Important Questions for CBSE Class 9 Science Chapter 10 Gravitation
Gravitation Class 9 Important Questions and Answers for CBSE Class 9 Science Chapter 10
Class 9 Science is very challenging for most students due to the complexity of the topics. Students always face problems in Physics as the subject requires a lot of understanding of basic concepts and formulas. It is very understandable that when the basics are strong, there are favourable chances of attempting tough questions in an exam and scoring higher marks.
Chapter 10 of Class 9 Science is about gravitation. The topic is the most interesting one for many teachers and students. The topic carries a considerable weightage in exams, so students need to understand the concept of gravitation thoroughly. The topic really requires a lot of hard work to get good marks. Important Questions Class 9 Science Chapter 10 is the best solution for all those students who find it difficult to study Physics.
Extramarks is the most trusted platform where students get solutions for all the essential topics in the most schematic fashion. Our team of experienced science teachers have prepared Important Questions Class 9 Science Chapter 10 according to the latest syllabus and CBSE guidelines.
All the solutions given in our Important Questions Class 9 Science Chapter 10 are explained in multiple steps with the possible best conceptual answer. The students are highly recommended to solve the questions given in our question bank before the exam. There are other links available on the website for CBSE sample papers, extra questions and past year question papers.
Get Access to CBSE Class 9 Science Important Questions 2022-23 with Solutions
Also, get access to CBSE Class 9 Science Important Questions for other chapters too:
CBSE Class 9 Science Important Questions
|Sr No||Chapters||Chapter Name|
|1||Chapter 1||Matter in Our Surroundings|
|2||Chapter 2||Is Matter Around Us Pure|
|3||Chapter 3||Atoms and Molecules|
|4||Chapter 4||Structure of Atom|
|5||Chapter 5||The Fundamental Unit of Life|
|7||Chapter 7||Diversity in Living Organisms|
|9||Chapter 9||Force and Laws of Motion|
|11||Chapter 11||Work and Energy|
|13||Chapter 13||Why Do We Fall ill|
|14||Chapter 14||Natural Resources|
|15||Chapter 15||Improvement in Food Resources|
Important Questions of Gravitation Class 9 Science Chapter 10 with Solutions
The Extramarks question bank for Important Questions Class 9 Science Chapter 10 is a complete approach for systematic preparation. Students will benefit from solving all types of questions like long, short, short answers, multiple choice, and fill-in-the-blanks. These solutions prepared by our expert team create a unique flow with which students can learn and apply the concept even to the most challenging questions.
Question 1. State the universal law of gravitation.
Answer 1. This law was given by Isaac Newton. The law states that the attractive force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them if A and B are two bodies having masses m1 and m2 whose centres are at a distance d from each other, then.
F m1m2………………. (1)
F 1/d2 ……………………(2)
Combing equations 1 and 2, we get
F = G m1m2/d2
Here G = 6.67 × 10 -11 N-m2/kg2
G is the universal gravitational constant.
Question 2. What do you mean by free fall?
Answer 2. When an object falls towards the earth under the effect of the earth’s gravitational force alone, then such a motion of a freely falling object is called free fall.
Question 3. Why is the weight of an object on the moon 1/6th its weight on the earth?
Answer 3. Weight = mass × acceleration due to gravity
W = mg
The value of acceleration due to gravity, g on earth is 9.8 m/s 2
Value of g on moon = 1.6 m/s 2
Value of g on moon/ value of g on earth = 1.6/9.8 = 1/6
= weight on moonweight on earth = 16
Weight on moon = 16 × weight on earth
Therefore, weight on the moon is 1/6 th the weight on earth as the value of g on the moon is 1/6th of that on the earth.
Also, the mass of an object m remains constant in all places. There is a change in acceleration due to gravity from place to place. So we can say that the body’s weight depends on the acceleration due to gravity. The acceleration due to gravity on the moon is 1/6 that of earth, so the weight of an object on the moon is 1/6 of the object on earth.
Question 4. (a) Name the balances used to measure mass and weight.
(b) Will the weight of an object be lesser or more in Antarctica than its weight in Delhi? Give a reason for your answer.
(a) Beam balance measures mass, and spring balance measures weight.
(b) Earth is flat at the poles.
Therefore, the distance of the body from its centre is less.
Since g 1/r2
Hence, the value of g in Antarctica will be more compared to Delhi.
Consequently, the weight will be more in Antarctica than in Delhi.
