JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Medium

1 of 1

Application +4 | -1

Moment of inertia of a circular loop of radius R about the axis of rotation parallel to horizontal diameter at a distance R/2 from it is

View answer and explanation

Correct option: D

\frac{3}{4}M{R^2}

The correct answer is 5 because it is calculated by equating the ratio of energies of photons produced during electron transitions in a hydrogen atom.

More Questions from SYSTEM OF PARTICLES

Practice All Chapter MCQs →

It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod

\mathrm{AB}

of length

\mathrm{L}

has linear mass density

\mu(x)=a+\frac{b x}{L}

, where

x

is measured from

\mathrm{A}

. If the

\mathrm{CM}

of the rod lies at a distance of

\left(\frac{7}{12}\right) \mathrm{L}

from

\mathrm{A}

, then a and

\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

8 \mathrm{~cm}

. Its moment of inertia will be The angular velocity of a particle is given by

\omega = {t^4} + 2{t^3}

then its angular acceleration of the particle at

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)