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NVAEasyJEE 2026Moment of Inertia & Radius of Gyration

JEE Physics 2026 Question with Solution

A circular disc has radius R1R_1 and thickness T1T_1. Another circular disc made of the same material has radius R2R_2 and thickness T2T_2. If the moments of inertia of both the discs are same and R1R2=2,thenT1T2=1α.\frac{R_1}{R_2} = 2, \quad \text{then} \quad \frac{T_1}{T_2} = \frac{1}{\alpha}. The value of α\alpha is _____.

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given: Two uniform circular discs are made of the same material, so their density ρ\rho is the same. Their moments of inertia about the central axis are equal. Also, R1R2=2\frac{R_1}{R_2} = 2.

Find: The value of α\alpha if T1T2=1α\frac{T_1}{T_2} = \frac{1}{\alpha}.

For a uniform circular disc about its central axis,

I=12MR2I = \frac{1}{2}MR^2

Since both discs are made of the same material,

M=ρ×Volume=ρπR2TM = \rho \times \text{Volume} = \rho \pi R^2 T

Therefore,

I=12ρπR4TI = \frac{1}{2}\rho \pi R^4 T

Now equating the moments of inertia of the two discs,

12ρπR14T1=12ρπR24T2\frac{1}{2}\rho \pi R_1^4 T_1 = \frac{1}{2}\rho \pi R_2^4 T_2

Cancelling common terms,

R14T1=R24T2R_1^4 T_1 = R_2^4 T_2

So,

T1T2=R24R14=(R2R1)4\frac{T_1}{T_2} = \frac{R_2^4}{R_1^4} = \left(\frac{R_2}{R_1}\right)^4

Given R1R2=2\frac{R_1}{R_2} = 2, hence

R2R1=12\frac{R_2}{R_1} = \frac{1}{2}

Therefore,

T1T2=(12)4=116\frac{T_1}{T_2} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}

Comparing with

T1T2=1α\frac{T_1}{T_2} = \frac{1}{\alpha}

we get

α=16\alpha = 16

the solution states α=8\alpha = 8, but this is inconsistent with the shown working. Therefore, the correct numerical value is 1616.

Common mistakes

  • Using I∝R2TI \propto R^2 T instead of I∝R4TI \propto R^4 T. This is wrong because the disc mass itself depends on R2R^2. First write M=ρπR2TM = \rho \pi R^2 T, then substitute into I=12MR2I = \frac{1}{2}MR^2.

  • Substituting R1R2=2\frac{R_1}{R_2} = 2 directly as T1T2=122\frac{T_1}{T_2} = \frac{1}{2^2}. This is wrong because the radius appears to the fourth power in the final relation. Use T1T2=(R2R1)4\frac{T_1}{T_2} = \left(\frac{R_2}{R_1}\right)^4.

  • Accepting the printed final answer without checking the algebra. The displayed working gives T1T2=116\frac{T_1}{T_2} = \frac{1}{16}, so comparing with 1α\frac{1}{\alpha} gives α=16\alpha = 16, not 88.

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