## Intext Questions 1

#### Question 1

State the universal law of gravitation.

**Answer**

The universal law of gravitation states that every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.

#### Question 2

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

**Answer**

F = G $\dfrac{\text{Mm}}{\text{d}^2}$

where,

F = force of attraction

G = constant of proportionality

M = mass of earth

m = mass of the object

d = distance between the earth's centre and object's centre

## Intext Questions 2

#### Question 1

What do you mean by free fall?

**Answer**

Earth attracts objects towards itself due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

#### Question 2

What do you mean by acceleration due to gravity?

**Answer**

Whenever an object falls towards the earth, an acceleration is involved due to the earth's gravitational force. So, this acceleration is called the acceleration due to the gravitational force of the earth (or acceleration due to gravity). It is denoted by g and its value is 9.8 ms^{-2}.

## Intext Questions 3

#### Question 1

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

**Answer**

Differences between the mass of an object and its weight:

Mass | Weight |
---|---|

Mass is the quantity of matter contained in the body. | Weight of an object is the force with which it is attracted towards the earth (or force of gravity acting on the body). |

It is the measure of inertia of the body. | It is the measure of gravity. |

It only has magnitude. | It has magnitude as well as direction. |

Mass is a constant quantity. | Weight is not a constant quantity. It is different at different places. |

Its SI unit is kilogram (kg). | Its SI unit is Newton (N). |

#### Question 2

Why is the weight of an object on the moon $\dfrac{1}{6}$th its weight on the earth?

**Answer**

The moon's gravitation force is determined by the mass and the size of the moon.

The mass of moon is $\dfrac{1}{100}$ times and its radius $\dfrac{1}{4}$ times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is $\dfrac{1}{6}$th its weight on the earth.

## Intext Questions 4

#### Question 1

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

**Answer**

Pressure is inversely proportional to the surface area on which the force acts. Incase of a strap made of a thin and strong string, the surface area of the strap in contact with the shoulder is less, hence, pressure on the shoulders is more which makes it difficult to hold the school bag.

#### Question 2

What do you mean by buoyancy?

**Answer**

The property of liquid to exert an upward force on a body immersed in it, is called buoyancy.

#### Question 3

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

**Answer**

An object floats or sinks in water because of the difference in densities of the substance and water. When the density of the substance is less than the density of water, then the upthrust of water is greater than the weight of the substance. Hence it floats.

When the density of substance is more than water, then the upthrust of water on the substance is less than the weight of the substance. Hence it sinks.

## Intext Questions 5

#### Question 1

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

**Answer**

When we stand on a weighing machine, the weight acts downwards while the upthrust due to air acts upwards.

As, true weight = (apparent weight + up thrust)

So our apparent weight becomes less than the true weight. This apparent weight is measured by the weighing machine and hence, the mass indicated is less than the actual mass. So our actual mass will be more than 42 kg.

#### Question 2

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 other. Can you say which one is heavier and why?

**Answer**

The cotton bag is heavier than an iron bar as it experiences a larger air thrust than the iron bar.

As, true weight = (apparent weight + up thrust)

The cotton bag's density is less than that of the iron bar, so, the volume of the cotton bag is more compared to the iron bar. Hence, the cotton bag experience more upthrust due to the presence of air. So, in the presence of air, the cotton bag's true weight is more compared to the true weight of the iron bar.

## Exercises

#### Question 1

How does the force of gravitation between two objects change when the distance between them is reduced to half?

**Answer**

We know,

F = G $\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}$

Where,

F = force of attraction

G = constant of proportionality

m_{1} = mass of first object

m_{2} = mass of second object

d = distance between the two objects.

When distance is reduced to half then, d' = $\dfrac{\text{d}}{2}$

So, substituting we get,

F' = G $\dfrac{\text{m}_1\text{m}_2}{\Big(\dfrac{\text{d}}{2}\Big)^2}$ = G $\dfrac{\text{m}_1\text{m}_2}{\dfrac{\text{d}^2}{4}}$ = 4G $\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}$ = 4F

Hence, when the distance between the objects is reduced to half, the gravitational force increases four times.

#### Question 2

Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

**Answer**

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). This acceleration remains constant and is independent of an object's mass. Hence, heavy objects do not fall faster than light objects.

