A spherical hollow is made in a lead sphere o | vedclass

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Question

Let $\rho$ be the density of lead

Then, $\mathrm{M}=\frac{4}{3} \pi \mathrm{R}^{3} \rho=$ mass of total sphere

$\mathrm{m}_{1}=\frac{4}{3} \pi\l

Vedclass · Accepted Answer

$\frac {R}{14}$

Vedclass · Answer

Let $ho$ be the density of lead

Then, $\mathrm{M}=\frac{4}{3} \pi \mathrm{R}^{3} ho=$ mass of total sphere

$\mathrm{m}_{1}=\frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}ight)^{3} ho=\mathrm{mass}$ of removed part $=\frac{\mathrm{M}}{8}$

$\mathrm{m}_{2}=\mathrm{M}-\frac{\mathrm{M}}{8}=\frac{7 \mathrm{M}}{8}=$ mass of remaining sphere

Choosing the centre of big sphere as the origin.

$\mathrm{X}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

$0=\frac{\frac{M}{8} 	imes \frac{R}{2}+\frac{7 M}{8} 	imes x_{2}}{M}$

Solving, we get $; x_{2}=\frac{-R}{14}$

i.e.., centre of mass of hollowed sphere be at a distance of $\mathrm{R} / 14$ on left on $\mathrm{O}$

i.e., shift in centre of mass $=\frac{\mathrm{R}}{14}$

A spherical hollow is made in a lead sphere of radius $R,$ such that its surface touches the outside surface of lead sphere and passes through the centre. What is the shift in the centre of mass of lead sphere due to the following ?

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