Question

A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2 cm and 2 cm, find the diameter of the third ball.

Answer 1

Volume of lead ball with radius \(\frac{3}{2}\) cm = \(\frac{4}{3}\)πr³

= \(\frac{4}{3}\) x π x (\(\frac{3}{2}\))³ 

Let, Diameter of first ball (d₁) = \(\frac{3}{2}\) cm 

Radius of first ball (r₁) = \(\frac{3}{4}\) cm 

Diameter of second ball (d₂) = 2 cm 

Radius of second ball (r₂) = \(\frac{2}{2}\) cm 

= 1 cm 

Diameter of third ball (d₃) = d 

Radius of third ball (r₃) = \(\frac{d}{2}\) cm 

Now,

So, diameter of third ball is 2.5 cm.