RBSE Class 9 Science Chapter 10 Gravity Solutions help students to get a proper grasp of the entire concepts of the chapter, thus preparing well for the exams. Students solve these solutions to revise the complete chapter thoroughly. They can practice these solutions and master the subject efficiently. Solving these questions from the RBSE Class 9 Solutions for Chapter 10 Science helps students to understand all the key highlights from the chapter. Students are encouraged to solve these chapter-wise important topics and questions from the RBSE Class 9 Science in order to prepare most competently for the Class 9 exams. In the meantime, students are advised to implement a strategic learning process so that they can improve their performance. Here, we have listed the important questions from Chapter 10, Gravity of RBSE Class 9 Science Textbook. Practising these questions help the students to score high marks. Students will also be able to get a proper foundation of the subject before moving to higher classes. This will also help them to excel in the academic year.
Rajasthan Board Class 9 Science Chapter 10- BYJU’S Important Questions & Answers
RBSE Class 9 Science Chapter 10 Objective Questions-Important Questions and Solutions
1. The Newton’s Law of Gravitation is universal because :
(a) It is always of attraction
(b) It is applicable on all the members and particles of the solar-system.
(c) It is applicable on all the masses, at all the distances and is not influenced by the medium.
(d) None of the above.
Answer: (c) It is applicable on all the masses, at all the distances and is not influenced by the medium.
2. Which is the force responsible for the circular motion of object :
(a) Gravitational force
(b) Frictional force
(c) Centripetal force
(d) None of the above
Answer: (a) Gravitational Force
3. The value of the Universal Constant G depends upon :
(a) Nature of the particles
(b) Medium present between the particles
(d) Does not depend on anything
Answer: (d) Does not depend on anything
4. The weight of a person on earth surface is 60 N then his weight on moon will be :
(a) 60 N
(b) 30 N
(c) 20 N
(d) 10 N
Answer: (d) 10 N
5. An object of mass m is taken into a very deep mine then :
(a) Its mass increases
(b) Its mass decreases
(c) Its weight decreases
(d) Its weight increases
Answer: (c) its weight decreases
6. On doubling the distance between two masses, the gravitational force between them :
(a) Remains unchanged
(b) will become one fourth
(c) will be reduced to half
(d) will double itself
Answer: (b) will become one fourth
RBSE Class 9 Science Chapter 10 Very Short Answer Type Questions-Important Questions and Solutions
1. From where does the satellite derive the centripetal force required by it to revolve round the planets?
Answer: Satellites derive the centripetal force required by it to revolve round the planets from the earth’s gravitational force.
2. Can the gravitational mass of a body be measured in an artificial satellite?
Answer: The mass of an artificial satellite is less and hence it applies negligible gravitational force and the object is in a state of weightlessness in its environment. Hence, the mass of a body cannot be determined in an artificial satellite.
3. What will be the change in the gravitational force between two masses if the distance between them is doubled?
Answer: If the distance between two masses is doubled then the gravitational force becomes 1/4th.
4. If the mass of a body is 10 Kg then what will be its weight on the earth surface?
Answer: If m=10kg and given that g=9.8m/s2, Then weight = mg= 10 x 9.8m/s2= 98N
5. Write the formula of the gravitational force between two bodies.
Answer: Suppose two bodies of mass M and m are placed at a distance ‘d’ from each other with G as the gravitational contestant. Then, the formula for the gravitational force F working between will be F= GMm/d2.
6. On which principle does a ball pen work?
Answer: Ball pen works on the principle of capillary action and surface tension.
7. A person can jump on the Moon for a greater height. Why?
Answer: The gravitational acceleration on the moon is 1.61 m/s2. This is 1/6th of
that present on the earth’s surface, whose gravitational acceleration is 9.8 m/s2. For this reason, the moon attracts an object with lesser force than the earth. Hence, it is possible for a person to jump higher on the moon than on earth.
8. Two bodies of 1 Kg each are at a distance of 1 meter from each other. Write the value of gravitational force between them.
Answer: The gravitational force acting between two objects of equal mass M and m is mathematically represented with the formula F= GMm/d2. The body of two masses M and m are 1 kg each. Hence, M=m=1kg, d=1m and G is the constant proportion with numerical value 6.67×10-11N. Here, F= GMm/d2= G(1 ×1)/12.= G × 1 = 6.67×10-11 x 1 = 6.67×10-11N.
