JEE Mains Previous Years Questions with Solutions

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1

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 $\rho$.

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

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

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 $T=ma$

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

$\Rightarrow \,\,\,a = {g \over 2}$
3

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.

4

JEE Main 2013 (Offline)

A hoop of radius $r$ and mass $m$ rotating with an angular velocity ${\omega _0}$ is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it cases to slip?
A
${{r{\omega _0}} \over 4}$
B
${{r{\omega _0}} \over 3}$
C
${{r{\omega _0}} \over 2}$
D
${r{\omega _0}}$

Explanation

From conservation of angular momentum at point of contact,

$m{r^2}{\omega _0} = mvr + m{r^2}\omega$

$m{r^2}{\omega _0} = mvr + m{r^2}\left( {{v \over r}} \right)$ [ as $v = r\omega$ ]

$m{r^2}{\omega _0} = mvr + mvr$

$m{r^2}{\omega _0} = 2mvr$

$v = {{{\omega _0}r} \over 2}$