Std 10th Science Part 1 – Chapter 1 : Gravitation Exercise Questions And Answers Maharashtra Board

Q1. Study the entries in the following table and rewrite them putting the connected items in a single row.

Column I	Column II	Column III
Mass	m/s2	Zero at the centre
Weight	kg	Measure of inertia
Acceleration due to gravity	Nm2/kg2	Same in the entire universe
Gravitational constant	N	Depends on height

Answer –

Column I	Column II	Column III
Mass	kg	Measure of inertia
Weight	N	Depends on height
Acceleration due to gravity	m/s2	Zero at the centre
Gravitational constant	Nm2/kg2	Same in the entire universe

Q2. Answer the following questions.

a. What is the difference between mass and weight of an object? Will the mass and weight of an object on the earth be same as their values on Mars? Why?

Answer –

• Mass is the amount of matter present in an object. Its SI unit is
  kilogram (kg). It is a scalar quantity , and its value is the same everywhere in the universe. It is a measure of inertia.
• Weight is the force with which the Earth (or any other planet/body) attracts the object. Its SI unit is Newton (N). It is a vector quantity , and its value changes from place to place because it depends on the acceleration due to gravity (g).
• Mass on Earth and Mars: The mass of an object will be the same on Earth and Mars, because mass is constant everywhere.
• Weight on Earth and Mars: The weight of an object will be different on Earth and Mars. This is because weight depends on the acceleration due to gravity (g), and the value of g is different for Earth and Mars (due to their different masses and radii).

b. What are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force?

Answer –

(i) Free Fall: A free fall occurs when an object moves under the influence of the force of gravity alone. In a true free fall (in a vacuum), the object’s initial velocity is zero and its velocity increases due to gravity’s acceleration.

(ii) Acceleration Due to Gravity (g): This is the acceleration produced in an object due to the Earth’s gravitational force. It is denoted by ‘
g’ , and its direction is always vertically downwards, towards the center of the Earth.

(iii) Escape Velocity (vesc​): This is the minimum initial velocity a body must have to completely escape the gravitational pull of a planet and never fall back down. For Earth, the escape velocity is 11.2 km/s.

(iv) Centripetal Force: This is the force that acts on any object moving along a circle and is directed towards the center of the circle. The term ‘centripetal’ means center seeking. For planets orbiting the Sun, this force is provided by gravity.

c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?

Answer –

Kepler’s First Law (Law of Orbits): The orbit of a planet is an ellipse with the Sun at one of the foci.

Kepler’s Second Law (Law of Areas): The line joining the planet and the Sun sweeps equal areas in equal intervals of time.

Kepler’s Third Law (Law of Periods): The square of its period of revolution (T2) around the Sun is directly proportional to the cube of the mean distance (r3) of a planet from the Sun. Mathematically:
r3T2​=constant=K.

How Kepler’s Laws Helped Newton:

• Newton used Kepler’s Third Law to establish the relationship between the centripetal force and distance.
• By combining the formula for centripetal force (
  F=rmv2​) with Kepler’s Third Law (r3T2​=K), Newton derived that the centripetal force (F) must be inversely proportional to the square of the distance (r2).
• Newton identified this centripetal force as the force of
  gravity and thus proposed the inverse square law of gravitation.

d. A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.

Part 1: Time taken to go up (tup​)

1. For upward motion, the final velocity (v) at the maximum height (h) is zero (v=0).
2. The acceleration (
   a) is due to gravity, acting downward, so a=−g.

Using Newton’s first equation of motion,
v=u+at:

Part 2: Time taken to come down (tdown​)

1. For downward motion from height h, the initial velocity (u) is zero (u=0).
2. The acceleration (
   a) is in the direction of motion, so a=+g.
3. The distance covered (s) is the maximum height, s=h.

e. If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?

• Difficulty in pulling an object along the floor is mainly determined by the force of friction.
• The force of friction is directly proportional to the Normal Force (N) pushing the object against the surface.
• For an object on a horizontal floor, the normal force (N) is equal to its weight (W).
• Weight is calculated as : W=mg.
• If the value of g suddenly becomes twice its value (2g), the weight (W) of the object will also become twice its value (W′=m(2g)=2W).
• Since the weight doubles, the normal force and consequently the force of friction doubles.
• Therefore, the force required to pull the object, which must overcome the increased friction, will also be roughly twice as large, making it two times more difficult.

