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 t/4 . Then the relation between the moment of inertia IX and IY is

The correct option is IY = 64 IX.

Step-by-step derivation:

The moment of inertia of a uniform circular disc of mass M and radius R about an axis passing through its center and perpendicular to its plane is given by the formula:

$I=\frac{1}{2}M{R}^2$

Since both discs are made from iron plates, they have the same mass density ($\rho$). The mass of a disc can be expressed in terms of its density, radius, and thickness ($t$):

$M=\text{density}\times \text{volume}=\rho \times \pi R^{2}t$

Substituting the expression for mass into the moment of inertia formula, we get:

$I=\frac{1}{2}\left(\rho \pi R^{2}t\right)R^2=\frac{1}{2}\rho \pi t{R}^4$

Now, let us write the expressions for the moments of inertia of both disc X and disc Y:

For disc X, the radius is R and the thickness is t. Thus, its moment of inertia is:

$I_X=\frac{1}{2}\rho \pi t{R}^4$

For disc Y, the radius is 4R and the thickness is $\frac{t}{4}$. Thus, its moment of inertia is:

$I_Y=\frac{1}{2}\rho \pi \left(\frac{t}{4}\right){\left(4R\right)}^4$

Simplifying the expression for disc Y:

$I_Y=\frac{1}{2}\rho \pi \left(\frac{t}{4}\right)\left(256R^4\right)$

$I_Y=\frac{256}{4}\left(\frac{1}{2}\rho \pi t{R}^4\right)$

$I_Y=64I_X$

Therefore, the relation between the moments of inertia is $I_Y=64I_X$.