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JEE Mains Previous Years Questions with Solutions

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1

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.
2

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

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

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

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

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

3

AIEEE 2007

MCQ (Single Correct Answer)
For the given uniform square lamina $$ABCD$$, whose center is $$O,$$
A
$${I_{AC}} = \sqrt 2 \,\,{I_{EF}}$$
B
$$\sqrt 2 {I_{AC}} = {I_{EF}}$$
C
$${I_{AD}} = 3{I_{EF}}$$
D
$${I_{AC}} = {I_{EF}}$$

Explanation


By perpendicular axes theorem,

$${I_x} = {I_x} + {I_y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$or,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_z} = 2{I_y}$$

( as $${I_x} = {I_y}$$ by symmetry of the figure)

$$\therefore$$ $${I_{EF}} = {{{I_z}} \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

By perpendicular axes theorem,

$${I_z} = {I_{AC}} + {I_{BD}} = 2{I_{AC}}$$

(As $${I_{AC}} = {I_{BD}}$$ by symmetry of the figure)

$$\therefore$$ $${I_{AC}} = {{{I_z}} \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$(i)$$ and $$(ii),$$ we get $${I_{EF}} = {I_{AC}}.$$
4

AIEEE 2007

MCQ (Single Correct Answer)
Angular momentum of the particle rotating with a central force is constant due to
A
constant torque
B
constant force
C
constant linear momentum
D
zero torque

Explanation

We know that $$\overrightarrow {{\tau _c}} = {{d\overrightarrow {{L_c}} } \over {dt}}$$
where $$\overrightarrow {{\tau _c}} $$ torque about the center of mass of the body and $$\overrightarrow {{L_c}} = $$ Angular momentum about the center of mass of the body.

Given that $$\overrightarrow {{L_c}} = $$ constant.

$$\therefore$$ $${{d\overrightarrow {{L_c}} } \over {dt}}$$ = 0

$$ \Rightarrow $$ $$\overrightarrow {{\tau _c}} = 0$$ [as $$\overrightarrow {{\tau _c}} = {{d\overrightarrow {{L_c}} } \over {dt}}$$]

Questions Asked from Rotational Motion

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