A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is αR from the centre of the bigger disc. The value of α is

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A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

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$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

If mass of bigger disc is M then mass of removed disc is M/4.

$\frac{M \times 0 - \frac{M}{4} R}{M - \frac{M}{4}} = \frac{R}{3} = α R \Rightarrow α = 1 / 3$

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