NCERT Exemplar Class 9 Science Chapter 10 Gravitation

NCERT Exemplar Class 9 Science Chapter 10 Gravitation are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 10 Gravitation.

NCERT Exemplar Class 9 Science Solutions Chapter 10 Gravitation

Multiple Choice  Questions

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
Solution:
(a) For the two /objects, of different masses falling freely near the surface of moon,
u = 0 and a = g_M
v = u + at =g_M^t
At any instant of time, both will have same velocity.

Question 2.
The value of acceleration due to gravity
(a) is same on equator and pole?
(b) is least on poles
(c) is least on equator
(d) increases from pole to equator
Solution:
(c) Acceleration due to gravity, \(g=G \frac{M}{R^{2}}\)
As the radius of the earth is smaller at the poles as compared to the equator, the value of g is greater at the poles and is least on equator.

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) 2 F
Solution:
(a): Let the masses of two objects be ml
and m2 placed at d distance apart.
So\(, F=G \frac{m_{1} m_{2}}{d^{2}}\)
When the masses of two objects are halved,
\(F^{\prime}=G \frac{\frac{m_{1}}{2} \cdot \frac{m_{2}}{2}}{d^{2}}=G \frac{m_{1} m_{2}}{4 d^{2}}=\frac{F}{4}\)

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
Solution:
(c) If the string breaks, the force that was causing it to move along a circular path, i.e., centripetal force is no longer there, so the stone will move along a straight line tangential to the circular path.

Question 5.
An object is put one by one in three liquids having different densities. The object floats 12 3
with 1/9,2/11 and 3/7 parts of their volumes outside
the liquid surface in liquids of densities d1r d2 and d3 respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₁>d₂< d₃
(c) d₁<d₂>d₃
(d) d₁<d₂<d₃
Solution:
(d) When an object floats in liquid, a buoyant force acts on it, which is equal to the weight of liquid displaced by that object. As more part of object’s volume floats outside the liq^O&ftore is the buoyant force and higher is the density of liquid.
So,\(d_{1}<d_{2}<d_{3}\left(\because \frac{1}{9}<\frac{2}{11}<\frac{3}{7}\right)\)

Question 6.
In the relation F=GM m/d2, 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
Solution:
(d) In the relation \(F=G \frac{M m}{d^{2}}\),the quantity G is universal constant of nature.

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
Solution:
(c) Law of gravitation gives the gravitational force between any two bodies having some mass as it is a universal law.

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
Solution:
(d) The value of quantity G in the law of gravitation is independent of mass and radius of the earth.

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) 1/4 times
(b) 4 times
(c) 1/2 times
(d) unchanged
Solution:
So, the gravitational force between the two particles will become 4 times.

Question 10.
The atmosphere is held to the earth by
(a) gravity
(b) wind
(c) clouds
(d) earth’s magnetic field
Solution:
(a) The atmosphere is held to the earth by gravity.

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
Solution:
(d) \(F=G \frac{m_{1} m_{2}}{d^{2}}\)
F = G, i.e., the force of attraction between two unit point masses separated by a unit distance is universal gravitational constant.

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/R2 times the weight at surface of the earth
Solution: (a): Acceleration due to gravity (g) is zero at the centre of earth. Therefore the weight of an object at the centre of earth = mg = m x 0 = 0.

Question 13.
An object weighs 10 N in air. When immeped fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2N
(b) 8 N
(c) 10N
(d) 12 N
Solution:
(a) Weight in air = 10 N Weight in water = 8 N
The weight of the liquid displaced by the object = Loss in weight of the object = 10N-8N = 2N.

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
Solution:
(b) As pressure=force/area=weight/area
Weight of girl is same in all the three cases. So, the pressure exerted by the brick will be maximum when area is least, i.e., when breadth and width rorm the base.

Question 15.
An apple falls from a tree because of gravitational attraction between the earth and apple. If F, is the magnitude of force exerted by the earth on the apple and F2 is the magnitude of force exerted by apple on earth, then
(a) F₁ is very much greater than F₂
(b)F₂ is very much greater than F₁
(c)F₁ is only a little greater than F₂
(d)F₁ and F₂ are equal
Solution:
(d) F₁ and F₂ are equal as Newton’s law of gravitation obeys the Newton’s third law of motion, i.e., if an object exerts a force on another object, then the second object exerts an equal and opposite force on the first object.

Short Answer Type Questions

Question 16.
What is the source of centripetal force that a planet requires to revolve around the sun? On i what factors does that force depend?
Solution:
A planet requires centripetal force to revolve around the sun which is provided by the gravitational force of sun on the planet.
As \(F=G \frac{M_{s} M_{p}}{r^{2}}\) the force depends on the mass of sun (M_s), mass of planet (M_P) and the distance between the two (r).

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?
Solution:
A stone is thrown from a height in a direction parallel to earth’s surface, i.e., the stone is given initial velocity in the horizontal direction.
For vertical motion of the stone, u = 0, a – g and s = h
Using \(s=u t+\frac{1}{2} a t^{2}\),
we get\(t=\sqrt{\frac{2 h}{g}}\),
Similarly, for the second stone, vertical motion is same as that of first. So, both the stones would reach the ground simultaneously.

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?
Solution:
The circular motion of the moon around earth is due to the centripetal force provided by gravitational force of earth. Therefore,
when gravity of earth suddenly becomes zero, the moon will begin to move in a straight line in the direction in which it was moving at that instant. That is the moon will move along the tangent to the circular orbit at that instant.

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.
Solution:
The value of acceleration due to gravity is greater at the poles than at the equator. So, 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 earth earlier.

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?
Solution:

Question 21.
Calculate the average density of the earth in terms of g, G and R.
Solution:

Question 22.
The earth is acted upon by gravitation of sun, even though it does not fall into the sun. Why?
Solution:
if The gravitational force of sun on earth is utilised in providing it centripetal force which is required for revolution around the sun. Hence, the earth does not fall in to the sun.

Long Answer Type Questions

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?
Solution:
Weight of an object of mass m,

Weight varies directly with respect to mass of earth, W∝ M and inversely with respect to radius of earth,

When diameter of the earth becomes half, radius of earth also becomes half and mass becomes 4 times of its initial value, then

i.e., weight will become 16 times its initial value.

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.
Solution:
From Newton’s law of gravitation, force of attraction between two bodies is directly proportional to the product of their masses, and inversely proportional to the square of distance between their centres i.e.,

Student’s hypothesis is wrong. As acceleration due to gravity is independent of the mass of the falling body, therefore, the two bricks tied together, falls with same speed as the single one to reach the ground at the same time under the action of gravity.

Question 25.
Two objects of masses m, and m2 having the same size are dropped simultaneously from heights h, and h2 respectively. 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.
Solution:

Yes, the ratio remains the same in both the cases as this ratio is independent of mass and size of the objects.

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 s^-3 is completely immersed in water of density 103 kg nrr3. Find the force of buoyancy on it. (Given g = 10 m s^-2 .)
Solution:
(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 =Vag (a is density of liquid) 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) Here, Mass of ball, m = 4 kg Density of ball, ρ= 4000 kg m^-3 Density of water σ=10³ kg m^-3

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