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JEE Mains Previous Years Questions with Solutions

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1

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

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

AIEEE 2007

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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, $$x$$ = $$\alpha R$$

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

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

Questions Asked from Rotational Motion

On those following papers in MCQ (Single Correct Answer)
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