JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Rotational Mechanics

Moment of inertia (M.I) of four bodies, having same mass and radius, are reported as:

I_{1} =

M.I. of thin circular ring about its diameter

I_{2} =

M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I_{3} =

M.I of solid cylinder about its axis and

I_{4} =

M.I of solid sphere about its diameter.
Then

Answer: (C)

Solution

For bodies with same mass

m

and radius

r

I_{1} = \frac{1}{2} m r^{2}

(thin ring about diameter)

I_{2} = \frac{1}{2} m r^{2}

(disc about perpendicular axis through center)

I_{3} = \frac{1}{2} m r^{2}

(solid cylinder about its axis)

I_{4} = \frac{2}{5} m r^{2}

(solid sphere about diameter)

Since

\frac{1}{2} > \frac{2}{5}

, we have:

I_{1} = I_{2} = I_{3} > I_{4}

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option