# Question-1

**State the universal law of gravitation.**

**Solution:**

According to the universal law of gravitation, every body in the universe attracts every other body with a force, which is directly proportional to products of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the centers of the two bodies.

Where F

_{grav }represtnts the force of gravity between two objests.

_{1}represents the mass of object1.

_{2}represents the mass of object2.

# Question-2

**Write the formulae to find the magnitude of gravitational force between the earth and an object on the surface of the earth.**

**Solution:**

Suppose the earth is a sphere of mass M and radius R and if an object of mass m is placed on the surface of the earth, then the magnitude of the gravitational force between the body and the earth will be

F = G

# Question-3

**What do you mean by free fall?**

**Solution:**

The motion of a body under the influence of the force of gravity alone is called a free

fall.

# Question-4

**What do you mean by acceleration due to gravity?**

**Solution:**

The acceleration produced in the motion of a body falling under the influence of the gravitational attraction of the earth is called as the acceleration due to gravity.

# Question-5

**What are the differences between the mass of an object and its weight?**

**Solution:**

Differences between mass and weight :

Mass |
Weight |

1. Mass is the quantity of matter contained in a body and is the measure of its inertia. |
1. Weight of a body is the force with which a body is attracted towards the centre of the earth. |

2. Its value remains constant at all places. |
2. Its value (W = mg) changes from place to place due to the change in the value of acceleration due to the gravity ‘g’. |

3. It is a scalar quantity. |
3. It is a vector quantity. |

4. It is measured by a pan balance. |
4. It is measured by a spring balance. |

5. Mass of a body is never zero. |
5. Weight of a body is zero at the centre of the earth because there, ‘g’ becomes zero. |

6. Its unit is kg. |
6. It units is Newton or kg wt. |

# Question-6

**Why is the weight of an object on the moon 1/6**

^{th}its weight on the earth?**Solution:**

The mass of the moon of times and its radius times that of the earth. As a result, the gravitational attraction on the moon is about one sixth when compared to that on the earth. Hence, the weight of an object on the moon is

^{th}of the weight on the earth.

# Question-7

**Why is it difficult to hold a school bag having a strap made of a thin and strong string?**

**Solution:**

The thin string exerts a large pressure on our hand due to its smaller area. Hence, we feel uncomfortable.

# Question-8

**What do you mean by buoyancy?**

**Solution:**

buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards. The magnitude of that force is proportional to the difference in the pressure between the top and the bottom of the column, and is also equivalent to the weight of the fluid that would otherwise occupy the column.

# Question-9

**When does an object float or sink when placed on the surface of water?**

**Solution:**

(i) An object sinks in water if its density is greater than that of water.

(ii) An object floats on water when its density is less than that of water.

# Question-10

**You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?**

**Solution:**

More than 42 kg, the weighing machine reads slightly less value due to the up-thrust of air acting on our body.

# Question-11

**You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality one is heavier than the other. Can you say which one is heavier and Why?**

**Solution:**

The cotton bag is heavier than the iron bar. The cotton bag experiences larger up thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

# Question-12

**Using expression g = G and knowing G, calculate the value of the acceleration due to gravity. Mass of the earth = 6 × 10**

^{24}kg and radius of earth = 6.4 × 10^{6}m and G = 6.7 × 10^{-11}Nm^{2}/kg^{2}.**Solution:**

We know that

g = G

**=**

= 9.8 m/s

^{2}.

# Question-13

**Using the data given at the end of the chapter, find first v**

^{2}/r for the moon where v is the velocity and r is the distance from the earth. The find the acceleration due to gravity at the distance of the moon using the inverse square law. Take acceleration due to gravity near the earth as 9.8 m/s^{2}.**Solution:**

Mass of the Moon, M = 7.3 × 10

^{22}kg

Distance from the Earth = 384000 km

= 384000 × 1000 m

= 3.84 × 10

^{8}m

Radius of the Moon, R = 1740 km

= 1740 × 1000 m

= 1.74 × 10

^{6}m

∴

**= 3.3 × 10**

=

=

^{-5}m/s

^{2}

Acceleration due to gravity g =

=

= 1.64 m/s

^{2}.

# Question-14

**Suppose you and your friend have mass 50 kg each. Suppose also that both of you are standing such that your centres of gravity are 1m apart. Calculate the force of gravitation between you and your friend. Calculate also the force of gravity acting on you.**

**Solution:**

Mass of each person m = 50 kg

Difference of distance in the centres of gravity = 1m

Radius of the earth = 1740 km

= 1740 × 1000 = 1.74 × 10

^{6}m

∴Force of attraction = G

=

= 1.7 × 10

^{-7}N

We know that, mg = G

∴ Force of gravity F =

= 5.5 × 10

^{-20}N.

