 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

The mass of a spaceship is $1000$ $kg.$ It is to be launched from the earth's surface out into free space. The value of $g$ and $R$ (radius of earth ) are $10\,m/{s^2}$ and $6400$ $km$ respectively. The required energy for this work will be:
A
$6.4 \times {10^{11}}\,$ Joules
B
$6.4 \times {10^8}\,$ Joules
C
$6.4 \times {10^9}\,$ Joules
D
$6.4 \times {10^{10}}\,$ Joules

Explanation

Potential energy at earth surface = $- {{GMm} \over R}$

and at free space potential energy = 0

Work done for this = $0 - \left( { - {{GMm} \over R}} \right)$ = ${{{GMm} \over R}}$

So the required energy for this work is

= ${{GMm} \over R}$

=${{{g{R^2}m} \over R}}$ [ as $g = {{GM} \over {{R^2}}}$ ]

= $mgR$

$= 1000 \times 10 \times 6400 \times {10^3}$

$= 6.4 \times {10^{10}}\,\,$ Joules
2

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

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

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

$\therefore$ ${{{a_1}} \over {{a_2}}} = {{{r_1}} \over {{r_2}}}$
3

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 $P,$

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

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

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

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

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

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

$\Rightarrow 3R = R + h$

$\therefore$ $h = 2R$