### JEE Mains Previous Years Questions with Solutions

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1

### 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 ${I_{EF}} = {I_{AC}}.$
2

### 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 $\overrightarrow {{\tau _c}} = {{d\overrightarrow {{L_c}} } \over {dt}}$]
3

### AIEEE 2007

A round uniform body of radius $R,$ mass $M$ and moment of inertia $I$ rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is
A
${{g\,\sin \theta } \over {1 - M{R^2}/I}}$
B
${{g\,\sin \theta } \over {1 + I/M{R^2}}}$
C
${{g\,\sin \theta } \over {1 + M{R^2}/I}}$
D
${{g\,\sin \theta } \over {1 - I/M{R^2}}}$

## Explanation

A uniform body of radius R, mass M and moment of inertia $I$ rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

$a = {{g\,\sin \,\theta } \over {1 + {I \over {M{R^2}}}}}$
4

### AIEEE 2007

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs coincide. The center of mass of the new disc is $\alpha R$ form the center of the bigger disc. The value of $\alpha$ is
A
$1/4$
B
$1/3$
C
$1/2$
D
$1/6$

## Explanation

Let the mass per unit area be $\sigma .$

Then the mass of the complete disc
$= \sigma \left[ {\pi {{\left( {2R} \right)}^2}} \right] = 4\pi \sigma {R^2}$

The mass of the removed disc $= \sigma \left( {\pi {R^2}} \right) = \pi \sigma {R^2}$

So mass of the remaining disc = $4\pi \sigma {R^2}$ - $\pi \sigma {R^2}$ = $3\pi \sigma {R^2}$

Let center of mass of $3\pi \sigma {R^2}$ mass is at x distance from origin O.

$\therefore$ ${{3\pi {R^2}\sigma .x + \pi {R^2}\sigma .R} \over {4\pi {R^2}\sigma }} = 0$

As center of mass of full disc is at Origin.

$\therefore$ $x = - {R \over 3}$

According to the question, $x$ = $\alpha R$

$\therefore$ $\alpha = - {1 \over 3}$

$\Rightarrow$ $\left| \alpha \right| = {1 \over 3}$

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