Height of a cylindrical iron rod is four times to its radius of the base. If it is melted and recast into spherical balls

Correct option is: B) 3

Given that h = 4 r

\(\therefore\) Volume of cylindrical iron rod = \(\pi r^2h\) = \(4 \pi r^3\) (\(\because\) h = 4r)

\(\because\) Cylindrical iron rod is melted and recast into n spherical balls.

Then n \(\times\) volume of one spherical ball = Volume of cylindrical iron rod

\(\Rightarrow\) n = \(\frac {Volume \, of \, cylindrical \, iron \, rod}{Volume \, of \, one \, spherical\, ball}\)  = \(\frac {4 \pi r^3}{\frac 43 \pi r^3 }\) (\(\because\) r is also radius of formed spherical ball)

= 3 

Hence, number of balls thus formed is 3.

Correct option is: B) 3