JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

From a circular disc of radius R and mass 4m, a small disc of radius

\frac{R}{2}

is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

View answer and explanation

Correct option: B

\frac{{15}}{8}m{R^2}

Option B (

\frac{{15}}{8}m{R^2}

) is correct because it's the difference between the moment of inertia of the full disc and the removed disc.

More Questions from SYSTEM OF PARTICLES

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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)