 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2002

Three identical blocks of masses $m=2$ $kg$ are drawn by a force $F=10.2$ $N$ with an acceleration of $0.6$ $m{s^{ - 2}}$ on a frictionless surface, then what is the tension (in $N$) in the string between the blocks $B$ and $C$? A
$9.2$
B
$3.4$
C
$4$
D
$7.8$

Explanation Assume tension between block A and B is T1 and block B and C is T2. Acceleration a = 0.6 m/s2.

For block A :

F - T1 = ma

For block B :

T1 - T2 = ma

By adding both those equation we get,

F - T2 = 2ma

T2 = F - 2ma

T2 = 10.2 - 2$\times$2$\times$0.6 = 10.2 - 2.4 = 7.8 N
2

AIEEE 2002

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}$$a Adding those equations we get a = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}} \therefore {g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}} \Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}} \Rightarrow {{{m_1}} \over {{m_2}}} + 1 = {8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)} \Rightarrow {{{m_1}} \over {{m_2}}} = {9 \over 7} 3 AIEEE 2002 MCQ (Single Correct Answer) Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are A 12N, 6N B 13N, 5N C 10N, 8N D 16N, 2N. Explanation Let the two forces be {F_1} and {F_2} and let {F_2} is smaller than {F_1} and assume R is the resultant force. Given {F_1} + {F_2} = 18 \,\,\,\,\,\, ....(i) From the right angle triangle, F_2^2 + {R^2} = F_1^2 or F_1^2 - F_2^2 = {R^2} or \left( {{F_1} + {F_2}} \right)$$\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 ${{F_2} = 5\,N}$ 4

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 ${F_1}.$
$i.e.$ $\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|$
$\therefore$ $\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}$