion?
Answer:
Coefficient of real expansion is defined as the ratio of the true rise in the volume of the liquid per degree rise in temperature to its unit volume. The SI unit of coefficient of real expansion is the K^-1.

Question 8.
What is the coefficient of apparent expansion?
Answer:
Coefficient of apparent expansion is defined as the ratio of the apparent rise in the volume of the liquid per degree rise in temperature to its unit volume.
The SI unit of the coefficient of apparent expansion is K^-1.

Class 10 Science Chapter 3  Numerical problems.

Question 1.
Find the final temperature of a copper rod whose area of cross section changes from 10 m² to 11 m² due to heating. The copper rod is initially kept at 90 K. (Coefficient of superficial expansion is 0.0021 /K).
Answer:
Change in area ΔA = 11 – 10 = 1 m²
Initial temperature T₁ = 90 K
Let Final temperature be T₂K
A₀ = 10 m²
Coefficient of superficial expansion is
α_A = 0.0021 / k
ΔAA0 = α_AΔT
110 = 0.0021 ΔT
∴ ΔT = 0.0021 × 10
= 0.021
T₂ – T₁ = 0.021
T₂ – 90 = 0.021
∴ Final temperature T₂ = 90.021 K

Question 2.
Calculate the coefficient of cubical expansion of a zinc bar. Whose volume is increased 0.25 m³ from 0.3 m³ due to the change in its temperature of 50 K.

Answer:

Calculation of Coefficient of Cubical Expansion

The coefficient of cubical expansion (β) is given by the formula:

β = ΔV / (V₀ × ΔT)

Given:

• Initial Volume, V₀ = 0.25 m³
• Final Volume = 0.30 m³
• Change in Volume, ΔV = 0.30 – 0.25 = 0.05 m³
• Temperature Change, ΔT = 50 K

Calculation:

Substitute the values into the formula:

β = 0.05 / (0.25 × 50)

β = 0.05 / 12.5

β = 0.004 K^-1

Result:

The coefficient of cubical expansion of the zinc bar is:

β = 0.004 K^-1

Class 10 Science Chapter 3 Answer in detail.

Question 1.
Derive the ideal gas equation.
Answer:
The ideal gas equation is an equation, which relates all the properties of an ideal gas. An ideal gas obeys Boyle’s law and Charles’s law and Avogadro’s law.
According to Boyle’s law, PV = constant ………. (1)
According to Charles’s law,
V/T = constant ……… (2)
According to Avogadro’s law,
V/T = constant …….. (3)
After combining equations (1), (2) and (3), you equation. can get the following
V/nT = constant ……. (4)
The above relation is called the combined law of gases. If you consider a gas, which contains µ moles of the gas, the number of atoms contained will be equal to µ times the Avogadro number, N₀.
i.e., n = µN_A
Using equation (5), in equation (4) can be written as
PV/µNAT = constant
The value of the constant in the above equation is taken to be K_B, which is called as Boltzmann constant (1.38 × 10^-23 JK^-1). Hence, we have the following equation:
PV/µNAT = K_B
PV = µN_AK_BT
µN_AK_B = R
which is termed as universal gas constant whose value is 8.31 J mol^-1 K^-1.
PV = RT
Ideal gas equation is also called as equation of state because it gives the relation between the state variables and it is used to describe the state of any gas.

Question 2.
Explain the experiment of measuring the real and apparent expansion of a liquid with a neat diagram.

Answer:

To start with, the liquid whose real and apparent expansion is to be determined is poured in a container up to a level. Mark this level as L₁. Now, heat the container and the liquid using a burner. Initially, the container receives the thermal energy and it expands. As a result, the volume of the liquid appears to have reduced. Mark this reduced level of liquid as L₂. On further heating, the thermal energy supplied to the liquid through the container results in the expansion of the liquid. Hence, the level of liquid rises to L₃. Now, the difference between the levels L₁ and L₃ is called as apparent expansion, and the difference between the levels L₂ and L₃ is called real expansion. The real expansion is always more than that of apparent expansion.
Real expansion = L₃ – L₂
Apparent expansion = L₃ – L₁

Class 10 Science Chapter 3  HOT Question

Question 1.
If you keep ice at 0°C and water at 0°C in either of your hands, in which hand you will feel more chillness? Why?
Answer:
The hand consisting of ice at 0°C would feel more chillness because, ice undergoes melting. More amount of energy (chillness) is transferred to hand. In addition ice has latent heat of fusion.

