### JEE Mains Previous Years Questions with Solutions

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1

### 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,

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

A particle is moving with velocity $\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$, where K is a constant. The general equation for its path is
A
y = x2 + constant
B
y2 = x + constant
C
xy = constant
D
y2 = x2 + constant

## Explanation

$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$ ........(1)

Also $\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$

$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$ ........(2)

Equating (1) and (2), we get

${{dx} \over {dt}} = ky\,\,\,\,\,\,$ .......(3)

and $\,\,\,\,\,{{dy} \over {dt}} = kx$ ......(4)

Dividing (3) and (4), we get

${{dy} \over {dx}} = {x \over y}$

$\Rightarrow ydy = xdx$

Integrating both sides of above equation, we get

$\int {ydy} = \int {xdx}$

$\Rightarrow {y^2} = {x^2} +$ constant