Question

An iron spherical ball of radius 21 cm is melted and is re-cast into hemispheres of radius 7 cm. The number of hemispheres formed will be :

• A. 27
• B. 64
• C. 54
• D. 16

Answer 1

The correct option is C 54

In the process of recasting, the volume will remain constant.

Let's assume $x$ such hemispherical balls will be formed.

Let R be the radius of spherical ball and r be the radius of each hemisphere.
Then, R = 21 cm and r = 7 cm.

$Volumeofsphere=\frac{4}{3}\pi R^3$
$Volumeofahemisphere=\frac{2}{3}\pi r^3$

$\therefore \frac{4}{3}\times \pi \times R^3=x\times \frac{2}{3}\times \pi \times r^3$

$\Rightarrow \frac{4}{3}\times \pi \times {21}^3=x\times \frac{2}{3}\times \pi \times 7^3$

$\Rightarrow 2\times {21}^3=x\times 7^3$

$\Rightarrow x=\frac{2\times 21\times 21\times 21}{7\times 7\times 7}$

$\Rightarrow x=54$