Question

The moment of inertia of a cylinder about its own axis is equal to its moment of inertia about as not passing through its center and normal to its length. The ratio length to radius is

• A. 2 : 1
• B. $\sqrt{3}:1$
• C. 3 : 1
• D. None

Answer 1

Answer: (B)

$\sqrt{3}:1$

$\begin{array}{c}\text{we know that -}\\ \text{Moment of Inertia in case}A\text{is}\\ I_A=\frac{M{R}^2}{2}\\ \text{And Moment of Inertia in case}B\text{is}\\ I_B=\frac{M{R}^2}{4}+\frac{M{L}^2}{12}=\frac{M}{12}(3R^2+L^2)\end{array}$

$\begin{array}{c}{\text{Sinu it is given I}}_A=I_B\\ \Rightarrow \frac{M{R}^2}{2}=\frac{M}{12}(3R^2+L^2)\\ \Rightarrow L^2=3R^2\\ \Rightarrow \frac{L}{R}=\frac{\sqrt{3}}{1}\end{array}$