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A uniform square plate has a side of length 2R. A circular piece of maximum possible area is cut and removed from one of the quadrants of the plate as shown in the figure. Shift in the centre of mass of the plate is,

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$\frac{R}{π (16 - π)}$

$\frac{πR}{\sqrt{2} (16 - π)}$

$\frac{R}{(16 - π)}$

answer is A.

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Detailed Solution

We know that,

$mass \propto area$

$oo' = d = \frac{\sqrt{2} R}{2} = \frac{R}{\sqrt{2}}$

Along OO',

$shift = \frac{m_{rem} (d)}{M_{total} - M_{rem}}$

$= \frac{[π {(\frac{R}{2})}^{2} σ] (\frac{R}{\sqrt{2}})}{4 R^{2} σ - π {(\frac{R}{2})}^{2} σ}$

$= \frac{πR}{\sqrt{2} (16 - π)}$

Hence the correct answer is $\frac{π R}{\sqrt{2} (16 - π)} .$

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