9. Why does the moon not fall down on experiencing the earth’s gravitational force?
Answer: Earth exerts an invisible force on every object, including the moon. Most objects fall towards the earth due to this gravitational force. However, the moon derives its centripetal force from the gravitational force. The movement of the moon around the earth is due to the presence of centripetal force. Hence, the moon does not fall down on experiencing the gravitational pull.
10. What is the main reason for oceanic tides?
Answer: Main cause of the oceanic tide is the Gravitational force exerted on Earth by the Moon.
RBSE Class 9 Science Chapter 10 Short Answer Type Questions-Important Questions and Solutions
1. Write the Universal Law of Gravitation.
Answer: Every particle (body) of the universe exerts a force of attraction towards itself on every other particle (body), which is known as the Gravitational Force. According to the Law of Gravitation “the force of attraction between two particles of matter or bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. The direction of this force is the same as the direction of the line joining the two particles”.
2. On the earth’s surface the value of g is maximum at what point and why?
Answer: When the change in an object’s velocity (acceleration) is the result of the earth’s force of gravitation, it is called the Gravitational Acceleration and is denoted by g. Given that M is the mass of earth and d is the distance of the object from the center of the earth. See that the value of g decreases as the value of d increases. If the radius of earth is R then the gravitational acceleration on the earth’s surface gs will be :
gs = GM / R2
Also, the radius of the earth is more at the equator than that at the poles. Hence, the value of g will be more at the poles, as compared to that at the equator.
3. Write the value of the universal constant of gravitation along with its unit.
Answer: The proportionality constant used in the Newton’s Law of Gravitation is the gravitational constant. The force of attraction between any two unit masses separated by a unit distance is called universal gravitational constant denoted by G measured in Nm2/kg2. Learn more about the value of gravitational constant here.
4. What is Gravitational Acceleration? Write its formula.
Answer: When the change in an object’s velocity i.e., acceleration, is due to the earth’s force of gravitation, it is called the Gravitational Acceleration, denoted by g and unit ms-2. Learn more about the Gravitational Acceleration formula here.
5. Clarify the difference between the mass and weight of an object.
Answer: Mass is the measure of the inertia of an object. Hence, if the mass of a body is greater its inertia will be more. The mass of a body remains the same on earth or in space. Meanwhile, on earth, the weight of an object is a type of a force that takes it towards the earth. According to the definition, the weight of a substance is the force by which it is attracted towards the earth.
Mathematically F = mg, hence, the weight of the body w = mg. Here F, the weight of the object is expressed in the unit of Newton, m is the mass of the object and g is the gravitational acceleration whose unit is m/s2. Given that the value of g at a given point is constant, the weight of an object is proportional to its mass. When the value of g changes, the mass remains the same. However, the weight changes along with the value of g.
6. What is meant by Free Fall? Give examples of free fall.
Answer: A free falling body is in a state of weightlessness as the reaction force on it is zero. Learn more about free fall motion with example.
7. What will be the change in weight of a person moving from the poles of the earth to its equator and why?
Answer: As man moves from the poles of the earth to its equator, the weight of the man will reduce.The value of g increases with the increase in latitude and the latitude at the poles is greater than that at equator. Hence, the value of g at poles is greater than that at equator. So, if a person moves from the equator to poles his weight decreases as the value of g increases.
8. What is weightlessness? Give two interesting examples of weightlessness.
Answer: When a body is in a free fall, it experiences the state of weightlessness. You experience weightlessness when there is no supporting force acting on your body. Zero gravity is also another term referring to this state. In case of weightlessness the reaction force.
(R) = 0. Now for some examples of experiencing weightlessness. If you have ever used a swing, you will see that you feel light when the swing comes down from up. Here, the notion of weight is indicated by the reaction force being applied on the object. Similarly, If the cable of a lift coming in downward direction breaks, there will be free fall of the lift and we will experience weightlessness, i.e. zero weight.
9. Write the three equations of motion for a freely falling body. Also, write the meaning of the Symbols.
Answer: Find out about the three equations of free fall formula. Read up for the answer.
10. An object is thrown in the upward direction with a velocity u so that it attains a height of h. Write the three equations for the movement of the object.
