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)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 23.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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
$X_{C M = \frac{m_{0} x_{0} - m x}{m_{0} - m}} = \frac{[π a^{2} \times 0 - \frac{a^{2}}{4} \times \frac{a}{2}] σ}{[π a^{2} - \frac{a^{2}}{4}] σ} = \frac{- a^{3} / 8}{(π - \frac{1}{4}) a^{2}} = \frac{- a}{2 (4 π - 1)} = \frac{- a}{8 π - 2} = - \frac{a}{23} ∴ x = 23$