Science Ncert Exemplar 2019 Solutions for Class 9 Science Chapter 10 - Gravitation

Share with your friends

Science Ncert Exemplar 2019 Solutions for Class 9 Science Chapter 10 Gravitation are provided here with simple step-by-step explanations. These solutions for Gravitation are extremely popular among class 9 students for Science Gravitation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science Ncert Exemplar 2019 Book of class 9 Science Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Science Ncert Exemplar 2019 Solutions. All Science Ncert Exemplar 2019 Solutions for class 9 Science are prepared by experts and are 100% accurate.

Page No 64:

Question 1:

Two objects of different masses falling freely near the surface of moon would
(a) have same velocities at any instant
(b) have different accelerations
(c) experience forces of same magnitude
(d) undergo a change in their inertia

Answer:

Two objects of different masses falling freely near the surface of moon would have same velocities at any instant because velocity at any instant depends on acceleration due to gravity and time. Both will be same for the two objects because acceleration due to gravity does not depend on the mass of the object.
Hence, the correct answer is option A.

Page No 64:

Question 2:

The value of acceleration due to gravity
(a) is same on equator and poles
(b) is least on poles
(c) is least on equator
(d) increases from pole to equator

Answer:

Acceleration due to gravity varies as the square of the distance between the surface of the earth and its centre.
$g \propto \frac{1}{R^{2}}$
This distance $(R)$ is maximum at the equator, therefore the value of g is least on equator.
Hence, the correct answer is option C.

Page No 64:

Question 3:

The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become
(a) F/4
(b) F/2
(c) F
(d) 2F

Answer:

Gravitational force between two bodies of mass m₁ and m₂kept at r distance apart is given by
$F = \frac{G m_{1} m_{2}}{r^{2}}$
If the masses are halved, then the new force:
$F' = G \frac{\frac{m_{1}}{2} \times \frac{m_{2}}{2}}{r^{2}} = \frac{1}{4} G \frac{m_{1} m_{2}}{r^{2}} \Rightarrow F' = \frac{F}{4}$
If mass of both the objects are halved then the gravitational force will become one fourth of its previous value.
Hence, the correct answer is option A.

Page No 64:

Question 4:

A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone
(a) will continue to move in the circular path
(b) will move along a straight line towards the centre of the circular path
(c) will move along a straight line tangential to the circular path
(d) will move along a straight line perpendicular to the circular path away from the boy

Answer:

If the string breaks, the stone will move along a straight line tangential to the circular path because centripetal force which was making it move in a circular path is no longer available.

Hence, the correct answer is option C.

Page No 64:

Question 5:

An object is put one by one in three liquids having different densities. The object floats with $\frac{1}{9}, \frac{2}{11}$ and $\frac{3}{7}$ parts of their volumes outside the liquid surface in liquids of densities d₁, d₂and d₃respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₁ > d₂ < d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃

Answer:

Fraction of the volume of the body outside the liquid surface is given as
$\frac{1}{9}, \frac{2}{11} and \frac{3}{7}$

LCM of the denominator of these fractions is 693. Now these fractions can be rewrittten as
$\frac{77}{693}, \frac{126}{693} and \frac{297}{693}$

These fractions can be arranged in this decreasing order
$\frac{297}{693}, \frac{126}{693} and \frac{77}{693}$

Higher the fraction outside the liquid, higher is the buoyant force. Buoyant force is directly proportional to the density of the liquid. The correct order of the density of the liquids is
d₃ > d₂ > d₁

Hence, the correct answer is option D.

Page No 65:

Question 6:

In the relation F = G $\frac{M m}{d^{2}}$ , the quantity G
(a) depends on the value of g at the place of observation
(b) is used only when the earth is one of the two masses
(c) is greatest at the surface of the earth
(d) is universal constant of nature

Answer:

The quantity G is universal Gravitational constant whose value is $6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}$ .
Its magnitude is independent of the mass of the bodies and distance between them.
Hence, the correct answer is option D.

Page No 65:

Question 7:

Law of gravitation gives the gravitational force between
(a) the earth and a point mass only
(b) the earth and Sun only
(c) any two bodies having some mass
(d) two charged bodies only

Answer:

Law of gravitation gives the gravitational force between any two bodies having some mass.
Hence, the correct answer is option C.