Answer: Given that an object is thrown up with a velocity “u” so that the height of the object is h. See the three equations of motion on throwing an object in the upward direction:
h= ut- ½ gt2
11. What are the difficulties faced by an astronaut due to weightlessness.
Answer: Artificial satellites continuously fall freely toward the center of the earth and all the things present in it including the organisms, are in a state of weightlessness. In such a condition, some very interesting results have been observed. It is seen that the water filled in the glass will not fall even if the glass is inverted as on tilting the glass, the water of the glass being heavy will float in the form of drops. Therefore an astronaut cannot drink the water of the glass. They even ingest food in the form of a paste directly from the tube by squeezing it. Everything inside the spacecraft will be afloat. Also, if an astronaut measures himself in the spacecraft with a spring balance, his weight will be zero.
12. Who were the ancient Bhartiya astronomy scientists prior to Kepler and Newton? Write their contributions in the field of gravitation.
Answer: In 5th century, Aryabhatt, the Bhartiya astronomer propounded the geocentric
model to understand the movement of planets. About 500 years before Newton and Kepler, Bhaskaracharya in his famous work Siddhantsiromani, discussed the power of earth and the planetary movements in detail. This was mentioned in the grah-ganit section of his work. Bhaskaracharya calculated the radius (R) and circumference (2pr) of the earth. Meanwhile, Copernicus (1473-1543), the western scientist on the basis of Aryabhatt’s vision propounded the model of the movement of celestial bodies.
13. The objects in an artificial satellite are in a condition of weightlessness while it is not so on a natural satellite. Why
Answer: Artificial satellites continue to fall freely toward the center of the earth and all the things present in it including the organisms, are in a state of weightlessness. In an artificial satellite everything experiences weightlessness, while the case is different on a natural satellite. Reason for this being that the natural satellite exerts gravitational force because of its mass, which is more. The mass of an artificial satellite, nevertheless, is less and so it applies negligible gravitational force and the object is in a state of weightlessness in its environment.
RBSE Class 9 Science Chapter 10 Essay Type Questions-Important Questions and Solutions
1. Find the value of gravitational force between two balls each having a weight of 10 Kg placed at a distance of 50 cm from each other.
Answer: Here M and m, the mass of two balls are given as 10 kg.
Hence, M=m=10kg and d = 50 cm= 50/100 = 0.5 m
The value for G is constant at 6.67×10–11
As per the Newton’s Gravitational Law F = GMm/d2
= G (10 x 10) / 0.5 2
= 6.67×10–11× 102 / 0.25
= 6.67×10–9 / 0.25 = 26.8 × 10–9= 2.68 × 10-8
Hence, the gravitational force between two balls, each having a weight of 10kg placed at a 50cm distance, is 2.68 x 10–8 N.
2. Calculate the gravitational force on a 40 Kg object placed on the earth’s surface. The radius of the earth R is 6400 Km and its mass is 6×1024 Kg.
Answer: Here, mass of the body (M) =40kg
Mass of the earth (m) = 6×1024 Kg
Radius of the earth = 6400 Km = 6400×1000 m = 6.4×106 m.
And Constant G = 6.67×10–11
Newton’s Gravitational Law states:
F= GMm/ R2 = 6.67×10–11× 40 × 6×1024/ (6.4×106)2
= 6.67 × 24 × 10-14 / (6.4)2× 1012
= [(6.67 × 24) / (6.4)2] × 102 = 390. 82 N
Hence, the gravitational force here is 390. 82 N.
3. A stone is thrown up in the air with an initial velocity of 20m/s. Taking g = 10m/s2 calculate the following :
(i) time taken by the stone to attain the maximum height
(ii) distance travelled by the stone.
Answer: Here, initial velocity is given as u = 20m/s
When the stone reaches the maximum height then the velocity is v=0.
Taking g= 10m/s2 , t= ?
(i) In the first scenario, v = u – gt
That is 0 = 20 – 10 × t
= 10t = 20.
Hence, t = 2 seconds
(ii) The equation applicable for the scenario is v2=u2-2gh
Given (0)2 = (20)2 – 2 × 10 × h
20h = (20) 2
4. Find the Gravitational Acceleration on the moon surface if the mass of the moon is 0.073×1024 Kg and its radius is 1738 Km.
Answer: Here, the mass of the moon is given as M = 0.073×1024 Kg
Radius of the moon (Rm)= 1738 Km
And Constant G = 6.67×10–11
Then find gravitational acceleration is (gm)= ?
Gm = GMm ./ (Rm.)
