JEE Main Solved Paper 2017 Question 7
Question: The moment of inertia of a uniform cylinder of length $ \ell $ and radius R about its perpendicular bisector is I. What is the ratio $ \ell /R $ such that the moment of inertia is minimum? JEE Main Solved Paper-2017
Options:
A) 1
B) $ \frac{3}{\sqrt{2}} $
C) $ \sqrt{\frac{3}{\sqrt{2}}} $
D) $ \frac{\sqrt{3}}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ I=\frac{m\ell }{12}+\frac{mR^{2}}{4} $ or $ I=\frac{m}{4}( \frac{{{\ell }^{2}}}{3}+R^{2} ) $ ?.. Also $ m=\pi R^{2}\ell \rho $
$ \Rightarrow $ $ R^{2}=\frac{m}{\pi \ell \rho } $ Put in equation $ I=\frac{m}{4}( \frac{{{\ell }^{2}}}{3}+\frac{m}{\rho \ell \rho } ) $ For maxima & minima $ \frac{dI}{d\ell }=\frac{m}{4}( \frac{2\ell }{3}-\frac{m}{\pi {{\ell }^{2}}\rho } )=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}{\sqrt{2}}} $