Question 5. Fill in the blanks:
- Gravitation is a __________ force unless it involves a large mass.
- __________ is the quantity of matter contained in the body.
- All objects experience __________ when they get immersed in a fluid.
- The force of gravity __________ with altitude.
- __________is the force acting on a body perpendicular to its surface.
- __________ is the state when the object does not weigh anything during free fall.
- __________ is responsible for formation of tides in the sea.
- The motion of the moon around the earth is due to __________
- The weight of the object at the centre of the earth is __________
- The weight of the body ____________ from pole to equator.
- Pressure is a __________ quantity.
- Gravitation is a weak force unless it involves a large mass.
- Mass is the quantity of matter contained in the body.
- All objects experience buoyancy when they get immersed in a fluid.
- The force of gravity decreases with altitude.
- Thrust is the force acting on a body perpendicular to its surface.
- Weightlessness is the state when the object does not weigh anything during free fall.
- The gravitational pull of the moon is responsible for the formation of tides in the sea.
- The motion of the moon around the earth is due to the centripetal force of the earth.
- The weight of the object at the centre of the earth is zero.
- The weight of the body decreases from pole to equator.
- Pressure is a scalar quantity.
Question 6. Why is G called a universal constant?
Answer 6. G is the universal constant because the value of G is constant at all the places in the universe.
Question 7. Why does an object float or sink when placed on the water surface?
Answer 7. Two forces act on the object’s surface when placed on the water surface. These are weight and upthrust. The weight acts vertically downwards and the upthrust works vertically upwards. Based on these two forces, the object will either float or sink. If the weight is greater, the object will sink. The object will float on the water’s surface if the upthrust is more than the weight.
Question 8. The relative density of gold is 19.5. The density of water is 1000 kg/m3. What will the density of gold be in the SI unit and in g/cc?
Density of gold = 19.5
Density of water = 1000 kg/m3
Density of gold = 1000 × 19.5 = 1.95 × 104 kg/m3
To calculate density in g/cc,
Density = 1.95 × 104 / 103 g/cm3 = 19.5 g/cm3
(a) Define pressure. State its SI unit.
(b) The dimensions of a metallic cuboid are 30 cm × 20cm × 15 cm, and its mass is 30 kg. If the acceleration due to gravity is 10 m/s 2, calculate the pressure exerted by the cuboid when it rests on the face with sides 20 cm × 15 cm on the table.
(c) In which of the following situations do we exert more pressure on the ground? Whether standing on the foot or standing on both feet? Justify the answer.
(a) The pressure is the thrust per unit area of a surface. The SI unit of pressure is pascal.
(b) It is given that mass = M = 30kg
Acceleration due to gravity = 10 m/s2
Area = A = 20 cm × 15 cm = 0.2 × 0.15 m = 0.03 m2
Thrust = M × g = 30 kg × 10 m/s2 = 300 N
Pressure = Thrust / Area = 300/0.03 = 104 Pascal
(c) When we stand on the ground, we exert pressure. This pressure is exerted more when we stand on one foot as all the weight is exerted on the area of one foot only, so the pressure increases because the pressure is inversely proportional to the area of contact.
Question 10. A sharp knife is more effective than a blunt knife. Why?
Answer 10: A sharp knife has a fragile edge and a small area compared to a blunt knife. When the area is less, pressure increases for a given amount of force, this enormous pressure cuts the object easily. That is why cutting tools have sharp edges.
Question 11. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer 11: A bag with a small string is difficult to hold. This is because the thin strap has a small surface area. Pressure is inversely proportional to the area on which the force acts. It can be said that these thin strings exert greater pressure on the student’s shoulder who is holding the bag.
(a) A metallic bar has a 200 g mass at poles. Does it change when it is taken to the equator?
(b) Is there any weight change when brought to the equator?
(c) What would happen when there is no acceleration due to gravity?
(d) Differentiate between acceleration due to gravity and universal gravitational constant.
(e) Derive a relation between G and g.
(a) Even when the metallic bar is taken to the equator, there will be no change in mass, and it will remain constant everywhere.
(b) When the bar is brought to the equator, its mass will decrease. This is due to lower value of ‘g’ at the equator as compared to the poles.
(c) All objects will move in a straight line with uniform velocity if there is no acceleration due to gravity.
(d) The universal gravitational constant ‘G’ equals the force between the two unit masses separated by unit distance. Its value of 6.67 × 10 -11 N m2/kg2 is constant at every point. It has a minimal value.