= 6.67×10–11× 0.073×1024 / (1.738 × 106)
= 6.67 × 0.073×10 / (1.738)2
= 4.8691/ 3.0206 = 1.62 m/s2
5. A stone is thrown from a 125 m high tower. Calculate
(i)time taken to reach down
(ii) final velocity of the stone (take g= 10m/s2)
Answer: Height of the tower denoted a h = 125m
Initial velocity (u) = 0
(i) h= ut + ½ gt2
125 = 0 × t + ½ × 10 × t2
1125 = 5t2
.25 = t2
Hence, taking the square root t = 5 seconds
(ii) Here equation is v = u + gt
So v= 0 + 10 × 5
v= 50 m/s
6. If the radius of the earth becomes half of the present radius then the mass will become 1/8th. What will be the g of this earth of half the shape?
Answer: Here, the diameter of the earth is ½ the radius and mass is ⅛ of its original value.
Here, R1 is the original radius of the earth and R2 is the new radius
Hence R2 = R1 /2 and M = m/ 8
So g = GM /R2
g= [G (m/8)] / (R1 /2)2.
g= (G . m × 4) / 8. R22
= 4 Gm / 8R22
= ½ . Gm/R22
= ½ . g
So, we conclude that the g of this earth is half its original value.
RBSE Class 9 Science Chapter 10 Additional Important Questions and Solutions
1. What is the initial velocity of a free falling object?
Answer: Initial velocity of a free falling object is zero
2. The mass of the earth is approximately 6×1024 Kg and that of the moon is approx. 7.4×1022 Kg. If the distance between the earth and the moon is 3.84×105 Km, then calculate the force exerted by the earth on the moon. Here, G = 6.7×10–11 Nm2/Kg2
Answer: Here, Mass of the earth (M) = 6×1024 Kg and
Mass of the moon (m) = 7.4×1022 Kg
Distance between the two denoted as d = 3.84×105 Km
G = 6.7×10-11
d = 3.84×105 Km = 3.84×105×1000 m
= 3.84×108 m
The force applied by the earth on the moon
F = GMm/ d2= (6.7×10-11× 6×1024 × 7.4×1022) / (3.84×108)2
= 20.17×1019 N
= 2.02×1020 N
3. A ball of 40 Kg mass experiences a gravitational force of 0.25×10-6 Kg weight by another ball of 80 Kg mass. The distance between the centers of the two balls is 30 cm. If the gravitational acceleration g = 9.8m/s2 then calculate the gravitational constant.
Answer: Given that m1 = 40 Kg
m2 = 80 Kg
W =F = mg = 0.25×10–6 ×9.8 Newton
r = 30 cm = 0.3 meter
F = Gm1m2 / r2
Hence, G= Fr2 /m1m2
= [0.25×10–6 ×9.8 × (0.3)2] / 40 × 80
=[ 225×98×10–11.] / 32 × 100
= 6.89×10-11 Nm2 / Kg2
4. Define gravitational acceleration?
Answer: The change in the velocity of an object under the influence of the gravitational force, is known as the gravitational acceleration.
5. Who is the famous seventeenth century scientist who gave the Laws of Motion and the Universal Law of Gravitation?
Answer: The famous seventeenth century scientist Isaac Newton (1642-1727) gave the Laws of Motion and the Universal Law of Gravitation.
6. What are the phenomena easily interpreted by the law of Gravitation?
Answer: Given below are the phenomena easily interpreted by the Law of Gravitation:
(1) The force that binds us to the earth
(2) Movement of planets around the sun
(3) Movement of the moon around the earth
(4) Occurrence of tides in the sea.
7. When a stone is thrown up, it moves up to a certain height then starts to fall down. What is the distance the stone moves dependent on?
Answer: This distance moved by the stone depends upon the force applied on it.
8. What is the weight of an object?
Answer: A type of a force that attracts it towards the earth is the weight of the object. Mathematically F = mg. Hence, the weight of the body w = mg. Also, given that the value of g at a given point is constant, the weight of an object is proportional to its mass.
9. Why is the gravitational acceleration of the planet Jupiter the maximum?
Answer: The largest planet of our solar system is Jupiter and hence its gravitational acceleration also is the maximum.
Mastering all the concepts from the textbook and revising the subject is the best way for the students to perform well in exams. They can also depend on these RBSE Class 9 Science Solutions to self-analyse their performance and know how to prepare for the exams. Additionally, extra resources that the students have been able to access in order to prepare for the exams include the RBSE textbooks , previous year papers and sample papers.
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