A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm.

Question:

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

Solution:

Radius of hemisphere = 9 cm

Volume of hemisphere $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi \times 9 \times 9 \times 9 \mathrm{~cm}^{3}$

Height of cone = 72 cm
Let the radius of the cone be r cm.

Volume of the cone $=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=\frac{1}{3} \pi \mathrm{r}^{2} \times 72 \mathrm{~cm}^{3}$

The volumes of the hemisphere and cone are equal. 
Therefore,

$\frac{1}{3} \pi r^{2} \times 72=\frac{2}{3} \pi \times 9 \times 9 \times 9$

$r^{2}=\frac{2 \times 9 \times 9 \times 9}{72}=\frac{81}{4}$

 

$r=\sqrt{\frac{81}{4}}=\frac{9}{2}=4.5 \mathrm{~cm}$

The radius of the base of the cone is 4.5 cm.

 

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