Question

Consider two uniform discs of the same thickness and different radii $R_1=R\text{and}{R}_2=\alpha R$, made of the same material. If the ratio of their moments of inertia $I_1$ and $I_2$, respectively, about their axes is $I_1:I_2=1:16$, then the value of $\alpha$ is :

• A. $2\sqrt{2}$
• B. $\sqrt{2}$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Let $\rho$ be the density of the discs and $t$ is their thickness.

Moment of inertia of the disc is given by,
$I=\frac{M{R}^2}{2}=\frac{\left[\rho \right(\pi R^2\left)t\right]R^2}{2}=\frac{\rho \pi t{R}^4}{2}$

$\Rightarrow I\propto R^4\because \rho \text{and}t\text{are same for the discs}$

$\therefore \frac{I_2}{I_1}={\left(\frac{R_2}{R_1}\right)}^4={\alpha }^4=\frac{16}{1}$

$\Rightarrow \alpha =2$

Hence, $\left(C\right)$ is the correct answer.