Question

$I_1$ is moment of inertia of a square plate of uniform thickness about an axis perpendicular to the plane of plate and passing through its centre. From this square plate, a circular plate of maximum radius is cut and removed. If $I_2$ is moment of inertia of this circular plate about an axis perpendicular to the plane of plate and passing through its centre then the ratio of $I_1$ to $I_2$ is:

• A. $56:33$
• B. $33:28$
• C. $28:11$
• D. $14:11$

Answer 1

The correct option is A $56:33$

Let the thickness of plate be t and density be $\rho$

Now $m=\rho l^{2}t$

So MI of square $I_1=\frac{m{l}^2}{6}$

Now the maximum possible radius of circular plate r=l/2,

Mass of circular plate $M=\pi (l/2{)}^{2}t\rho =\pi m/4$

So $I_2=M(l/2{)}^2/2=\frac{\pi m{l}^2}{32}$

So $\frac{I_1}{I_2}=56:33$, using $\pi =22/7$