 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be
A
$20\sqrt 2$ m
B
10 m
C
$10\sqrt 2$ m
D
20 m

Explanation

We know, $R = {{{u^2}{{\sin }2}\theta } \over g}$ and $H = {{{u^2}{{\sin }^2}\theta } \over {2g}};$

${H_{\max }}\,\,$ is possible when $\theta = 90$$^\circ$

${H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2$

As $R = {{{u^2}\sin 2\theta } \over g}$

Range is maximum when projectile is thrown at an angle $45^\circ$.

$\Rightarrow {R_{\max }} = {{{u^2}} \over g}$

${R_{\max }} = {{10 \times g \times 2} \over g} = 20$ meter
2

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

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

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

$\Rightarrow t = 2\,sec$
3

AIEEE 2011

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :
A
$\pi {{{v^4}} \over {{g^2}}}$
B
${\pi \over 2}{{{v^4}} \over {{g^2}}}$
C
$\pi {{{v^2}} \over {{g^2}}}$
D
$\pi {{{v^2}} \over g}$

Explanation

Maximum range of water coming out of fountain,

${R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}$

Total area around fountain,

$A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}$
4

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

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

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

$= \sqrt {144 + 51.84}$

$= \sqrt {195.84}$

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