# Question-15

**A particle is dropped from a tower 180 m height. How long does it take to reach the ground? What is its velocity when it touches the ground? Make a table showing the distance covered by the particle, the velocity and acceleration at the end of each second. Plot distance-time, velocity-time and acceleration-time graphs.**

**Solution:**

Height h = 180 m

Initial velocity u = 0

We know that,

v

^{2}– u

^{2}= 2as

v

^{2}– (0)

^{2}= 2 × 10 × 180

v

^{2}= 3600

v = 60 m/s

We know that

v = u + gt

60 = 0 + 10 × t

∴ Time taken t = = 10 s.

# Question-16

**A ball moving on a table reaches the edge and falls. Sketch the path it will follow while falling.**

**Solution:**

A ball dropped from the edge of the table bounces off progressively decreasing heights.

# Question-17

**Imagine that u are visiting planet Mars. You want to record your weight in a note-book on this planet. If your weight on the Earth is 450 N, what would you record your weight on the Mars? You will find the required data at the end of this chapter. Take g = 10 m/s**

^{2}.**Solution:**

Weight on the earth = 450 N

Acceleration due to gravity on Earth, g = 10 m/s

^{2}

∴ We know that

W = m × g

450 = m × 10

∴ m = = 45 kg

Mass on Mars = 45 kg

Mass of the Mars = 6 × 10

^{24}kg

Radius of the Mars = 4.3 × 10

^{6}m

∴ F = G

= 9.8 N.

# Question-18

**In the following five questions choose the correct answer: The force of gravitation between two bodies varies with 'r' as**

(a) r

(c) (d)

(a) r

^{2}(b) r(c) (d)

**.**

**Solution:**

(d) .

# Question-19

**At which of the following locations the value of g is the largest.**

(i) On the top of Mount Everest.

(ii) On top of Qutub Minar

(iii) A place on the equator

(iv) A camp site in Antarctica.

(i) On the top of Mount Everest.

(ii) On top of Qutub Minar

(iii) A place on the equator

(iv) A camp site in Antarctica.

**Solution:**

(iv) A camp site in Antarctica.

# Question-20

**A stone is dropped from the top of the tower. Its speed after it has fallen 20m is (take g = 10 m/s**

(i) –10 m/s (ii) 10 m/s

(iii) –20 m/s (iv) 20 m/s.

^{2}).(i) –10 m/s (ii) 10 m/s

(iii) –20 m/s (iv) 20 m/s.

**Solution:**

v

^{2}–u

^{2}= 2as

v

^{2}– (0)

^{2}= 2 × 10 × 20

v

^{2}= 400

v = 20 m/s

Hence (iv) is correct.

# Question-21

**A ball is thrown vertically upwards. The acceleration due to gravity:**

(i) is in the direction opposite to the direction of its motion.

(ii) is in the same direction as the direction of its motion

(iii) increases as it comes down

(iv) becomes zero at the highest point.

(i) is in the direction opposite to the direction of its motion.

(ii) is in the same direction as the direction of its motion

(iii) increases as it comes down

(iv) becomes zero at the highest point.

**Solution:**

(i) is in the direction opposite to the direction of its motion.

# Question-22

**At the top of its path, a projectile:**

(i) has no acceleration opposite to the direction

(ii) is in the same direction as the direction of its motion

(iii) increases as it comes down

(iv) becomes zero at the highest point.

(i) has no acceleration opposite to the direction

(ii) is in the same direction as the direction of its motion

(iii) increases as it comes down

(iv) becomes zero at the highest point.

**Solution:**

(i) has no acceleration opposite to the direction

# Question-23

**In what sense does the moon fall towards the earth? Why does it not actually fall on the earth’s surface?**

**Solution:**

At each point of its orbit the moon falls towards the earth instead of going straight.

The reason why the moon does not actually fall on the Earth’s surface is because, the gravitational attraction of Earth provides the necessary centripetal force to the moon for its orbital motion around the earth. Due to which the Moon is revolving around the earth.

# Question-24

**The Earth attracts an apple. Does the apple also attract the Earth? If it does, why does the Earth not move towards the apple?**

**Solution:**

Yes, the apple also attracts the earth. The apple has a very small mass and the Earth has a very large mass. Therefore, the gravitational force acting between the apple and the Earth produces a large acceleration in the apple due to its very small mass. The same gravitational force however, produces very small (negligible) acceleration in the earth due to its very large mass. It is due to large acceleration produced in the apple that we can see the apple moving towards the earth. The acceleration produced in the Earth is so small that the displacement of the Earth is almost zero. That is why the Earth does not move towards the apple.

