Question

A circular disc X of radius R is made from an iron plate of thickness t and another disc Y of radius 4R is made from an iron plate of thickness $\frac{t}{4}$ Then the relation between the moment of inertia about their natural axis $I_X$ and $I_Y$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is (D)

Given,

The radius of the disc = R

The thickness of the disc = t

The radius of the second disc = 4R

The thickness of the second disc = $\frac{t}{4}$

Now,

Moment of Inertia of disc 'X', $I_x=\frac{M_x{R}_x^2}{2}$

Where $M_x=\rho \left(\pi R_x^2\right)t_x$ = $\rho \pi R^{2}t$

Moment of Inertia of disc 'Y', $I_y=\frac{M_y{R}_y^2}{2}$

Where $M_y=\rho \pi (4R{)}^2\frac{t}{4}=4\rho \pi R^{2}t$

Ratio of the moment of inertia of both the discs

$\therefore \frac{I_x}{I_y}=\frac{\left(\rho \pi R^{2}t\right)R^2}{2}\times \frac{2}{\left(4\rho \pi R^{2}t\right)16R^2}$

After solving

$\Rightarrow \frac{I_x}{I_y}=\frac{1}{64}\Rightarrow I_y=64I_x$

Hence, the moment of inertia of Y is 64 times the moment of inertia of X