Question

Find the weight of a spherical shell of copper having external diameter $\mathbf{30}\mathbf{}\mathbf{cm}$ and thickness of $\mathbf{2}\mathbf{}\mathbf{cm}$. Density of copper is $\mathbf{10}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$.

Answer 1

Step 1: Given information.

External diameter of the spherical shell $\left(D\right)=30\mathrm{cm}$.

Thickness of ball$\left(t\right)=2\mathrm{cm}$.

Density of copper$=10\mathrm{g}/{\mathrm{cm}}^3$.

Step 2: Calculate the volume of the spherical shell.

The external radius of the spherical shell $\left(R\right)=\frac{30}{2}=15\mathrm{cm}$.

The inner diameter of the spherical shell $\left(d\right)=30-2t=30-4=26\mathrm{cm}$.

So, the inner radius $\left(r\right)=\frac{26}{2}=13\mathrm{cm}$.

$$\begin{gathered} \therefore \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{spherical}\mathrm{shell}=\frac{4}{3}\mathrm{\pi }{R}^3-\frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow =\frac{4}{3}\mathrm{\pi }(R^3-r^3) \\ \Rightarrow =\frac{4}{3}\times \frac{22}{7}\times \left({15}^3-{13}^3\right) \\ \Rightarrow =\frac{4}{3}\times \frac{22}{7}\times (3375-2197) \\ \Rightarrow =\frac{4}{3}\times \frac{22}{7}\times 1178 \\ \Rightarrow =4936.38{\mathrm{cm}}^3 \end{gathered}$$

Step 3: Calculate the weight of the spherical shell.

The density of copper is $10\mathrm{g}/{\mathrm{cm}}^3$.

$$\begin{gathered} \therefore \mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{a}\mathrm{spherical}\mathrm{shell}=4936.38\times 10 \\ \Rightarrow =49363.8\mathrm{grams} \\ \Rightarrow =\frac{49363.8}{1000} \\ \Rightarrow =49.364\mathrm{kg} \end{gathered}$$

Hence, the weight of the spherical shell is $49.364\mathrm{kg}$.