GSEB Solutions for Class 9 Science and Technology – Gravitation (English Medium)
GSEB SolutionsMathsScience
Exercise 42:
Solution 1.1:
B. will move downwards with accelerated motion
When an object is released freely from a certain height without applying any external force, the object moves downwards with accelerated motion due to the force of gravity acting on it.
Solution 1.2:
B. rotates
The Earth rotates about its imaginary axis in space.
Solution 1.3:
A. revolve
The Earth and other planets revolve around the Sun due to its gravitational force.
Solution 1.4:
D. retarded motion
The motion of an object vertically upwards is retarded motion. Here, the retardation is 9.8 m/s^{2}.
Solution 1.5:
C. slightly less than
The value of acceleration due to gravity (g) at the equator is less than that at the poles. Hence the weight (= mg) of an object at the equator is slightly less than the weight of an object at the poles.
Exercise 43:
Solution 1.6:
B. remains constant
The mass of an object is the amount of matter it contains, and hence, it remains constant throughout the universe.
Solution 1.7:
B. 1000 kg/m^{3
}The density of water is 1000 kg/m^{3}.
Solution 1.8:
D. Gravitational force
The direction of the weight of an object is in the direction where gravitational force acts on it.
Solution 1.9:
A. sink in water
If the density of an object is more than that of water, then that object will sink in water.
Solution 1.10:
D. is unitless
Relative density is the ratio of similar quantities and hence has no units.
Solution 1.11:
B. 9.8 m/s
Freely falling bodies move down with a constant acceleration of g = 9.8 m/s^{2}. Thus, the increase in velocity per second is 9.8 m/s.
Solution 1.12:
C. 6 kg
The mass of an object always remains constant.
Solution 1.13:
B. Cavendish
The value of G for the first time was measured by practically by Cavendish.
Solution 1.14:
D. N/m^{2
}The S.I. unit of pressure is Pascal or N/m^{2}.
Solution 1.15:
C. Zero
As a freely falling object falls from rest, the initial velocity is taken as zero.
Solution 2.1:
Gravitational force depends on:
 The masses of the two bodies involved
 The distance between the two bodies
Solution 2.2:
The Moon revolves around the Earth in circular motion due to the centripetal gravitational force of the Earth.
Solution 2.3:
Newton’s universal law of gravitation states that every object of the universe attracts every other object. The force of attraction between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The direction of this force is along the line joining the centres of the objects.
Solution 2.4:
At any place in the universe and at any time, the value of G is found to be constant for any two bodies. Thus, G is called the universal constant of gravitation.
Solution 2.5:
When an object is released (freely) from a certain height, without applying any external force, it moves downwards under the effect of the gravitational force. The motion of such an object is described as free fall.
Solution 2.6:
A freely falling body undergoes accelerated motion under the effect of gravitational force.
Solution 2.7:
Gravitational force acting on a body at a given location is its weight at that location.
Weight of an object is not constant; it depends on local value of g.
Mathematically,
Weight = Mass × Acceleration due to gravity
Solution 2.8:
Pressure is the force acting per unit area, in a direction perpendicular to the surface.
Pressure = Force/Area
Solution 2.9:
Archimedes’ principle: When an object is partially or completely immersed in a liquid, the buoyant force acting on it is equal to the weight of the liquid displaced by the object.
Solution 2.10:
The value of gravitational acceleration on the Moon is one sixth as that compared to the gravitational acceleration on the Earth.
g_{moon} = (1/6) g _{earth
}g_{moon} = 1.63 m/s^{2}^{ }
Exercise 44:
Solution 2.11:
The buoyant force acts in the upward direction on an object immersed in liquid.
Solution 2.12:
The man’s mass will be 60 kg as mass of an object remains constant.
Solution 2.13:
The decrease in velocity of an object thrown upwards will be 9.8 m/s.
Solution 2.14:
The velocity of a stone thrown upwards will be zero when it reaches its maximum height, as the stone momentarily comes to rest at its maximum height.
Solution 3.1:
The universal law of gravitation explained the following phenomena:
 The force which binds us to the Earth.
 The revolution of the Moon around the Earth.
 The revolution of planets around the Sun.
 Tide arising due to the Moon and the Sun.
Solution 3.2:
The property of a liquid to push the immersed object in the upward direction is known as buoyancy.
The net upward force acting on a body partially or completely immersed in a liquid is known as buoyant force.
Solution 3.3:
Illustration: Take a plastic bottle and fill it with pure water up to a certain height. Draw a mark on the bottle indicating the upper surface of the water level in it and measure the mass of the filled bottle.
Next mix some salt/sugar with the water and fill the bottle up to the same mark with this solution. Measure the mass of the filled bottle. We observe that the mass of the bottle increases. This happens because the addition of solutes (salt/sugar) increases the density of water.
Sometimes the density of an object is compared with the density of water.
Thus, the ratio of density of an object and the density of the water is known as the relative density or specific density of the object.
Solution 3.4:
Equations of uniformly accelerated motion are as follows:
For a freely falling body from height h, we have
s = h, initial velocity (u) = 0, a = g.
