### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### 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 $a$

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

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

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

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

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

### 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
3

### AIEEE 2004

A uniform chain of length $2$ $m$ is kept on a table such that a length of $60$ $cm$ hangs freely from the edge of the table. The total mass of the chain is $4$ $kg.$ What is the work done in pulling the entire chain on the table?
A
$12$ $J$
B
$3.6$ $J$
C
$7.2$ $J$
D
$1200$ $J$

## Explanation

Mass of hanging part $(m') = {4 \over 2} \times \left( {0.6} \right)kg$ = 1.2 kg

Let at the surface $PE=0$

Center of mass of hanging part $=0.3$ $m$ below the surface of the table

${U_i} = - m'gx = - 1.2 \times 10 \times 0.30$ = - 3.6 J

$\Delta U = m'gx = 3.6 J =$ Work done in putting the entire chain on the table.
4

### AIEEE 2004

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to
A
$x$
B
${e^x}$
C
${x^2}$
D
${\log _e}x$

## Explanation

Given that, retardation $\propto$ displacement

$\Rightarrow$ $a=-kx$

But we know $a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$

$\therefore$ ${{vdv} \over {dx}} = - kx$

$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx}$

$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$

$\Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$

$\therefore$ Loss in kinetic energy is proportional to ${x^2}$.

$\therefore$ $\Delta K \propto {x^2}$

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET

Class 12