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# NCERT book Solutions class-9 Gravitation (Part 2)

This page contains NCERT book Solutions class -9 Gravitation starting from page 143 at the end of the chapter. I have also been writing the notes for this chapter to learn more about notes you can follow this

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[standout-css3-button href=”https://physicscatalyst.com/article/ncert-solutions-class9-gravitation/”]Class 9 gravitation NCERT book in text questions answers[/standout-css3-button]
If you want to get the solutions of first 11 questions follow this

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For assignment on gravitation follow this

[standout-css3-button href=”https://physicscatalyst.com/Class9/gravitation_questions.php”]gravitation assignment(FA)[/standout-css3-button]

Question 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 newtons of a 10 kg object on the moon and on the earth?

Weight of any object on moon=$\frac{1}{6} \times$ weight of object on earth

Also, Weight = mass $\times$ acceleration

Acceleration due to gravity $g=9.8 ms^{2}$

So, weight of an object of mass 10 Kg on earth = $10 \times 9.8=98N$

And weight of the same object on moon = $\frac{1}{6} \times 98=16.3N$

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.

(i) According to question

Initial velocity of the ball (u)=$49ms^{-1}$

final velocity of the ball (v)=$0ms^{-1}$

Downward gravity (g) = $9.8ms^{-2}$

Upward gravity (g) = $-9.8ms^{-2}$

Now we have to find the maximum height $h$ attained by the ball

From equation of motion we have

$v^{2}-u^{2}=2gh$

putting respective values in above equation we get

$(0)^{2}-(49)^{2}=2\times (-9.8)\times h$

$h=\frac{49 \times 49}{2 \times 9.8}=122.5m$

(ii) let $t$ be the time taken by the ball to reach the height 122.5 m , then according to the equation of motion $v=u+gt$

we get,

$0=49+t \times (-9.8)$

$9.8t=49$

or,

$t=\frac{49}{9.8}=5s$

But, time of ascent=time of descent

Therefore , total time taken by ball to return back = 5+5 = 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.

According to question

initial velocity of stone = $0 ms^{-1}$

acceleration due to gravity (g)=$9.8 ms^{-2}$

height (h) = $-19.6 m$

From equation of motion of object under gravity we have

$v^{2}-u^{2}=2gh$

putting respective values in above equation we get

$v^{2}-(0)^{2}=2 \times 9.8 \times 19.6$

$v^{2}=2 \times 9.8 \times 19.6$

$v^{2}=( 19.6)^{2}$

or

$v=19.6 ms^{-1}$

Hence velocity of  the stone just before touching the ground is $19.6 ms^{-1}$

Question 15

A stone is thrown vertically upward with an initial velocity of 40 m/s. 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?

Equation of motion of object under gravity we have

$v^{2}-u^{2}=2gh$               (1)

From values given in question we have

initial velocity u = 40 m/s and acceleration due to gravity is given as $g = 10 ms^{2}$.

Let $h$ be the maximum height attained and final velocity $v=o$

Rearranging equation (1) we get

$h= \frac{v^{2}-u^{2}}{2g}$

putting respective values we have

$h=\frac{40 \times 40}{20}=80m$

Total distance traveled would be = $2 \times h = 160m$

Since the stone returns back to the original position displacement of the stone during its total upward and downward journey  = $80 +(-80)=0$

Question 16

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

As given in the question

Mass of earth  $M_{e}= 6 \times 10^{24} kg$

Mass of sun $M_{s}= 2 \times 10^{30} kg$

distance between the two is $R=1.5 \times 10^{11} m$

From Universal law of gravitation

$F = G\frac{{M \times m}}{{{R^2}}}$

Therefore, putting all the values given in question in above equation we get

$F = 6.67\times10^{-11}\frac{{(6 \times 10^{24}) \times (2 \times 10^{30})}}{{{(1.5 \times 10^{11})^2}}}=3.56\times10^{22}N$

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

Let $t$ be the point at which two stones meet and let $h$ be their height from the ground. It is given in the question that height of the tower is $H=100m$

Now first consider the stone which falls from the top of the tower. So distance covered by this stone at time $t$ can be calculated using equation of motion

$x-x_{0}=u_{0}t+\frac{1}{2}gt^{2}$

Since initial velocity $u=0$ so we get

$100-x=\frac{1}{2}gt^{2}$                           (1)

The distance covered by the same stone that is thrown in upward direction from ground is

$x=u_{0}t-\frac{1}{2}gt^{2}$

In this case initial velocity is 25 m/s . So

$x=25t-\frac{1}{2}gt^{2}$                                  (2)

Combining equations 1 and 2 we get

$100=25t$

this gives $t=4s$

This gives

$x=25 \times 4-\frac{1}{2} \times 9.8 \times (4)^{2}=100-78.4=21.6m$

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.

(a) Here we have to find the velocity with which the ball is thrown in upwards direction.

Now, Acceleration due to gravity ,$g=-9.8 ms^{-2}$

It is given in the question that total time taken by the ball in going upward and in comming back is 6 s.

Therefore time taken in upward journey = t=6/2=3s

Final velocity v= 0 m/s

we have to find the initial velocity u

We know that from first equation of motion

$v=u+gt$

so,

putting all the values we gat

$0=u+(9.8) \times 3$

calculating we get

$u=29.4 m/s$

So the velocity with which ball is thrown in the upward direction is 29.4 m/s

(b) Now we have to find the maximum height say $h_{max}$ reached by the ball

From second equation of motion we have

$s=ut+\frac{1}{2}gt^{2}$

here s (distance) is the maximum height $h_{max}$ reached

putting all the values in above equation we have

$h_{max}=29.4 \times 3 +\frac{1}{2}(-9.8)(3)^{2}=44.1 m$

(c) Ball reaches its maximum height after 3 s . After reaching this height ball started to move in downward direction. So in this case

initial velocity u=0

Position of the ball after 4 s of the throw is given by the distance at which ball reaches while going downward direction in $4-3=1$ $s$

Again from second equation of motion we have

$s=ut+\frac{1}{2}gt^{2}$

putting all the values in above equation and calculating we find $s=4.9m$

here $h_{max}=44.1 m$

So, ball is 39.2 m (44.1 m- 4.9 m) above the ground after 4 s

Question 19

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

Buoyant force in this case will acts in vertically upwards direction.

Question 20

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

For an object immersed in water two force acts on it

1. gravitational force which tends to pull object in downward direction

2. buoyant force that pushes the object in upward direction

here in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water. Note:- this only happens in case where density of block is less than that of water.

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?

If the density of this object is greater then the density of water then it will sink and if it is less than water then t will float in the water. So we calculate the density of this substance

$Density=\frac{mass}{volume}=\frac{50}{20}=25 g$ $cm^{-3}$

Since density of this substance is greater then the density of water so it 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?

Density of packet = (mass of packet)/(volume of packet)

From the question we have

mass of packet = 500 g

volume of packet =350 $cm^{3}$

so density of packet =$\frac{500}{350}=1.428g$ $cm^{-3}$

The density of this substance is greater then the density of water. Hence it will sink in water. The mass of water displaced by the packet is equal to the volume of packet that is 350 gm

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