Gravitation Test

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Gravitation Test - 58

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Question 1
1 / -0

Find the force of gravitational attraction between two neutrons whose centres are $$10^{-12}$$m apart. Given $$G=6.67\times 10^{-11}Nm^{2}kg^{-2}$$, mass of neutron =$$1.67\times 10^{-27}kg$$

A
$$1.8\times 10^{-42}N$$

B
$$1.8\times 10^{-29}N$$

C
$$1.8\times 10^{-40}N$$

D
$$1.07\times 10^{-13}N$$

Solution
Question 2
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Find the gravitational force between two protons kept at a separation of $$1$$ femtometre ($$1$$ femtometre$$=10^{-15}m$$). The mass of a protons is $$1.67\times 10^{-27}kg$$

A
$$1.8\times 10^{-42}N$$

B
$$1.8\times 10^{-29}N$$

C
$$1.8\times 10^{-34}N$$

D
$$1.86\times 10^{-36}N$$

Solution

Given
$$m_{p} = 1.67\times 10^{-27} kg$$; mass of the proton
Distance between two proton
$$r = 10^{-15} m$$
We know that
$$F = \dfrac {Gm_{1}m_{2}}{r^{2}}$$
$$\therefore F = \dfrac {6.67\times 10^{-11} \times (1.67 \times 10^{-27})^{2}}{(10^{-15})^{2}}$$
$$\Rightarrow F = 1.8\times 10^{-34} N$$
So, option (C) is correct.
Question 3
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If the distance between two bodies is reduced to half, the force of gravitation between them is

A
decreased by two times

B
increased by four times

C
increased by two times

D
decreased by four times

Solution

According to newton's law of gravitation,
Force between two masses $$F = \dfrac{GMm}{r^2}$$
If distance is halved $$r = r/2$$

$$F' = \dfrac{GMm}{\left(\dfrac{r}{2}\right)^2} = \dfrac{4GMm}{r^2}=4F$$
Hence, the distance between mass reduces to half then gravitation force increases by four times.
Thus, option (B) is correct
Question 4
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The value of universal gravitational constant depend upon:

A
Nature of material of two bodies

B
Heat constant of two bodies

C
Acceleration of two bodies

D
None of these

Solution

Every object in the universe attracts every another 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 force is along the line joining the centers of two objects.

$$F\propto \dfrac{m_1m_2}{r^2}$$

$$F=G \dfrac{m_1m_2}{r^2}$$
where $$G$$ is the constant of proportionality and is called the universal gravitation constant. $$G$$ does not depend on anything.
Question 5
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In a rocket of mass 1000 kg fuel is consumed at a rate of 40 kg/s. The velocity of the gases ejected from the rocket is $$5 \times {10^4}\;m/s.$$ the thrust on the rocket is.

A
$$2 \times {10^3}\;N.$$

B
$$2.5\;m/{s^2}.$$

C
$$2 \times {10^6}\;N.$$

D
$$2 \times {10^9}\;N.$$

Solution

Thrust is a reaction force. When a system accelerates a mass in a particular direction, the accelerated mass will cause a force of equal magnitude and opposite direction on that system. Thrust is defined as,
$$F_{thrust}=u\dfrac{dm}{dt}$$
$$=(5\times 10^4)(40)$$
$$=2\times 10^6\, N$$
Question 6
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Choose the correct statement:

A
The dimensional formula for $$G$$ is $${M}^{-1}{L}^{3}{T}^{-2}$$

B
$$G$$ is independent of medium

C
$$F=G\cfrac{{m}_{1}{m}_{2}}{{r}^{2}}$$

D
All of the above

Solution
Question 7
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Find the false statement

A
Gravitational force acts along the line joining the two interacting particles

B
Gravitational force is independent of medium

C
Gravitational force forms an action-reaction pair

D
Gravitational force does not obey the principle of superposition
Question 8
1 / -0

When a body is weighted in a liquid, the loss in its weight depends upon:

A
volume of the fluid displaced by the body.

B
mass of the body

C
shape of the body

D
CG of the body

Solution

When a body is placed in a liquid, it experiences an upward thrust. This upward thrust is known as the buoyant force. According to the Archimedes' principle, this force is equal to the weight of the fluid that the body displaces.

Hence, the net force experienced by the body will be
$$F_{net} = F_{weight}-F_{upthrust}$$

Hence, the loss in its weight depends upon the buoyant force acting on the body.

Since, buoyant force is given by , $$F_b = \rho_l Vg$$, where $$V$$ is the volume of the fluid displaced.

Hence, loss in its weight depends upon the volume of the fluid displaced
Question 9
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The mass of a planet is twice and its radius is three times of the earth. The weight of a body, which has a mass of $$5kg$$, on that planet will be:

A
$$11.11N$$

B
$$10.88N$$

C
$$9.88N$$

D
$$20.99N$$

Solution

Given:
The mass of the planet ($$M_p$$) = $$2\times $$ the mass of the earth ($$M_e$$)
The radius of the planet ($$R_p$$) = $$3\times $$ the radius of the earth ($$R_e$$)
The mass of the body $$m=5\ kg$$
The gravitational acceleration of the earth is $$g=\dfrac{GM_e}{R_e^2} = 9.8\ m/s^2$$

The gravitational acceleration of the earth is $$g'=\dfrac{GM_p}{R_p^2}$$

$$\therefore \dfrac{g'}{g} = \dfrac{M_p}{M_e} \times \dfrac{R_e^2}{R_p^2} = 2\times \dfrac{1}{9}=\dfrac{2}{9}$$
$$\Rightarrow g'=\dfrac{2g}{9}=\dfrac{2\times 9.8}{9} = 2.18\ m/s^2$$

The weight of the body of mass $$5\ kg$$ on the planet will be $$W'=m\times g'=5\times 2.18 = 10.88\ N$$
Question 10
1 / -0

The distance between two objects is reduced to half. So, the force of gravitation between the two objects become

A
twice of the initial force of gravitation.

B
four times of the initial force of gravitation

C
one-fourth of the initial force of gravitation

D
half of the initial force of gravitation

Solution