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Class 10 Science Guide

Class 10 Science

Chapter 2: Qptics – Questions & Answers

Class 10 Science Chapter 2 Choose the Correct Answer

Question 1.

The refractive index of four substances A, B, C and D are 1.31,1.43,1.33, 2.4 respectively. The speed of light is maximum in:

(a) A
(b) B
(c) C
(d) D

Answer:
(a) A

Question 2.

Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens:

(a) f
(b) 2f
(c) infinity
(d) between f and 2f

Answer:
(b) 2f

Question 3.

Where should an object be placed so that a real and inverted image of the same size is obtained by a convex lens ______.

(a) f
(b) 2f
(c) infinity
(d) between f and 2f.

Answer:
(b) 2f

Question 4.

Magnification of a convex lens is _____.

(a) positive
(b) negative
(c) either positive or negative
(d) zero.

Answer:
(b) negative

Question 5.

A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at:

(a) focus
(b) infinity
(c) at 2f
(d) between f and 2f

Answer:
(b) infinity

Question 6.

Power of a lens is -4D, then its focal length is:

(a) 4 m
(b) -40 m
(c) -0.25 m
(d) -2.5 m

Answer:
(d) -2.5 m

Question 7.

In a myopic eye, the image of the object is formed _____.

(a) behind the retina
(b) on the retina
(c) in front of the retina
(d) on the blind spot.

Answer:
(c) in front of the retina

Question 8.

The eye defect ‘presbyopia’ can be corrected by:

(a) convex lens
(b) concave lens
(c) convex mirror
(d) Bi focal lenses

Answer:
(d) Bi focal lenses

Question 9.

Which of the following lens would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 5 cm
(b) A concave lens of focal length 5 cm
(c) A convex lens of focal length 10 cm
(d) A concave lens of focal length 10 cm

Answer:
(d) A concave lens of focal length 10 cm

Question 10.

If V_B, V_G, V_R be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?

(a) V_B = V_G = V_R
(b) V_B > V_G > V_R
(c) V_B < V_G < V_R
(d) V_B < V_G > V_R

Answer:
(c) V_B < V_G < V_R

Class 10 Science Chapter 2

Fill in the Blanks

1. The path of the light is called as ray
2. The refractive index of a transparent medium is always greater than unity
3. If the energy of incident beam and the scattered beam are same, then the scattering of light is called as scattering elastic
4. According to Rayleigh’s scattering law, the amount of scattering of light is inversely proportional to the fourth power of its wavelength
5. Amount of light entering into the eye is controlled by iris

Answer:

1. ray
2. unity
3. elastic
4. wavelength
5. iris

Class 10 Science Chapter 2

True or False. If false correct it.

1. Velocity of light is greater in denser medium than in rarer medium. – False 
2. The power of lens depends on the focal length of the lens. – True
3. Increase in the converging power of eye lens cause ‘hypermetropia’ – True
4. The convex lens always gives small virtual image. – False 

Answer:

1. False – Velocity of light is greater in rarer medium than in denser medium.
2. True
3. True
4. False – The convex lens does not give small virtual image always.

Class 10 Science Chapter 2

Match the following.

Column – I	Column – II
1. Retina	a. Path way of light
2. Pupil	b. Far point comes closer
3. Ciliary muscles	c. near point moves away
4. Myopia	d. Screen of the eye
5. Hypermetropia	e. Power of accommodation

Answer:
1. d
2. a
3. e
4. b
5. c

Class 10 Science Chapter 2

Assertion and reasoning type.

Mark the correct choice as-
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
1. Assertion: If the refractive index of the medium is high (denser medium) the velocity of the light in that medium will be small
Reason: Refractive index of the medium is inversely proportional to the velocity of the light.

2. Assertion: Myopia is due to the increase in the converging power of eye lens.
Reason: Myopia can be corrected with the help of concave lens.
Answer:
1. (a)
2. (a)

Class 10 Science Chapter 2

Answer Briefly.

