# Gravitation

## NCERT Solution

### Part 2

Question 14: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

**Answer:** This can be calculated by following formula:

`v^2 - u^2 = 2gs`

Or, `v^2 - 0 = 2 xx 9.8 xx 19.6`

Or, `v^2 = 19.6 xx 19.6`

Or, `v = 19.6 m//s`

Question 15: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

**Answer:** This can be calculated by following formula:

`v^2 - u^2= 2gs`

Here, final velocity will be zero;

Or, `0 – 40^2 = 2 xx – 10 xx s`

Or, `1600 = 20 xx s`

Or, `s = 1600/20 = 80 m`

The stone will finally return on the starting point, i.e. on the ground. So, net displacement is zero.

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:** The force of gravitation between the earth and the sun can be calculated as follows:

`F=G(Mm)/(R^2)`

`=6.67xx10^-11xx(6xx10^24\xx10^30)/(1.5xx10^11)^2`

`=(6.67xx6xx2xx10^43)/(2.25xx10^22)`

`=3.58xx10^22N`

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

**Answer:** Let us assume that the stones meet at a height of x from the ground. The first stone will travel 100 – x to reach this point. The time taken to reach this distance by the first stone can be calculated as follows:

`s=ut+1/2\g\t^2`

Or, `100-x=0+1/2\xx9.8xxt^2`

Or, `100-x=4.9xxt^2`

Or, `x=100-4.9xxt^2`

Let this be the equation (1).

Now, time taken by second stone to reach the distance x can be calculated as follows:

`x=ut+1/2\gxxt^2`

Or, `x=25t-4.9xxt^2`

Let it be the equation (2)

From equation (1) and (2), we get;

25t = 100

Or, t = 4 s

Now, putting the value of t, in equation (2) we get;

`x = 25 xx 4 – 4.9 xx 4^2`

Or, `x = 100 – 4.9 xx 16`

Or, `x = 100 – 78.4 = 21.6 m`

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 4 s.

**Answer:**Given; initial velocity u = ?, final velocity v = 0, t = 3s, s = ?

Initial velocity can be calculated as follows:

`v = u + gxxt`

Or, `0 = u - 9.8 xx 3`

Or, `u = 9.8 xx 3 = 29.4 m//s`

Now, maximum height can be calculated as follows:

`s=ut+1/2\gxxt^2`

`=29.4xx3-1/2xx9.8xx3^2`

`=88.2-44.1=44.1m`

Distance after 4 s can be calculated as follows:

`s=ut+1/2\gxxt^2`

`=29.4xx4-1/2xx9.8xx16`

`=117.6-78.4=39.2m`

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

**Answer:** Upward direction

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

**Answer:** Because of buoyancy

Question 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?

**Answer:** Density = mass/volume

Hence, density of object = 50g/20 cm^{3}

= 2.5 g cm^{-3}

Here; density of object is more than the density of water.

Hence, the object will sink in water.

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:** Density = mass/volume

Hence, density of object = 500g/350 cm^{3}

= 1.43 g cm^{-3}

Here; density of object is more than the density of water.

Hence, the object will sink in water.