NCERT Class 9 Science Chapter 10 Gravitation Extra Questions and Answers
Class 9 Science Chapter 10 Extra Inside Questions and Answers – Gravitation. Here in this Page Class IX Students can Learn Extra Questions & Answer 10th Chapter Science fully Inside.
We Provided Here Gravitation Science Chapter 10 Long Answer Type Question, Multiple Choice Questions & Answer, Short Answer Type Questions (2 or 3 marks), and Very Short Answer Type Question (1 marks) Solution.
Class 9 Science Chapter 10 Extra Question with Answer – Gravitation
Science Chapter 10 Gravitation Class 9 Inside 5 Marks, 3 marks, 2 Marks & And 1 Marks Important Questions and Answers.
Very Short Answer Type questions:
1.) Is gravitational force depend on separation between objects?
Answer: Yes, gravitation force is inversely proportional to square of distance between the objects.
2.) Why does the ball moves towards earth when it dropped from top of tower?
Answer: The ball dropped from top of tower moves towards earth is due to gravitational force acting towards earth.
3.) State the direction of centripetal force.
Answer: Centripetal force always acts on object towards the centre of circle.
4.) State the centripetal force which binds earth in its orbit.
Answer: The gravitational force acting towards the centre of sun is responsible for binding earth in its orbit.
5.) Who did discover the value of G?
Answer: Henry Cavendish discovered the value of G.
6.) Is mass changes with altitude from earth’s surface?
Answer: No, Mass does not changes with altitude.
7.) On what factors does the weight depend?
Answer: Weight of the object is depend on mass and value of acceleration due to gravity.
8.) Who sink into water?
Cork, nail, empty plastic bottle, 2 g stone.
Answer: Nail and 2 g stone sink into water.
9.) State the name of country from which Archimedes belongs.
Answer: Archimedes belongs from Italy.
10.) What is the meaning of ‘Eureka’?
Answer: Eureka means ‘I have got it’.
11.) State the principle on which ancient people measure the mass of elephant?
Answer: Ancient people measure the mass of elephant using Archimedes principle.
12.) Is relative density of platinum greater than water?
Answer: Yes, relative density of platinumis greater than water.
13.) What is the relative density of gold?
Answer: The value of relative density of gold is 19.3.
14.) State the quantity which depend on force and area.
Answer: Pressure is depend on force and area.
15.) Which force exerts on bottle in upward direction when it trying to push into water?
Answer: The force acting on bottle in upward direction by water is called buoyant force.
Short Answer Type Questions:
1.) Give the examples of centripetal force.
Answer: Centripetal force is defined as the force exerted on the object towards the centre of circle which it performing circular motion.
Examples-
- The motion of electron around proton. In this type of motion, electrostatic force binds the orbit of electron
- The motion of moon around earth. In this example, gravitational force binds the orbit of moon.
- The motion of car along circular track. In this type of motion, Frictional force provides sufficient centripetal force.
2.) If the distance between two object becomes 4 times then calculate gravitational force between them.
Answer: As we know that, gravitational force is inversely proportional to square of distance between two objects.
If d is the separation between two objects then the magnitude of force is,
F ∝ 1/d² ………..1
Now the distance becomes 4 time that of initial then,
F’ ∝ [1/(4d)²]
F’ ∝1/16d² ………..2
Divide equation 2 by equation 1 we get,
F’/F ∝ [(1/16d²)/(1/d²)]
F’ ∝F/16
Final force becomes 1/16 times initial force.
If distance increases by 4 times then force decreases by 16 times.
3.) Calculate gravitational force on the object having mass 100 kg.
Answer: Given, Mass of object = 100 kg.
Acceleration due to gravity = g =9.8 m/s².
According to second law of motion,
Force = mass × acceleration
Force = 100× 9.8
Force = 980 N.
4.) Find the unit of universal gravitational constant.
Answer: Let the mass of first object = m kg,
Mass of second object = M kg,
Separation between them = r,
Universal gravitational constant = G then the force of attraction between them is calculated by,
F= G × m × M / r²
G = (F × r²)/ m × M ——–1
We know the units of each term in right hand side,
F- Newton
r – meter.
M – Kilogram.
m – Kilogram.
Put unit of each physical quantity in equation 1.
G = (N × m²)/(kg × kg)
G = Nm²/kg²
The unit of G is Nm²/kg².
5.) If a force of 20 N is exerted on an object having mass 4 kg then calculate acceleration produced in it.
Answer: Given, Mass of the object = 4 kg, Force = 20 N.
According to second law of motion,
Force = mass × acceleration
Acceleration = force / mass.
Acceleration = 20 / 4.
Acceleration = 5 m/s².
6.) An object is thrown vertically upward with initial velocity of 20 m/s. Calculate total time it takes to return at ground. (take g = 10 m/s²)
Answer: Given, initial velocity = u = 20 m/s.
When the object is thrown vertically upward then it’s acceleration will be negative.
g = -10 m/s².
