### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2004

A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to
A
R
B
${1 \over R}$
C
${1 \over {{R^2}}}$
D
${R^2}$

## Explanation

Range is same for angle of projection $\theta ,$ and ${90^ \circ } - \theta$

${T_1} = {{2u\sin \theta } \over g},\,\,{T_2} = {{2u\cos \theta } \over g}$

${T_1}{T_2} =$ ${{4{u^2}\sin \theta \cos \theta } \over {{g^2}}}$

= ${2 \over g} \times \left( {{{{u^2}\sin 2\theta } \over g}} \right)$

= ${{2R} \over g}$

(as $R = $${{{{u^2}\sin 2\theta } \over g}} ) Hence, {T_1}{T_2} is proportional to R. 2 ### AIEEE 2004 MCQ (Single Correct Answer) If \overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A , then the angle beetween A and B is A {\pi \over 2} B {\pi \over 3} C \pi D {\pi \over 4} ## Explanation \overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A \overrightarrow A \times \overrightarrow B - \overrightarrow B \times \overrightarrow A = 0 \Rightarrow \overrightarrow A \times \overrightarrow B + \overrightarrow A \times \overrightarrow B = 0 \therefore \overrightarrow A \times \overrightarrow B = 0 \Rightarrow AB\sin \theta = 0 \theta = 0,\pi ,\,\,$$2\pi$ ........

from the given options, $\theta = \pi$
3

### 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

$= h - {h \over 9} = {{8h} \over 9}$
4

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

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

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

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