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A square shaped hole of side $l = \frac{a}{2}$ is curved out at a distance $d = \frac{a}{2}$ from the centre ‘O’ of a uniform circular disk of radius a, If the distance of the centre of mass of the remaining portion from O is $- \frac{a}{X}$ , value of O (to the nearest integer is)

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answer is 23.

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Detailed Solution

Let $σ$ be the mass density of circular disc.
Original mass of the disc, $m_{0} = π a^{2} σ$
Removed mass, $m' = (π a^{2} - \frac{a^{2}}{4}) σ = a^{2} (\frac{4 π - 1}{4}) σ$
Question Image
New position of centre of mass
$X_{C M = \frac{m_{0} x_{0} - m x}{m_{0} - m}} = \frac{[π a^{2} \times 0 - \frac{a^{2}}{4} \times \frac{a}{2}] σ}{[π a^{2} - \frac{a^{2}}{4}] σ} = \frac{- a^{3} / 8}{(π - \frac{1}{4}) a^{2}} = \frac{- a}{2 (4 π - 1)} = \frac{- a}{8 π - 2} = - \frac{a}{23} ∴ x = 23$

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