A square hole is punched out of a circular lamina, the diagonal of the square coinciding with the radius of circle. If ‘a’ be the diameter of the circle, the distance of the centre of mass of reminder from previous centre of mass. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{4 (2 π - 1)}{a}$
B
$\frac{a}{(2 π - 1)}$
C
$\frac{a}{4 (2 π - 1)}$
D
no change

Solution:

$∵ x_{shift} = \frac{m_{removed} d}{(M - m_{removed})}$
Mass is proportional to area of plate.

Let d be the side of the square, then
$M \propto [{πR}^{2}] and d \sqrt{2} = R$
$m_{removed} \propto d^{2} \Rightarrow m_{removed} \propto \frac{R^{2}}{2} and a = \frac{R}{2}$

$x_{shift} = \frac{\frac{R^{2}}{2} . \frac{R}{2}}{[{πR}^{2} - \frac{R^{2}}{2}]} = \frac{R}{2 [2 π - 1]}$

$x_{shift} = \frac{a}{4 (2 π - 1)}$

Solution:

Related content