 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2021 (Online) 31st August Morning Shift

Angular momentum of a single particle moving with constant speed along circular path :
A
changes in magnitude but remains same in the direction
B
remains same in magnitude and direction
C
remains same in magnitude but changes in the direction
D
is zero

## Explanation $$\left| {\overrightarrow L } \right|$$ = mvr

And direction will be upward & remain constant.

Option (b)
2

### JEE Main 2021 (Online) 31st August Morning Shift

A body of mass M moving at speed V0 collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles $$\theta$$1 and $$\theta$$2 with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles $$\theta$$1 and $$\theta$$2 will be equal, is :
A
4
B
1
C
3
D
2

## Explanation Given $$\theta$$1 = $$\theta$$2 = $$\theta$$

from momentum conservation

in x-direction MV0 = MV1 cos$$\theta$$ + mV2 cos$$\theta$$

in y-direction 0 = MV1 sin$$\theta$$ $$-$$ mV2 sin$$\theta$$

Solving above equations

$${V_2} = {{M{V_1}} \over m}$$, V0 = 2V1 cos$$\theta$$

From energy conservation

$${1 \over 2}MV_0^2 = {1 \over 2}MV_1^2 + {1 \over 2}MV_2^2$$

Substituting value of V2 & V0, we will get

$${M \over m} + 1 = 4{\cos ^2}\theta \le 4$$

$${M \over m} \le 3$$

Option (c)
3

### JEE Main 2021 (Online) 27th August Morning Shift

Moment of inertia of a square plate of side l about the axis passing through one of the corner and perpendicular to the plane of square plate is given by :
A
$${{M{l^2}} \over 6}$$
B
$${M{l^2}}$$
C
$${{M{l^2}} \over {12}}$$
D
$${2 \over 3}M{l^2}$$

## Explanation

According to perpendicular Axis theorem. Ix + Iy = Iz

Iz $$\Rightarrow$$ $${{m{l^2}} \over 3} + {{m{l^2}} \over 3}$$

$$= {{2m{l^2}} \over 3}$$
4

### JEE Main 2021 (Online) 26th August Evening Shift

A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that $$r = {L \over {\sqrt 2 }}$$. The speed of particle will be :
A
$${\sqrt {rg} }$$
B
$${\sqrt {2rg} }$$
C
$${2\sqrt {rg} }$$
D
$${\sqrt {{{rg} \over 2}} }$$

## Explanation $$r = {l \over {\sqrt 2 }}$$

$$\sin \theta = {r \over l} = {l \over {\sqrt 2 }}$$

$$\theta$$ = 45$$^\circ$$

$$T\sin \theta = {{m{v^2}} \over r}$$

$$T\cos \theta = mg$$

$$\tan \theta = {{{v^2}} \over {rg}} \Rightarrow v = \sqrt {rg}$$

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