A uniform semicircular ring having mass m and radius r is hanging at one of its ends freely as shown in figure. The ring is slightly disturbed so that it oscillates in its own plane. The time period of oscillation of the ring is

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$2 π \sqrt{\frac{r}{g (1 + 1 / π^{2})}}$

$2 π \sqrt{\frac{2 r}{g (1 + 4 / π)^{1 / 2}}}$

$2 π \sqrt{\frac{r}{g {(1 - 4 / π^{2})}^{1 / 2}}}$

$2 π \sqrt{\frac{r}{g {(1 - 2 / π^{2})}^{1 / 2}}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{I_{0}}{m g d}} \\ I_{0}^{'} = I_{C} + m r^{' 2} \Rightarrow m r^{2} = I_{C} + m r^{' 2} \\ \Rightarrow I_{C} = m (r^{2} - r'^{2}) \\ I_{0} = I_{C} + m d^{2} = m (r^{2} - r^{' 2}) + m (r^{2} + r^{' 2}) = 2 m r^{2} \\ d = \sqrt{r^{2} + {(\frac{2 r}{π})}^{2}} = r \sqrt{1 + \frac{4}{π^{2}}} \\ \Rightarrow T = 2 π \sqrt{\frac{2 r}{g {(1 + \frac{4}{π})}^{1 / 2}}} \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET