### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2012

Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is
A
m1r1 : m2r2
B
m1 : m2
C
r1 : r2
D
1 : 1

## Explanation

We know, $a = r\,{w^2} = r \times {\left( {{{2\pi } \over T}} \right)^2}$

Given, ${T_1} = {T_2} = T$

${a_1} = {r_1} \times {\left( {{{2\pi } \over T}} \right)^2}$

${a_2} = {r_2} \times {\left( {{{2\pi } \over T}} \right)^2}$

$\therefore$ ${{{a_1}} \over {{a_2}}} = {{{r_1}} \over {{r_2}}}$
2

### 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 $P,$

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

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

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

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

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

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

$\Rightarrow 3R = R + h$

$\therefore$ $h = 2R$
4

### 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.