 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2005

The change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $d$ below the surface of earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct?
A
$d = {{3h} \over 2}$
B
$d = {h \over 2}$
C
$d = h$
D
$d = 2\,h$

Explanation

At height h acceleration due to gravity, ${g_h} = g\left[ {1 - {{2h} \over R}} \right];$

At depth d acceleration due to gravity, ${g_d} = g\left[ {1 - {d \over R}} \right]$

According to the question,

${g_h} = {g_d}$

$\therefore$ $g\left[ {1 - {{2h} \over R}} \right]$ = $g\left[ {1 - {d \over R}} \right]$

$\Rightarrow {{2hg} \over R} = {{dg} \over R}$

$\Rightarrow$ $d=2h$
2

AIEEE 2005

A particle of mass $10$ $g$ is kept on the surface of a uniform sphere of mass $100$ $kg$ and radius $10$ $cm.$ Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $G$ $= 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}$)
A
$3.33 \times {10^{ - 10}}\,J$
B
$13.34 \times {10^{ - 10}}\,J$
C
$6.67 \times {10^{ - 10}}\,J$
D
$6.67 \times {10^{ - 9}}\,J$

Explanation

We know, Work done = Difference in potential energy

$\therefore$ $W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]$

$\Rightarrow$$W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}$

$= 6.67 \times {10^{ - 10}}J$
3

AIEEE 2004

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to
A
${R^n}$
B
${R^{\left( {{{n - 1} \over 2}} \right)}}$
C
${R^{\left( {{{n + 1} \over 2}} \right)}}$
D
${R^{\left( {{{n - 2} \over 2}} \right)}}$

Explanation

For moving a planet around the sun in the circular orbit,

The necessary centripetal force = Gravitational force exerted on it

$\therefore$ ${{m{v^2}} \over r} = {{GMm} \over {{R^n}}}$

$\Rightarrow$ $v = \sqrt {{{GM} \over {{R^{n - 1}}}}}$

We know, $T = {{2\pi R} \over v}$

$= 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}}$

= $2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}}$

= $2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}$

$\therefore$ $T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$
4

AIEEE 2004

The time period of an earth satellite in circular orbit is independent of
A
both the mass and radius of the orbit
B
C
the mass of the satellite
D
neither the mass of the satellite nor the radius of its orbit

Explanation

For satellite, gravitational force = centripetal force

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

$x=$ height of satellite from earth surface

$m=$ mass of satellite

$\Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$ or $v = \sqrt {{{GM} \over {R + x}}}$

We know, $T = {{2\pi } \over \omega }$

$T = {{2\pi \left( {R + x} \right)} \over v}$ [ as $\omega = {v \over r}$ ]

$= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}$

which is independent of mass of satellite