This post all important questions from chapter 10 Gravitation Class 9 Science. Gravitation is a very important chapter from class 9 physics. Students must prepare this chapter thoroughly to build a strong foundation in physics. These questions are hand-picked by CBSE Guidance. Hence students who are looking to score maximum in their exam, must prepare these questions before appearing in the exams.
Gravitation Important Questions
Q. No. 1) A boy is whirling a stone tied with a string in a 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 center 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
Ans. Option (c).
Q. No. 2) Write the SI unit of the given quantities.
i. Universal Gravitation Constant (G)
ii. Acceleration due to gravity (g)
iii. Weight (W)
iv. Thrust
v. Pressure
vi. Density
vii. Relative density
Ans. i. N m2 kg-2
ii. m s-2
iii. N
iv. N
v. N m-2
vi. Kg m-3
vii. No unit of relative density
Q. No. 3) The gravitational force between two objects is F. If the masses of both objects are halved without changing the distance between them, then the gravitational force would become
a. F/4
b. F/2
c. F
d. 2F
Ans. Option (a).
F = G M'm'/R2 = G (M/2)(m/2)/R2 = ¼ GMm/R2 = ¼ F
Q. No. 4) The weight of an object at the center 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 on the surface of the earth
Ans. At the center of the earth, R = 0
F = G Mm/R2 = G Mm/0 = ∞
Option (b).
Q. No. 5) Define the weight of an object on the Moon.
Ans. The force with which the Moon attracts the object is known as the weight of the object on the Moon.
| Â Also See: Class 9 Science Important Questions |
Q. No. 6) 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
Ans. Buoyant force = 10 – 8 = 2 N
Therefore, by Archimedes Principle, the weight of the liquid displaced by the object = buoyant force = 2 N.
Q. No. 7) Weight of a girl is 294 N. Find her mass.
Ans. W = mg
⇒ 294 = m x 9.8
⇒ m = 30 kg
Q. No. 8) What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?
Ans. Gravitational force.
This force depends on the product of the masses of the planet and the Sun and the distance between them.
Q. No. 9) Evaluate what happens to the force between two objects, if
i. The mass of one object is doubled?
ii. The distance between the objects is tripled?
iii. The masses of both objects are doubled?
Ans.
Q. No. 10) It is seen that a falling apple is attracted towards the earth. Does the apple also attract the earth? If so, we do not see the earth moving towards the apple. Why?
Ans. As per Newton’s third law of motion, the apple also attracts the earth with equal and opposite force. But according to Newton’s second law of motion, for a given force, acceleration is inversely proportional to the mass of an object. The mass of an apple is negligibly small compared to that of the earth. So, we do not see the earth moving towards the apple.
Q. No. 11) The density of gold is 19300 kg m-3 and that of water is 1000 kg m-3. What is the relative density of gold?
Ans. Relative density of gold = Density of gold/Density of water = 19300/1000 = 19.3
Q. No. 12) State the universal law of gravitation. Why this law is called universal? What is the importance of the universal law of gravitation?
Ans. The universal law of gravitation: Every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
The law is universal in the sense that it is applicable to all bodies, whether the bodies are big or small, whether they are celestial or terrestrial.
Importance of the Universal law of gravitation:
The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:
i. The force that binds us to the earth;
ii. The motion of the moon around the earth;
iii. The motion of planets around the Sun;
iv. The tides due to the moon and the Sun.
Q. No. 13) What do you mean by the acceleration due to gravity?
Ans. Whenever an object falls towards the earth, acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore, this acceleration is called acceleration due to gravity.
Q. No. 14) Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Ans. F = G Mm/R2
F = Gravitational force between the earth and the object
G = Universal gravitation constant = 6.673 x 10 -11 N m2 kg-2
M = Mass of earth
m = mass of the object
R = radius of the earth
Q. No. 15) With the help of the Second Law of Motion and the Universal Law of Gravitation derive an expression for acceleration due to gravity ‘g’.
Ans.
Q. No. 16) What are the differences between the mass of an object and its weight?
Ans.
| Mass | Weight |
| 1. The quantity of matter contained in a body is known as mass. | 1. Force with which the earth attracts an object is known as the weight of the object. |
| 2. It has only magnitude (scalar quantity). | 2. It has both magnitude and direction (vector quantity). |
| 3. It is constant everywhere in the universe. | 3. The weight of an object is different at different places (depending on the value of g). |
| 4. Its SI unit is kg. | 4. Its SI unit is N. |
Q. No. 17) A force of 20 N is distributed uniformly over an area of 20 cm2, find pressure in Pascal.
Ans. Area = 20 cm2 = 20/(100 x 100) = 0.002 m2
Pressure = Force/Area = 20 / 0.002 = 10000 N/m2 = 10000 Pa
Q. No. 18)Â The weight of any person on the moon is about 1/6 times that of 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?
Ans. Acceleration due to gravity on earth (ge) = g
Acceleration due to gravity on moon (gm) = g/6
Force applied on earth = F = m ge= 15 g
Now,
Force applied on moon = m gm
⇒ F = m gm
⇒ 15 g = m g/6 (Same force applied)
⇒ m = 15 x 6 = 90 kg
Q. No. 19) Why are the ends of tools like knives, pins, and nails pointed?
Ans. The ends of tools such as pins, knives, and nails are made pointed to increase the effect of the force. Because the pointed ends have much smaller areas. As result for a certain amount of force applied, the pressure becomes very large. That is why sharp edges need a very small force to cut fruits or vegetables.
Q. No. 20) The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in a saturated salt solution, if the density of the solution is 1.2 g cm-3? What will be the mass of the solution displaced by this packet?
Ans. Density of sealed packet = m/V = 500/350 = 1.4 g cm-3
As the density of the packet is greater than that of the saturated salt solution, the packet will sink.
Volume of solution displaced = 350 cm3
Density of salt solution = Mass of salt solution/Volume of salt solution
⇒ 1.2 = Mass of salt solution / 350
⇒ Mass of salt solution = 1.2 x 350 = 420 g.
Q. No. 21) Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the sheet of paper is more than the resistance offered to the crumpled paper ball.
Q. No. 22) Two bodies of mass 10 kg and 100 kg respectively dropped from the same height of 4.9 m in a vacuum. Calculate and compare their final velocities just before hitting the ground.
Ans. Since they are falling in a vacuum, their velocities will be equal (due to the absence of upthrust).
We know, v2= u2 + 2aS
⇒ v2= 0 + 2 x 9.8 x 4.9
⇒ v2 = 2 x 2 x 4.9 x 4.9
⇒ v = 2 x 4.9 = 9.8 m/s
Q. No. 23) Why centripetal force is called the center-seeking force?
Ans. This force keeps the body moving along the circular path and is directed towards the center.
Q. No. 24) A particle is thrown up vertically with a velocity of 50 m/s.
a. What will be its velocity at the highest point of its journey?
b. How high would the particle rise?
c. What time would it take to reach the highest point?
Ans. a. At the highest point, the velocity will be zero.
b. Here, u = 50 m/s; a = - 9.8 ms-2; v = 0 m/s
We know,
v2= u2 + 2aS
⇒ 02= 502 + 2 x (-9.8) x S
⇒ S = 2500/19.6 = 127.5 m
c. We know, v = u + at
⇒ 0 = 50 + (-9.8) t
⇒ t = 50/9.8 = 5.1 s
Q. No. 25) A ball weighing 4 kg of density 4000 kg m-3 is completely immersed in water of density 103 kg m-3. Find the force of buoyancy on it. ( Take g = 10 ms-2)
Ans. Volume of ball = mass/density = 4/4000
Mass of water displaced (m) = Density of water x volume of displaced water = 1000 x 4/4000 = 1
Buoyant force = Weight of the liquid displaced = mg = 1 x 10 = 10 N
Q. No. 26) i. Why does an object float or sink when placed on the surface of the water? What are the factors on which buoyant force depends?
ii. Why is it easier to lift a mass inside water than in the air?
Ans. i. If the density of an object is greater than the density of water then it will sink, if the density of the object is less than the density of water then it will float. The magnitude of this buoyant force depends on the density of the fluid.
ii. The force needed to lift an object in water = Weight of object – Buoyant Force offered by the water.
Q. No. 27) State Archimedes’ principle. Mention its applications.
Ans. Archimedes’ Principle: When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.
Applications:
i. It is used in designing ships and submarines.
ii. Lactometers and hydrometers are based on this principle.
Q. No. 28) Identical packets are dropped from two airplanes, one above the equator and the other above the north pole, both at height ‘h’. Assuming all conditions are identical, will these packets take same time to reach the surface of the earth? Justify your answer.
Ans. We know that the value of ‘g’ at the equator of the earth is less than that at the poles. Therefore, the packet falls slowly at the equator in comparison to the poles. Thus, the packet will remain in the air for a longer time interval, when it is dropped at the equator.
Q. No. 29) Define relative density. Relative densities of two substances A and B are 2.5 and 0.9 respectively. Find densities of A and B. Also, find whether they will sink or float in water. (Density of water = 1000 kg/m3)
Ans. Relative Density: Relative density of a substance is the ratio of its density to that of water.
Relative density of A = Density of A / Density of water
⇒ 2.5 = Density of A/1000
⇒ Density of A = 2.5 x 1000 = 2500 kg/m3
Also
Relative density of B = Density of B / Density of water
⇒ 0.9 = Density of B/1000
⇒ Density of B = 0.9 x 1000 = 900 kg/m3
Since the density of A is greater than that of water, it will sink. And the density of B is lesser than that of water, it will float.
In addition to the above questions also prepare the Examples and Exercises given in your NCERT Textbook.
Also, watch the detailed explanation of Gravitation chapter class 9 here:
Need an answer to the following question on class 9 gravitation
A solid of density D is floating in a liquid of density d. If V is the volume of solid submerged in the liquid, and V1 is the total volume of the solid, then V1 /V will be d/D or D/ d
Buoyant force = Weight of water displaced by solid
mass of solid (m) x acceleration due to gravity (g) = Mass of water displaced (M) x acceleration due to gravity (g)
mass of solid (m) = Mass water displaced (M)
Density of solid (D) x Volumne of solid(V1) = Density of water (d) x volume of water displaced (V)
D x V1 = d x V
V1/V = d/D