Question 1.
What is refractive index?
Answer:
Refractive index gives us an idea of how fast or how slow light travels in a medium.

Question 2.
State Snell’s law.
Answer:
The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
sinisinr = µ2µ1

Question 3.
Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
Answer:

Question 4.
Define dispersion of light.
Answer:
When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as ‘dispersion of light’.

Question 5.
State Rayleigh’s law of scattering.
Answer:
Rayleigh’s scattering law states that, “The amount of scattering of light is inversely proportional to the fourth power of its wavelength”.
Amount of scattering ‘S’ ∝1λ4

Question 6.
Differentiate convex lens and concave lens.
Answer:

S.No	Convex Lens	Concave Lens
1	A convex lens is thicker in the middle than at edges.	A concave lens is thinner in the middle than at edges.
2	It is a converging lens.	It is a diverging lens.
3	It produces mostly real images.	It produces virtual images.
4	It is used to treat hypermeteropia.	It is used to treat myopia.

 

Question 7.
What is the power of accommodation of the eye?
Answer:

• The ability of the eye lens to focus nearby as well as the distant objects is called the power of accommodation of the eye.
• This is achieved by changing the focal length of the eye lens with the help of ciliary muscles.

Question 8.
What are the causes of ‘Myopia’?
Answer:

1. The lengthening of eye ball.
2. The focal length of eye lens is reduced.
3. The distance between eye lens and retina increases.
4. The far point will not be at infinity.
5. The far point comes closer.

Question 9.
Why does the sky appear blue in colour?
Answer:
When sunlight passes through the atmosphere, the blue colour (shorter wavelength) is scattered to a greater extent than the red colour (longer wavelength). This scattering causes the sky to appear blue in colour.

Question 10.
Why are traffic signals red in colour?
Answer:

• Red light has the highest wavelength.
• It is scattered by atmospheric particles.
• So red light is able to travel the longest distance through a fog, rain etc.

Class 10 Science Chapter 2

Give the answer in detail.

Question 1.
List any five properties of light?
Answer:

• Light is a form of energy.
• Light always travels along a straight line.
• Light does not need any medium for its propagation. It can even travel through a vacuum.
• The speed of light in vacuum or air is, c = 3 × 10⁸ ms^-1
• Since light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (v), which are related by the following equation: c = vλ (c = velocity of light).
• Different coloured light has a different wavelength and frequency.

Question 2.
Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.
Answer:
Rule-1: When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation.

Rule-2: When rays parallel to the principal axis strikes a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus.

Rule-3: When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis.

Question 3.
Differentiate the eye defects: Myopia and Hypermetropia.
Answer:

S.No	Myopia	Hypermetropia

1

It is due to the lengthening of the eyeball.

It is due to the shortening of the eyeball.

2

With this defect, distant objects cannot be seen clearly.

With this defect, nearby objects cannot be seen clearly.

3

The focal length of the eye lens is reduced.

The focal length of the eye lens is increased.

4

The far point will not be at infinity.

The near point will not be at 25 cm.

5

The far point has come closer.

The near point has moved further.

6

The image of distant objects is formed before the retina.

The image of nearby objects is formed behind the retina.

7

It can be corrected by using a concave lens.

It can be corrected by using a convex lens.

8

This defect is known as myopia.

This defect is known as hypermetropia.

Question 4.

Explain the construction and working of a ‘Compound Microscope’.
Answer:
Construction : A compound microscope consists of two convex lenses. The lens with the shorter focal length is placed near the object, and is called as ‘objective lens’ or ‘objective piece’. The lens with larger focal length and larger aperture placed near the observer’s eye is called as ‘eye lens’ or ‘eye piece’. Both the lenses are fixed in a narrow tube with adjustable provision.

Working : The object (AB) is placed at a distance slightly greater than the focal length of objective lens (u > F₀). A real, inverted and magnified image (A’B’) is formed at the other side of the objective lens. This image behaves as the object for the eye lens. The position of the eye lens is adjusted in such a way, that the image (B’B’) falls within the principal focus of the eye piece. This eye piece forms a virtual, enlarged and erect image (A”B”) on the same side of the object.

Compound microscope has 50 to 200 times more magnification power than simple microscope.

Class 10 Science Chapter 2

Numerical Problems.