This acceleration opposes the motion so the velocity at maximum height becomes zero.
v = 0 m/s.
We calculate time for reach maximum height.
According to 1st equation of motion
v = u + at
0 = 20 + (-10)t
10t = 20
t = 20/10
t = 2 seconds.
So I will achieve maximum height in 2 seconds.
Also it fall on ground after 2 seconds from maximum height.
Thus,
Total time = time required for upward motion + time required for downward motion
Total time = 2 + 2
Total time = 4 seconds.
The object will fall on ground after 4 seconds.
7.) A wooden block of mass 15 kg is places on table. Calculate force exerted by table on wooden block. (Take g = 10 m/s²)
Answer: A wooden block is at rest on table. Thus according to first law of motion, Net force on the wooden block is zero.
We can calculate force exerted on wooden block by table using third law.
According to 3rd law of motion, Force exerted by table on wooden block is equal to force exerted by wooden block on table.
We understand this using following diagram.
It is to calculate force exerted by wooden block on table.
The mass of wooden block is 15 kg. g = 10 m/s².
According to second law of motion,
Force = mass × acceleration
Force = 15 × 10
Force = 150 N.
Thus the magnitude of force exerted by wooden block on table is 150 N. According to 3rd aw of motion we states that the magnitude of force exerted by floor on wooden block is also 150 N.
8.) Mass of elephant is 5000 kg. What is its weight on earth?
Answer: Given,
Mass of elephant = 5000 kg,
We know that the value of g on earth,
g on earth = 9.8 m/s².
According to second law of motion,
Weight = mass × acceleration due to gravity
Weight = 5000 × 9.8
Weight = 49000 N.
The weight of elephant on earth is 49000 N
9.) Why does wooden block not sink into water?
Answer: We know the condition of sink objects into water. If the density of object is less than density of water then that objects does not sink into water.
The density of wooden block is less than water. Thus wooden block does not sink into water.
10.) A piece of paper takes greater time to reach at ground from certain height than crumpled paper. Explain.
Answer: When an object falls on ground then air resistance opposes this motion. This air resistance depends on the area. If area is greater then air resistance also greater.
The area of paper is greater than area of crumpled paper. So due to maximum oppose the paper takes greater time to reach at ground than crumpled paper.
11.) Why does brick sink into water.
Answer: We know the condition of sink objects into water. If the density of object is less than density of water then that objects does not sink into water.
The density of brick is greater than water. Brick sink into water.
12.) State the applications of Archimedes principle.
Answer: Archimedes principle is widely used in purposes. Some applications are as,
- Archimedes principle is used to design submarines and ships.
- We calculate the purity of milk using lactometer. The lactometer is based on Archimedes principle.
- It is used to find the purity of metals.
13.) Define weight. What is the SI unit of weight?
Answer: Weight- The gravitational force acting on the object is called weight.
SI unit of weight is kg m/s² or simple Newton.
In case you are missed :- Previous Chapter Extra Questions
14.) A man has mass 70 kg then calculate its weight on moon.
Answer:
- According to second law of motion, weight is the product of mass and acceleration due to gravity.
- The value of acceleration due to gravity on earth is 9.8 m/s².
- The value of acceleration due to gravity on moon is 6 times less than value of acceleration due to gravity on earth.
- Thus the value of acceleration due to gravity on moon becomes,
Acceleration due to gravity on moon = 1/6 × acceleration due to gravity on earth.
= 1/6 × (9.8)
= 1.633 m /s².
Given, Mass of man = 70 kg.
g on moon = 1.633 m/s².
According to second law of motion,
Weight = mass × g
Weight = 70 × 1.633
Weight = 114.31 N.
15.) A ball having mass 200 gm moves with 60 m/s catches by a player. If the ball takes 0.2 seconds in hands of player before stop then calculate force exerted by that player to stop the ball.
Answer: Given mass of ball = 200 g = 0.2 kg,
Initial speed = u = 60 m/s,
Final speed = v = 0 m/s.
Time = 0.2 seconds.
Second law of motion states that rate of change of momentum is equal to applied force.
Force = (final momentum- initial momentum )/time
Force = (mv – mu)/time
Force = (0.2 × 0 –0.2 ×60)/0.2
Force = -12/0.2
Force = – 60 N
Minus sign indicates that force opposes the motion.
The force exerted by player to stop the ball is 60 N.
Long Answer Type Questions
1.) (a) State the universal law of gravitation. Give its mathematical form. Explain each keywords.
(b) An abject having mass 90 kg attracts another object of mass 50 kg. If the separation of them is 2 m then calculate gravitational force between them.
Answer:
(a) Universal law of gravitation is used to calculate the gravitational force between two objects. It holds everywhere. This law states that,
Every object in the universe attracts another object with a force, that force is directly proportional to product of their masses and universally proportional to square of distance between them.
It is mathematically wrote as,
F = (G ×m× M) /r²
Where,
G –6.67 × 10-11 N m²/kg²- universal gravitation constant
m- mass of first object
M – Mass of second object.
r – speciation between them.
