 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

Average density of the earth
A
is a complex function of $g$
B
does not depend on $g$
C
is inversely proportional to $g$
D
is directly proportional to $g$

Explanation

Mass of earth = Volume $\times$ Density of earth($\rho$)

$\therefore$ M = ${{4 \over 3}\pi {R^3}} $$\times \rho We know, g = {{GM} \over {{R^2}}} \Rightarrow g = {{G \times \rho \times {4 \over 3}\pi {R^3}} \over {{R^2}}} \Rightarrow g = {4 \over 3}\rho \pi GR \therefore g \propto \rho or \rho \propto g 2 AIEEE 2005 MCQ (Single Correct Answer) A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take G = 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}) A 3.33 \times {10^{ - 10}}\,J B 13.34 \times {10^{ - 10}}\,J C 6.67 \times {10^{ - 10}}\,J D 6.67 \times {10^{ - 9}}\,J Explanation We know, Work done = Difference in potential energy \therefore W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right] \Rightarrow$$W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}$

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

AIEEE 2004

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
4

AIEEE 2004

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

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