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A lamina lies in the $x - y$ plane. Its mass distribution is symmetrical about the $y$ axis and its moment of inertia about the axis $X X'$ is $I = y^{2} - \frac{4 r}{3 π} y + \frac{r^{2}}{4}$ . The coordinates of the centre of mass of the lamina are

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$(\frac{8 r}{3 π}, 0)$

$(0, \frac{2 r}{3 π})$

$(0, \frac{4 r}{3 π})$

$(\frac{r}{3}, \frac{8 r}{3 π})$

answer is B.

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Detailed Solution

The moment of inertia about the axis XX’ is minimum if XX’ passes through the centre of mass

$d I / d y = 2 y - 4 r / (3 π) = 0,$
Which gives $y = y_{c} = 2 r / (3 π) .$

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