 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2011

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :
${{dv} \over {dt}} = - 2.5\sqrt v$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :
A
2 s
B
4 s
C
8 s
D
1 s

Explanation

Given ${{dv} \over {dt}} = - 2.5\sqrt v$

$\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt$

On integrating, $\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}}} \,dv = - 2.5\int_0^t {dt}$

$\Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}}} \over {\left( {{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t$

$\Rightarrow - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}} = - 2.5t$

$\Rightarrow t = 2\,sec$
2

AIEEE 2010

A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length $s = {t^3} + 5,$ where $s$ is in metres and $t$ is in seconds. The radius of the path is $20$ $m.$ The acceleration of $'P'$ when $t=2$ $s$ is nearly. A
$13m/{s_2}$
B
$12m/{s^2}$
C
$7.2m{s^2}$
D
$14m/{s^2}$

Explanation

Given $s = {t^3} + 5$

$\Rightarrow$ Speed,$\,\,\,v = {{ds} \over {dt}} = 3{t^2}$

Tangential acceleration ${a_t} = {{dv} \over {dt}} = 6t$

Radial acceleration ${a_c} = {{{v^2}} \over R} = {{9{t^4}} \over R}$

At $\,\,\,\,t = 2s,\,\,{a_t} = 6 \times 2 = 12\,\,m/{s^2}$

${a_c} = {{9 \times 16} \over {20}} = 7.2\,\,m/{s^2}$

$\therefore$ Net acceleration

$= \sqrt {a_t^2 + a_c^2}$

$= \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}}$

$= \sqrt {144 + 51.84}$

$= \sqrt {195.84}$

$= 14\,m/{s^2}$
3

AIEEE 2010

A small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity ${v_0}$ in the $x$-$y$ plane as shown in the figure. At a time $t < {{{v_0}\sin \theta } \over g},$ the angular momentum of the particle is ................, where $\widehat i,\widehat j$ and $\widehat k$ are unit vectors along $x,y$ and $z$-axis respectively.
A
$- mg\,{v_0}{t^2}\cos \theta \widehat j$
B
$mg\,{v_0}t\cos \theta \widehat k$
C
$- {1 \over 2}mg\,{v_0}{t^2}\cos \,\theta \widehat k$
D
${1 \over 2}mg\,{v_0}{t^2}\cos \theta \widehat i$

Explanation

Position vector of the particle from the original at any time t is

$\overrightarrow r = {v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j$

As Velocity vector, $\overrightarrow v = {{d\overrightarrow r } \over {dt}}$

$\therefore$ $\overrightarrow v = {d \over {dt}}\left( {{v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j} \right)$

$= {v_0}\cos \theta \widehat i + \left( {{v_0}\sin \theta - gt} \right)\widehat j$

Angular momentum of the particle about the origin is

$\overrightarrow L = m\left( {\overrightarrow r \times \overrightarrow v } \right)$

\eqalign{ & \overrightarrow L = m\left[ {{v_0}\,\cos \theta t\widehat i + \left( {{v_0}\,\sin \theta t - {1 \over 2}g{t^2}} \right)\widehat j} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left[ {{v_0}\,\cos \,\theta \widehat i + \left( {{v_0}\,\sin \theta - gt} \right)\widehat j} \right] \cr}

$= m{v_0}\cos \theta t\left[ { - {1 \over 2}gt} \right]\widehat k$

$= - {1 \over 2}mg{v_0}{t^2}\cos \theta \widehat k$
4

AIEEE 2010

For a particle in uniform circular motion the acceleration $\overrightarrow a$at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)
A
$- {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j$
B
$- {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j$
C
$- {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$
D
${{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j$

Explanation

For a particle in uniform circular motion,

${a_c} = {{{v^2}} \over R}$ towards the center of the circle

From figure, $\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$

$= {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$ 