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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2011

MCQ (Single Correct Answer)
A thin horizontal circular disc is rotating about a vertical axis passing through its center. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.
A
continuously decreases
B
continuously increases
C
first increases and then decreases
D
remains unchanged

Explanation

Here no external force is applied on the disc so Torque ($$\tau $$) = 0.

So angular momentum is conserved.

That means $${I_1}{\omega _1} = {I_2}{\omega _2}$$

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

$$\therefore$$ Angular speed is inversely proportional to Moment of inertia.

For disc $$I = {1 \over 2}M{R^2}$$

$$\therefore$$ Moment of Inertia is proportional to Mass.

As insect moves along a diameter, the effective mass of disc first decreases then increases and hence the moment of inertia first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.
2

AIEEE 2009

MCQ (Single Correct Answer)
A thin uniform rod of length $$l$$ and mass $$m$$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $$\omega $$. Its center of mass rises to a maximum height of:
A
$${1 \over 6}\,\,{{l\omega } \over g}$$
B
$${1 \over 2}\,\,{{{l^2}{\omega ^2}} \over g}$$
C
$${1 \over 6}\,\,{{{l^2}{\omega ^2}} \over g}$$
D
$${1 \over 3}\,\,{{{l^2}{\omega ^2}} \over g}$$

Explanation


The moment of inertia of the rod about $$O$$ is $${1 \over 2}m{\ell ^2}.$$ The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this conditions is $${1 \over 2}I{\omega ^2}$$ where $$I$$ is the moment of inertia of the rod about $$O.$$ when the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the center of mass is raised through $$h$$, so the increase in potential energy is $$mgh$$. This is equal to kinetic energy $${1 \over 2}I{\omega ^2}$$.

$$\therefore$$ $$mgh = {1 \over 2}I{\omega ^2} = {1 \over 2}\left( {{1 \over 3}m{l^2}} \right){\omega ^2}$$.

$$ \Rightarrow h = {{{\ell ^2}{\omega ^2}} \over {6g}}$$
3

AIEEE 2008

MCQ (Single Correct Answer)
A thin rod of length $$'L'$$ is lying along the $$x$$-axis with its ends at $$x=0$$ and $$x=L$$. Its linear density (mass/length) varies with $$x$$ as $$k{\left( {{x \over L}} \right)^n},$$ where $$n$$ can be zero or any positive number. If the position $${X_{CM}}$$ of the center of mass of the rod is plotted against $$'n',$$ which of the following graphs best approximates the dependence of $${X_{CM}}$$ on $$n$$?
A
B
C
D

Explanation

Given The linear mass density $$\lambda = k{\left( {{x \over L}} \right)^n}$$



$${x_{CM}} = {{\int\limits_0^L {x{\mkern 1mu} dm} } \over {\int\limits_0^L {dm} }}$$

$$ = {{\int\limits_0^L {x\left( {\lambda {\mkern 1mu} dx} \right)} } \over {\int\limits_0^L {\lambda {\mkern 1mu} dx} }}$$

$$ = {{\int\limits_0^L {k{{\left( {{x \over L}} \right)}^n}} .xdx} \over {\int\limits_0^L {k{{\left( {{x \over L}} \right)}^n}{\mkern 1mu} dx} }}$$

= $${{k\left[ {{{{x^{n + 2}}} \over {\left( {n + 2} \right){L^n}}}} \right]_0^L} \over {\left[ {{{k{\mkern 1mu} {x^{n + 1}}} \over {\left( {n + 1} \right){L^n}}}} \right]_0^L}}$$

$$ = {{L\left( {n + 1} \right)} \over {n + 2}}$$



For $$n=0,$$ $${x_{CM}} = {L \over 2};n = 1,$$

$${x_{CM}} = {{2L} \over 3};n = 2,\,{x_{CM}} = {{3L} \over 4};\,...$$

From here you can see only option (A) can be the right answer.
4

AIEEE 2008

MCQ (Single Correct Answer)
Consider a uniform square plate of side $$' a '$$ and mass $$'m'$$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
A
$${5 \over 6}m{a^2}$$
B
$${1 \over 12}m{a^2}$$
C
$${7 \over 12}m{a^2}$$
D
$${2 \over 3}m{a^2}$$

Explanation


Moment of inertia for the square plate through O, perpendicular to the plate is

$${I_{nn'}} = {1 \over {12}}m\left( {{a^2} + {a^2}} \right) = {{m{a^2}} \over 6}$$

Also, $$DO = {{DB} \over 2} = {{\sqrt 2 a} \over 2} = {a \over {\sqrt 2 }}$$

According to parallel axis theorem

$${{\mathop{\rm I}\nolimits} _{mm'}} = {I_{nn'}} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}$$

$$ = {{m{a^2}} \over 6} + {{m{a^2}} \over 2} $$

$$= {{m{a^2} + 3m{a^2}} \over 6} $$

$$= {2 \over 3}m{a^2}$$

Questions Asked from Rotational Motion

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