Question

A cylindrical rod of iron rod whose height is eight times its radius is melted and cast into spherical balls each of half of the radius of the cylinder, the number of spherical balls is:
A. $12$
B. $16$
C. $24$
D. $48$

Answer 1

To solve this problem, we need to find the volume of the rod and each spherical ball. Since, the volume of the rod is equal to the sum of the volume of all spherical balls,equating this we get a number of spherical balls.

Complete step-by-step answer:
It is given that; the height of the cylindrical iron rod is eight times its radius. It is melted and cast into some spherical balls. The radius of the spherical balls is half of the radius of the cylindrical iron rod.
We have to find out the number of spherical balls.
Let us consider, the radius of the cylindrical iron rod is $2r$.
Since, the height of the cylindrical iron rod is eight times of its radius, the height of the cylindrical iron rod is $2 \times 8r = 16r$.
We know that the volume of the cylinder with $r$ as radius and $h$ as height is $\pi {r^2}h$.
So, the volume of the cylindrical rod is $\pi {(2r)^2}(16r)$
Now, the radius of the spherical balls is half of the radius of the cylindrical iron rod.
So, the radius of the spherical ball is $\dfrac{{2r}}{2} = r$.
We know that the volume of a sphere with radius $r$ is $\dfrac{4}{3}\pi {r^3}$.
Dividing volume of cylindrical rod with volume of spherical ball,we get number of spherical balls
So, the number of spherical balls is $\dfrac{{\pi {{(2r)}^2}(16r)}}{{\dfrac{4}{3}\pi {r^3}}}$
Simplifying we get,
The number of spherical balls is $\dfrac{3}{4} \times \dfrac{{\pi {{(2r)}^2}(16r)}}{{\pi {r^3}}} = 48$
Hence, the number of spherical balls is $48$.

So, the correct answer is “Option D”.

Note: We know that, the volume of the cylinder with $r$ as radius and $h$ as height is $\pi {r^2}h$ and the volume of a sphere with radius $r$ is $\dfrac{4}{3}\pi {r^3}$.Students should remember these formulas for solving these types of questions.