Question

A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. What is the number of such spherical balls ?

Answer 1

Let the radius of the base of the cylinder be r units. Height = 8r units 

Its volume = πr² × 8r = 8πr³ cu. units 

Radius of sphere = \(\frac{r}{2}\)units  ∴ Its volume = \(\frac43π\big(\frac{r}{2}\big)^3 = \frac{πr^3}{6}\) cu units.

Number of spherical balls = \(\frac{\text{Volume of cylinder}}{\text{Volume of sphere}} = \frac{8πr^3}{πr^3}\times6\) = 48.