 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

If $\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A$, then the angle beetween A and B is
A
${\pi \over 2}$
B
${\pi \over 3}$
C
$\pi$
D
${\pi \over 4}$

Explanation

$\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A$

$\overrightarrow A \times \overrightarrow B - \overrightarrow B \times \overrightarrow A = 0$

$\Rightarrow \overrightarrow A \times \overrightarrow B + \overrightarrow A \times \overrightarrow B = 0$

$\therefore$ $\overrightarrow A \times \overrightarrow B = 0$

$\Rightarrow AB\sin \theta = 0$

$\theta =$ $0,\pi ,\,\,$$2\pi$ ........

from the given options, $\theta = \pi$
2

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

$= h - {h \over 9} = {{8h} \over 9}$
3

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

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

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

$= 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}}$
4

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