Question

Moment of inertia of a cylinder of mass $M,$ length $L$ and radius $R$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $I=M(\frac{R^2}{4}+\frac{L^2}{12})$. If such a cylinder to be made for a given mass of a material, the ratio $\frac{L}{R}$ for it to have minimum possible $I$ is-

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

The correct option is C $\sqrt{\frac{3}{2}}$

Given, $I=\frac{M{R}^2}{4}+\frac{M{L}^2}{12}$

$\because V=\pi R^{2}L\Rightarrow R^2=\frac{V}{\pi L}$

$\Rightarrow I=\frac{M}{4}\times \frac{V}{\pi L}+\frac{M{L}^2}{12}=\frac{MV}{4\pi L}+\frac{M{L}^2}{12}$

For $I$ to be minimum, $\frac{dI}{dL}=0$

$\frac{dI}{dL}=-\frac{MV}{4\pi L^2}+\frac{M\times 2L}{12}=0$

$\Rightarrow {\left(\frac{R}{L}\right)}^2=\frac{2}{3}$

$\therefore \frac{L}{R}=\sqrt{\frac{3}{2}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.