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If a square of side R/2 is removed from a uniform circular disc of radius R as shown in the figure, the shift in centre of mass is

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$\frac{R}{3 (4 π - 1)}$

$\frac{R}{4 π - 1}$

$\frac{R}{2 (4 π - 1)}$

$\frac{R}{4 (4 π - 1)}$

answer is D.

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Detailed Solution

$Δ_{X_{c m}} = \frac{(m_{r e m o v e d}) d}{M - m_{r e m o v e d}}$

$σ= \frac{M}{π {(\frac{R}{2})}^{2}}$

2a = R

$a = \frac{R}{2}$

$m_{removed} = {(\frac{R}{2})}^{2} \times \frac{M}{{πR}^{2}} = \frac{R^{2} M}{4 {πR}^{2}} = \frac{M}{4 π}$

$m_{reminding} = M - \frac{M}{4 π} = \frac{M}{4 π} (4 π - 1)$

$X_{cm} = \frac{\frac{M}{4 π} \times \frac{R}{4}}{\frac{M}{4 π} (4 π - 1)} = \frac{R}{4 (4 π - 1)}$

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