A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is half the radius of the rod. Find the number of balls.

Volume of rod = $\mathrm{\pi r}2h$ =  $\mathrm{\pi r}2\times r$ =  $\mathrm{\pi r}3$  

Volume of one spherical ball =r/2 

Volume of the spherical ball =  $\frac{\frac{4}{3}}{\mathrm{\pi r}}$  

Number of balls=Volumeof rod/volumeof one ball