JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

A circular disc of radius \(b\) has a hole of radius \(a\) at its centre (see figure). If the mass per unit area of the disc varies as \(\left( {\frac{{{\sigma _0}}}{r}} \right)\), then the radius of gyration of the disc about its axis passing through the centre is

A \(\frac{{a + b}}{3}\)
B \(\sqrt {\frac{{{a^2} + {b^2} + ab}}{3}} \)
C \(\frac{{a + b}}{2}\)
D \(\sqrt {\frac{{{a^2} + {b^2} + ab}}{2}} \)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt {\frac{{{a^2} + {b^2} + ab}}{3}} \)

Step-by-step Solution

Detailed explanation

\(dI = \left( {dm} \right){r^2}\) \( = \left( {\sigma dA} \right){r^2}\) \( = \left( {\frac{{{\sigma _0}}}{r}2\pi dr} \right){r^2} = \left( {{\sigma _0}2\pi 0{r^2}dr} \right)\) \(I = \int {DI = \int\limits_a^b {{\sigma _0}2\pi {r^2}dr} } \)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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