Question 1.

An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image.

Answer:

Distance of an object u = 20 cm
Focal length of a convex lens f = 10 cm
Let the image distance be v
We know

1/u + 1/v = 1/f

1/20 + 1/v = 1/10

1/v = 1/10 – 1/20

1/v = (20 – 10)/200

=10 / 200 = 1/20

v = 20 cm
Magnification m = v/u = 20/20 = 1

Hence a real image of same size is formed at 20 cm.
Image distance = 20 cm

Question 2.

An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image.

Answer:

Concave Lens Image Calculation

Given:

• Object height (h) = 3 cm
• Object distance (u) = -10 cm (since the object is placed in front of the lens)
• Focal length (f) = -15 cm (for a concave lens)

Step 1: Use the Lens Formula

The lens formula is:

        1/f = 1/v - 1/u
    

Rearranging for image distance (v):

        1/v = 1/f + 1/u
    

Substitute the given values:

        1/v = 1/(-15) + 1/(-10)
    

Calculating gives:

        v ≈ -6 cm
    

Step 2: Calculate the Magnification

The magnification (M) is given by:

        M = v / u
    

Substitute the values:

        M = -6 / -10 = 0.6
    

The height of the image (h’) is:

        h' = M × h = 0.6 × 3 = 1.8 cm
    

Final Answer:

• Image distance (v) ≈ -6 cm (indicating the image is virtual and located on the same side as the object).
• Image height (h’) ≈ 1.8 cm (indicating the image is upright and smaller than the object).

Class 10 Science Chapter 2

Higher order thinking (HOT) questions.

Question 1.
While doing an experiment for the determination of focal length of a convex lens, Raja Suddenly dropped the lens. It got broken into two halves along the axis. If he continues his experiment with the same lens,
(a) can he get the image?
(b) Is there any change in the focal length?
Answer:
(a) He can get the image.
(b) The focal length of the lens will be doubled.

Question 2.
The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?
Answer:

• The large pupil opens wider and allows the maximum amount of light to enter the eye in the dark.
• Their lens is large and situated near the retina. This also allows a lot of light to register on the retina. The retina contains 2 types of light-sensing cells rods and cones.
• Cones are responsible for the coloured vision and require bright, focused light.
• Rods are extremely sensitive to light and have a photosensitive pigment called rhodopsin which plays a vital role in night vision.

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Class 10 Science Guide

Class 10 Science Chapter 1

Fill in the Blanks

1. To produce a displacement force is required.
2. Passengers lean forward when the sudden brake is applied in a moving vehicle. This can be explained by inertia
3. By convention, the clockwise moments are taken as negative and the anticlockwise moments are taken as positive
4. Accelerator is used to change the speed of the car.
5.  A man of mass 100 kg has a weight of 980 N at the surface of the Earth.

Answer:

1. force
2. inertia
3. negative, positive
4. Accelerator
5. Weight = m × g = 100 × 9.8 = 980 N

Class 10 Science Chapter 1

State whether the following statements are true or false. Correct the statement if it is false

1. The linear momentum of a system of particles is always conserved. – True
2. Apparent weight of a person is always equal to his actual weight. – False 
3. Weight of a body is greater at the equator and less at the polar region. – False 
4. Turning a nut with a spanner having a short handle is so easy than one with a long handle. – False 
5. There is no gravity in the orbiting space station around the Earth. So the astronauts feel weightlessness. – False 

Answer:

1. True
2. False – Apparent weight of a person is not always equal to his actual weight.
3. False – Weight of a body is minimum at the equator. It is maximum at the poles.
4. False – Turning a nut with a spanner having a longer handle is so easy than one with a short handle.
5. False – Astronauts are falling freely around the earth due to their huge orbital velocity.

Class 10 Science Chapter 1  Match the following

Column I	Column II
a. Newton’s I law	i. propulsion of a rocket
b. Newton’s II law	ii. Stable equilibrium of a body
c. Newton’s III law	iii. Law of force
d. Law of conservation of Linear momentum	iv. Flying nature of bird

Answer:
A. (ii)
B. (iii)
C. (iv)
D. (i)

Class 10 Science Chapter 1  Assertion and Reasoning

Mark the correct choice as:

(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

1. Assertion: The sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Reason: The principle of conservation of momentum is valid if the external force on the system is zero.

