• # question_answer A spherical hollow cavity is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through the center. What is the shift in the center of mass of the lead sphere due to hollowing? A)  $\frac{R}{7}$                       B)   $\frac{R}{14}$                      C) $\frac{R}{2}$                        D)    $R$

Let $\rho$by the density of lead. Then M$M=\frac{4}{3}\pi {{r}^{3}}p=$mass of total sphere ${{m}_{1}}=\frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}}\rho =$ Mass of removed part $=\frac{M}{8}$ ${{m}_{2}}=M-\frac{M}{8}=\frac{7M}{8}=$Mass of remaining sphere choosing the centre of the sphere as the origin. ${{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$or, $o=\frac{\frac{M}{8}\times \frac{R}{2}+\frac{7M}{8}{{x}_{2}}}{M}$or ${{x}_{2}}=-\frac{R}{14}\,\,\therefore$Shift in centre of mass$=R/14.$ Hence, the correction option is (b).