 JEE Mains Previous Years Questions with Solutions

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1

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
2

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
3

AIEEE 2002

Speeds of two identical cars are $u$ and $4$$u$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is
A
$1:1$
B
$1:4$
C
$1:8$
D
$1:16$

Explanation

Given that initial speed of two cars $u$ and $4u$ and final speed $v$ is 0 for both car. Both car is gradually slowing down so acceleration = $(-a)$

So formula becomes, 0 = ${u^2}$ - 2$a$s.

$\Rightarrow {u^2} = 2as$

For first car ${u^2} = 2a{s_1}$ ..........(i)

For second car ${\left( {4u} \right)^2} = 2a{s_2}$ ..........(ii)

Dividing $(i)$ and $(ii),$

${{{u^2}} \over {16{u^2}}} = {{2a{s_1}} \over {2a{s_2}}}$

$\Rightarrow {1 \over {16}} = {{{s_1}} \over {{s_2}}}$
4

AIEEE 2002

From a building two balls A and B are thrown such that A is thrown upwards and B downwards ( both vertically with the same speed ). If vA and vB are their respective velocities on reaching the ground, then
A
${v_B} > {v_A}$
B
${v_A} = {v_B}$
C
${v_A} > {v_B}$
D
their velocities depend on their masses.

Explanation Assume the initial velocity of each particle is = u

And height of building = h

If final velocity of A is vA then vA2 = u2 + 2(-g)(-h) = u2 + 2gh

If final velocity of B is vB then vB2 = u2 + 2gh

$\therefore$ vA = vB

Sign Rule : Take the direction of initial velocity positive opposite direction as negative.

Here for ball A initial velocity u is upward so upward is positive and downward is negative. That is why gravity is = - g and height = - h

And for ball B initial velocity u is downward so downward is positive and upward is negative. That is why gravity is = + g and height = + h