Q73 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
A circular disc has radius $R_{1}$ and thickness $T_{1}$. Another circular disc made of the same material has radius $R_{2}$ and thickness $T_{2}$. If the moment of inertia of both discs are same and $\frac{R_{1}}{R_{2}}=2$ then $\frac{T_{1}}{T_{2}}=\frac{1}{\alpha}$. The value of $\alpha$ is $\_\_\_\_$.
Correct answer: 16
Solution
For a circular disc made of material with density $\rho$, moment of inertia is $I = \frac{1}{2}MR^2 = \frac{1}{2}\rho \pi R^2 T \cdot R^2 = \frac{1}{2}\rho \pi R^4 T$.
For disc 1: $I_1 = \frac{1}{2}\rho \pi R_1^4 T_1$
For disc 2: $I_2 = \frac{1}{2}\rho \pi R_2^4 T_2$
Given $I_1 = I_2$:
$R_1^4 T_1 = R_2^4 T_2$
With $\frac{R_1}{R_2} = 2$:
$(2R_2)^4 T_1 = R_2^4 T_2$
$16 R_2^4 T_1 = R_2^4 T_2$
$\frac{T_1}{T_2} = \frac{1}{16}$
Therefore $\alpha = 16$.
For disc 1: $I_1 = \frac{1}{2}\rho \pi R_1^4 T_1$
For disc 2: $I_2 = \frac{1}{2}\rho \pi R_2^4 T_2$
Given $I_1 = I_2$:
$R_1^4 T_1 = R_2^4 T_2$
With $\frac{R_1}{R_2} = 2$:
$(2R_2)^4 T_1 = R_2^4 T_2$
$16 R_2^4 T_1 = R_2^4 T_2$
$\frac{T_1}{T_2} = \frac{1}{16}$
Therefore $\alpha = 16$.
Hence, the correct answer is 16.