### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed?
A
The velocity vector is tangent to the circle.
B
The acceleration vector is tangent to the circle.
C
The acceleration vector points to the centre of the circle.
D
The velocity and acceleration vectors are perpendicular to each other.

## Explanation

Only option $(b)$ is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle.
2

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

from the given options, $\theta = \pi$
4

### 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

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