A disc (of radius r cm) of uniform thickness and uniform density `sigma` has a square hole with sides of length `l = (r )/(sqrt(2)) cm`. One corner of the hole is located at the centre of the disc and centre of the hole lies y-axis as shown. Then the y-coordinate of position of centre of mass of disc with hole (in cm) is.
.

Options

A. `-(r)/(2(pi - ^1//_4)`
B. `-(r)/(4(pi - ^1//_4)`
C. `-(r)/(4(pi - ^1//_2)`
D. `-(3r)/(4(pi - ^1//_4)`

Answer

C

Explanation

( c) This disc can be assumed to be made of a complete uniform disc and a square plate with same negative mass density.
`Y_(cm) =(m_1 y_1 + m_2 y_2)/(m_1 + m_2) = ((pi r^2) sigma(0) + l^2 (- sigma)(r//2))/(pi r^2 sigma + l^2 (-sigma))`
=`(-l^2 r)/(2(pi r^2 - l^2)) =(-(r^2)/(2))/(2(pi r^2 -(r^2)/(2))) =(-r)/(4(pi -(1)/(2)))`.