#### Question 3

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10^{24} kg and radius of the earth is 6.4 × 10^{6} m.)

**Answer**

Mass of the body (m) = 1 kg

Mass of the earth (M) = 6 × 10^{24} kg

Radius of the Earth (R) = 6.4 × 10^{6} m

G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

According to the formula for gravitational force we know,

F = G $\dfrac{\text{M}\text{m}}{\text{R}^2}$

Substituting we get,

$\text{F} = \dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} \\[1em] = \dfrac{6.67 \times 10^{13}}{6.4 \times 6.4 \times10^{12}} \\[1em] = 9.77 \text{ N} \approx 9.8 \text{ N}$

Hence, **magnitude of the gravitational force = 9.8 N**

#### Question 4

The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

**Answer**

According to universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions.

This force is given by: F = G $\dfrac{\text{M}\text{m}}{\text{R}^2}$

Hence, the earth attracts the moon with the same force with which the moon attracts the earth.

#### Question 5

If the moon attracts the earth, why does the earth not move towards the moon?

**Answer**

According to the universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions. Hence, moon and earth both attract each other with the same force. Therefore, the earth does not move towards the moon.

#### Question 6

What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

**Answer**

(i) We know,

F = G $\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}$

Where,

F = force of attraction

G = constant of proportionality

m_{1} = mass of first object

m_{2} = mass of second object

d = distance between the two objects.

When mass of one object is doubled.

m' = 2m_{1}

So, substituting we get,

F' = G $\dfrac{\text{2m}_1\text{m}_2}{\text{d}^2}$ = 2G $\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}$ = 2F

Hence, **when the mass of one object is doubled, the gravitational force increases two times.**

(ii) When distance is doubled then, d' = 2r

So, substituting we get,

F' = G $\dfrac{\text{m}_1\text{m}_2}{(2d)^2}$ = G $\dfrac{\text{m}_1\text{m}_2}{4d^2}$ = $\dfrac{\text{F}}{4}$

Hence, **when the distance between the objects is doubled, the gravitational force becomes one fourth of its original force.**

Now, if it's tripled

d' = 3r

So, substituting we get,

F' = G $\dfrac{\text{m}_1\text{m}_2}{(3d)^2}$ =G $\dfrac{\text{m}_1\text{m}_2}{9d^2}$ = $\dfrac{\text{F}}{9}$

Hence, **when the distance between the objects is tripled, the gravitational force becomes one ninth of its original force.**

(iii) If masses of both the objects are doubled, then

m_{1}' = 2m_{1}

and

m_{2}' = 2m_{2}

So, substituting we get,

F' = G $\dfrac{\text{2m}_1\text{2m}_2}{\text{d}^2}$ = 4G $\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}$ = 4F

Hence, **when the masses of both the objects are doubled, the gravitational force becomes four times the original force.**

#### Question 7

What is the importance of universal law of gravitation?

**Answer**

The universal law of gravitation successfully explained several phenomena that were believed to be unconnected:

- The force that binds us to the earth
- The motion of the moon around the earth
- The motion of planets around the sun
- The tides due to the moon and the sun.

#### Question 8

What is the acceleration of free fall?

**Answer**

The acceleration of free fall is denoted by g and its value on the surface of the earth is 9.8 ms^{-2}.

#### Question 9

What do we call the gravitational force between the earth and an object?

**Answer**

Gravitational force between the earth and an object is known as the weight of the object.

#### Question 10

Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

**Answer**

The weight of a body on the earth's surface (W) = mg (where m = mass of the body and g = acceleration due to gravity)

Given, value of g is larger at poles when compared to the equator. So, for the same mass of gold, weight will be less at the equator as compared to the poles. Therefore, Amit's friend will not agree with the weight of gold bought.

#### Question 11

Why will a sheet of paper fall slower than one that is crumpled into a ball?

**Answer**

A sheet of paper has a larger surface area in comparison to a crumpled paper ball. The larger surface area will suffer greater air resistance. Hence, a sheet of paper falls slower than the crumpled ball.

#### Question 12

Gravitational force on the surface of the moon is only $\dfrac{1}{6}$ as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?

