JEE Physics 2025 Question with Solution

Given: Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ with moments of inertia $I_1$ and $I_2$. Also, $R_2 = 2R_1$.

Find: The value of $x$ if

$$\frac{I_1}{I_2} = \frac{1}{x}$$

For a solid disc about its central axis,

$$I = \frac{1}{2}MR^2$$

Since both discs are made of the same material and have negligible thickness, mass is proportional to area. Therefore,

$$M \propto R^2$$

So,

$$I \propto MR^2 \propto R^2 \cdot R^2 = R^4$$

Hence,

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

Using $R_2 = 2R_1$,

$$\frac{I_1}{I_2} = \left(\frac{R_1}{2R_1}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$

Comparing with

$$\frac{I_1}{I_2} = \frac{1}{x}$$

we get

$$x = 16$$

Therefore, the value of $x$ is $16$.

Given: Two solid iron discs of the same material and negligible thickness, with $R_2 = 2R_1$.

Find: The value of $x$ in

$$\frac{I_1}{I_2} = \frac{1}{x}$$

Because the discs have the same thickness and material, mass varies as area, so $M \propto R^2$. Since for a disc

$$I = \frac{1}{2}MR^2$$

it follows immediately that

$$I \propto R^2 \times R^2 = R^4$$

Therefore doubling the radius makes moment of inertia increase by

$$2^4 = 16$$

So,

$$I_2 = 16I_1$$

and hence

$$\frac{I_1}{I_2} = \frac{1}{16}$$

Thus, $x = 16$.