Hence, the equations for a freely falling body become:
Solution 3.5:
Equations of uniformly accelerated motion are as follows:
For a body moving in the upward direction, equations of motion can be obtained by putting u = u, v = 0 (as the body momentarily comes to rest at the highest point), a = g and s = h in equations for accelerated motion.
Thus, we get:
Solution 3.6:
The gravitational constant is the force of attraction between two bodies of unit mass separated by unit distance. The value of G in the S.I. system is 6.67 × 10^{11 }Nm^{2}kg^{2} and it always remains constant, whereas the rate at which the velocity of a freely falling body increases is called acceleration due to gravity. Its value varies from place to place. It is the acceleration caused by the gravitational force of the Earth on the object.
Solution 3.7:
Pressure is the force applied (thrust) per unit area of the surface.
Its S.I. unit is ‘newton per metre^{2}‘ or ‘pascal’.
If the area increases, pressure will decrease: The base of a dam is made broad, so that pressure exerted on it can be reduced.
If the area decreases, pressure will increase: In a knife, the edge (or blade) is very small and thin. So, applying a small force produces pressure needed for cutting.
Solution 3.8:
Universal law of gravitation: Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, the direction of the force being along the line joining the masses.
Derivation of mathematical form:
Let two objects of mass m_{1} and m_{2} lie at a distance r from each other. According to Newton’s law of gravitation, these objects will attract each other with a force, acting along the line joining the centres of the two objects.
Let, F_{12} = Gravitational force acting on object 1 due to object 2.
F_{21} = Gravitational force acting on object 2 due to object 1.
According to Newton’s third law of motion, these forces are equal and opposite.
i.e., F_{12} = – F_{21
}The magnitude of gravitational force acting between the two objects is
F_{12} = F_{21} = F
According to universal law of gravitation,
At any place and at any given time in the universe, the value of G is found to be constant and is given as, G = 6.67 × 10^{11} Nm^{2}/kg^{2}.
Solution 3.9:
Mass  Weight 
1. It is the quantity of matter contained in a body.  1. It is the gravitational force acting on a body. 
2. It is a scalar quantity.  2. It is a vector quantity. 
3. Mass of a body = m  3. Weight of a body = mg 
4. Its S.I. unit is kilogram.  4. Its S.I. unit is newton. 
5. It is measured using a beam balance.  5. It is measured using a spring balance. 
6. Mass of a body (at rest) remains constant at all locations.  6. Weight of a body is not constant. It depends on the local value of g. 
Solution 3.10:
Density is the ratio of the mass of an object and its volume. It is a scalar quantity and its S.I. unit is kg/m^{3}. Density of a given material is always the same.
Relative density is the ratio of density of an object to the density of water. Since it is a ratio of two similar quantities, it has no unit.
Solution 4.1:
Solution 4.2:
Solution 4.3:
Solution 4.4:
Solution 4.5:
Solution 4.6:
Solution 4.7:
Weight = mass × acceleration due to gravity
Here, mass = 10 kg
Acceleration due to gravity = 9.8 m/s^{2
}Therefore,
Weight = 10 × 9.8 = 98 N
Solution 4.8:
We know that,
Weight = mass × acceleration due to gravity
Here, mass = 6 kg
Acceleration due to gravity on the Earth = 9.8 m/s^{2
}Acceleration due to gravity on the Moon = 1.63 m/s^{2}
Therefore,
Weight on the Earth = 6 × 9.8 = 58.8 N
Solution 4.9:
Given:
Weight on Earth (W)= 245 N
Acceleration due to gravity on the Earth (g) = 9.8 m/s^{2
}Therefore,
Mass = W /g = (245/9.8) = 25 kg
Solution 4.10:
Given,
Weight = 392 N
Therefore, Mass = Weight/Acceleration due to gravity
Or, Mass = 392/ 9.8 = 40 kg
We know that,
Force (F) = Mass (m) x Acceleration (a)
Given,
Mass = 40 kg
Force = 80 N
Therefore,
Acceleration = F/m = 80/40 = 2 m/s^{2}
Solution 4.11:
We know that,
Density = Mass/Volume
And,
Making use of the above two equations, we can calculate the density and relative density for a beaker filled with different volumes of water.
Mass of beaker m_{1} (kg) 
Volume of beaker
V (m^{3}) 
Volume of water filled in beaker
v (m^{3}) 
Mass of water filled in beaker m_{2}= v x 1000
(kg) 
Mass of beaker with water m= m_{1} + m_{2}
(kg) 
Density of beaker with water
ρ = m/V (kg/m^{3}) 
Relative density of beaker with water ρ/1000

0.25  5 x10^{4}  0  0  0.25  500  0.5 
0.25  5 x10^{4}  1 x10^{4}  0.1  0.35  700  0.7 
0.25  5 x10^{4}  2 x10^{4}  0.2  0.45  900  0.9 
0.25  5 x10^{4}  3 x10^{4}  0.3  0.55  1100  1.1 
The beaker will sink in water when the density of the beaker filled with water is more than the density of water i.e. 1000 kg/m^{3}. Thus, the beaker will sink when it is filled with 300 ml of water.
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