# Question-25

**If the force of gravity somehow vanished today, why would we be sent flying into space?**

**Solution:**

If the force of gravity is vanished today, we would be sent flying in space due to the centrifugal force.

# Question-26

**Suppose the mass of the Earth some how increases by 10% without any change in its size. What would happen to your weight? Suppose the radius of the Earth becomes twice of its present radius without any change in its mass, what will happen to your weight?**

**Solution:**

We know that

g =

When the mass of the Earth increases by 10%, without any change in its size, the value of g also increases. Hence, the weight will increase.

When the radius of the Earth becomes twice of its present radius without any change in its mass, the value of g will decrease. Hence, the weight will decrease.

# Question-27

**Suggest a method for calculating the mass of the moon.**

**Solution:**

Let M

_{1}be the mass of the Moon and M

_{2}be the mass of Earth.

Both Moon and Earth revolve round a common centre of mass O.

Taking the movements about O, we get

M_{1}r_{1} = M_{2}r_{2} -------(1)

where r_{1} and r_{2} are the distances of the Moon and the Earth from O respectively and r = r_{1} + r_{2}

From equation (1), we get

or

(since r_{1} + r_{2} = r)

r_{1} =

Similarly r_{2} =

If v_{1} and v_{2} are the magnitudes of the velocities about the centre of mass and T the

time period, then

v_{1} = and

v_{2} =

=

or

Substituting r_{2} = , we get

M_{1} + M_{2} =

Neglecting M_{2},

M_{1} =

From the above formula mass of the moon can be calculated.

# Question-28

**At what height above the Earth’s surface would the value of acceleration due to gravity be half of what it is on the surface?**

**Solution:**

If h is the height of location above the surface of earth, then

g′ = g

If the acceleration due to gravity is half of what it is on the surface, then

g′ = g/2

∴

Hence

R = R + h

(-1)R = h

where R is the radius of the earth.

# Question-29

**You must have seen two types of balances. A grocer has two platforms, or pans and a needle moving on a scale, while junk dealers (Kabadi walas) may be using a spring balance to weigh used newspapers. Suppose the two balances give the same measure for a given body on the Earth. Now take the balance to Moon. Would the two balances give the same measure? Explain your answer.**

**Solution:**

The two balances would not give the same measure. The spring balance measures the weight of

the body.

Weight = mass × acceleration due to gravity

The acceleration due to gravity of the Moon is 1/6

^{th}of the Earth. The mass of the newspaper is

the same as on the earth but its weight on the moon is one-sixth of that on the Earth. Therefore,

the weight of the newspaper (reading on the spring balance) on the Moon will be 1/6

^{th}that on the

surface of the Earth.

# Question-30

**You buy a bag of sugar of weight w at a place on the equator. You take this to Antarctica. Would its weight be the same there? If not, would it increase or decrease?**

**Solution:**

The weight would not be the same at Antarctica. The value of ‘g’ at Antarctica is larger than at the equator. Therefore, its weight would increase at Antarctica that at the equator.

# Question-31

**Find the percentage decrease in the weight of the body when taken to a height of 16 km above the surface of Earth. Radius of the Earth is 6400 km.**

**Solution:**

Height, h = 16 km

Radius, R = 6400 km

We know that

g′ = g

= g -

∴ g′ - g = -

Therefore the percentage decrease in the weight of the body is given by

= = 0.5 %.

# Question-32

**Compare the gravitational forces exerted by the Sun and the Moon on the Earth. Which exerts greater force and by how much?**

**Solution:**

Mass of the Earth, M

_{e}= 6 × 10

^{24}kg

Mass of the Moon, M

_{m}= 7.4 × 10

^{22}kg

Mass of the Sun, M

_{s}= 2 × 10

^{30}kg

Distance between the Earth and the Moon, R

_{1}= 3.84 × 10

^{8}m

Distance between the Earth the Sun, R

_{2}= 1.5 × 10

^{11}m

Gravitational force between the Earth and the Moon,

F

_{1}=

=

= 2 × 10

^{20}N

Similarly gravitational force between the Earth and the Sun

F

_{2}=

=

**= 3.6 × 10**

^{22}N

∴ = 180

The Sun exerts 180 times greater force.

