 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

From a solid sphere of mass $M$ and radius $R$ a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is:
A
${{4M{R^2}} \over {9\sqrt {3\pi } }}$
B
${{4M{R^2}} \over {3\sqrt {3\pi } }}$
C
${{M{R^2}} \over {32\sqrt {2\pi } }}$
D
${{M{R^2}} \over {16\sqrt {2\pi } }}$

Explanation From the diagram you can see diagonal of the cube is equal to the diameter of the sphere.

$\therefore$ $\sqrt 3 a = 2R$

$\Rightarrow$ $a = {2 \over {\sqrt 3 }}R$

Let M is the mass of sphere and M' is the mass of the cube.

So ${M \over {M'}} = {{{4 \over 3}\pi {R^3}} \over {{a^3}}}$

$= {{{4 \over 3}\pi {R^3}} \over {{{\left( {{2 \over {\sqrt 3 }}R} \right)}^3}}} = {{\sqrt 3 } \over 2}\pi$

$\therefore$ $M' = {{2M} \over {\sqrt 3 \pi }}$

Moment of inertia of the cube about the axis passing through its center and perpendicular to one of its face,

$I = {{M'{a^2}} \over 6} = {{{{2M} \over {\sqrt 3 \pi }} \times \left( {{2 \over {\sqrt 3 }}R} \right)} \over 6}$

$= {{4M{R^2}} \over {9\sqrt 3 \pi }}$
2

JEE Main 2015 (Offline)

Distance of the center of mass of a solid uniform cone from its vertex is $z{}_0$. If the radius of its base is $R$ and its height is $h$ then $z{}_0$ is equal to :
A
${{5h} \over 8}$
B
${{3{h^2}} \over {8R}}$
C
${{{h^2}} \over {4R}}$
D
${{3h} \over 4}$

Explanation Let the density of solid cone $\rho$.

$dm = \rho \pi {r^2}dy$

${y_{cm}} = {{\int {ydm} } \over {\int {dm} }}$
$= {{\int\limits_0^h {\pi {r^2}} dy\rho \times y} \over {{1 \over 3}\pi {R^2}h\rho }}$
$= {{3h} \over 4}$
3

JEE Main 2014 (Offline)

A mass $'m'$ is supported by a massless string wound around a uniform hollow cylinder of mass $m$ and radius $R.$ If the string does not slip on the cylinder, with what acceleration will the mass fall or release? A
${{2g} \over 3}$
B
${{g} \over 2}$
C
${{5g} \over 6}$
D
$g$

Explanation Here string is not slipping over pulley.

From figure,

Acceleration $a = R\alpha \,\,\,\,...(i)$

Applying Newton's second law on the block we get,

$mg-T=ma$ $...(ii)$

Torque on cylinder due to tension in string about the fixed point,

$T \times R = I\alpha$

$T \times R = m{R^2}\alpha = m{R^2}\left( {{a \over R}} \right)$

[ as $I$ = MR2 for hollow cylinder]

or $T=ma$

$\Rightarrow \,\,mg - ma = ma$

$\Rightarrow \,\,\,a = {g \over 2}$
4

JEE Main 2014 (Offline)

A bob of mass $m$ attached to an inextensible string of length $l$ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $\omega \,rad/s$ about the vertical. About the point of suspension:
A
angular momentum is conserved
B
angular momentum changes in magnitude but not in direction.
C
angular momentum changes in direction but not in magnitude.
D
angular momentum changes both in direction and magnitude.

Explanation Torque working on the bob of mass $m$ is, $\tau = mg \times \ell \,\sin \,\theta .$ (Direction of torque by the weight is parallel to plane of rotation of the particle)

As $\tau$ is perpendicular to the angular momentum $\overrightarrow L$ of the bob, so the direction of $L$ changes but magnitude remains same.