 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2008

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

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

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

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 $\overrightarrow {{\tau _c}} = {{d\overrightarrow {{L_c}} } \over {dt}}$]