# Question-33

**A ball is dropped from the top of a tower 40 m tall. What is its velocity when it has covered 20 m. What would be its velocity when it hits the ground?**

**Solution:**

Height of the tower

**=**40 m

Initial

**Velocity of the ball, u = 0**

Acceleration due to gravity a = 10 m/s

Distance covered s = 20 m

We know that

v^{2} – u^{2} = 2as

v^{2} – 0 = 2 × 10 × 20

v^{2} = 400

∴ v = 40 m/s

Velocity of the ball when it reaches the ground v =

=

= 28.3 m/s.

# Question-34

**A helicopter is ascending with a velocity of 2 m/s at a height of 24 m, when it drops a mail packet. The packet contains materials, which can be damaged if it hits ground with a velocity greater than 72 km/h. Was the packet damaged? Explain your answer.**

**Solution:**

Initial velocity u = 2 m/s

Height h = 24 m

We know that

v

^{2}= u

^{2}+ 2as

= (2)

^{2}+ 2 × 10 × 24

= 484

∴v = 22 m/s

It is given that the packet hits the ground with a velocity greater than 72 km/h =

= 20 m/s

Therefore since the packet hits the ground with a velocity greater than 72 km/h, the mail packet is damaged.

# Question-35

**A ball is dropped from the jump board of swimming pool, which is at a height of 20 m. A second ball is thrown from the same board after one second with initial velocity u. Both the balls hit the water together. What was the initial velocity with which the second ball was thrown? Do they hit water with the same velocity? Explain your answer.**

**Solution:**

Height of the jump board h = 20 m

Time taken for the first ball to fall freely will be t =

= 2s

Thus the second ball thrown after 1s with velocity u should cover a distance of 20 m in (2 - 1) = 1s

Considering the second ball, we have

s = ut + at

^{2}

20 = u × 1 + × 10 × (1)

^{2}

∴ u = 20 – 5

= 15 m/s

They will not hit the water with the same velocity. The second ball hits the water with a greater velocity.

# Question-36

**A rocket is launched to travel vertically upwards with a constant velocity of 20 m/s. After travelling for 35 s the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. What is the total height achieved by it? After what time of its launch will it come back to the Earth?**

**Solution:**

Initial velocity u = 20 m/s

Final velocity v = 0

Acceleration due to gravity g = -10 m/s

^{2}

For the vertically upward motion, the equation of motion is

v

^{2}= u

^{2}+ 2gh

0 = (20)

^{2}+ 2 × (-10) × h

0 = 400 - 20h

∴ Height achieved by the rocket h = = 20 m

For a freely falling body

v

^{2}= u

^{2}+ 2gh

v

^{2}= 0 + 2 × 10 × 20

v

^{2}= 400

∴ v = 20 m/s

Now u = 0, v = 20 m/s, g = 10 m/s

^{2}

v = u + gt

20 = 0 + 10 × t

∴ Time taken t = = 2s

Therefore the rocket will come back to Earth after (35 +2) = 37 s.

# Question-37

**A helicopter is on mission to drop food for people stranded on a boat. It is at a height of 20 m and moving with a steady horizontal velocity of 2 m/s when it spots the nearest end of the boat immediately below it. It then drops the packets . If the length of the boat is 5m, will the people in the boat receive the packets?**

**Solution:**

Height h = 20 m

Initial velocity u = 2 m/s

Acceleration due to gravity g = 10 m/s

^{2}

Horizontal range R =

= 4m

Length of the boat = 5m

Therefore people in the boat will receive the packets.

# Question-38

**A man is standing at the top of a 60 m high tower. He throws a ball vertically upwards with a velocity of 20 m/s. After what time will the ball pass him going downwards? How long after its release will the ball reach the round?**

**Solution:**

Height of the tower h = 60 m

Initial velocity u = -20 m/s

Final velocity v = 0

Time of ascent = Time of descent

∴ After 4s, the ball will pass him going downwards

Time taken to cover 60 m.

Initial velocity = 20 m/s

We know that

s = ut + at

^{2}

60 = 20 × t + × 10 × t

^{2}

60 = 20t + 5t

^{2}

5t

^{2}+ 20t –60 = 0

t

^{2}+ 4t – 12 = 0

t(t + 6) – 2(t + 6) = 0

∴ t = 2 s

Hence the total time taken to reach the ground after its release = 4 + 2 = 6s.

# Question-39

**A coconut is hanging on a tree at a height of 15 m from the ground. A boy launches a projectile vertically upwards with a velocity of 20 m/s. After what time will projectile pass by the coconut? Explain the two answers that you get in this problem.**

**Solution:**

Height of the coconut hanging on the tree = 15 m

(i) In 1s the projectile passes the coconut going upwards.

(ii) In 3s the projectile passes the coconut coming downwards.