JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2002

MCQ (Single Correct Answer)
A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)
A
solid sphere
B
hollow sphere
C
ring
D
all same

Explanation

Each bodies is sliding along the frictionless inclined plane and there is no rolling, therefore the acceleration of all the bodies is same $\left( {g\,\sin \,\theta } \right).$
2

AIEEE 2002

MCQ (Single Correct Answer)
Initial angular velocity of a circular disc of mass $M$ is ${\omega _1}.$ Then two small spheres of mass $m$ are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?
A
$\left( {{{M + m} \over M}} \right)\,\,{\omega _1}$
B
$\left( {{{M + m} \over m}} \right)\,\,{\omega _1}$
C
$\left( {{M \over {M + 4m}}} \right)\,\,{\omega _1}$
D
$\left( {{M \over {M + 2m}}} \right)\,\,{\omega _1}$

Explanation

When two small spheres of mass $m$ are attached gently, the external torque, about the axis of rotation, is zero.

So, ${{d\overrightarrow L } \over {dt}} = \overline z$ = 0

$\overrightarrow L$ = conserved

So the angular momentum about the axis of rotation is conserved.

$\therefore$ ${I_1}{\omega _1} = {I_2}{\omega _2}$

$\Rightarrow {\omega _2} = {{{I_1}} \over {{I_2}}}{\omega _1}$

Here Moment of inertia of Disc ${I_1} = {1 \over 2}M{R^2}$ and

After adding two sphere Moment of Inertia of disc and two sphere,

${I_2} = {1 \over 2}M{R^2} +$$2\left( {{1 \over 2}m{R^2} + {1 \over 2}m{R^2}} \right)$

$\therefore$ ${\omega _2} = {{{1 \over 2}M{R^2}} \over {{1 \over 2}MR + 2m{R^2}}} \times {\omega _1} = {M \over {M + 4m}}{\omega _1}$

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