### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2002

When forces ${F_1},\,\,{F_2},\,\,{F_3}$ are acting on a particle of mass $m$ such that ${F_2}$ and ${F_3}$ are mutually perpendicular, then the particle remains stationary. If the force ${F_1}$ is now removed then the acceleration of the particle is
A
${F_1}/m$
B
${F_2}{F_3}/m{F_1}$
C
$\left( {F{}_2 - {F_3}} \right)/m$
D
${F_2}/m$

## Explanation

When ${F_1},{F_2}$ and ${F_3}$ are acting on a particle then the particle remains stationary. This means that the resultant of ${F_1},{F_2}$ and ${F_3}$ is zero. When ${F_1}$ is removed then particle will start moving due to the force ${F_2}$ and ${F_3}$ in the resultant of ${F_2}$ and ${F_3}$ and it should be equal and opposite to ${F_1}.$
$i.e.$ $\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|$
$\therefore$ $\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}$
2

### AIEEE 2002

A lift is moving down with acceleration $a.$ A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively
A
$g,g$
B
$g-a, g-a$
C
$g-a, g$
D
$a, g$

## Explanation

Let acceleration of ball = ${\overrightarrow a _b}$ and acceleration of man is = ${\overrightarrow a _m}$

With respect to the man standing in the lift, the acceleration of the ball

${\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}$

$\Rightarrow {a_{bm}} = g - a$

Where $a$ is the acceleration of the man as the acceleration of the lift is $a$.

With respect to the man standing on the ground the acceleration of the ball

${\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}$

$\Rightarrow {a_{bm}} = g - 0$

$= g$