Gravitation Test

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Gravitation Test - 44

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Question 1
1 / -0

S.I. Unit of universal gravitational constant $$G$$ is-

A
$$\dfrac{Nm^2}{Kg}$$

B
$$\dfrac{Nm^2}{Kg^2}$$

C
$$\dfrac{Nm}{Kg^2}$$

D
$$\dfrac{Nm}{Kg}$$

Solution

Gravitational force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance $$r$$, $$F = \dfrac{Gm_1 m_2}{r^2}$$
$$\implies$$ $$G = \dfrac{F r^2 }{m_1 m_2}$$
SI unit of universal gravitational constant $$G$$ is $$\dfrac{Nm^2}{Kg^2}$$.
Question 2
1 / -0

If the radius of the earth shrinks by $$1.5$$% (mass remaining same), then the value of acceleration due to gravity changes by :

A
$$1$$%

B
$$2$$%

C
$$3$$%

D
$$4$$%

Solution

Acceleration due to gravity on the earth's surface is $$g=\dfrac{GM}{R^2}$$
When the earth shrinks $$1.5$$ %, the radius becomes $$R'=R-R(1.5/100)=(9.85/100)R$$ or $$R'/R=98.5/100$$
After shrinks, the acceleration due to gravity becomes, $$g'=\dfrac{GM}{R'^2}$$
So, $$\dfrac{g'}{g}=(R/R')^2=(100/98.5)^2=1.03$$
Thus, percentage change in acceleration $$=\dfrac{g'-g}{g}\times 100=[g'/g -1]100=[1.03-1]100=3$$ %
Question 3
1 / -0

Variation of acceleration due to gravity $$\left(g\right)$$ with distance $$x$$ from the centre of the earth is best represented by : ($$R \rightarrow$$ Radius of the earth)

A

B

C

D

Solution

Acceleration due to gravity at a depth $$d$$ from the earth surface $$g = g_o\bigg(1-\dfrac{d}{R}\bigg)$$
Thus distance from earth's centre $$x = R-d$$ i.e. $$d = R-x$$
$$\implies$$ $$g = g_o\dfrac{x}{R}$$
Thus acceleration due to gravity increases linearly with the increase in distance from centre of earth.
Acceleration due to gravity at a height $$h$$ from the earth surface $$g = g_o\bigg(\dfrac{R}{R+h} \bigg)^2$$
Thus distance from earth's centre $$x = R+h$$ i.e. $$h = x-R$$
$$\implies$$ $$g = g_o\bigg(\dfrac{R}{x} \bigg)^2$$
Thus acceleration due to gravity decreases as $$\dfrac{1}{x^2}$$ with the increase in distance from centre of earth.
Hence, option D is correct.
Question 4
1 / -0

At the centre of the earth acceleration due to gravity is:

A
zero

B
infinity

C
9.8

D
98

Solution

gravitational field $$g$$ at distance $$r$$ where $$r<R$$( radius of earth) is
$$g=\frac{G\rho r}{3}$$ ($$\rho$$ is density of earth)
So at centre $$r=0$$
hence $$g=0$$
So at the centre of earth acceleration due to gravity is zero
Question 5
1 / -0

Dimensional formula of universal gravitational constant $$G$$ is-

A
$$M^{-1}L^3T^{-2}$$

B
$$M^{-1}L^2T^{-2}$$

C
$$M^{-2}L^3T^{-2}$$

D
$$M^{-2}L^2T^{-2}$$

Solution
Question 6
1 / -0

The minimum speed of a particle projected from earth's surface so that it will never return is/are:

A
$$\dfrac{GM}{R}$$

B
$$22.1 km/s$$

C
$$(4g_oR)$$

D
none of above

Solution

An object can be thrown up with a certain minimum initial velocity so that, the object goes beyond the earth's gravitational field and escape from earth, this velocity escape velocity of the earth.
potential energy at surface is $$-\frac{GMm}{R}$$, so kinetic energy that has to give to reach particle to infinite ( zero energy) is equal to potential energy.
$$\frac{1}{2}mv_e^2=\frac{GMm}{R}=\sqrt{\frac{2GM}{R}}$$.
so none ofthe above give this expression so the answer is D.
Question 7
1 / -0

A particle hanging from a massless spring stretches it by $$2\ cm$$ at earth's surface. How much will the same particle stretch the spring at height $$2624\ km$$ from the surface of earth? ( Radius of earth $$=6400\ km$$)

A
$$1\ cm$$

B
$$2\ cm$$

C
$$3\ cm$$

D
$$4\ cm$$

Solution

$$x=2cm=0.02m$$
$$h=2624km$$
$$R=6400km$$
$$\therefore$$ $$\cfrac { { x }^{ \prime } }{ x } =\cfrac { { g }^{ \prime } }{ g } =\cfrac { { 6400 }^{ 2 } }{ { \left( 6400+2624 \right) }^{ 2 } } \approx 0.5\\ { x }^{ \prime }=0.5\times x=0.5\times 2cm=1cm$$
Question 8
1 / -0

The SI unit of gravitational potential is

A
$$J$$

B
$$Jkg^{-1}$$

C
$$Jkg$$

D
$$Jkg^{-2}$$

Solution

It is a very simple idea. Gravitational potential is the potential energy per kilogram at a point in a field. So the units are joule per kilogram.
so the best possible answer is option B.
Question 9
1 / -0

How the gravitational constant will change if a brass plate is introduced between two bodies?

A
No change

B
Decreases

C
Increases

D
No sufficient data

Solution

Universal Gravitational Constant (G) remains constant irrespective of the medium between two bodies. It has a constant value of $$6.67 \times 10^{-11} Nm^{2}kg^{-2}$$
Question 10
1 / -0

The minimum energy required to launch a $$m$$ kg satellite from earth's surface in a circular orbit at an altitude of $$2R$$ where $$R$$ is the radius of earth, will be:

A
$$3mgR$$

B
$$\dfrac{5}{6} mgR$$

C
$$2mgR$$

D
$$\dfrac{1}{5} mgR$$

Solution

The kinetic energy at altitude $$2R$$ is $$= \dfrac{G M m} {6 R}$$
The gravitational potential energy at altitude $$2R$$ is $$=\dfrac{– G M m}{ 3 R}$$
Total energy $$=$$ Kinetic energy $$+$$ potential energy
$$= \dfrac{G M m} {6 R}+\dfrac{– G M m}{ 3 R}$$

Potential energy at the surfce is $$= \dfrac{G M m} { R}$$

Therefore, required kinetic energy $$= -\dfrac{G M m} {6 R}+ \dfrac{G M m} { R}= \dfrac{5G M m} { 6R}$$
hence the best possible answer is option B.