JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Rotational Mechanics

I_{C M}

is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc.

I_{A B}

is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance

\frac{2}{3} R

from center. Where

R

is the radius of the disc. The ratio of

I_{A B}

and

I_{C M}

x : 9

. The value of

x

is ______.

Answer: 17

Solution

A uniform disc has

I_{C M} = \frac{m R ^{2}}{2}

about its central perpendicular axis. The parallel axis AB lies a distance

d = \frac{2 R}{3}

away, so

I_{A B} = I_{C M} + m d^{2} = \frac{m R ^{2}}{2} + m (\frac{2 R}{3})^{2} = \frac{m R ^{2}}{2} + \frac{4 m R ^{2}}{9} = \frac{17 m R ^{2}}{18}

. The ratio is

\frac{I _{A B}}{I _{C M}} = \frac{17 m R ^{2} /18}{m R ^{2} /2} = \frac{17}{9}

, so against

x : 9

the value is

x = 17

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer