### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:
A
2gH = n2u2
B
gH = (n - 2)2u2
C
2gH = nu2(n - 2)
D
gH = (n - 2)u2

## Explanation

Time taken to reach highest point is $t = {u \over g}$

Time taken by the particle to reach the ground = $nt = {nu \over g}$

Speed on reaching ground $v = \sqrt {{u^2} + 2gH}$

Now, $v = u + at$

$\Rightarrow \sqrt {{u^2} + 2gH} = - u + gt$

$\Rightarrow t = {{u + \sqrt {{u^2} + 2gH} } \over g} = {{nu} \over g}$ (from question)

$\Rightarrow 2gH = n\left( {n - 2} \right){u^2}$
2

### 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

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

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

$\therefore$ $y = 2x - {1 \over 2}g{x^2} = 2x - 5{x^2}$
3

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

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

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

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

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

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

$\Rightarrow v_{max} = {{{F_0}} \over {mb}}$
4

### 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