NCERT Solutions for Science Class 9 Chapter 9

CHAPTER-9

Intex Excercise 1:

Question 1: State the universal law of gravitation

Answer: Every object in the universe attracts every other object with some force which is directly proportional to the product of their masses and inversely
proportional to the square of the distance between them. The force acts along the line joining the centres of two objects.

Let the two objects A and Bof masses Mand m lie at a distance d from each other. Let the force of attraction between two objects be F.

\(F=\frac{GMm}{f^2}\)

Where, G is the universal gravitation constant which is given by:

\(G=6.67\times10^{-11}Nm^2kg^{-2}\)

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: Let the mass of the Earth be M and the mass of an object on its surface be m. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force ( F) that acts between the Earth and the object can be given by the relation:

\(F=\frac{GMm}{R^2}\)

Intext Exercise 2

Question 1: What do you mean by free fall?
Answer: Each object is drawn towards the centre of the Earth by its gravity. When any object is released from a certain height, under the impact of gravitational force, it falls to the Earth’s surface. The movement of the object is said to be in free fall.
Question 2: What do you mean by acceleration due to gravity?
Answer: When any object falls freely from a certain height towards the earth’s surface, its velocity changes with respect to time. This change in velocity causes acceleration. This acceleration is known as the acceleration due to gravity ( g ). The value of acceleration due to gravity is \(9.8\space ms^{-2}\)

Intext Exercise 3

Question 1: What are the differences between the mass of an object and its weight?

Answer: The difference between mass of an object and its weight is given in the table below:

Question 2: Why is the weight of an object on the moon \(\frac{1}{6}\)th its weight on the earth?

Answer: Let the mass of the Earth be \(M_E\) and the mass of an object on the surface of earth \(=m\) and the radius of earth \(R_E\).

According to the Universal law of gravitation, weight \(W_E\) of the object on the surface of the earth is given by,

\(W_E=\frac{GM_Em}{R_E^2}\)

Let \(M_M\) and \(R_M\) be the mass and radius of the moon. Then, according to the universal law of gravitation, weight \(W_4\) of the object on the surface of the moon is given by:

\(W_M=\frac{GM_Mm}{R_M^2}\)

So, ratio of weight of object on moon to weight on earth is

\(\frac{W_M}{W_E}=\frac{M_MR_E^2}{M_ER_M^2}\)

Where, \(M_E=5.98\times10^{24}kg\)

\(M_M=7.36\times100^{22}kg\)

\(R_E=6.4\times10^6m\)

\(R_M=1.74\times10^6m\)

Substituting the values in the ratio,

\(⇒\(\frac{W_M}{W_E}=\frac{7.36\times10^{22}\times(6.37\times10^6)^2}{5.98\times10^{24}\times(1.74\times10^6)^2}\)

\(⇒\frac{W_M}{W_E}=0.165≈\frac{1}{6}\)

Intext Exercise 4

Question 1: Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer: Pressure can be given by the formula,

\(P=\frac{F}{A}\)

Pressure is inversely proportional to the surface area on which the force is acting. The smaller is the surface area, the larger will be the pressure on the surface on which the force is being acted upon. In the case of a thin strap of the school bag, the contact surface area is very less. Hence, the pressure exerted on the shoulder is very high. Therefore, it becomes difficult to hold a  school bag with a thin strap.

Question 2: What do you mean by buoyancy?
Answer: The liquid exerts an upward force on any object when it is immersed in a liquid. The tendency of the liquid to exert such an upward force on the object is called buoyancy, and the upward force which is exerted on the object by the liquid is called the buoyant force.

Question 3: Why does an object float or sink when placed on the surface of water?
Answer: If the density of an object is greater than the density of liquid, it will sink into the liquid. This is due to the buoyant force which is acted by the object is less than the force of gravity.
On the contrary, if the density of object is less than the density of the liquid, it floats on the liquid’s surface. This is because the force that is acting on the object is greater than the force of gravity.

Intext Exercise 5

Question 1: You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42kg?
Answer: An upward force acts on our body when we weigh our body while standing on a weighing machine. The buoyant force is this upward force which is acting. Consequently, the body is pushed up slightly, resulting in the weighing machine showing less reading than the real value.

Question 2: You have a bag of cotton and an iron bar, each indicating a mass of 100kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is  eavier and why?
Answer: Weight measured = Actual weight = buoyant force Therefore, Actual weight = Weight measured = buoyant force As the surface area of a cotton bag is greater than the iron bar, more buoyant force acts on the bag than that on the iron bar. Hence, the mass of the cotton bag is more than that of the iron bar.

