### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

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.
2

### AIEEE 2003

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.
3

### AIEEE 2003

The time period of satellite of earth is $5$ hours. If the separation between the earth and the satellite is increased to $4$ times the previous value, the new time period will become
A
$10$ hours
B
$80$ hours
C
$40$ hours
D
$20$ hours

## Explanation

According to kepler's law,

${T^2} \propto {R^3}$

$\therefore$ ${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$

$\Rightarrow$ ${T_2} = {T_1}{\left( {{{{R_2}} \over {{R_1}}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 3} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}} = 5 \times {\left[ {{{4R} \over R}} \right]^{{\raise0.5ex\hbox{\scriptstyle 3} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$

$= 5 \times {2^3} = 40\,\,$ hour
4

### AIEEE 2002

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$ is
A
${{GMm} \over {12{R^2}}}$
B
${{GMm} \over {3{R^2}}}$
C
${{GMm} \over {8R}}$
D
${{GMm} \over {6R}}$

## Explanation

Gravitational potential energy E = $- {{GMm} \over r}$

where M = mass of earth

m = mass of body

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$

$=$ (Potential energy of the Earth-mass system when mass is at distance $3R$ ) $-$ (Potential energy of the Earth-mass system when mass is at distance $2R$)

$= {{ - GMm} \over {3R}} - \left( {{{ - GMm} \over {2R}}} \right)$

$= {{ - GMm} \over {3R}} + {{GMm} \over {2R}}$

$= {{ - 2GMm + 3GMm} \over {6R}} = {{GMm} \over {6R}}$

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