A) 1
B) \[\frac{3}{\sqrt{2}}\]
C) \[\sqrt{\frac{3}{\sqrt{2}}}\]
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: C
Solution :
\[I=\frac{m\ell }{12}+\frac{m{{R}^{2}}}{4}\] or \[I=\frac{m}{4}\left( \frac{{{\ell }^{2}}}{3}+{{R}^{2}} \right)\] ?.. Also \[m=\pi {{R}^{2}}\ell \rho \] \[\Rightarrow \] \[{{R}^{2}}=\frac{m}{\pi \ell \rho }\]Put in equation \[I=\frac{m}{4}\left( \frac{{{\ell }^{2}}}{3}+\frac{m}{\rho \ell \rho } \right)\] For maxima & minima \[\frac{dI}{d\ell }=\frac{m}{4}\left( \frac{2\ell }{3}-\frac{m}{\pi {{\ell }^{2}}\rho } \right)=0\] \[\Rightarrow \] \[\frac{2\ell }{3}=\frac{m}{\pi {{\ell }^{2}}\rho }\Rightarrow \frac{2\ell }{3}=\frac{\pi {{R}^{2}}\ell \rho }{\pi {{\ell }^{2}}\rho }\] Or \[\frac{2\ell }{3}=\frac{{{R}^{2}}}{\ell }\] \[\Rightarrow \] \[\frac{{{\ell }^{2}}}{{{R}^{2}}}=\frac{3}{2}\] \[\frac{{{\ell }^{2}}}{R}=\sqrt{\frac{3}{2}}\]
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