**Answer**

Given,

Acceleration due to earth's gravity g = 9.8 ms^{-2}

Object's mass, m = 10 kg

Acceleration due to moon gravity = g_{m}

Weight on the earth = W_{e}

Weight on the moon = W_{m}

Weight = mass x gravity

g_{m} = $\dfrac{1}{6}$ g_{e} (given)

So, W_{m} = m g_{m} = m x $\dfrac{1}{6}$ x g_{e}

W_{m} = 10 x $\dfrac{1}{6}$ x 9.8 = 16.34 N

W_{e} = m x g_{e} = 10 x 9.8

W_{e} = 98 N

Hence, **weight on moon and earth = 16.34 and 98 N respectively.**

#### Question 13

A ball is thrown vertically upwards with a velocity of 49 m/s.

Calculate

(i) The maximum height to which it rises,

(ii) The total time it takes to return to the surface of the earth.

**Answer**

Given,

Initial velocity (u) = 49 ms^{-1}

Final velocity v at maximum height = 0

Acceleration due to earth gravity g = -9.8 ms^{-2} (negative as ball is thrown up).

Let H be the maximum height to which the ball rises.

By third equation of motion,

2gH = v^{2} - u^{2}

Substituting we get,

2 × (-9.8) × H = 0 - (49)^{2}

-19.6 H = -2401

H = $\dfrac{-2401}{-19.6}$ = 122.5 m

Hence, **the maximum height to which it rises = 122.5 m**

(ii) Total time T = Time to ascend (T_{a}) + Time to descend (T_{d})

According to the first equation of motion,

v = u + gt

Substituting we get,

0 = 49 + (-9.8) x T_{a}

T_{a} = $\dfrac{49}{9.8}$ = 5 s

Also, T_{d} = 5 s

∴ T = T_{a} + T_{d} = 5 + 5 = 10 s

Hence, **total time it takes to return to the surface of the earth = 10 s**

#### Question 14

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

**Answer**

Given,

Initial velocity (u) = 0

Tower height = total distance = 19.6 m

g = 9.8 ms^{-2}

Final velocity (v) = ?

According to the third equation of motion,

v^{2} = 2gs + u^{2}

Substituting we get,

v^{2} = (2 x 9.8 x 19.6) + 0

v = $\sqrt{384.16}$

v = 19.6 ms^{-1}

Hence, **final velocity = 19.6 ms ^{-1}**

#### Question 15

A stone is thrown vertically upward with an initial velocity of 40 ms^{-1}. Taking g = 10 ms^{-2}, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

**Answer**

Given,

Initial velocity (u) = 40 ms^{-1}

Final velocity (v) = 0

g = 10 ms^{-2}

Max height = ?

According to the third equation of motion,

v^{2} = u^{2} - 2gs

Note: [negative g as the object goes up]

Substituting we get,

0 = (40)^{2} - 2 x 10 x s

⇒ 1600 = 20s

⇒ s = $\dfrac{1600}{20}$ = 80 m

Total Distance = s_{a} + s_{d} = 80 + 80 = 160 m

where, s_{a} and s_{d} are distance going up and coming down, respectively.

Displacement = 0 [as the stone comes to the same point from where it was thrown ].

Hence, **total distance covered = 160 m and displacement = 0, as the first point is the same as the last point**

#### Question 16

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10^{24} kg and of the Sun = 2 × 10^{30} kg. The average distance between the two is 1.5 × 10^{11} m.

**Answer**

Given,

Mass of the sun (m) = 2 × 10^{30} kg

Mass of the earth (M) = 6 × 10^{24} kg

G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

Average distance R = 1.5 × 10^{11} m

According to the formula for gravitational force we know,

F = G $\dfrac{\text{M}\text{m}}{\text{R}^2}$

Substituting we get,

F = $\dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 2 \times 10^{30} }{(1.5 \times 10^{11})^2}$ = 3.56 x 10^{22} N

Hence, **magnitude of the gravitational force = 3.56 x 10 ^{22} N**

#### Question 17

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 ms^{-1}. Calculate when and where the two stones will meet.

