A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal. In order to reduce its weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of metal A to the volume of the metal B in the solid.

Given,

Radius of cone (R) = 6 cm

Height of cone (H) = 10 cm

Radius of conical hole (r) = $\dfrac{6}{2}$ = 3 cm

Height of conical hole (h) = 4 cm.

Volume of complete cone = $\dfrac{4}{3}πR^2H$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 6^2 \times 10 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 36 \times 10 \\[1em] = \dfrac{31680}{21} \\[1em] = 1508.57 \text{ cm}^3.$

Volume of conical hole = $\dfrac{4}{3}πr^2h$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 3^2 \times 4 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 9 \times 4 \\[1em] = \dfrac{3168}{21} \\[1em] = 150.85 \text{ cm}^3.$

Volume of cone (with only metal A) = Volume of complete cone - Volume of conical hole

= 1508.57 - 150.85

= 1357.71 cm³.

$\text{Ratio} = \dfrac{\text{Vol. of cone (with only metal A)}}{\text{Vol. of conical hole}} \\[1em] = \dfrac{1357.71}{150.85} \\[1em] = \dfrac{9}{1} \\[1em] = 9 : 1$

Hence, ratio of the volume of metal A to the volume of the metal B in the solid = 9 : 1.