### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2007

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is
A
v0 + g/2 + f
B
v0 + 2g + 3f
C
v0 + g/2 + f/3
D
v0 + g + f

## Explanation

Given that, v = v0 + gt + ft2

We know that, $v = {{dx} \over {dt}}$

$\Rightarrow dx = v\,dt$

Integrating, $\int\limits_0^x {dx} = \int\limits_0^t {v\,dt}$

or $\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t$

or $\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}$

At $t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.$
2

### AIEEE 2007

A particle is projected at $60^\circ$ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is
A
K/2
B
K
C
Zero
D
K/4

## Explanation

Let $u$ be the velocity with which the particle is thrown and $m$ be the mass of the particle. Then

$KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

At the highest point the velocity is $u$ $\cos \,{60^ \circ }$ (only the horizontal component remains, the vertical component being zero at the top-most point).

Therefore kinetic energy at the highest point,

${\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ$

$\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}$ [ From eq $(1)$ ]
3

### AIEEE 2006

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity 'v' that varies as $v = \alpha \sqrt x$. The displacement of the particle varies with time as
A
t2
B
t
C
t1/2
D
t3

## Explanation

Given $v = \alpha \sqrt x$

$\therefore$ ${{dx} \over {dt}} = \alpha \sqrt x$

$\Rightarrow {{dx} \over {\sqrt x }} = \alpha \,dt$

On integrating we get,

$\int\limits_0^x {{{dx} \over {\sqrt x }} = \alpha \int\limits_0^t {dt} }$

$\Rightarrow \left[ {{{2\sqrt x } \over 1}} \right]_0^x = \alpha \left[ t \right]_0^t$

$\Rightarrow 2\sqrt x = \alpha t$

$\Rightarrow x = {{{\alpha ^2}} \over 4}{t^2}$

So $x \propto {t^2}$
4

### AIEEE 2005

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
A
${1 \over 2}m{s^{ - 2}}$ towards north
B
${1 \over {\sqrt 2 }}m{s^{ - 2}}$ towards north-east
C
${1 \over {\sqrt 2 }}m{s^{ - 2}}$ towards north-west
D
zero

## Explanation

Average acceleration

$= {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{\rm int}} erval}}$

$= {{\Delta \overrightarrow v } \over t}$

${\overrightarrow v _1} = +5\widehat i$, towards east direction.

$\overrightarrow {{v_2}} = +5\widehat j$, towards north direction

$\Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}}$ = $5\widehat i - 5\widehat j$

$\therefore$ $\,\,\,\,\overrightarrow a = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}$

$\therefore$ $\,\,\,\, a = {{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}$

$\tan \theta = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1$

$\therefore$ $\,\,\,\,\theta = {45^ \circ }$

Therefore the direction is North-west.