NCERT Exercise
Question 1: How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer: According to thee universal law of gravitation, the gravitational force (F) acting between two objects of mass \(m_1\) and \(m_2\), separated by a distance ‘r’ is given by

\(F=\frac{Gm_1m_2}{r^2}\)

Where \(m_1\) and \(m_2\) are the masses of two bodies and \(r\) is the distance between them, \(G\) is the universal gravitational constant.

When the distance is reduced to hald, i.e., \(r’=\frac{r}{2}\)

\(⇒F=\frac{Gm_1m_2}{(\frac{r}{2})^2}\)

\(⇒F=\frac{Gm_1m_2}{\frac{r^2}{4}}\)

\(⇒F=\frac{4Gm_1m_2}{r^2}\)

Hence, if the distance is reduced to half, then the gravitational forcec becomes four times that of the previuos value.

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 the objects fall towards the ground with constant acceleration, called acceleration due to gravity (if there is no air resistance present). It is constant and independent of the mass of the object. Hence, heavy objects do not fall faster than light objects.

Question 3: What is the magnitude of the grtavitational force between the earth and a 1 kg object on its surface? (Mass of the earth is \(6\times10^{24}\space kg\) and radius of the earth is (\(6.4\times10^6\space m\)).

Answer: According to the Universal law of gravitation, the gravitational force exerted on an object of mass mis given by:

\(F=\frac{GMm}{r^2}\)

Where,

Mass of Earth, \(M=6\times10^{24}kg\)

Mass of object, \(m=1\space kg\)

Universal gravitational constant, \(G=6.7\times10^{-11}Nm^2kg^{-2}\)

Since the object is on the surface of the Earth, \(r=\) radius of the Earth (R)

\(r=R=6.4\times10^6m\)

Gravitational force,

\(F=\frac{GMm}{r^2}\)

\(⇒F=\frac{6.7\times10^{-11}\times6\times10^{24}\times1}{(6.4\times10^6)^2}=9.8 N\)

The magnitude of the gravitational force between the earth and a 1kg object on its surface is 9.8N.

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 the Universal law of gravitation, two objects attract each other and according to Newton’s third law of motion, the force of attraction between two objects is the same but acts in the opposite direction. Thus, the earth attracts the moon with the same force as the moon exerts on earth but the force acts in the opposite direction.
Question 5: If the moon attracts the earth, why does the earth not move towards the moon?
Answer: The Earth and the moon experience equal gravitational forces acting towards each other.

By Newton’s Second Law, \(F=ma\)

\(⇒a=\frac{F}{m}\)

For a certain force, acceleration is inversely proportional to the mass of an object.

\(a∝\frac{F}{m}\)

Mass of the Earth >> Mass of the moon.

Hence, the acceleration experienced by earth due to the gravitational pull of the moon is very small when compared to that experienced by the moon due to earth. That is why the Earth does not move towards the moon.

Question 6: What happens to the force between two objects, if
(a) the mass of one object is doubled?

Answer: According to the universal law of gravitation, the force of gravitation between two objects is given by: \(F=\frac{GMm}{r^2}\)

\(F\) is directly proportional to the product of masses of the two objects.

\(F∝Mm\)

If the mass of one object is doubled, then the gravitational force will also change to double the original.

(b) The distance between the objects is doubled an tripled?

Answer: \(F\) is inversely proportional to the square of the distance between the objects.

If the distance between the objects is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance is tripled, then the gravitational force becomes one-ninth of its original value. 
(c) the masses of both objects are doubled?
Answer: F is directly proportional to the product of masses of the objects.
\(F∝Mm\)
If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.

Question 7: What is the importance of the universal law of gravitation?

Answer: The universal law of gravitation states that every object in the universe attracts every other object.
The force of gravitation binds us to the earth.
It is the cause for the motion of the moon around the earth and planets around the sun.

It results in the formation of tides due to the moon and the Sun. High tide occurs at the side where the moon pulls towards itself.
Question 8: What is the acceleration of free fall?

Answer: A free-falling object is an object that is falling due to gravity without any air resistance. When it falls, there is a variation in velocity with respect to time, that is, there is an acceleration associated with it.
Acceleration of free fall is denoted by g and its value on the surface of earth is \(9.8\space ms^{-2}\), which is constant for all objects (irrespective of their masses.)

Question 9: What do we call the gravitational force between the Earth and an object?
Answer: The gravitational force between the earth and an object is called the weight of that object. It is equal to the product of acceleration due to gravity and mass of the object.

Question 10: Amit buys a 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: Weight of a body on the Earth is given by:
W=mg
Where,
m=Mass of the body
g =Acceleration due to gravity
The shape of Earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator. Since the value of g is greater at the poles than the equator.
Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.
Question 11: Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer: When a sheet of paper is crumbled into a ball, then its surface area becomes much lesser than the surface area of a plain non-crumpled sheet of paper.

Hence, the upward force exerted by air on the sheet is greater as compared to the one exerted on the ball. Hence the sheet falls slower as compared to a paper ball.

