## Exercise-10.5 Page: 142 ( Gravitation )

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

**Solution:**

When weighing our body, it is acting by an upward force. The buoyant force is this upward force. As a result, the body is pushed up slightly, resulting in the weighing machine showing less reading than the actual value.

**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?**

**Solution:**

The bag of cotton is heavier than the bar of iron. The cotton bag has a larger air thrust than the iron bar. The weighing machine therefore indicates a smaller cotton bag weight than its actual weight.

#### Exercises-10.6 Page: 143 ( Gravitation )

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

**Solution:**

Consider the Universal law of gravitation,

According to that law, the force of attraction between two bodies is

Where,

m_{1} and m_{2} are the masses of the two bodies.

G is the gravitational constant.

r is the distance between the two bodies.

Given that the distance is reduced to half then,

r = 1/2 r

Therefore,

F = 4F

Therefore once the space between the objects is reduced to half, then the force of gravitation will increase by fourfold the first force.

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

**Solution:**

All objects fall on the bottom with constant acceleration called acceleration thanks to gravity (g). It’s constant and therefore the price of ‘g’ doesn’t depend on the mass of associate object. So serious objects don’t fall quicker than light-weight objects provided there’s no air resistance.

**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.)**

**Solution:**

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by

**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?**

**Solution:**

The earth attracts the moon with associate degree equal force with that the moon attracts the planet however these forces are in opposite directions. By universal law of gravitation, the force between moon and also the sun can be,

Where,

d = distance between the earth and moon.

m_{1} m_{2} = masses of earth and moon respectively.

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

**Solution:**

According to universal law of gravitation and Newton third law, we all know that the force of attraction between 2 objects is the same, however in wrong way. So the planet attracts the moon with the identical force because the moon exerts on earth however in opposite directions. Since earth is far larger in size than moon, that the acceleration cannot be detected on earth surface.

**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?**

**Solution:**

(i)

According to universal law of gravitation, the force between 2 objects (m_{1} and m_{2}) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

If the mass is doubled for one object.

F = 2F, so force is also doubled.

(ii)

If the distance between the objects is doubled and tripled

If it’s doubled

Hence,

F = 4F, Force thus becomes one-fourth of its initial force.

If it’s tripled

Hence,

F = 9F, Force thus becomes one-ninth of its initial force.

(iii)

If masses of both the objects are doubled, then

F = 4F, Force will therefore be four times greater than its actual value.

**7. What is the importance of universal law of gravitation?**

**Solution:**

The universal law of gravitation explains many phenomena that were believed to be unconnected:

(i) The motion of the moon round the earth

(ii) The force that binds North American nation to the world

(iii) The tides because of the moon and therefore the Sun

(iv) The motion of planets round the Sun

**8. What is the acceleration of free fall?**

**Solution:**

When anybody is in free fall, the sole force functioning on the article is that the earth’s field of force. By Newton’s second law of motion all the forces manufacture acceleration, therefore all the objects accelerate toward the world’s surface thanks to attraction of the earth.

This acceleration is thought as acceleration thanks to gravity close to earth’s surface. It’s denoted by ‘g’ and its worth is 9.8m/s^{2} and it’s constant for all objects close to earth’s surface (irrespective of their masses).

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

**Solution:**

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

**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.]**

**Solution:**

The weight of a body on the earth’s surface;

W = mg wherever (m = mass of the body and g= acceleration thanks to gravity)

The value of g is a lot of at poles as compared to equator. So gold can weigh less at the equator as compared to poles.

Therefore, Amit’s friend won’t believe the load of the gold bought.

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

**Solution:**

A sheet of paper has a lot of area as compared to a crumpled paper ball. A sheet of paper must face a lot of air resistance. As result a sheet of paper falls slower than the crumpled ball.

**12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton’s of a 10 kg object on the moon and on the earth?**

**Solution:**

Given data:

Acceleration due to earth’s gravity = g_{e} or g = 9.8 m/s^{2}

Object weight 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} = (1/6) g_{e} (given)

So W_{m} = m g_{m} = m x (1/6) g_{e}

W_{m} = 10 x (1/6) x 9.8 = 16.34 N

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

W_{e} = 98N

**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.**

**Solution:**

Given data:

Initial velocity u = 49m/s

Final speed v at maximum height = 0

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

By third equation of motion,

v^{2} = u^{2} – 2gs

Substitute all the values in the above equation

Total time T = Time to ascend (Ta) + Time to descend (Td)

V = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5s

Also, Td = 5s

Therefore T = Ta + Td

T = 5 + 5

T = 10s

**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.**

**Solution:**

Given data:

Initial velocity

u = 0

Tower height = total distance = 19.6m

g = 9.8 m/s^{2}

Consider third equation of motion

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

v^{2 }= 0 + 2 × 9.8 × 19.6

v^{2} = 384.16

v = √(384.16)

v = 19.6m/s

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

**Solution:**

Given data:

Initial velocity u = 40m/s

g = 10 m/s^{2}

Max height final velocity = 0

Consider third equation of motion

v^{2} = u^{2} – 2gs [negative as the object goes up]

0 = (40)^{2} – 2 x 10 x s

s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

Total displacement = 0 (The first point is the same as the last point)

**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.**

**Solution:**

Given data:

Mass of the sun m_{s} = 2 × 10^{30} kg

Mass of the earth m_{e} = 6 × 10^{24} kg

Gravitation constant G = 6.67 x 10^{-11} N m^{2}/ kg^{2}

Average distance r = 1.5 × 10^{11} m

Consider Universal law of Gravitation

**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 m/s. Calculate when and where the two stones will meet.**

**Solution:**

Given data:

(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

From equations (a) and (b)

5t^{2} = 100 -25t + 5t^{2}

t = (100/25) = 4sec.

After 4sec, two stones will meet

From (a)

x = 5t^{2} = 5 x 4 x 4 = 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

This means that after 4sec, 2 stones meet a distance of 20 m from the ground.

**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.**

**Solution:**

Given data:

g = 10m/s^{2}

Total time T = 6sec

T_{a} = T_{d} = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u – gt_{a}

u = v + gt_{a}

= 0 + 10 x 3

= 30m/s

The velocity with which stone was thrown up is 30m/s.

(b) From second equation of motion

The maximum height stone reaches is 45m.

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

Distance travelled in another 1sec = s’

The distance travelled in another 1sec = 5m.

Therefore in 4sec, the position of point p (45 – 5)

= 40m from the ground.

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

**Solution:**

The buoyant force on an object that is immersed in a liquid will be in a vertically upward direction.

**20. Why a block of plastic does released under water come up to the surface of water?**

**Solution:**

The density of plastic is a smaller amount than that of water, therefore the force of buoyancy on plastic block are going to be bigger than the load of plastic block displaced. Hence, the acceleration of plastic block are going to be in upward direction, and comes up to the surface of water.

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

**Solution:**

To find the Density of the substance the formula is

Density = (Mass/Volume)

Density = (50/20) = 2.5g/cm^{3}

Density of water = 1g/cm^{3}

Density of the substance is greater than density of water. So the substance will sink.

**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?**

**Solution:**

Density of sealed packet = 500/350 = 1.42 g/cm^{3}

Density of sealed packet is greater than density of water

Therefore the packet will sink.

Considering Archimedes Principle,

Displaced water volume = Force exerted on the sealed packet.

Volume of water displaced = 350cm^{3}

Therefore displaced water mass = ρ x V

= 1 × 350

Mass of displaced water = 350g.