 JEE Mains Previous Years Questions with Solutions

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1

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
2

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.
3

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 $a$

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

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

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

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

$= m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$
4

AIEEE 2004

A force $\overrightarrow F = \left( {5\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right)N$ is applied over a particle which displaces it from its origin to the point $\overrightarrow r = \left( {2\overrightarrow i - \overrightarrow j } \right)m.$ The work done on the particle in joules is
A
$+10$
B
$+7$
C
$-7$
D
$+13$

Explanation

Work done when the particle displaces from the origin,

$W = \overrightarrow F .\overrightarrow x$

$= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)$

$=10-3=7$ J