### JEE Mains Previous Years Questions with Solutions

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1

### 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 $(1)$ ]
2

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

The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is
A
2bv3
B
-2abv2
C
2av2
D
-2av3

## Explanation

Given $t = a{x^2} + bx;$

Differentiate with respect to time $(t)$

${d \over {dt}}\left( t \right) = a{d \over {dt}}\left( {{x^2}} \right) + b{{dx} \over {dt}}$

$= a.2x{{dx} \over {dt}} + b.{{dx} \over {dt}}$

$1 = 2axv + bv = v\left( {2ax + b} \right)$

[as ${{dx} \over {dt}} = v$]

$\Rightarrow 2ax + b = {1 \over v}.$

Again differentiating,

$2a{{dx} \over {dt}} + 0 = - {1 \over {{v^2}}}{{dv} \over {dt}}$

$\Rightarrow {{dv} \over {dt}} = f = - 2a{v^3}$

(as ${{dv} \over {dt}} = f$ = Acceleration )