Page No 65:

Question 8:

The value of quantity G in the law of gravitation
(a) depends on mass of earth only
(b) depends on radius of earth only
(c) depends on both mass and radius of earth
(d) is independent of mass and radius of the earth

Answer:

Since G is universal gravitation constant, it does not depend on mass or radius of the earth.
Hence, the correct answer is option D.

Page No 65:

Question 9:

Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be

(a) $\frac{1}{4}$ times

(b) 4 times

(c) $\frac{1}{2}$ times

(d) unchanged

Answer:

Since force between two particles is directly proportional to the product of the masses of two particles. Therefore, if the mass of both the particles is doubled then the gravitational force will become 4 times.
Hence, the correct answer is option B.

Page No 65:

Question 10:

The atmosphere is held to the earth by
(a) gravity
(b) wind
(c) clouds
(d) earth’s magnetic field

Answer:

The atmosphere is held to the earth by gravity.
Hence, the correct answer is option A.

Page No 65:

Question 11:

The force of attraction between two unit point masses separated by a unit distance is called
(a) gravitational potential
(b) acceleration due to gravity
(c) gravitational field
(d) universal gravitational constant

Answer:

We know
$F = \frac{G m_{1} m_{2}}{r^{2}}$

If m₁ = 1 unit m₂ = 1 unit r = 1 unit
then, F = G
Hence, the correct answer is option D.

Page No 65:

Question 12:

The weight of an object at the centre of the earth of radius R is
(a) zero
(b) infinite
(c) R times the weight at the surface of the earth
(d) 1/R²times the weight at surface of the earth

Answer:

Since g is zero at the centre of the earth and we know
Weight (W) = mg = 0
Hence, the correct answer is option A.

Page No 66:

Question 13:

An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N

Answer:

According to Archimedes principle, when an object is immersed in a liquid it displaces some liquid whose weight is equal to the reduction in the apparent weight of the body i.e. equal to buoyant force acting on the body.
Apparent reduction in the weight of the body is 10 N - 8 N = 2 N. Thus, the weight of the liquid displaced by the object will be 2 N.
Hence, the correct answer is option A.

Page No 66:

Question 14:

A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base
(b) maximum when breadth and width form the base
(c) maximum when width and length form the base
(d) the same in all the above three cases

Answer:

Pressure is defined as the force acting per unit area.
P = $\frac{Force}{Area}$
Lesser the area, greater the pressure.
Area in I case = 60 × 40 = 2400 cm²
Area in II case = 40 × 20 = 800 cm²
Area in III case = 60 × 20 = 1200 cm²
Area will be minimum when breadth and width form the base.
Hence, the correct answer is option B.

Page No 66:

Question 15:

An apple falls from a tree because of gravitational attraction between the earth and apple. If F₁is the magnitude of force exerted by the earth on the apple and F₂is the magnitude of force exerted by apple on earth, then
(a) F₁is very much greater than F₂
(b) F₂is very much greater than F₁
(c) F₁is only a little greater than F₂
(d) F₁and F₂are equal

Answer:

Both F₁ and F₂ are equal as per Newton's third law of motion.
The direction of F₁ and F₂ will be opposite to each other.
Hence, the correct answer is option D.

Page No 66:

Question 16:

What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?

Answer:

Gravitational force is the source of centripetal force required to revolve around the sun. This force depends on the distance between the planet and the sun along with their masses. If this force becomes zero as a result of the absence of centripetal force, the planet would shift to moving tangentially outwards to the circular route.

Page No 66:

Question 17:

On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?

Answer:

The stone thrown from the height has zero initial velocity in the vertical direction. It has the horizontal velocity only.

For the vertical motion of the stone

u = 0 m/s
a = g
s = h

using the second equation of motion $s = u t + \frac{1}{2} a t^{2}$

The time t is calculated as $t = \sqrt{\frac{2 h}{g}}$

This time period is similar for the stone thrown in the vertically downward direction, So both the stones will reach the ground simultaneously.

Page No 66:

Question 18:

Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?

Answer:

The centripetal force provided by the earth’s gravitational force is responsible for the moon’s circular motion around the earth. When the earth’s gravity suddenly disappears (g = 0) the moon will start to move in a straight line in the same direction it was moving at the time. At that time, the moon will be moving along the tangent to the circular orbit.

Page No 66:

Question 19:

Identical packets are dropped from two aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming all conditions are identical, will those packets take same time to reach the surface of earth. Justify your answer.

Answer:

The value of acceleration due to gravity is greater at the poles than at the equator.