2. Assertion: The value of ‘g’ decreases as height and depth increases from the surface of the Earth.
Reason: ‘g’ depends on the mass of the object and the Earth.

Answer:
1. (b)
2. (c)

Class 10 Science Chapter 1  Answer Briefly.

Question 1.
Define inertia. Give its classification.
Answer:
The inherent property of a body to resist any change in its state of rest or the state of uniform motion, unless it is influenced upon by an external unbalanced force, is known as ‘inertia’.
Classifications:

1. Inertia of rest
2. Inertia of motion
3. Inertia of direction

Question 2.
Classify the types of force based on their application.
Answer:
Based on the direction in which the forces act, they can be classified into two types as:

1. Like parallel forces: Two or more forces of equal or unequal magnitude acting along the same direction, parallel to each other are called like parallel forces.
2. Unlike parallel forces: If two or more equal forces or unequal forces act along with opposite directions parallel to each other, then they are called, unlike parallel forces.

Question 3.
If a 5 N and a 15 N forces are acting opposite to one another. Find the resultant force and the direction of action of the resultant force.
Answer:
F₁ = 5 N
F₂ = 15 N
∴ Resultant force F_R = F₁ – F₂
= 5 – 15 = -10 N
It acts in the direction of the force of 15 N (F₂).

Question 4.
Differentiate mass and weight.
Answer:
Ratio of masses of planets is
m₁ = m₂ = 2 : 3
Ratio of radii
R₁ = R₂ = 4 : 7
We know

Question 5.
Define the moment of a couple.
Answer:
When two equal and unlike parallel forces applied simultaneously at two distinct points constitute a couple. A couple results in causes the rotation of the body. This rotating effect of a couple is known as the moment of a couple.

Question 6.
State the principle of moments.
Answer:
Principle of moments states that if a rigid body is in equilibrium on the action of a number of like (or) unlike parallel forces then the algebraic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anticlockwise direction.

Question 7.
State Newton’s second law.
Answer:
The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force.

Question 8.
Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
Answer:
When a spanner is having a long handle, the turning effect of the applied force is more when the distance between the fixed edge and the point of application of force is more. Hence a spanner with a long handle is preferred to tighten screws in heavy vehicles.

Question 9.
While catching a cricket ball the fielder lowers his hands backwards. Why?
Answer:
While catching a cricket ball the fielder lowers his hands backwards, so increase the time during which the velocity of the cricket ball decreases to zero. Therefore the impact of force on the palm of the fielder will be reduced.

Question 10.
How does an astronaut float in a space shuttle?
Answer:
Astronauts are not floating but falling freely around the earth due to their huge orbital velocity. Since spaceshuttle and astronauts have equal acceleration, they are under free fall condition. (R = 0) Hence, both the astronauts and the space station are in the state of weightlessness.

Class 10 Science Chapter 1  Solve the given problems.

Question 1.
Two bodies have a mass ratio of 3 : 4 The force applied on the bigger mass produces an acceleration of 12 ms². What could be the acceleration of the other body, if the same force acts on it.
Answer:
Ratio of masses m₁ : m₂ = 3 : 4
Acceleration of m₂ is a₂ = 12 m/s²
Force acting of m₂ is F₂ = m₂a₂
F₂ = 4 × 12 = 48N
but F₂ = F₁
∴ Force acting on m₁ is F₁ = 48N
∴ Acceleration of m₁ = a₁ = F1m1
a₁ = 483
= 16 m/s²
Acceleration of the other body ax = 16 m/s²

Question 2.
A ball of mass 1 kg moving with a speed of 10 ms-1 rebounds after a perfect elastic collision with the floor. Calculate the change in linear momentum of the ball.
Answer:
Given mass = 1 kg, speed = 10 ms^-1
Initial momentum = mu = 1 × 10 = 10 kg ms^-1
Final momentum = -mu = -10 kg ms^-1
Change in momentum = final momentum – initial momentum = -mu – mu
Change in momentum = -20 kg ms^-1

