NCERT Solutions Class 9 Science Chapter 10

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

Ans-

\begin{array}{l}\text{Gravitational force}\left(\text{F}\right)\text{acting between two objects is inversely proportional to the square of the distance}\\ \left(\text{r}\right)\text{between them}\text{.}\\ \text{Mathematically,}\\ \text{F}\propto \frac{1}{{\text{r}}^{\text{2}}}\\ \text{When distance them becomes}\frac{r}{2}\text{then,}\\ \text{F}\propto \frac{1}{{\left(\frac{\text{r}}{2}\right)}^2}\propto \frac{4}{r^2}\\ \text{Thus, if the distance is reduced to half, the gravitational force weill become four times larger}\text{.}\end{array}

Q.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?

Ans-

All the objects fall towards the ground under the acceleration due to gravity and this acceleration is constant and does not depend upon the mass of the falling object. Thus, a heavy object does not fall faster than a light object.

Q.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 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)

Ans-

$\begin{array}{l}\text{Given, Mass of the object, m = 1 kg}\\ {\text{Mass of the Earth, M = 6 × 10}}^{\text{24}}\text{kg}\\ \text{According to universal law of gravitation,}\\ \text{Force, F =}\frac{\text{GMm}}{{\text{r}}^{\text{2}}}...\left(\text{i}\right)\\ \text{Where, G = universal gravitational constant}\\ {\text{= 6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{Since, the object is on the earth surface hence,}\\ {\text{r = radius of earth = 6.4×10}}^{\text{6}}\text{m}\\ \text{On putting these values in}\left(\text{i}\right)\text{, we have}\\ {\text{= 6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\text{×}\frac{\left({\text{6×10}}^{\text{24}}\text{kg}\right)\left(\text{1 kg}\right)}{{\left({\text{6.4 ×10}}^{\text{6}}\text{m}\right)}^{\text{2}}}\\ \text{= 9.8 N}\end{array}$

Q.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?

Ans-

The universal law of gravitation states that two objects attract each other with equal forces, but in opposite directions. Hence, the Earth attracts the moon with an equal force with which the moon attracts the earth.

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

Ans-

The Earth and the moon exert equal gravitational forces to each other. Due to heavier mass of Earth, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. Therefore, the Earth does not move towards the moon.

Q.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?

Ans-

(i) According to universal law of gravitation, force is directly proportional to the product of the masses. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) According to universal law of gravitation, force is inversely proportional to the square of the distance between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance gets tripled, then the gravitational force becomes one-ninth of its original value.

(iii) Force is directly proportional to the product of the masses. When masses of both the objects are doubled, then the gravitational force becomes four times.

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

Ans-

Importance of the universal law of gravitation

(i) It binds us to the earth.

(ii) It explains the motion of moon around the earth.

(iii) It explains the motion of planets around the sun.

Q.8 What is the acceleration of free fall?

Ans-

An object falling towards the earth under the effect of gravitation is said to be in free fall. Acceleration of free fall also called acceleration due to gravity is 9.8 ms⁻² near the surface of earth. It is constant for all the objects.

Q.9 What do we call the gravitational force between the Earth and an object?

Ans-

Gravitational force acting between the earth and an object is called the weight of the object.

Q.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.]

Ans-

As the gold purchased at the equator weighs less than at the poles so the value of acceleration due to gravity (g) for an object is greater at poles than at the equator. Therefore, Amit’s friend will not agree with the weight of the gold bought.

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

Ans-

When the paper is crumbled in the shape of a ball, its surface area decreases. Thus, air does not exert too much resistance to its motion and it falls faster than the sheet of paper.

Q.12 Gravitational force on the surface of the moon is only MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0le9yqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaqGXaaabaGaaeOnaaaaaaa@376B@

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?

Ans-

\begin{array}{l}\text{Let,}{\text{Weight of an object on the moon = W}}_{\text{m}}\\ \text{Weight}\text{of}{\text{an object on the Earth = W}}_{\text{E}}\\ \text{Hence,}\\ {\text{W}}_{\text{m}}\text{=}\frac{\text{1}}{\text{6}}{\text{× W}}_{\text{E}}\\ \text{Also, Weight = Mass × Acceleration}\\ \text{Acceleration due to gravity, g = 9}{\text{.8 ms}}^{\text{-2}}\\ \text{Hence, weight of a 10 kg object on the Earth}\\ \text{= 10 kg × 9}{\text{.8 ms}}^{\text{-2}}\text{= 98}\text{N}\\ \text{And,}\text{weight}\text{of the same object on the moon}\\ \text{=}\frac{\text{1}}{\text{6}}\text{× 98}\text{N =16}\text{.3}\text{N}\end{array}

Q.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.

