### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2002

MCQ (Single Correct Answer)
A light string passing over a smooth light pulley connects two blocks of masses ${m_1}$ and ${m_2}$ (vertically). If the acceleration of the system is $g/8$, then the ratio of the masses is
A
$8:1$
B
$9:7$
C
$4:3$
D
$5:3$

## Explanation

Assume that, mass m1 is greater than mass m2, so the heavier mass m1 is accelerating downward and the lighter mass m2 is accelerating upwards.

For mass ${m_1}$ the equation will be

${m_1}$$g-T=$${m_1}$$a For mass {m_2} the equation will be T-$${m_2}$$g=$${m_2}$$a Adding those equations we get a = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}} \therefore {g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}} \Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}} \Rightarrow {{{m_1}} \over {{m_2}}} + 1 = {8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)} \Rightarrow {{{m_1}} \over {{m_2}}} = {9 \over 7} 2 ### AIEEE 2002 MCQ (Single Correct Answer) Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are A 12N, 6N B 13N, 5N C 10N, 8N D 16N, 2N. ## Explanation Let the two forces be {F_1} and {F_2} and let {F_2} is smaller than {F_1} and assume R is the resultant force. Given {F_1} + {F_2} = 18 \,\,\,\,\,\, ....(i) From the right angle triangle, F_2^2 + {R^2} = F_1^2 or F_1^2 - F_2^2 = {R^2} or \left( {{F_1} + {F_2}} \right)$$\left( {{F_1} - {F_2}} \right)$ = ${R^2}$

or $\left( {18} \right)\left( {{F_1} - {F_2}} \right)$ = ${\left( {12} \right)^2}$ = 144

or $\left( {{F_1} - {F_2}} \right) = 8$ $\,\,\,\,\,\,$ ....$(ii)$

By solving equation $(i)$ and $(ii)$ we get,

${{F_1} = 13\,N}$ and ${{F_2} = 5\,N}$
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### AIEEE 2002

MCQ (Single Correct Answer)
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 ${F_1}.$
$i.e.$ $\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|$
$\therefore$ $\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}$
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### AIEEE 2002

MCQ (Single Correct Answer)
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

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

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

$= g$

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