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A uniform sphere has radius ‘R’. A sphere of diameter ‘R’ is cut from its edge so that shift in centre of mass is maximum. Then the distance of centre of mass of the remaining portion from the centre of mass of original sphere is

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$\frac{R}{14}$

$\frac{2 R}{7}$

$\frac{R}{7}$

$\frac{R}{18}$

answer is B.

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Detailed Solution

$\begin{array}{l} d = \frac{R}{2} \\ m_{1} = \frac{4}{3} {πR}^{3} ρ \\ m_{2} = \frac{4}{3} π {(\frac{R}{2})}^{3} ρ \\ {Δx}_{cm} = \frac{m_{2} d}{m_{1} - m_{2}} \end{array}$

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