Question

Consider two uniform discs of the same thickness and different radii $R_1=R$ 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. $\sqrt{2}$
• B. $2$
• C. $2\sqrt{2}$
• D. $4$

Answer 1

The correct option is B

$2$

Step 1. Given data

Radii of the uniform disc, $1$$=R$

Radii of the uniform disc, $2=\alpha R$

The ratio of Moment of inertia, $I_1:I_2=1:16$

Step 2. Moment of inertia of the uniform disc

Moment of inertia of the uniform disc, $I=\frac{M{R}^2}{2}$ [Where $M$ is the mass]

$I=\frac{\left[\rho \left({\mathrm{\pi R}}^2\right)t\right]R^2}{2}$ [Where $\rho$is the uniform density and $t$ is the thickness]

Step 3. Finding the value of $\alpha$

By using the above formula,

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

$\frac{1}{16}={\left(\frac{R}{\alpha R}\right)}^4$

$\alpha =2$

Hence, the correct option is $\left(B\right)$.