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 ( 0 r), then the radius of gyration of the disc about its axis passing through the centre is

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

JEE Mains · Physics · STD 11 - 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

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Verified Solution

Correct Answer

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

Detailed explanation

\(dI = \left( {dm} \right){r^2}\)

\( = \left( {\sigma dA} \right){r^2}\)

\( = \left( {\frac{{{\sigma _0}}}{r}2\pi dr} \right){r^2} = \left( {{\sigma _0}2\pi 0{r^2}dr} \right)\)

\(I = \int {DI = \int\limits_a^b {{\sigma _0}2\pi {r^2}dr} } \)

\( = {\sigma _0}2\pi \left( {\frac{{{b^3} - {a^3}}}{3}} \right)\)

\(m = \int {dm = \int {\sigma dA} } \)

\( = {\sigma _0}2\pi \int\limits_a^b {dr} \)

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