Question 3.
A mechanic unscrew a nut by applying a force of 140 N with a spanner of length 40 cm. What should be the length of the spanner if a force of
40 N is applied to unscrew the same nut?
Answer:
Force acting on the screw F₁ = 140 N
Length of a spanner d₁ = 40 × 10^-2 m
Second force applied to the screw F₂ = 40 N
Let the length of spanner be d₂
According to the Principle of moments,
F₁ × d₁ = F₂ × d₂
= 140 × 40 = 40 × d₂
∴ d₂ = 140×4040
= 140 × 10^-2 m
Length of a spanner = 140 × 10^-2 m

Question 4.
The ratio of masses of two planets is 2 : 3 and the ratio of their radii is 4 : 7. Find the ratio of their accelerations due to gravity.
Answer:
Ratio of masses of two planets is
m₁ : m₂ = 2 : 3
Ratio of their radii,
R₁ : R₂ = 4 : 7
We know g
Img 2
∴ g₁ : g₂ = 49 : 24

Class 10 Science Chapter 1  Answer in Detail.

Question 1.
What are the types of inertia? Give an example for each type.
Answer:
Types of Inertia:
(i) Inertia of rest: The resistance of a body to change its state of rest is called inertia of rest.
E.g.: When you vigorously shake the branches of a tree, some of the leaves and fruits are detached and they fall down (Inertia of rest).

(ii) The inertia of motion: The resistance of a body to change its state of motion is called inertia of motion.
E.g.: An athlete runs some distance before jumping. Because this will help him jump longer and higher. (Inertia of motion)

(iii) Inertia of direction: The resistance of a body to change its direction of motion is called inertia of direction.
E.g.: When you make a sharp turn while driving a car, you tend to lean sideways, (Inertia of direction).

Question 2.
State Newton’s laws of motion.
Answer:
(i) Newton’s First Law : States that “every body continues to be in its state of rest or the state of uniform motion along a straight line unless it is acted upon by some external force”.

(ii) Newton’s Second Law : States that “the force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.

(iii) Newton’s third law : States that “for every action, there is an equal and opposite reaction. They always act on two different bodies”.

Question 3.
Deduce the equation of a force using Newton’s second law of motion.
Answer:
Let, ‘m’ be the mass of a moving body, moving along a straight line with an initial speed V. After a time interval of ‘t’, the velocity of the body changes to v due to the impact of an unbalanced external force F.
Initial momentum of the body P_i = mu
Final momentum of the body P_f = mv
Change in momentum Δp = P_i – P_f – mv – mu
By Newton’s second law of motion,
Force, F ∝ rate of change of momentum
F ∝ change in momentum / time
F ∝ mv−mut
F = km(v−u)t
Here, k is the proportionality constant.
k = 1 in all systems of units. Hence,
F = m(v−u)t
Since,
acceleration = change in velocity/time,
a = (v – u)/t.
Hence, we have F = m × a
Force = mass × acceleration

Question 4.
State and prove the law of conservation of linear momentum.

Answer:
Proof:
Let two bodies A and B having masses m₁ and m₂ move with initial velocity u₁ and u₂ in a straight line. Let the velocity of the first body be higher than that of the second body, i.e,, u₁ > u₂. During an interval of time t second, they tend to have a collision. After the impact, both of them move along the same straight line with a velocity v₁ and v₂ respectively.
Force on body B due to A,
F_B = m₂(v₂ – u₂)/t
Force on body A due to B,
F_A = m₁(v₁ – u₁)/t
By Newton’s III law of motion,
Action force = Reaction force
F_A = -F_B
m₁(v₁ – u₁)/t = -m₂ (v₂ – u₂)/t
m₁ v₁ + m₂ v₂ = m₁ u₁ + m₂ u₂
The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
Hence the law of conservation of linear momentum is proved.

Question 5.
Describe rocket propulsion.
Answer:

1. Propulsion of rockets is based on the law of conservation of linear momentum as well as Newton’s III law of motion.
2. Rockets are filled with fuel (either liquid or solid) in the propellant tank. When the rocket is fired, this fuel is burnt and hot gas is ejected with high speed from the nozzle of the rocket, producing a huge momentum.
3. To balance this momentum, an equal and opposite reaction force is produced in the combustion chamber, which makes the rocket project forward.
4. While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out.
5. Since there is no net external force acting on it, the linear momentum of the system is conserved.
6. The mass of the rocket decreases with altitude, which results in the gradual increase in the velocity of the rocket.
7. At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity.

