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Using formula for physical pendulum but unable to solve.


8 months ago

## Answers : (2)

Arun
24742 Points
							Moment of inertia for disc -$I=\frac{1}{2} MR^{2}$ $=\frac{MR^{2}}{2}-[\frac{3}{32}MR^{2}]$$I_{remain}=\frac{13}{32}MR^{2}$ - whereinAbout an axis perpendicular to the plane of disc & passing through its centre .  $I=I_{remain}+I_\frac{R}{2}$$I_{remain}=I-I_\frac{R}{2}$$=\frac{MR^{2}}{2}- \left[ \frac{ \frac{M}{4} \left( \frac{R}{2} \right )^{2} }{2} + \frac{M}{4} \left( \frac{R}{2} \right )^{2} \right ]$$=\frac{MR^{2}}{2}-[\frac{MR^{2}}{32}+\frac{MR^{2}}{16}]$$= \frac{MR^2}{2}-\frac{3MR^2}{32}$$I_{remaining}=\frac{13MR^2}{32}$

8 months ago
Vikas TU
12276 Points
							Dear student Please refer the link below for more information,https://www.askiitians.com/iit-study-material/iit-jee-physics/oscillation/simple-pendulum/Good Luck

8 months ago
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