Acceleration due to gravity ‘g’ is equal to the acceleration experienced by a body of any mass. It has a large magnitude which changes from point to point.
- We know that, by Newton’s law of gravitation,
F = GMm/ r2 ……………….. (1)
Where, M is the mass of earth
m is the mass of object
r is the distance between the centre of earth and object.
According to Newton’s second law,
F = ma
Here a is the acceleration due to gravity
F = mg ……………… (2)
Equating 1 and 2
mg = GMm/r2
g = GM/r2
g is 9.8 m/s 2 on the surface of the earth.
Question 13. What happens to the magnitude of the force of gravitation between two objects if
(a) the distance between the object is tripled?
(b) Mass of both objects is doubled?
(c) Mass of both objects as well as distance between them is doubled?
- Let us suppose that R is the distance
When the distance is tripled, R = 3R.
According to Newton’s law,
F = (13)2
So, we can say that F is one-ninth of the original value.
- Mass is doubled
M = 2M
M1 = 2M1
M2 = 2M2
F ∝ M1M2
So F is 4 times of the original value
- R = 2R
M1 = 2M1
M2 = 2M2
F ∝ M1M2/R2
F = GM1M2/R2
On changing mass and distance,
F1 = G (2M1) (2M2) / (2R)2 = GM1M2/R2 = F
It is proved that F is unchanged.
Question 14: What is meant by the statement that acceleration due to gravity is 9.8 m/s2?
Answer 14. Acceleration due to gravity is 9.8 m/s 2 means that a freely falling body accelerates at 9.8 m/s2 towards the earth’s centre during its fall.
Question 15. If the earth’s gravity suddenly becomes zero, in which direction will the moon begin to move if no other celestial body affects it?
Answer 15: If the earth’s gravity suddenly becomes zero, the moon will move in a tangential direction along a straight line at a point where gravitational force ceases to exist.
Question 16. Find the weight of an 80 kg man on the moon’s surface. Calculate his mass on the earth and the moon?
ge = 9.8 m/s 2
gm = 1.63 m/s 2
Mass on the earth = mass on the moon = 80 kg
Weight on moon = 1/63 m/s2 × 80 = 130.4 N
Question 17. Moon does not have an atmosphere. Why?
Answer 17. The moon does not have a gravity strong enough to hold atmospheric gases.
Question 18. Multiple choice questions:
- Law of gravitation gives the gravitational force between
(a) the earth and a point mass only,
(b) the earth and the sun only
(c) any two bodies having some mass
(d) two charged bodies only
Explanation: The law of gravitation applies to all the bodies having some mass. F = Gm1m2/r2
In the equation F = force of attraction between two bodies m1 and m2 = mass of two bodies.
- The value of G in the law of gravitation depends upon
(a) mass of the earth only
(b) radius of the earth only
(c) both mass and radius of the earth
(d) independent of mass and radius of the earth.
Explanation: G is independent of the mass and radius of the earth. This is because it is the constant of proportionality and is the universal gravitational constant.
- An object weighs 10 N in air. When immersed fully in water, it weighs 8N. The weight of the liquid displaced by the object will be:
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N
Explanation: weight in water is 8 N
Weight in water is 10 N
Weight of liquid displaced = loss in weight
= 10 N – 8 N = 2 N
Question 19. Why does the earth not move towards the moon if the moon attracts the earth?
Answer 19. This is not possible because the mass of the earth is greater than the moon and the acceleration due to the earth’s gravity towards the moon is less.
Question 20. Your mass is 42 kg on a weight machine. Is your mass more or less than 42 kg?
Answer 20: The weight of the person is slightly less than the real weight. This is because there is a loss in weight of the person due to the upthrust which is exerted by the air.
Question 21. State the importance of the universal law of gravitation.
Answer 21. The universal law of gravitation is responsible for the following phenomena:
- Binding us to the earth
- The moon’s motion around the earth
- Motion of planets around the sun
- Occurrence of tides
- Flow of water in rivers
Question 22. Explain Kepler’s laws of planetary motion.
Answer 22. Johannes Kepler gave three laws of planetary motion. They are:
- Kepler’s first law states that the path of a planet around the sun in orbit follows the shape of an ellipse at one of the foci with the sun.
- Kepler’s second law states that an imaginary line from the sun to the planet sweeps out equal areas in equal time intervals.