Question 6.
State the universal law of gravitation and derive its mathematical expression.
Answer:
Newton’s universal law of gravitation states that every particle of matter in this universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centres of these masses. The direction of the force acts along the line joining the masses.

Force between the masses is always attractive and it does not depend on the medium where they are placed.

Let, m₁ and m₂ be the masses of two bodies A and B placed r metre apart in space
Force
F ∝ m₁ × m₂
F ∝ 1/r²
On combining the above two expressions
F ∝ m1×m2r2
F = Gm1m2r2
Where G is the universal gravitational constant. Its value in SI unit is 6.674 × 10^-11 N m² kg^-2.

Question 7.
Give the applications of gravitation.
Answer:

1. Dimensions of the heavenly bodies can be measured using the gravitation law. Mass of the Earth, the radius of the Earth, acceleration due to gravity, etc. can be calculated with higher accuracy.
2. Helps in discovering new stars and planets.
3. One of the irregularities in the motion of stars is called ‘Wobble’ lead to the disturbance in the motion of a planet nearby. In this condition, the mass of the star can be calculated using the law of gravitation.
4. Helps to explain germination of roots is due to the property of geotropism, which is the property of a root responding to the gravity.
5. Helps to predict the path of the astronomical bodies.

Class 10 Science Chapter 1

HOT questions

Question 1.
Two blocks of masses 8 kg and 2 kg respectively lie on a smooth horizontal surface in contact with one other. They are pushed by a horizontally applied force of 15 N. Calculate the force exerted on the 2 kg mass.

Answer:
Mass of first block m₁ = 8 kg
Mass of second block m₂ = 2 kg
Total mass M = 8 + 2 = 10 kg
Force applied F = 15 N
∴ Acceleration a = FM
1510 = 1.5 m/s²
Force exerted on the 2 kg mass,
F = ma
= 2 × 1.5 = 3 N

Question 2.
A heavy truck and bike are moving with the same kinetic energy. If the .mass of the truck is four times that of the bike, then calculate the ratio of their momenta. (Ratio of momenta = 1 : 2)

Answer:

Ratio of Momentas: Truck and Bike

We are given:

• The mass of the truck is 4 times the mass of the bike: m_t = 4m_b
• The kinetic energy of the truck equals the kinetic energy of the bike: KE_t = KE_b

Step 1: Express velocities in terms of kinetic energy

From the formula for kinetic energy:

KE = 1/2 * m * v²

We can solve for velocity:

v = √(2KE/m)

For the bike:

vb = √(2KEb/mb)

For the truck:

vt = √(2KEt/mt) = √(2KE/4mb) = 1/2 * √(2KE/mb) = 1/2 * vb

Thus, the truck’s velocity is half that of the bike’s velocity.

Step 2: Calculate the ratio of momenta

The momentum (p) of each vehicle is:

• For the bike: p_b = m_b * v_b
• For the truck: p_t = m_t * v_t = 4m_b * (1/2 * v_b) = 2m_b * v_b

The ratio of their momenta is:

pb : pt = mb * vb : 2mb * vb = 1 : 2

Conclusion:

The ratio of the truck’s momentum to the bike’s momentum is 2 : 1.

Question 3.
“Wearing helmet and fastening the seat belt is highly recommended for safe journey” Justify your answer using Newton’s laws of motion.
Answer:
(i) According to Newton’s Second Law, when you fall from a bike on the ground with a force equal to your mass and acceleration of the bike.
According to Newton’s Third Law, an equal and opposite reacting force on the ground is exerted on your body. When you do not wear a helmet, this reacting force can cause fatal head injuries. So it is important to wear helmet for a safe journey.

(ii) Inertia in the reason that people in cars need to wear seat belts. A moving car has inertia, and so do the riders inside it. When the driver applies the brakes, an unbalanced force in applied to the car. Normally the bottom of the seat applies imbalanced force friction which slows the riders down as the car slows. If the driver stops the car suddenly, however, this force is not exerted over enough time to stop the motion of the riders. Instead, the riders continue moving forward with most of their original speed because of their inertia.

 

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