Question 12: gravitational force on the surface of the moon is only \(\frac{1}{6}\) as strong as gravitational force on the Earth. What is the weight newtons of a 10 kg object on the moon and on the Earth?

Answer: it is provided that,

Weight of an object on the moon \(=\frac{1}{6}\times\) Weight of an object on the Earth

Alsom

Weight = Mass \(\times\) Acceleration, \(g=9.8\space ms^{-2}\)

Therefore, the weight of a 10 kg object on the Earth \(=10\times9.8N=98N\)

Weight of the same object on the moon \(=\frac{1}{6}\times9.8N=16.3N\)

Question 13: A ball is thrown vertically upwards with a velocity of \(49\space ms^{-1}\).

Calculate

(a) The maximum height to which it rises.

Answer: According to the equation of motion under gravity:

\(v^2-u62=3gh\)

Where,

u = Initial velocity of the ball
v = Final velocity of the ball
h =Height achieved by the ball
g =Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., \(v=0\)

\(u=49\space ms^{-1}\)

During upward motion, \(g=9.8\space ms^{-2}\)

Let \(h\) be the maximum height attained by the ball.

Hence,

\(⇒(0)^2-(49)^2=2\times(-9.8)\times h\)

\(⇒h=\frac{49\times49}{2\times9.8}=122.5\)

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

Answer: 

Let \(t\) be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

\(v=u+gt\)

Substituting the values and solving,

\(⇒0=49+t\times(-9.8)\)

\(⇒9.8t=49\)

\(⇒t=\frac{49}{9.8}=5s\)

However,

Time of ascent = Time of descent

Therefore, total time taken by the ball to return is \(5+5=10s\).

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: According to the equation of motion under gravity:

\(v^2-u^2=2gh\)

Where,

u= Initial velocity of the stone = 0

v= Final velocity of the stone

s= Height of the stone=9.6 m

g=Acceleration due to gravity = \(9.8ms^{-2}\)

\(⇒v^2-0^2=2\times9.8\times19.62\)

\(⇒v^22\times9.8\times19.6=(19.6)^2\)

\(v=19.6,s^{-1}\)

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

Question 15: A stone is thrown vertically upward with an initial velocity of \(40ms^{-1}\). Taking \(g=10ms^2\), find the maximum height reached by the stone. What is the net disoaplacement and the total distance covered by the stone?

Answer: According to the equation of motion under gravity:

\(v^2-u^2=3gh\)

Where,

u= Intital velocity of the stone \(=40ms^{-1}\)

v=Final velocity of the stone =0

s= Height of the stone

g= Acceleration due to gravity \(=-10ms^{-2}\)

Let \(h\) be the maximum height attained by the stone.

Therefore,

\(⇒0-(40)^2=2\times h\times (-10)\)

\(⇒h=\frac{40\times40}{20}=80\space m\)

Therefore, total distance covered by the stone during its upward and downward journey is \(80+80=160\space m\).

Net displacement of the stone during its upward and downward journey is \(80+(-80)=0\space m\).

Question 16: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth \(=6\times10^{24}kg\) and of the Sun \(=2\times10^{30}kg\). The average distance between the two is \(1.5\times10^{11}m\).

Answer: According to the Universal l law of gravitation, the force of attraction between the Earth and the Sun is given by:

Given, 

\(M_{Sun}=\) Mass of the Sun \(=2\times10^{30}kg\)

\(M_{Earth}=\) Mass of the Earth \(=6\times10^{24}kg\)

\(R=\) Average distance between the Earth and the Sun \(=1.5\times10^{11}m\)

\(G=\) Universal gravitational constant \(=6.7\times10^{-11}Nm^2Kg^{-2}\)

\(F=\frac{GM_{Sun}M{Earth}}{R^2}\)

\(⇒F=\frac{6.7\times10^{-11}\times2\times10^{30}\times6\times10^{24}}{(1.5\times10^{11})^2}\)

\(⇒F=3.57\times10^{22}N\)

Hence, the force of gravitation between the earth and the sun is \(3.57\times10^{22}N\).

Question 17: A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of \(25\space ms^{-1}\). Calculate when and where the two stones will meet.

Answer: Let the two stones meet after time \(t\) from the start.

(a) Fos the stone dropped from the tower:

Intial velocity, \(u=0\)

Let the displacement of the stone in time \(t\) from the top of the tower ve \(s\).

Acceleration due to gravity, \(g=9.8\space ms^{-2}\)

From the equation of motion,

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

\(⇒s=0\times t=\frac{1}{2}\times9.8\times t^2\)

\(⇒s=4.9t^2\)   ….(1)

(b) For the stone thrwon upwards:

Intial velocity, \(u=25ms^{-1}\)

Let the displacement of the stone from the ground in time \(t\) be \(s’\).