Q3. Explain why the value of g is zero at the centre of the earth.

The value of the acceleration due to gravity (g) at a depth inside the Earth depends on the following factors:

1. As you go deeper, the distance (r) from the center decreases.
2. However, the mass (M) of the Earth that contributes to the gravitational force (the mass inside your position) also decreases.

As you move toward the center, the outer shell of the Earth no longer exerts a gravitational force on you. Only the mass closer to the center matters.

At the very center of the Earth, the radius r is effectively zero , and most importantly, the entire mass of the Earth is all around the point. The net gravitational force exerted by all the matter in every direction cancels out completely. Since the gravitational force (F) is zero, and F=mg, the acceleration due to gravity (g) must also be zero at the center of the Earth.

Q4. Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8​T.

Given:

• Initial distance (r1​) = R
• Initial period (T1​) = T
• New distance (r2​) = 2R
• New period (T2​) = ?

Proof using Kepler’s Third Law: Kepler’s third law states:

5. Solve the following examples.

a. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?

Given:

• Initial velocity (u) = 0 (since it falls from a height)
• Distance (s) = 5 m
• Time (t) = 5 s
• Acceleration due to gravity (g) = ?

b. The radius of planet A is half the radius of planet B. If the mass of A is MA​, what must be the mass of B so that the value of g on B is half that of its value on A?

Given:

• Radius of A (RA​)
• Radius of B (RB​) = 2R_A​ (since R_A​= ​R_B​)
• Mass of A (M_A​)
• Acceleration due to gravity on B (g_B​) = ½​g_A

c. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6^th of that on the earth.

Answer –

d. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g=10 m/s².

Answer –

Part 1: Initial Velocity (u)

Given:

• Distance (s) = 500 m
• Final velocity at peak (v) = 0
• Acceleration (a) = −g=−10 m/s²
• Initial velocity (u) = ?

Part 2: Total time to come back to Earth (T )

• The time to go up (t_up​) is equal to the time to come down (t_down​). The total time is
  T=2×t_up​.
• We can find

Answer: The initial velocity was 100 m/s, and the total time to come back is 20 s.

e. A ball falls off a table and reaches the ground in 1 s. Assuming g=10 m/s², calculate its speed on reaching the ground and the height of the table.

Given:

• Initial velocity (u) = 0 (since it falls off)
• Time (t) = 1 sec
• Acceleration (a) = +g = +10 m/s²

Part 1: Speed on reaching the ground (v)

Answer: The speed on reaching the ground is 10 m/s and the height of the table is 5 m.

f. The masses of the earth and moon are 6×10²⁴ kg and 7.4×10²² kg, respectively. The distance between them is 3.84×10⁵ km. Calculate the gravitational force of attraction between the two? Use G=6.7×10−11 N⋅m²/kg²

Given:

• Mass of Earth (ME​) = 6×10²⁴ kg
• Mass of Moon (MM​) = 7.4×10²² kg
• Distance (r) = 3.84×10⁵ km
  • Convert to meters: r=3.84×10⁵×10³ m=3.84×10⁸ m
• Gravitational constant (G) = 6.7×10⁻¹¹ N⋅m²/kg²

g. The mass of the earth is 6×10²⁴ kg. The distance between the earth and the Sun is 1.5×10¹¹ m. If the gravitational force between the two is 3.5×10²² N, what is the mass of the Sun? Use G=6.7×10⁻¹¹ N⋅m²/kg².

Given:

• Mass of Earth (ME​) = 6×10²⁴ kg
• Distance (r) = 1.5×10¹¹ m
• Gravitational Force (F) = 3.5×10²² N
• Gravitational constant (G) = 6.7×10⁻¹¹ N⋅m²/kg²
• Mass of Sun (MS​) = ?

If you want to download the all above questions and answers in pdf format the click on below download button

Download PDF