Ans-

\begin{array}{l}{\text{Given, u = 49 ms}}^{\text{-1}}\text{,}\\ \text{Here, v = final velocity of the ball = 0}\\ \text{Let, s = maximum height achieved by the ball}\\ \text{On using third equation of motion,}\\ {\text{V}}^{\text{2}}{\text{=u}}^{\text{2}}\text{+ 2 as}\\ \text{Where, a = g = acceleration due to gravity}\\ \text{It would be negative, as the ball has thrown upward}\\ \text{On putting these values, we have,}\\ \text{0 =}\frac{\left({\text{49 ms}}^{\text{-1}}\right)\text{×}\left({\text{49 ms}}^{\text{-2}}\right)}{\text{2×9}{\text{.8 ms}}^{\text{-2}}}\text{= 122}\text{.5 m}\\ \text{Let, t = time taken by the ball to reach at height of 122}\text{.5 m}\text{.}\\ \text{On using the first equation of motion under gravity ,}\\ \text{v = u + at}\\ \text{On putting the values,}\\ {\text{0 = 49 ms}}^{\text{-1}}\text{+ t ×}\left(\text{-9}\text{.8}\right)\\ \text{t =}\frac{{\text{49 ms}}^{\text{-1}}}{\text{9}{\text{.8 ms}}^{\text{-2}}}\text{= 5 s}\\ \text{Here,}\\ \text{Time of ascent = Time of descent}\\ \text{Hence, total time taken by the ball of return}\\ \text{= 5 s + 5 s = 10 s}\end{array}

Q.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.

Ans-

\begin{array}{l}\text{Given,}\text{inital}\text{velocity,}\text{u}\text{=}\text{0}\\ \text{height}\text{of}\text{the}\text{stone,}\text{h}\text{= 19}\text{.6}\text{m}\\ \text{acceleration}\text{due}\text{to}\text{gravity,}\text{g}\text{=}\text{9}\text{.8}{\text{ms}}^{\text{-2}}\\ \text{Let,}\text{final}\text{velocity}\text{of}\text{the}\text{stone,}\text{v}\text{=}\text{?}\\ \text{On}\text{using}\text{the}\text{third}\text{equation}\text{of}\text{motion}\text{under}\text{gravity,}\\ {\text{v}}^{\text{2}}\text{=}{\text{u}}^{\text{2}}\text{+}\text{2gh}\\ \text{On}\text{putting}\text{the}\text{values,}\\ {\text{v}}^{\text{2}}\text{=}\text{0}\text{+}\text{2}\text{(9}\text{.8}{\text{ms}}^{\text{-2}}\text{)(19}\text{.6}\text{m)}\\ {\text{v}}^{\text{2}}\text{=}{\left(\text{19}{\text{.6ms}}^{\text{-1}}\right)}^{\text{2}}\\ \text{v}\text{=}\text{19}\text{.6}{\text{ms}}^{\text{-1}}\end{array}

Q.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?

Ans-

\begin{array}{l}{\text{Given, inital velocity, u = 40 ms}}^{\text{-1}}\\ {\text{final velocity, v = 0 ms}}^{\text{-1}}\\ \text{gravitational acceleration for upward motion,}\\ {\text{g = -10 ms}}^{\text{-2}}\\ \text{Let h be the height attained by the stone}\\ \text{Using third equation of motion under gravity,}\\ {\text{v}}^{\text{2}}{\text{= u}}^{\text{2}}\text{+2gh}\\ \text{On putting the values,}\\ \text{0 =}{\left({\text{40 m s}}^{\text{-1}}\right)}^{\text{2}}\text{+2}\left({\text{-10 m s}}^{\text{-2}}\right)\text{h}\\ \therefore \text{h =}\frac{{\text{1600 m}}^{\text{2}}{\text{s}}^{\text{-2}}}{{\text{20 m s}}^{\text{-2}}}\text{=80 m}\\ \text{Now, total distance covered by the stone after reaching}\\ \text{on the earth = 80 m + 80 m = 160 m.}\\ \text{The net displacement of the stone when it reaches to the}\\ \text{ground = 80 m – 80 m = 0.}\end{array}

