I is moment of inertia of a thin circular ring about an axis perpendicular to the plane of ring and passing through its centre. The same ring is folded into 2 turns coil. The moment of inertia of circular coil about an axis perpendicular to the plane of coil and passing through its centre is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{I}{2}$

$\frac{I}{4}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

New radius $= \frac{r}{2}$

$Mass of each new rings = \frac{m}{2}$

$\frac{I^{'}}{I} = 2 \frac{m^{'}}{m} {[\frac{r^{'}}{r}]}^{2} = \frac{1}{4} (∵ I^{'} = 2 \times m^{'} r^{'^{2}})$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test