Question

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.
A) 3
B) 4
C) 5
D) 6

Answer 1

Hey.


Here is the answer.


Given : The radius of the base of the cylindrical rod = height of the rod

i .e., r = h

Also, radius of the spheres formed is R = r/2


Solution :


We know that Volume of a cylinder

= π r^2 h

So, volume of the rod here will be

= π r^3 __________{as h = r }


Also the volume of sphere is given by

= 4/3 π R^3

So, volume of one such sphere will be

= 4/3 π (r/2)^3

= 4/3 π r^3/8

= 1/6 π r^3


Hence, the no. of balls will be

= (Volume of rod) / (volume of sphere)

= (π r^3) / (1/6 π r^3)

= 6


Hence, the no. of such balls is 6.


Thanks