Acceleration due to gravity, \(g=9.8ms^{-2}\)

Equation of motion,

\(s’=ut+\frac{1}{2}gt^2\)

\(⇒s’=25t-\frac{1}{2}\times9.8\times t^2\)

\(⇒s’=25t-4.9t^2\)  …(2)

The combined displacement of both the stones at the meeting point is equal to the height of the towe \(100\space m\)

\(s+s’=100\)   ….(3)

Substituting equation (1) and (2) in (3),

\(4.9t^2+25t-4.9t^2=100\)

\(⇒25t=100\)

\(⇒t=\frac{100}{25}=4s\)

In \(4\space s\), the falling stone has covered a distance given by equation (1) as 

\(s=\frac{1}{2}\times9.8\times4^2=78.4\space m\)

Therefore, the stones will meet after \(4\space s\) at a height \((100-78.4)=21.6\space m\) from the ground.

Question 18: A baal thrown up vertically return to the thrower after \(6\space s\). Find 

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

Answer: Time of ascent is equal to the time of descent. The ball takes a total of \(6s\) for its upward and downward journey.

Hence, time taken for upward journey, \(t=\frac{6}{2}=3s\)

Final velocity of the ball at the maximum height, \(v=0\)

Acceleration due to gravity, \(g=9.8\space ms^{-2}\)

Equation of motion, \(v=u+gt\) will give,

\(⇒0=u+(-9.8\times3)\)

\(⇒u=9.8\times3=29.4\space ms^{-1}\)

Hence, the ball was thrwon upwards with a velocity of \(29.4\space ms^{-1}\).

(b) The maximum height it reaches

Answer: Let the maximum height attained by the ball be \(h\).

Initial velocity during the upward journey, \(u=29.4\space ms^{-1}\)

Final veloxity, \9v=0\)

Acceleration due to gravity, \(g=-9.8\space ms^{-2}\)

\9s=ut=\frac{1}{2}at^2\)

From the equation of motion,

\(h=29.4\times3+\frac{1}{2}\times(-9.8)\times(3)^2=44.1\space m\)

(c) its position after \(4s\).

Answer: Ball attains the maximum height after \(3s\). After attaining this height, it will start falling downwards.

in this case, Initial velocity, \(u=0\)

Position of the ball after \(4s\) of the throw is given by the distance travelled by it during downward journey in \(4s-3s=1s\)

Equation of motion,

\(s=ut+\frac{1}{2}at^2\)

Will give,

\(s=0\times t+\frac{1}[2}\times9.8\times1^2=4.9m\)

Total height \(=44.1m\)

This means that the ball is \(44.1m-4.9m=39.2m\) above the ground after 4 seconds.

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

Answer: An object immersed in a liquid is acted upon by the buoyant force in the vertically upward direction.

Question 20: Why does a block of plastic released under water come up to the surface of water?
Answer: Number of forces acting on a certain item in water are two. The first one is the gravitational force pulling down the object, and the other is the buoyant force pushing up the object. If the buoyant force acting in the upward direction is higher than the gravitational force that is acting downward, then the object goes up to the water’s surface as quickly as it is released into water. That is why, a block of plastic released under the water comes up to the surface of the water.

Question 21: The volume of \(50g\) of a substance is \(20\space cm^3\). If the density of water is \(1\space gcm^{-3}\), will the substance float or sink?

Answer: If the density of an object is more than the density of a liquid, then it sinks in the liquid. If the density of an object is less than the density of a liquid, then it floats

Density of the substance \(=\frac{Mass\space of\space the\space substance}{Volume\space of \space the \space substance}\)

⇒Density of the substance \(=\frac{50}{20}\)

⇒ Density of the substance \(=2.5g\space cm^{-3}\)

The density of the substance > The density of water \((1g\space cm^{-3})\)

Hence, the substance will sink in water.

Question 22: The volume of a \(500g\) sealed packet is \(350\space cm^3\). Will the packet float or sink in water if the density of water is \(1\space cm^{-3}\)? What will be the volume of the water displaced by this packet?

Answer: If the density of an object is greater than the density of a liquid, then the object will sink in the liquid. If the density of an object is less than the density of a liquid, then it will float on the surface of the liquid.

Density of the 500 g sealed packet \(=\frac{Mass\space of\space the\space packet}{Volume\space of \space the \space packet}\)

⇒Density of the 500 g sealed packet \(=\frac{500}{350}\)

⇒Density of the 500 g sealed packet \(=1.428g\space cm^{-3}\)

The density of the substance is more than the density of water \((1g\space cm^{-3})\).

Hence, the object will sink in water.

Clearly, the mass of water displaced by the packet can be considered equal to the volume of the packet \(=0.350g\).