Question

From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid. [CBSE 2011]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cylinder},H=14\mathrm{cm}, \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{cylinder},R=\frac{7}{2}\mathrm{cm}, \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{each}\mathrm{conical}\mathrm{holes},r=2.1\mathrm{cm}\mathrm{and} \\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{each}\mathrm{conical}\mathrm{holes},h=4\mathrm{cm} \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{solid}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylinder}-\mathrm{Volume}\mathrm{of}2\mathrm{conical}\mathrm{holes} \\ =\mathrm{\pi }{R}^{2}H-2\times \frac{1}{3}\mathrm{\pi }{r}^{2}h \\ =\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 14-\frac{2}{3}\times \frac{22}{7}\times 2.1\times 2.1\times 4 \\ =539-36.96 \\ =502.04{\mathrm{cm}}^3 \end{gathered}$$

So, the volume of the remaining solid is 502.04 cm³.