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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$\frac{{πb}^{2}}{(a^{2} - c^{2})}$

$\frac{{ca}^{2}}{(c^{2} - b^{2})}$

$\frac{{cb}^{2}}{(a^{2} - b^{2})}$

$\frac{{πc}^{2}}{(a^{2} - b^{2})}$

answer is B.

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Detailed Solution

Consider the mass of the removed portion as negative.

Mass of original disc $= M = σ π a^{2}$

Mass of removed portion $= - m = - σ π b^{2}$

$Position of centre of mass = \frac{M . x_{M} + (- m) x_{m}}{M - m}$

$x_{c . o . m} = \frac{(- {σπb}^{2}) . c}{{σπa}^{2} - {σπb}^{2}} = \frac{{cb}^{2}}{a^{2} - b^{2}}$

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