### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2011

Two bodies of masses $m$ and $4$ $m$ are placed at a distance $r.$ The gravitational potential at a point on the line joining them where the gravitational field is zero is:
A
$- {{4Gm} \over r}$
B
$- {{6Gm} \over r}$
C
$- {{9Gm} \over r}$
D
zero

## Explanation

Let the gravitational field at $P,$ distant $x$ from mass $m,$ be zero.

$\therefore$ ${{Gm} \over {{x^2}}} = {{4Gm} \over {{{\left( {r - x} \right)}^2}}}$

$\Rightarrow 4{x^2} = {\left( {r - x} \right)^2}$

$\Rightarrow 2x = r - x$

$\Rightarrow x = {r \over 3}$

Gravitational potential at point $P,$

$V = - {{Gm} \over {{r \over 3}}} - {{4Gm} \over {{{2r} \over 3}}} = -{{9Gm} \over r}$
2

### AIEEE 2009

The height at which the acceleration due to gravity becomes ${g \over 9}$ (where $g=$ the acceleration due to gravity on the surface of the earth) in terms of $R,$ the radius of the earth, is:
A
${R \over {\sqrt 2 }}$
B
$R/2$
C
$\sqrt 2 \,\,R$
D
$2\,R$

## Explanation

Given that, at height h from ground the acceleration due to gravity becomes ${g \over 9}$.

We know acceleration at earth surface due to gravity g = ${{GM} \over {{R^2}}}$

and acceleration at height h due to gravity g' = ${{GM} \over {{{\left( {R + h} \right)}^2}}}$

So ${{g} \over 9}$ = ${{GM} \over {{{\left( {R + h} \right)}^2}}}$

$\Rightarrow$ ${{g} \over 9}$ = ${{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} \right)}^2}}}$

= $g.{\left( {{R \over {R + h}}} \right)^2}$

$\Rightarrow {1 \over 9} = {\left( {{R \over {R + h}}} \right)^2}$

$\Rightarrow {R \over {R + h}} = {1 \over 3}$

$\Rightarrow 3R = R + h$

$\therefore$ $h = 2R$
3

### AIEEE 2008

This question contains Statement - $1$ and Statement - $2$. of the four choices given after the statements, choose the one that best describes the two statements.

Statement - $1$:

For a mass $M$ kept at the center of a cube of side $'a'$, the flux of gravitational field passing through its sides $4\,\pi \,GM.$

Statement - 2:

If the direction of a field due to a point source is radial and its dependence on the distance $'r'$ from the source is given as ${1 \over {{r^2}}},$ its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.
A
Statement - $1$ is false, Statement - $2$ is true
B
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is a correct explanation for Statement - $1$
C
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is not a correct explanation for Statement - $1$
D
Statement - $1$ is true, Statement - $2$ is false

## Explanation

Gravitational field $\overrightarrow g$ = $- {{GM} \over {{r^2}}}$

where, $M=$ mass enclosed in the closed surface

Gravitational flux through a closed surface is given by

${\left| {\overrightarrow g .d\overrightarrow S } \right|}$ = $4\pi {r^2}.{{GM} \over {{r^2}}}$ = $4\pi GM$

So Statement - 1 is correct.

Statement - 2 is also correct because when the shape of the earth is spherical, area of the Gaussian surface is $4\pi {r^2}$. This proves inverse square law.
4

### AIEEE 2008

A planet in a density solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11\,\,km\,{s^{ - 1}},$ the escape velocity from the surface of the planet would be
A
$1.1\,\,km\,{s^{ - 1}}$
B
$100\,\,km\,{s^{ - 1}}$
C
$110\,\,km\,{s^{ - 1}}$
D
$0.11\,\,km\,{s^{ - 1}}$

## Explanation

Let Me is mass of earth then mass of planet Mp = 10Me.

And let Re is radius of earth then radius of planet Rp = ${{{R_e}} \over {10}}$

Escape velocity of earth, ${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}}$

Escape velocity of planet, ${v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}}$

$\therefore$ ${{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}$

$= \sqrt {{{{M_p}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}}$

$= \sqrt {{{10{M_e}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{{\mathop{\rm R}\nolimits} _e}/10}}} = 10$

$\therefore$ ${{v_p}} = 10 \times {{v_e}}$

$= 10 \times 11 = 110\,km/s$