### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2004

MCQ (Single Correct Answer)
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
${1 \over 4}mgR$
B
$2mgR$
C
${1 \over 2}mgR$
D
$mgR$

## Explanation

Gravitational potential energy on the earth surface of a body

U = $-{{GmM} \over R}$

And at the height h from the earth surface the potential energy

${U_h} = - {{GmM} \over {R + h}}$ = $- {{GmM} \over {2R}}$ [ as h = R ]

So the gain in the potential energy

$\Delta U = {U_h} - U$

$\therefore$ $\Delta U = {{ - GmM} \over {2R}} + {{GmM} \over R};$

$\Rightarrow$ $\Delta U = {{GmM} \over {2R}}$

Now ${{GM} \over {{R^2}}} = g;$ $\,\,\,$ $\therefore$ ${\mkern 1mu} {{GM} \over R} = gR$

$\therefore$ $\Delta U = {1 \over 2}mgR$
2

### AIEEE 2004

MCQ (Single Correct Answer)
A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
A
${{g{R^2}} \over {R + x}}$
B
${{gR} \over {R - x}}$
C
${gx}$
D
${\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}$

## Explanation

Gravitational force applied on the satellite,

= ${{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$

For satellite, gravitational force = centripetal force

$\therefore$ ${{m{v^2}} \over {\left( {R + x} \right)}} = {{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$

where $v$ is the orbital speed of satellite.

also $\,\,g = {{GM} \over {{R^2}}}$ $\Rightarrow GM = g{R^2}$

$\therefore$ ${v^2} = {{g{R^2}} \over {R + x}}$

$\Rightarrow v = {\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}$
3

### AIEEE 2003

MCQ (Single Correct Answer)
Two spherical bodies of mass $M$ and $5M$ & radii $R$ & $2R$ respectively are released in free space with initial separation between their centers equal to $12R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is
A
$2.5$ $R$
B
$4.5$ $R$
C
$7.5$ $R$
D
$1.5$ $R$

## Explanation

Let $t$ be the time taken for the two masses to collide and ${x_{5M,}}\,{x_M}$ be the distance travelled by the mass $5M$ and $M$ respectively.

The gravitational force acting between two sphere when the distance between them (12R - x) where x is a variable,

$F = {{GM \times 5M} \over {{{\left( {12R - x} \right)}^2}}}$

Acceleration of mass M, ${a_M} = {{G \times 5M} \over {{{\left( {12R - x} \right)}^2}}}$

Acceleration of mass 5M, ${a_{5M}} = {{GM} \over {{{\left( {12R - x} \right)}^2}}}$

For mass $5M$

$u = 0,\,\,S = {x_{5M}},\,\,t = t,\,\,a{ = a_{5M}}$

$S = ut + {1 \over 2}a{t^2}$

$\therefore$ ${x_{5M}} = {1 \over 2}{a_{5M}}{t^2}\,\,\,\,\,\,\,\,\,\,...\left( {i} \right)$

For mass $M$

$u = 0,\,\,s = {x_M},\,\,t = t,\,\,a = {a_M}$

$\therefore$ $s = ut + {1 \over 2}a{t^2} \Rightarrow$

${x_M} = {1 \over 2}{a_M}{t^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Dividing $(ii)$ by $(iii)$

${{{x_{5M}}} \over {{x_M}}} = {{{1 \over 2}{a_5}_M{t^2}} \over {{1 \over 2}{a_M}{t^2}}}$

$= {{{a_{5M}}} \over {{a_M}}} = {1 \over 5}$

$\therefore$ $5{x_{5M}} = {x_M}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left....( {iii} \right)$

From the figure,

${x_{5M}} + {x_M} = 12R - 2R - R = 9R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( iv \right)$

From $(iii)$ and $(iv)$

${{{x_M}} \over 5} + {x_M} = 9R$

$\therefore$ $6{x_M} = 45R$

$\therefore$ ${x_M} = {{45} \over 6}R = 7.5R$

So two sphere collide when the sphere of mass M covered the distance of 7.5R.
4

### AIEEE 2003

MCQ (Single Correct Answer)
The escape velocity for a body projected vertically upwards from the surface of earth is $11$ $km/s.$ If the body is projected at an angle of ${45^ \circ }$ with the vertical, the escape velocity will be
A
$11\sqrt 2 \,\,km/s$
B
$22$ $km/s$
C
$11$ $km/s$
D
${{11} \over {\sqrt 2 }}km/s$

## Explanation

We know, Escape velocity, ${v_e} = \sqrt {2gR}$

So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.

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