 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

The upper half of an inclined plane with inclination $\phi$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by
A
$2\,\cos \,\,\phi$
B
$2\,sin\,\,\phi$
C
$\,\tan \,\,\phi$
D
$2\,\tan \,\,\phi$

Explanation

Let the length of the inclined plane is = $l$. So only ${l \over 2}$ part will have friction.

According to work-energy theorem, $W = \Delta k = 0$ (Since initial and final speeds are zero)

$\therefore$ Work done by friction + Work done by gravity $=0$

i.e., $- \left( {\mu \,mg\,\cos \,\phi } \right){\ell \over 2} + mg\ell \,\sin \,\phi = 0$

or ${\mu \over 2}\cos \,\phi = \sin \phi$

or $\mu = 2\,\tan \,\phi$
2

AIEEE 2005

A bullet fired into a fixed target loses half of its velocity after penetrating $3$ $cm.$ How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
A
$2.0$ $cm$
B
$3.0$ $cm$
C
$1.0$ $cm$
D
$1.5$ $cm$

Explanation

Let $K$ be the initial kinetic energy and $F$ be the resistive force. Then according to work-energy theorem, $$W = \Delta K$$

i.e., $3F = {1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2}...\left( 1 \right)$

Let the bullet will penetrate x cm more before coming to rest.

$\therefore$ $Fx = {1 \over 2}m{\left( {{v \over 2}} \right)^2} - {1 \over 2}m{\left( 0 \right)^2}...\left( 2 \right)$

Dividing eq. $(1)$ and $(2)$ we get,

${x \over 3} = {1 \over 3}$ or x = 1 cm
3

AIEEE 2004

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that
A
its kinetic energy is constant
B
is acceleration is constant
C
its velocity is constant
D
it moves in a straight line

Explanation

Work done by such force is always zero when a force of constant magnitude always at right angle to the velocity of a particle when the motion of the particle takes place in a plane.

$\therefore$ From work-energy theorem, $\Delta K = 0$

$\therefore$ $K$ remains constant.
4

AIEEE 2004

A body of mass $' m ',$ acceleration uniformly from rest to $'{v_1}'$ in time ${T}$. The instantaneous power delivered to the body as a function of time is given by
A
${{m{v_1}{t^2}} \over {{T}}}$
B
${{mv_1^2t} \over {T^2}}$
C
${{m{v_1}t} \over {{T}}}$
D
${{mv_1^2t} \over {{T}}}$

Explanation

Assume acceleration of body be $a$

$\therefore$ ${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$

$\therefore$ $v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$

${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v$

$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$

$= m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$