JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline)

A projectile is given an initial velocity of $\left( {\widehat i + 2\widehat j} \right)$ m/s, where ${\widehat i}$ is along the ground and ${\widehat j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is:
A
y = x - 5x2
B
y = 2x - 5x2
C
4y = 2x - 5x2
D
4y = 2x - 25x2

Explanation

$\overrightarrow u = \widehat i + 2\widehat j = {u_x}\widehat i + {u_y}\widehat j$

$\Rightarrow u\cos \theta = 1,u\sin \theta = 2$

Also $x = {u_x}t$ and

$y = {u_y}t - {1 \over 2}g{t^2}$

$\Rightarrow$ $y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {u_x^2}}$

$\therefore$ $y = 2x - {1 \over 2}g{x^2} = 2x - 5{x^2}$
2

AIEEE 2012

A particle of mass $m$ is at rest at the origin at time $t=0.$ It is subjected to a force $F\left( t \right) = {F_0}{e^{ - bt}}$ in the $x$ direction. Its speed $v(t)$ is depicted by which of the following curves?
A
B
C
D

Explanation

Given that $F\left( t \right) = {F_0}{e^{ - bt}}$

$\Rightarrow m{{dv} \over {dt}} = {F_0}{e^{ - bt}}$

${{dv} \over {dt}} = {{{F_0}} \over m}{e^{ - bt}}$

$\Rightarrow \int\limits_0^v {dv} = {{{F_0}} \over m}\int\limits_0^t {{e^{ - bt}}} \,dt$

$v = {{{F_0}} \over m}\left[ {{{{e^{ - bt}}} \over { - b}}} \right]_0^t = {{{F_0}} \over {mb}}\left[ { - \left( {{e^{ - bt}} - {e^{ - 0}}} \right)} \right]$

$\Rightarrow v = {{{F_0}} \over {mb}}\left[ {1 - {e^{ - bt}}} \right]$

$\Rightarrow v_{max} = {{{F_0}} \over {mb}}$
3

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
4

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

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

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

$\Rightarrow t = 2\,sec$