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A circular plate has a uniform thickness and has a diameter 56 cm. A circular disc of diameter 42 cm is removed from one edge of the plate. The distance of centre of mass of the remaining portion from the centre of the circular plate is

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18 cm

9 cm

27 cm

4.5 cm

answer is B.

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Detailed Solution

$R = \frac{56}{2} = 28 cm$
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$r = \frac{42}{2} = 21 cm; d = R - r = 7 cm x_{shift} = \frac{r^{2} d}{R^{2} - r^{2}} = \frac{(21)^{2} \times 7}{(28)^{2} - (21)^{2}} = 9 cm$

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