Question : A hemisphere of lead of radius 4 cm is cast into a right circular cone of height 72 cm. What is the radius of the base of the cone?
Option 1: 1.63 cm
Option 2: 1.35 cm
Option 3: 1.33 cm
Option 4: 1.45 cm
Correct Answer: 1.33 cm
Solution :
Volume of hemisphere = $\frac{ 2 \pi \times \text{Radius}^3}{3}$
Volume of cone = $\frac{\pi \times \text{Radius}^2 \times \text{Height}}{3}$
Given, Radius = 4 cm
Volume of hemisphere = $\frac{ 2 \pi \times 4^3}{3}$ = $\frac{ 128 \pi}{3}$ cm
3
Height of the cone = 72 cm
Let the radius of the base of the right circular cone be $R$ cm.
Volume of the cone = $\frac{\pi \times R^2 \times 72}{3}$
According to the question,
$\frac{ 128 \pi}{3} = \frac{\pi \times R^2 \times 72}{3}$
$⇒R^2 = \frac{16}{9}$
$⇒R = \frac{4}{3}$
$\therefore R= 1.33$
$\therefore$ The radius of the base of the cone is 1.33 cm.
Hence, the correct answer is 1.33 cm.
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