• # question_answer A sphere of radius R has a concentric cavity of radius r. The relative density of the material of the sphere is a. It just floats when placed in a tank full of water. The ratio $\frac{R}{r}$ is: A)  ${{\left( \frac{\sigma -1}{\sigma } \right)}^{1/3}}$               B)  ${{\left( \frac{\sigma }{\sigma -1} \right)}^{1/3}}$ C)  ${{\left( \frac{\sigma +1}{\sigma } \right)}^{2}}$                        D)  ${{\left( \frac{\sigma }{\sigma +1} \right)}^{3}}$

Weight of the hollow sphere $=\frac{4}{3}\pi ({{R}^{3}}-{{r}^{3}})\sigma g=\frac{4}{3}\pi R_{1}^{3}g$ Weight of water displaced $=\frac{4}{3}\pi {{R}^{3}}\times l\times g$ Now according to law of floatation $\frac{4}{3}\pi ({{R}^{3}}-{{r}^{3}})\sigma g=\frac{4}{3}\pi R_{1}^{3}g$                 $\frac{R}{r}={{\left( \frac{\sigma }{\sigma -1} \right)}^{1/3}}$