Physics - Rotation Question with Solution | TestHub

$m=\rho \left(\pi R^{2}l\right) ...\left(1\right)$

$dm=\rho \pi \left(2R.dRl+R^{2}dl\right)$

$0=\rho \pi \left(2RldR+R^{2}dl\right)$

$2Rl dR=-R^{2}dl ...\left(2\right)$

$I=\frac{M{R}^2}{4}+\frac{M{l}^2}{12}$

$I=\frac{M}{12} \left(3R^2+l^2\right)$

$dI=\frac{M}{12}\left(6R\left(dR\right)+2l dl\right)=0 ... \left(3\right)$

Putting$\left(1\right), \left(2\right)$in$\left(3\right)$

$6R\left(\frac{-R^{2}dl}{2Rl}\right)+2l dl=0$

$\left(\frac{-3R^2}{l}+2l\right)dl=0$

$-3R^2+2l^2=0$

$2l^2=3R^2$

$\frac{l}{R}= \sqrt{\frac{3}{2}}$