Question

a spherical ball with 10 cm external diameter, weighs 1789 1/3 g. find the thickness of the shell if the density is 7 g/cm³.

Answer 1

$$\begin{gathered} \mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{shell}=\frac{5368}{3}\mathrm{g} \\ \mathrm{density}=7\mathrm{g}/{\mathrm{cm}}^3 \\ \mathrm{So}\mathrm{volume}=\frac{\mathrm{mass}}{\mathrm{density}}=\frac{\frac{5368}{3}}{7}=255.61{\mathrm{cm}}^3 \\ \mathrm{Now}\mathrm{external}\mathrm{diameter}=10\mathrm{cm} \\ \mathrm{So}\mathrm{external}\mathrm{radius}=5\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{internal}\mathrm{radius}=\mathrm{r} \\ \mathrm{So}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \frac{4}{3}\mathrm{\pi }\left({\mathrm{R}}^3-{\mathrm{r}}^3\right)=255.61 \\ \Rightarrow 5^3-{\mathrm{r}}^3=\frac{255.61\times 3}{4\times 3.14}=14.58 \\ \Rightarrow {\mathrm{r}}^3=125-14.58 \\ \Rightarrow {\mathrm{r}}^3=110.42 \\ \Rightarrow \mathrm{r}=4.79\mathrm{cm} \\ \mathrm{So}\mathrm{thickness}=\mathrm{R}-\mathrm{r}=5-4.79=0.21\mathrm{cm} \end{gathered}$$