Gravitation MCQ Questions & Answers in Basic Physics | Physics
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71.
A central particle $$M$$ is surrounded by a square array of other particles, separated by either distance $$d$$ or distance $$\frac{d}{2}$$ along the perimeter of the square. The magnitude of the gravitational force on the central particle due to the other particles is
72.
A solid sphere of mass $$'M\,'$$ and radius $$'a\,'$$ is surrounded by a uniform concentric spherical shell of thickness $$2a$$ and mass $$2M.$$ The gravitational field at distance $$'3a\,'$$ from the centre will be:
A
$$\frac{{2GM}}{{9{a^2}}}$$
B
$$\frac{{GM}}{{9{a^2}}}$$
C
$$\frac{{GM}}{{3{a^2}}}$$
D
$$\frac{{2GM}}{{3{a^2}}}$$
Answer :
$$\frac{{GM}}{{3{a^2}}}$$
$${E_P} = \frac{{GM}}{{{{\left( {3a} \right)}^2}}} + \frac{{G\left( {2M} \right)}}{{{{\left( {3a} \right)}^2}}} = \frac{{GM}}{{3{a^2}}}$$
For a part on the surface of a spherical uniform charge distribution the whole mass acts as a point mass kept at the centre.
73.
Two concentric uniform shells of mass $${M_1}$$ and $${M_2}$$ are as shown in the figure. A particle of mass $$m$$ is located just within the shell $${M_2}$$ on its inner surface. Gravitational force on $$'m'$$ due to $${M_1}$$ and $${M_2}$$ will be
A
zero
B
$$\frac{{G{M_1}m}}{{{b^2}}}$$
C
$$\frac{{G\left( {{M_1} + {M_2}} \right)m}}{{{b^2}}}$$
D
None of these
Answer :
$$\frac{{G{M_1}m}}{{{b^2}}}$$
Gravitational force will be due to $${{M_1}}$$ only.
74.
The period of revolution of the planet $$A$$ round the sun is 8 times that of $$B.$$ The distance of $$A$$ from the sun is how many times greater than that of $$B$$ from the sun?
A
5
B
4
C
3
D
2
Answer :
4
According to Kepler's third law $${T^2} \propto {r^3}$$
where, $$T =$$ Time period of revolution
$$r =$$ Semi major axis
$$\eqalign{
& \therefore \frac{{T_A^2}}{{T_B^2}} = \frac{{r_A^3}}{{r_B^3}} \cr
& \therefore \frac{{{r_A}}}{{{r_B}}} = {\left( {\frac{{{T_A}}}{{{T_B}}}} \right)^{\frac{2}{3}}} = {\left( 8 \right)^{\frac{2}{3}}} = {2^{3 \times \frac{2}{3}}} = 4 \cr
& {\text{or}}\,\,{r_A} = 4{r_B} \cr} $$
75.
The gravitational field due to a mass distribution is $$E = \frac{K}{{{x^3}}}$$ in the $$x$$-direction. ($$K$$ is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance $$x$$ is
A
$$\frac{K}{x}$$
B
$$\frac{K}{{2x}}$$
C
$$\frac{K}{{{x^2}}}$$
D
$$\frac{K}{{2{x^2}}}$$
Answer :
$$\frac{K}{{2{x^2}}}$$
$$V = - \int_\infty ^x E dx = \int_\infty ^x K {x^{ - 3}}dx = \frac{K}{{2{x^2}}}.$$
76.
If the earth were to rotates faster than its present speed, the weight of an object will
A
increase at the equator but remain unchanged at the poles
B
decrease at the equator but remain unchanged at the poles
C
remain unchanged at the equator but decrease at the poles
D
remain unchanged at the equator but increase at the poles
Answer :
decrease at the equator but remain unchanged at the poles
At the equator, $$g' = g - R{\omega ^2}$$
As $$\omega $$ increases, $${g'}$$ decreases and hence weight decreases.
At the pole, $$g' = g.$$
So weight remain unchanged.
77.
If $$g$$ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth, is-
A
$$\frac{1}{2}\,mgR$$
B
$$2\,mgR$$
C
$$mgR$$
D
$$\frac{1}{4}mgR$$
Answer :
$$\frac{1}{2}\,mgR$$
$${U_i} = - \frac{{GMm}}{R} = $$ Initial potential energy of the system.
$${U_f} = - \frac{{GMm}}{{2R}} = $$ Final P.E. of the system.
$$\eqalign{
& \therefore \Delta U = {U_f} - {U_i} \cr
& = - GMm\left[ {\frac{1}{{2R}} - \frac{1}{R}} \right] = \frac{{GMm}}{{2R}}\,.....(i) \cr
& {\text{But }}g = \frac{{GM}}{{{R^2}}} \cr
& \therefore GM = g{R^2}\,.....(ii) \cr
& {\text{From (i) and (ii)}} \cr
& \Delta U = \frac{{g{R^2}m}}{{2R}} = \frac{{mgR}}{2} \cr} $$
78.
