### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in ${T \over 3}$ seconds?
A
${{8h} \over 9}$ meters from the ground
B
${{7h} \over 9}$ meters from the ground
C
${h \over 9}$ meters from the ground
D
${{7h} \over {18}}$ meters from the ground

## Explanation

We know that equation of motion, $s = ut + {1 \over 2}g{t^2},\,\,$

Initial speed of ball is zero and it take T second to reach the ground.

$\therefore$ $h = {1 \over 2}g{T^2}$

After $T/3$ second, vertical distance moved by the ball

$h' = {1 \over 2}g{\left( {{T \over 3}} \right)^2}$

$\Rightarrow h' = {1 \over 2} \times {{8{T^2}} \over 9}$

$= {h \over 9}$

$\therefore$ Height from ground

$= h - {h \over 9} = {{8h} \over 9}$
2

### AIEEE 2003

The co-ordinates of a moving particle at any time 't' are given by x = $\alpha$t3 and y = βt3. The speed to the particle at time 't' is given by
A
$3t\sqrt {{\alpha ^2} + {\beta ^2}}$
B
$3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}}$
C
${t^2}\sqrt {{\alpha ^2} + {\beta ^2}}$
D
$\sqrt {{\alpha ^2} + {\beta ^2}}$

## Explanation

Given that $x = \alpha {t^3}\,\,\,\,$ and $\,\,\,\,y = \beta {t^3}$

$\therefore$ ${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$

and$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$

$\therefore$ $v = \sqrt {v_x^2 + v_y^2}$

$= \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}}$

$= 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}}$
3

### AIEEE 2003

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $30^\circ$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $\left[ {g = 10m/{s^2},\sin 30^\circ = {1 \over 2},\cos 30^\circ = {{\sqrt 3 } \over 2}} \right]$
A
5.20 m
B
4.33 m
C
2.60 m
D
8.66 m

## Explanation

From the figure it is clear that maximum horizontal range

$R = {{{u^2}\sin 2\theta } \over g}$

$= {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}$

$= 5\sqrt 3$ = 8.66 m
4

### AIEEE 2003

A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is
A
12 m
B
18 m
C
24 m
D
6 m

## Explanation

For case 1 :
u = 50 km/hr = ${{50 \times 1000} \over {3600}}$ m/s = ${{125} \over 9}$ m/s, v = 0, s = 6 m, $a$ = ?

$\therefore$ 02 = u2 + 2$a$s

$\Rightarrow$ $a = - {{{u^2}} \over {2s}}$

$\Rightarrow$ $a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}$ = $-$16 m/s2

For case 2 :
u = 100 km/hr = ${{100 \times 1000} \over {3600}}$ m/s = ${{250} \over 9}$ m/s, v = 0, $a$ = $-$16, s = ?

$\therefore$ 02 = u2 + 2$a$s

$\Rightarrow$ $s = - {{{u^2}} \over {2a}}$

$\Rightarrow$ $s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}$ = 24 m