A circular disc of radius b has a hole of radius a at its centre … - Vidyadip

MCQ

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
$\frac{{a + b}}{3}$
✓
$\sqrt {\frac{{{a^2} + {b^2} + ab}}{3}} $
C
$\frac{{a + b}}{2}$
D
$\sqrt {\frac{{{a^2} + {b^2} + ab}}{2}} $

✓

Answer

Correct option: B.

$\sqrt {\frac{{{a^2} + {b^2} + ab}}{3}} $

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

$ = \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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