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
MCQ (Single Correct Answer)
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$$\theta$$
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
MCQ (Single Correct Answer)
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 :
