Question

A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of $\frac{3}{2}$ cm and its depth is $\frac{8}{9}$ cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape. [CBSE 2015]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cylinder},R=3\mathrm{cm}, \\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cylinder},H=5\mathrm{cm}, \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{conical}\mathrm{hole},r=\frac{3}{2}\mathrm{cm}\mathrm{and} \\ \mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{conical}\mathrm{hole},h=\frac{8}{9}\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylinder},V=\mathrm{\pi }{R}^{2}H \\ =\mathrm{\pi }\times 3^2\times 5 \\ =45\mathrm{\pi }{\mathrm{cm}}^3 \\ \mathrm{Also}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{removed}\mathrm{from}\mathrm{the}\mathrm{cylinder},v=\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ =\frac{\mathrm{\pi }}{3}\times {\left(\frac{3}{2}\right)}^2\times \left(\frac{8}{9}\right) \\ =\frac{2\mathrm{\pi }}{3}{\mathrm{cm}}^3 \\ \mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{metal}\mathrm{left}\mathrm{in}\mathrm{the}\mathrm{cylinder},V\text{'}=V-v \\ =45\mathrm{\pi }-\frac{2\mathrm{\pi }}{3} \\ =\frac{133\mathrm{\pi }}{3}{\mathrm{cm}}^3 \\ \therefore \mathrm{The}\mathrm{required}\mathrm{ratio}=\frac{V\text{'}}{v} \\ =\frac{\left(\frac{133\mathrm{\pi }}{3}\right)}{\left(\frac{2\mathrm{\pi }}{3}\right)} \\ =\frac{133}{2} \\ =133:2 \end{gathered}$$

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.