A hemisphere of lead of radius of 8cm is recast into a right circular cone of height of 64cm. Find the radius of the base of the cone.(Use $$\pi= \frac{22}{7}$$​)

Correct option is B

Given:

Radius of hemisphere = 8cm

Height of cone =64cm

Formula Used:

Volume of hemisphere =$$\frac{2}{3} \pi r^{3}$$​​

Volume of cone = $$\frac{1}{3}\times\pi \times r^{2}\times h$$​​

Solution:

If hemisphere is recast into cone then the volume remains same:

​$$\frac{2}{3} \pi r^{3} = \frac{1}{3}\times\pi \times r^{2}\times h$$​​

​$$\frac{2}{3} \pi 8^{3} = \frac{1}{3}\times\pi \times r^{2}\times 64$$​​

​$$2 \times 8^{3} = r^{2}\times 64$$​​

​$$2 \times 8 = r^{2}$$​​

​$$r= 4cm$$​​