(b) Given,
Mass of first object = m = 90 kg,
Mass of second object = M = 50 kg,
G – 6.67 × 10-11 N m²/kg²
Separation = r = 2 m.
F = (G ×m× M) /r²
F = (6.67 × 10-11 × 90 × 50 ) /2²
F = 30015× 10-11 /4
F = 7503.75 × 10-11
F = 74.04 × 10-9 N
The attractive force between these two objects is 74.04 × 10-9 N.
2.) (a) Write three equation of motions for free fall. Explain each keywords.
(b) If the object is released from 80 m then calculate its velocity at the ground. Also calculate time required to reach the object at ground.
Answer:
(a) We know the three equations of kinematics. These equations are as,
v = u + at
s = ut + ½ at²
v² = u² +2 as
Where,
v – final velocity,
u – initial velocity,
a – acceleration,
t – time,
s – displacement,
When the objects falls under gravitational force only then such motion is called as free fall. In free the acceleration arises due to gravitational force only, thus acceleration is called acceleration due to gravity. The value of this acceleration is constant.
The value of g on earth = 9.8 m/s².
So we can write these equations by replacing a by g.
v = u + gt
s = ut + ½ gt²
v² = u² +2 gs
Where,
v – final velocity,
u – initial velocity,
g – acceleration,
t – time,
s – displacement,
we take the value of g is positive when objects falls down and negative when object moves up.
(b) Given, height = s = 80 m,
It is dropped from some height so initial velocity = 0 m/s
Acceleration = g = 9.8 m/s².
According to third equation,
v² = u² +2 gs
v² = 0² + 2 × 9.8 ×80
v² = 0 + 1556
Taking square root on both side,
v = 39.45 m/s.
Now we calculate time using first equation,
v = u + gt
39.45 = 0 + 9.8 ×t
t = 39.45/9.8
t = 4 seconds.
Thus the final velocity becomes 39.45 m/s and it requires 4 seconds to reach at ground.
3.) A ball is thrown vertically upwards with a velocity of 60m/s. Calculate
Maximum height, and time required to gain maximum height.
Answer: Given, initial velocity =u = 60 m/s,
The value of acceleration in upwards journey = -9.8 m/s².
- We know that the velocity of ball becomes zero when it reach at maximum height.
v = 0 m/s.
The value of acceleration in upwards journey = -9.8 m/s².
We can calculate maximum height using third equation of motion,
v² = u² +2as
0² = 60² + 2 × (-9.8 )× s
19.6 s = 3600
s = 3600/19.6
s = 183.7 m.
The maximum height gained by ball is 183.7 m.
- Now we calculate time required to gain maximum height.
According to 1st equation of motion,
v = u + at
0 = 60+ (-9.8)× t
9.8t = 60
t = 60/9.8
t = 6.12 seconds
Thus, the maximum height gain by the object is 183. 7 m and time required to gain this height is 6.12 seconds.
4.) A metal cylinder having mass 80 kg is placed on the table. If the radius of cylinder is 1.4 m then calculate pressure exerted by cylinder on table.
Answer: Given, radius of cylinder = r = 1.4 m,
Mass of cylinder = m = 80 kg,
Firstly we calculate, Force exerted by cylinder on table.
As we know that, gravitational force acting on the object is called weight.
We calculate weight using second law of motion,
Weight = mass × acceleration due to gravity.
Weight = 80 × 9.8
Weight = 784 N.
Now we calculate pressure.
As we know,we must know the area to find pressure.
The shape of lower part is a circular.
Hence area = π r²,
Area = (22/7 ) × 1.4²
Area = 3.14 × 1.96
Area = 6.02 m²,
Now we calculate pressure,
Pressure = perpendicular force/ area.
Pressure = 784/ 6.02
Pressure = 130.23 N/m².
The pressure exerted by cylinder on table is 130.23 N/m².
5.) The density of brick is 15000 kg/m³. If the brick is placed on table as shown below then calculate pressure exerted by brick on table. The dimensions of brick are as,
Answer: Given, Density of brick = 15000 kg/m³.
Length = l = 1 m, breadth =0.5 m,
Height = 1.5 m.
Volume of brick = l × b × h
Volume = 1 × 0.5 × 1.5
Volume = 0.75 m³.
We know,
Density = mass/ volume
Mass = Density × volume
Mass = 15000 ×0.75
Mass = 11250 kg.
Now we calculate force exerted by brick on table.
We know that,
Force = weight.
Force = mass × acceleration due to gravity
Force = 11250 × 9.8
Force = 110250 N.
We must know area of lower side of brick to calculate pressure,
As area = length × breadth
Area = 1 ×0.5
Area = 0.5 m².
As we know,
Pressure = force /area
Pressure = 110250/0.5
Pressure = 220500 N /m².
The pressure exerted by brick on table is 220500 N /m².
In case you are missed :- Next Chapter Extra Questions