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

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))`
`=((pir^(2))sigma(0)+l^(2)(r//2))/(pi r^(2)sigma+l^(2)(-sigma))`
`=(-l^(2)r)/(2(pir^(2)-l^(2)))=(-(r^(2))/(2))/(2(pir^(2)-(r^(2))/(2)))=(-r)/(4(r-(1)/(2)))`