 JEE Mains Previous Years Questions with Solutions

4.5
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1

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 $2R$)

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

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

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

AIEEE 2002

The escape velocity of a body depends upon mass as
A
${m^0}$
B
${m^1}$
C
${m^2}$
D
${m^3}$

Explanation

Escape velocity,

${v_e} = \sqrt {2gR} = \sqrt {{{2GM} \over R}} \Rightarrow {V_e}\, \propto \,{m^0}$

Where $M,R$ are the mass and radius of the planet respectively. In this expression the mass of the body $(m)$ is not present. The escape velocity is independent of the mass m or it depends on m0.

3

AIEEE 2002

If suddenly the gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the satellite will
A
continue to move in its orbit with same velocity
B
move tangentially to the original orbit with the same velocity
C
become stationary in its orbit
D
move towards the earth

Explanation

When gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the centripetal force becomes zero. So the satellite will move tangentially to the original orbit with the same velocity as it has at the instant when gravitational force becomes zero.
4

AIEEE 2002

The kinetic energy needed to project a body of mass $m$ from the earth surface (radius $R$) to infinity is
A
$mgR/2$
B
$2mgR$
C
$mgR$
D
$mgR/4$

Explanation

The required velocity is called escape velocity (${v_e}$) to leave the earth surface of a body.

${v_e} =$ escape velocity $= \sqrt {2gR}$

Kinetic Energy $K.E = {1 \over 2}mv_e^2$

$\therefore$ $K.E = {1 \over 2}m \times 2gR = mgR$