**Answer**

Given,

(a) In the case, when the stone falls from the top of the tower,

Initial velocity u = 0

Distance travelled = x

Time taken = t

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$gt^{2}

⇒ S = 0 x t + $\dfrac{1}{2}$gt^{2}

⇒ S = $\dfrac{1}{2}$gt^{2} **[Eq 1]**

(b) When another stone is projected vertically upwards,

Initial velocity u = 25 ms^{-1}

Distance travelled = (100 - x)

Time taken = t

Using S = ut + $\dfrac{1}{2}$gt^{2}

we get,

S' = 25t - $\dfrac{1}{2}$gt^{2} **[Eq 2]**

From equations (1) and (2)

S + S' = $\dfrac{1}{2}$gt^{2} + 25t - $\dfrac{1}{2}$gt^{2}

100 = $\dfrac{1}{2}$gt^{2} + 25t - $\dfrac{1}{2}$gt^{2}

100 = 25t

t = $\dfrac{100}{25}$ = 4 sec.

Hence, **after 4 secs, the two stones will meet.**

From (1)

S = $\dfrac{1}{2}$ x 10 x 4^{2} = 5 x 16 = 80m.

Hence, **after 4 sec, 2 stones meet a distance of 80 m from the top.**

#### Question 18

A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4s.

**Answer**

Given,

Ball returns to thrower in 6s. So, time_{up} + time_{down} = 6 s hence, time_{up} = 3 s.

g = 9.8 ms^{-2}

(a) Final velocity (v) = 0

From first equation of motion

v = u – gt_{up}

⇒ u = v + gt_{up}

⇒ u = 0 + (9.8 x 3)

⇒ u = 29.4 ms^{-1}

Hence, **the velocity with which it was thrown up = 29.4 ms ^{-1}.**

(b) According to the second equation of motion,

S = ut - $\dfrac{1}{2}$gt^{2}

Substituting we get,

S = (29.4 x 3) - ($\dfrac{1}{2}$ x 9.8 x 3 x 3)

= 88.2 - 44.1 = 44.1 m

Hence, **maximum height stone reaches is 44.1 m**

(c) In 3 sec, it reaches the maximum height.

Distance travelled in another 1 sec = s'

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$gt^{2}

g is +ve as ball is going down

Substituting we get,

S = 0 + ($\dfrac{1}{2}$ x 9.8 x 1 x 1)

= 4.9 m

Hence, **in 4 sec the ball will be 4.9 m from the top or 39.2 m (i.e., 44.1 - 4.9) from the bottom.**

#### Question 19

In what direction does the buoyant force on an object immersed in a liquid act?

**Answer**

The buoyant force acts vertically upwards on the object that is immersed in a liquid.

#### Question 20

Why a block of plastic when released under water come up to the surface of water?

**Answer**

When plastic is immersed in water, two forces act on it:

- Gravitational force or its weight in vertically downward direction
- Buoyant force in vertically upward direction.

As the density of plastic is less than that of water, the buoyant force is greater than its weight. Hence, the plastic comes up to the surface.

#### Question 21

The volume of 50 g of a substance is 20 cm^{3}. If the density of water is 1 gcm^{-3}, will the substance float or sink?

**Answer**

Given,

Volume of substance = 20 cm^{3}

Mass of substance = 50 g

Density of water = 1 gcm^{-3}

Density of the substance = ?

Density = $\dfrac{\text{Mass}}{\text{Volume}}$ = $\dfrac{50}{20}$ = 2.5 gcm^{-3}

As density of the substance is greater than density of water. Hence, **the substance will sink.**

#### Question 22

The volume of a 500 g sealed packet is 350 cm^{3}. Will the packet float or sink in water if the density of water is 1 g cm^{-3}? What will be the mass of the water displaced by this packet?

**Answer**

Given,

Volume of sealed packet = 350 cm^{3}

Mass of sealed packet = 500 g

Density of water = 1 gcm^{-3}

Density of the sealed packet = ?

Density = $\dfrac{\text{Mass}}{\text{Volume}}$ = $\dfrac{500}{350}$ = 1.42 gcm^{-3}

As density of the packet is greater than density of water. Hence, **the sealed packet will sink.**

According to Archimedes Principle,

Displaced water volume = Volume of packet

Volume of water displaced = 350 cm^{3}

Mass of water displaced = ρ x V = 1 × 350 = 350 g

Hence, **mass of displaced water = 350 g**