Q.16 Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Ans-

\begin{array}{l}{\text{Given, Mass of the Sun, m}}_{\text{1}}{\text{= 2 × 10}}^{\text{30}}\text{kg}\\ {\text{Mass of the Earth, m}}_{\text{2}}{\text{= 6 × 10}}^{\text{24}}\text{kg}\\ \text{Average distance between the Earth and the Sun, r = 1}{\text{.5 × 10}}^{\text{11}}\text{m}\\ \text{Universal gravitational constant, G = 6}\text{.67}\times {\text{10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{As per universal law of gravitation,}\\ \text{Force, F = G}\frac{{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}}{{\text{r}}_{\text{2}}}\\ \text{On putting the given values,}\\ \text{=}\left(\text{6}\text{.67}\times {\text{10}}^{\text{11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\right)\frac{\left(2\times {10}^{30}\text{kg}\right)\times \left(6\times {10}^{24}\text{kg}\right)}{{\left(1.5\times {10}^{11}\text{m}\right)}^2}\\ \text{= 3}{\text{.57 × 10}}^{\text{22}}\text{N}\end{array}

Q.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.

Ans-

\begin{array}{l}\end{array} \begin{array}{l}\text{Let, two stones meet at point O in time t}\text{.}\\ \text{For the stone dropped from the tower}\text{.}\\ \text{Initial velocity, u = 0}\\ \text{Let the displacement of the stone in time t from the top of the tower = s}\\ {\text{g = 10 ms}}^{\text{-2}}\\ \text{On using the second equation of motion under gravity,}\\ \text{s}\text{=}\text{ut +}\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}}\\ \text{or,}\text{=}\text{0}\text{+}\frac{\text{1}}{\text{2}}\left({\text{10 ms}}^{\text{-2}}\right){\text{t}}^{\text{2}}\\ \text{or,}\text{s =}{\text{5t}}^{\text{2}}\mathrm{\dots }\text{(i)}\\ \text{For the stone thrown upward}\\ {\text{Initial velocity, u = 25 ms}}^{\text{-1}}\\ \text{Let the displacement of the stone form the ground in time t = s’}\\ \text{Acceleration}\text{due}\text{to}\text{gravity,}\text{g}\text{=}\text{-10}{\text{ms}}^{\text{-2}}\\ \text{On using the second equation of motion under gravity,}\\ \text{s = ut +}\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}}\\ \text{or,}\text{= 25t +}\frac{\text{1}}{\text{2}}\left({\text{-10 ms}}^{\text{-2}}\right){\text{t}}^{\text{2}}\\ \text{or,}\text{s’ =}\left({\text{25 ms}}^{\text{-1}}\right){\text{t – 5t}}^{\text{2}}\mathrm{\dots .}\left(\text{ii}\right)\\ \text{Now,}\text{from}\text{the}\text{figure,}\\ \text{s+s’ = 100}m\mathrm{..}\left(\text{iii}\right)\\ \therefore {\text{5t}}^{\text{2}}{\text{+25t – 5t}}^{\text{2}}\text{=100 m}\\ \Rightarrow \text{t =}\frac{\text{100 m}}{{\text{25 ms}}^{\text{-1}}}\text{= 4}\text{s}\\ \text{Hence, Both the stones will meet after 4 s}\text{.}\\ \text{Now, from}\left(\text{1}\right)\text{, distance covered by first stone in 4 s}\\ {\text{s = 5t}}^{\text{2}}\text{= 5}{\left(\text{4 s}\right)}^{\text{2}}\text{= 80 m}\\ \text{From}\left(\text{iii}\right)\text{, distance covered by second stone in 4 s,}\\ \text{s’ = 100 m – 80 m = 20 m}\end{array}

