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.

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$\frac{4 (2 π - 1)}{a}$

$\frac{a}{(2 π - 1)}$

$\frac{a}{4 (2 π - 1)}$

$\frac{2 a}{(2 π - 1)}$

answer is C.

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Detailed Solution

Let initial mass = m

Mass of removed square portion

$= \frac{m}{π R^{2}} \times (\frac{R^{2}}{2}) = \frac{m}{2 π}$

Let centre of circular lamina is origin.

Then C.O.M of square is at $(\frac{a}{4}, 0)$

$X_{s h i f t} = \frac{m_{1} x_{1} - m_{2} x_{2}}{m_{1} - m_{2}}$

$= \frac{m (0) - \frac{m}{2 π} (\frac{a}{4})}{m - \frac{m}{2 π}}$

$= \frac{m a}{4 m (2 π - 1)}$

$= \frac{a}{4 (2 π - 1)}$

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