A hollow sphere of internal and external diameters 4 cm and 8 cm is melted into cone of base diameter 8 cm. The height of the cone is:

Correct option is B

Given:
Internal diameter of the hollow sphere, $$d_1 = 4 \, \text{cm}$$​​
External diameter of the hollow sphere $$d_2 = 8 \, \text{cm}$$​​
Diameter of the base of the cone,  d = 8 cm
Formula Used:
Volume of a hollow sphere = Volume of the outer sphere - Volume of the inner sphere.
Volume of a sphere =$$\frac{4}{3} \pi r^3$$​​
Volume of a cone =$$\frac{1}{3} \pi r^2 h$$​​
Since the material is melted and reshaped, the volume remains the same.
Solution:
Internal radius of the hollow sphere $$r_1 = \frac{d_1}{2} = \frac{4}{2} = 2 \, \text{cm}$$​

External radius of the hollow sphere $$r_2 = \frac{d_2}{2} = \frac{8}{2} = 4 \, \text{cm}$$​
Radius of the base of the cone,$$r = \frac{d}{2} = \frac{8}{2} = 4 \, \text{cm}$$​
Calculate the volume of the hollow sphere:
$$\text{Volume of the outer sphere} = \frac{4}{3} \pi r_2^3 = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \, \text{cm}^3$$​​

$$\text{Volume of the inner sphere} = \frac{4}{3} \pi r_1^3 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \, \text{cm}^3$$​​

​$$\text{Volume of the hollow sphere} = \frac{256}{3} \pi - \frac{32}{3} \pi = \frac{224}{3} \pi \, \text{cm}^3$$​​

$$\text{Volume of the cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (4)^2 h = \frac{1}{3} \pi (16) h = \frac{16}{3} \pi h \, \text{cm}^3$$​​

​$$\frac{224}{3} \pi = \frac{16}{3} \pi h$$​​

224 = 16 h

​$$h = \frac{224}{16} = 14 \, \text{cm}$$​​
The height of the cone is 14 cm.