Q.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

Ans-

$\begin{array}{l}\text{(a)\hspace{0.17em}Time of ascent of the ball = Time of descent of the ball}\\ \text{As total time of journey = 6 s}\\ \therefore \text{\hspace{0.17em}Time of ascent = 3 s}\\ {\text{For upward motion, v = 0, g = -9.8 ms}}^{\text{-2}}\\ \text{Now, by using the relation, v = u + gt}\\ {\text{or, 0 = u + (-9.8 ms}}^{\text{-2}}\text{) × 3\hspace{0.17em}s}\\ {\text{or,\hspace{0.17em}u = 29.4 ms}}^{\text{-1}}\\ {\text{Therefore,\hspace{0.17em}the ball was thrown upwards with a velocity of 29.4 ms}}^{\text{-1}}.\\ \left(\text{b}\right)\\ {\text{Here,\hspace{0.17em}Initial velocity, u = 29.4 ms}}^{\text{-1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Final velocity, v = 0}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Acceleration due to gravity, g = -9.8 ms}}^{\text{-2}}\\ \text{Let, the maximum height = h}\\ \text{on\hspace{0.17em}using the relation}\\ \text{h = ut +}\frac{\text{1}}{\text{2}}{\text{\hspace{0.17em}gt}}^{\text{2}}\\ \text{On putting the given values,}\\ {\text{h = 29.4 ms}}^{\text{-1}}\text{× 3 s +}\frac{\text{1}}{\text{2}}\left({\text{-9.8\hspace{0.17em}m\hspace{0.17em}s}}^{\text{-2}}\right){\left(\text{3\hspace{0.17em}s}\right)}^{\text{2}}\text{= 44.1\hspace{0.17em}m}\\ \text{(c)}\\ \text{After 3 s,\hspace{0.17em}the ball attains the maximum height}\\ {\text{At maximum height, initial velocity, u = 0 m s}}^{\text{-1}}\\ \text{Position of the ball after 4 s of the throw is given by the}\\ \text{distance travelled by it during its downward}\\ \text{journey in (4s – 3s) = 1 s}\\ \text{On\hspace{0.17em}using the relation for downward motion,}\\ \text{s\hspace{0.17em}=\hspace{0.17em}ut +}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ \text{s\hspace{0.17em}=\hspace{0.17em}0\hspace{0.17em}×\hspace{0.17em}t\hspace{0.17em}+}\frac{\text{1}}{\text{2}}\text{×\hspace{0.17em}}\left({\text{9.8\hspace{0.17em}ms}}^{\text{-2}}\right){\left(\text{1s}\right)}^{\text{2}}\\ \text{=\hspace{0.17em}4.9\hspace{0.17em}m}\\ \text{As total height = 44.1 m}\\ \text{Therefore, the\hspace{0.17em}}\mathrm{h}\text{eight\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}ball\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}ground\hspace{0.17em}after 4\hspace{0.17em}s would\hspace{0.17em}be\hspace{0.17em}}\\ \text{=\hspace{0.17em}}\left(\text{44.1\hspace{0.17em}m – 4.9\hspace{0.17em}m}\right)\text{= 39.2\hspace{0.17em}m.}\end{array}$

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

Ans-

The buoyant force acts in the upward direction for an object immersed in a liquid.

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

Ans-

A plastic block under the water experiences two forces. One is the gravitational force, which pulls the block downwards and the other is the buoyant force which pushes the block upwards. If the upward buoyant force is greater than the downward gravitational force, then the block comes up to the surface of the water as soon as it is released within water.

Q.21 The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm⁻³, will the substance float or sink?

Ans-

\begin{array}{l}\text{Density}\text{of}\text{the}\text{substance}=\frac{\text{Mass}\text{of}\text{the}\text{substance}}{\text{Volume}\text{of}\text{the}\text{substance}}\\ =\frac{\mathrm{50\; g}}{20{\text{cm}}^3}=\text{2}.\text{5}{\text{gcm}}^{-3}\end{array}

if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid. Conversely, the density of an object is more than the density of a liquid, then it sinks in the liquid.

Here, the density of the substance is more than the density of water i.e., 1 g cm⁻³. Therefore, the substance will sink into the water.

Q.22 The volume of a 50 g sealed packet is 350 cm³. 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?

Ans-

\begin{array}{l}\text{Density of packet =}\frac{\text{Mass}}{\text{Volume}}\text{=}\frac{\text{500 g}}{{\text{350 cm}}^{\text{3}}}\text{=1}{\text{.428 gcm}}^{\text{-3}}\\ \text{Here, the density of the packet is more than the density of water}\text{.}\\ \text{Therefore, it will sink into the water}\text{.}\\ \text{The mass of water displaced}\\ \text{= volume of packet × density of water}\\ {\text{= 350 cm}}^{\text{3}}{\text{× 1 gcm}}^{\text{-3}}\text{=350g}\end{array}