Question

The moment of inertia of a cylinder about its own axis is equal to its M.I. about an axis passing through its centre and perpendicular to its length. The ratio of of length to radius is

• A. 1 : 2
• B. $\sqrt{2}:1$
• C. $\sqrt{3}:1$
• D. $\frac{1}{\sqrt{3}}:1$

Answer 1

Answer: (C)

$\sqrt{3}:1$

$\begin{array}{c}\text{Let, the length of cylinder}=l\\ \text{and radius}=r\text{.}\end{array}$

$\text{m.o.I about axis-}A\text{(its own axis)}=\frac{m{r}^2}{2}\dots .\text{(1)}$

$\text{Mass of elementary element}\left(dm\right)=\frac{m}{l}\cdot dx$

$\begin{array}{c}dI=[\frac{\left(dm\right)r^2}{4}+(dm\left)x^2\right]\begin{array}{c}\text{(parallel axistheorem)}\end{array}\\ \Rightarrow \int dI={\int }_{\frac{l}{2}}^{1/2}\frac{m}{l}(\frac{r^2}{4}+x^2)dx=\frac{2m}{\ell }{\int }_0^{1/2}(\frac{r^2}{4}+x^2)dx\end{array}$

$\text{Thus,}{I}_B=m(\frac{l^2}{12}+\frac{r^2}{4})\cdots -\text{(2)}$

$\begin{array}{cc}\text{Given that,}{I}_A=I_B& \Rightarrow \frac{m{r}^2}{2}=m(\frac{l^2}{12}+\frac{r^2}{4})\\ & \Rightarrow \frac{l}{r}=\frac{\sqrt{3}}{1}\end{array}$