A point $$P$$ lies on the axis of a fixed ring of mass $$M$$ and radius $$R,$$ at a distance $$2R$$ from its centre $$O.$$ A small particle starts from $$P$$ and reaches $$O$$ under gravitational attraction only. Its speed at O will be
A
zero
B
$$\sqrt {\frac{{2GM}}{R}} $$
C
$$\sqrt {\frac{{2GM}}{R}\left( {\sqrt 5 - 1} \right)} $$
D
$$\sqrt {\frac{{2GM}}{R}\left( {1 - \frac{1}{{\sqrt 5 }}} \right)} $$
Gravitational potential at $$P,{V_p} = \frac{{ - GM}}{{\sqrt 5 R}}$$
Gravitational potential at $$O,{V_O} = \frac{{ - GM}}{R}$$
By work energy theorem,
$$\eqalign{
& W = \Delta K \Rightarrow m\left[ {{V_p} - {V_O}} \right] = \frac{1}{2}m{v^2} \cr
& m\left[ {\frac{{GM}}{R} - \frac{{GM}}{{\sqrt 5 R}}} \right] = \frac{1}{2}m{v^2}\,{\text{or}}\,V = \sqrt {\frac{{2GM}}{R}\left( {1 - \frac{1}{{\sqrt 5 }}} \right)} \cr} $$
79.
The acceleration due to gravity on the planet $$A$$ is 9 times the acceleration due to gravity on the planet $$B.$$ A man jumps to a height of $$2\,m$$ on the surface of $$A.$$ What is the height of jump by the same person on the planet $$B$$ ?
A
$$6\,m$$
B
$$\frac{2}{3}\,m$$
C
$$\frac{2}{9}\,m$$
D
$$18\,m$$
Answer :
$$18\,m$$
It is given that, acceleration due to gravity on planet $$A$$ is 9 times the acceleration due to gravity on planet $$B$$ i.e.
$${g_A} = 9{g_B}\,......\left( {\text{i}} \right)$$
From third equation of motion, $${v^2} = 2gh$$
At planet $$A,$$ $${h_A} = \frac{{{v^2}}}{{2{g_A}}}\,......\left( {{\text{ii}}} \right)$$
At planet $$B,$$ $${h_B} = \frac{{{v^2}}}{{2{g_B}}}\,......\left( {{\text{iii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we have
$$\frac{{{h_A}}}{{{h_B}}} = \frac{{{g_B}}}{{{g_A}}}$$
From Eq. (i), $${g_A} = 9{g_B}$$
$$\eqalign{
& \therefore \frac{{{h_A}}}{{{h_B}}} = \frac{{{g_B}}}{{9{g_B}}} = \frac{1}{9} \cr
& {\text{or}}\,\,{h_B} = 9{h_A} = 9 \times 2 = 18\;m\,\,\left( {\because {h_A} = 2\;m} \right) \cr} $$
80.
Two particles of equal mass $$'m'$$ go around a circle of radius $$R$$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is
A
$$\sqrt {\frac{{Gm}}{{4R}}} $$
B
$$\sqrt {\frac{{Gm}}{{3R}}} $$
C
$$\sqrt {\frac{{Gm}}{{2R}}} $$
D
$$\sqrt {\frac{{Gm}}{R}} $$
Answer :
$$\sqrt {\frac{{Gm}}{{4R}}} $$
Here, centripetal force will be given by the gravitational force between the two particles.
$$\eqalign{
& \frac{{G{m^2}}}{{{{\left( {2R} \right)}^2}}} = m{\omega ^2}R \cr
& \Rightarrow \frac{{Gm}}{{4{R^3}}} = {\omega ^2} \Rightarrow \omega = \sqrt {\frac{{Gm}}{{4{R^3}}}} \cr} $$
If the velocity of the two particles with respect to the centre of gravity is $$v$$ then $$v = \omega R$$
$$v = \sqrt {\frac{{Gm}}{{4{R^3}}}} \times R = \sqrt {\frac{{Gm}}{{4R}}} $$