Question

The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the
figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{glass},h=16\mathrm{cm}\mathrm{and} \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{cylinder}=\mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{hemisphere},r=\frac{7}{2}\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{apparent}\mathrm{capacity}\mathrm{of}\mathrm{the}\mathrm{glass}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylinder} \\ =\mathrm{\pi }{r}^{2}h \\ =\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 16 \\ =616{\mathrm{cm}}^3 \\ \mathrm{Also}, \\ \mathrm{The}\mathrm{actual}\mathrm{capacity}\mathrm{of}\mathrm{the}\mathrm{glass}=\mathrm{Volume}\mathrm{of}\mathrm{cylinder}-\mathrm{Volume}\mathrm{of}\mathrm{hemisphere} \\ =616-\frac{2}{3}\mathrm{\pi }{r}^3 \\ =616-\frac{2}{3}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times \frac{7}{2} \\ =616-\frac{539}{6} \\ =\frac{3157}{6}{\mathrm{cm}}^3 \\ \approx 526.17{\mathrm{cm}}^3 \end{gathered}$$