- Kepler’s third law states that:
The cube of the mean distance of a planet from the sun is directly proportional to the square of its orbital period T.
r 3 T2 = r3 = k × T2
k = r3/T2
T is the period of the planet.
k is Kepler constant.
r is radius as mean distance of the planet from the sun.
Question 23. Why do astronauts in space feel weightless?
Answer 23. Gravity is absent in space, and astronauts do not exert any weight or force on their spaceship.
Question no 24. An astronaut carried a pot of soil weight 60 N from the earth to the moon’s surface. He kept it there, and just before returning from moon to earth, he weighed the soil there on the moon’s surface and found that it was only 10 N. What about the rest of the soil, and how much soil mass was lost? g earth = 10 m/s 2 and g moon = g earth/6
weight of soil on earth = 60 N
g earth = 10 m/s 2
mass on the earth = m1 = 60/10 = 6 kg
weight of soil on moon = 10 N
g moon = g earth/6 = 10/6 m/s2
mass on moon = m2 = 60/10 = 6 kg
m1 and m2 are equal.
Therefore, it is stated that there is no loss of mass of soil on the moon’s surface. The difference in gravity decreases weight.
Question 25. A stone is released from the top of a tower of 19.6 m. Calculate its velocity just before touching the ground.
Answer: Height of the tower = h = 19.6 m
Initial velocity = u = 0 m/s
Final velocity , v = ?
g = 9.8 m/s2
v2 = u2 + 2 gh
v2 = 0 + 2 × 9.8 × 19.6 = 19.6 × 19.6
v = 19.6 m/s
Question 26. A stone is dropped from a height of 10 m on an unknown planet having g = 20 m/s2. Calculate the speed of the stone when it hits the surface of the planet. Also, calculate the time it takes to fall through this height.
h = 10m, g = 20 m/s2
V2 = u2 + 2gh
v2 = 0 + 2 × 20 × 10 =400
v = 20 m/s
v = u + gt
20 = 0 + 20t
t = 1 sec
Question 27. Mention a few applications of Archimedes’ principle.
Answer 27: Important applications of Archimedes’ principle includes:
- Designing ships and submarines
Question 28. A 100 N force acts on a surface of an area 25 square cm. Calculate thrust and pressure. Calculate the changed pressure if the force is now reduced to 25 N.
Force = F = 100 N
Area = A = 25 cm2 = 25 × 10 -4 m2
Thrust = force = 100 N
Pressure = Thrust/ Area = 100 / 25 × 10 -4 = 40000 Pa
If force = 25 N i.e, 1/4th then,
Pressure also becomes ¼ th = 40000/4 = 10000 Pa
Question 29. The value of g depends on which factors?
Answer 29: The value of g depends on the latitude of the place and the mass of the earth. The value of g at the poles is maximum and minimum at the equator.
Question 30. What is buoyancy?
Answer 30. The upward force by a fluid on an object on immersion is buoyancy. When buoyancy is greater than the weight, it floats. When buoyancy is less than the weight of an object, it sinks.
Question 31. Find the ratio of the pressure exerted by a block of 400 N when placed on a tabletop along its two sides with dimensions 20 cm × 10 cm and 10 cm × 15 cm.
Thrust = Weight = 400 N
Area 1 = A1 = 20 cm × 10 cm = 0.2 × 0.1 m = 0.02 m2
Pressure = P1 = W/A1 = 400/0.02 = 2× 10 4 Pa
Area 2 = 10 cm × 15 cm = 0.1 m × 0.15 m = 0.015 m2
Pressure = P2 = w/A2 = 400/0.015 = 2.6 × 104 Pa
To calculate the ratio,
P1 : P2 = 2 × 10 4/ 2.6 × 104 = 10/13
P1:P2 = 10:13
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Q.1 Harry drives a car weighing 1500 kg. He travels at a speed of 25 m/s and stops at a distance of 40 m decelerating uniformly.
(i) What is the force exerted on the car by the brakes?
(ii) What is the work done by the brakes?
Q.2 An object of mass 10 kg is moving with a speed of 4 m/s. What is the kinetic energy of the object?
m = 10 kg; v = 4 m/s
Kinetic energy, EK = (1/2 )mv2
= (1/2) — 10 kg — (4 m/s)2
= 80 J
Q.3 The energy transformation, which takes place when an object is dropped from top of building hits the floor, is
Potential energy to kinetic energy
Electrical energy to kinetic energy
Kinetic energy to potential energy
Muscular energy to kinetic energy
Potential energy to kinetic energy