Due to this difference in the acceleration due to gravity the packet dropped at north pole from a height h, will accelerate more than the packet dropped at equator from the same height and hence will reach the surface of the earth earlier.

Page No 66:

Question 20:

The weight of any person on the moon is about 1/6 times that on the earth. He can lift a mass of 15 kg on the earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon?

Answer:

The value of acceleration due to gravity on the surface of the Moon is 1/6 the value of acceleration due to gravity that on the surface of the earth.

â€‹Force applied by the person to lift a mass of 15 kg on Earth is $F_{E} = 15 g_{E}$

Let m be the maximum mass that can be lifted on the surface of the moon = $\frac{m g_{E}}{6}$

On comparing the two equations we get

$15 g_{E} = \frac{{mg}_{E}}{6} m = 90 kg$

â€‹

Page No 66:

Question 21:

Calculate the average density of the earth in terms of g, G and R.

Answer:

Let the radius of the sphere be R and M be the mass, V be the Volume.

The density is given by $ρ = \frac{Mass}{Volume} = \frac{M}{\frac{4}{3} π R^{3}}$

$g = \frac{G M}{R^{2}} M = \frac{g R^{2}}{G}$

Using the value of mass M in the equation for density gives $ρ = \frac{g R^{2}}{G} \times \frac{3}{4 π R^{3}} = \frac{3 g}{4 π R G}$

$ρ = \frac{3 g}{4 π R G}$

Page No 66:

Question 22:

The earth is acted upon by gravitation of Sun, even though it does not fall into the Sun. Why?

Answer:

The gravitational force exerted by the sun on the earth provides the necessary centripetal force required by the earth to revolve around the sun. Hence, the earth does not fall into the sun.

Page No 67:

Question 23:

How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?

Answer:

Weight of an object is given by mass m times the acceleration due to gravity g which can further be expressed as $\frac{G M m}{R^{2}}$

The radius of the earth is halved as the diameter is halved.

â€‹Mass of the earth becomes four times its initial value.

So, the value of the weight of the object can be written as $\frac{16 G M m}{R^{2}}$

So, the weight becomes 16 times of its original value.
â€‹

Page No 67:

Question 24:

How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment.

Answer:

Newton's law of gravitation states that the force of attraction between body is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Mathematically, $F = G \frac{m_{1} m_{2}}{R^{2}}$

â€‹The acceleration due to gravity (g) is independent of the mass of the falling body.
$g = G \frac{M}{R^{2}}$ Here, M is mass of earth and R is radius of earth.
Therefore, the two bricks tied together falls with the same speed as the single one.

Hence the hypothesis of the boy is not correct.
â€‹

Page No 67:

Question 25:

Two objects of masses m₁and m₂having the same size are dropped simultaneously from heights h₁and h₂respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case. Give reason.

Answer:

Time taken by a body to reach the ground is given by $t = \sqrt{\frac{2 h}{g}}$ where h is the height through which the body falls

Time taken by a body of mass m₁is $t_{1} = \sqrt{\frac{2 h_{1}}{g}}$

Similarly, time taken by a body of mass m₂ $t_{2} = \sqrt{\frac{2 h_{2}}{g}}$

The ratios of the time taken by the body is given by $\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$

The ratio remains the same as we can see that the ratio is independent of the mass of the body.

Page No 67:

Question 26:

(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water. Give reason for each case.
(b) A ball weighing 4 kg of density 4000 kg m^–3is completely immersed in water of density 10³kg m^–3Find the force of buoyancy on it. (Given g = 10 m s^–2.)

Answer:

â€‹â€‹(a)
The cube will experience a greater buoyant force in saturated salt solution than in water as the density of saturated salt solution is greater than that of water.
As buoyant force = weight of liquid displaced by the object = $ρ V g$
As each side of the cube is reduced to 4 cm from 5 cm, so volume of cube decreases and hence the buoyant force also decreases.

(b)
â€‹â€‹Mass of ball, m = 4 kg
Density of ball, ρ = 4000 kg m^-3
Density of water, σ =10³ kg m^-3

Volume of the Ball = $\frac{Mass of the ball}{Density of the Ball} = \frac{4}{4000} = 10^{- 3} m^{3}$

The buoyant force exerted by the water of density 10³kg m^-3is given by = $σ \times V \times g = 10^{- 3} \times 10^{3} \times 10 = 10 N$
â€‹

View NCERT Solutions for all chapters of Class 9