A uniform circular disc of radius a is taken. A circular portion of radius b has been removed from it as shown in the figure. If the centre of hole is at a distance c from the centre of the disc, the distance x2 of the centre of mass of the remaining part from the initial centre of mass O is given by

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πb2(a2-c2)

cb2(a2-b2)

πc2(a2-b2)

ca2(c2-b2)

answer is B.

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Detailed Solution

Centre of mass of remaining portion was at point O2. Hence, x2 (area of remaining portion) = c (area of removeddisc)∴ x2πa2-πb2=cπb2⇒x2=cb2a2-b2

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