A cylindrical rod of iron whose height is twelve times its radius, is melted and cast into spherical balls, each of one-third the radius of the cylinder. Find the number of spherical balls.

The correct option is 243.

Step-by-step Explanation:

1. Define the dimensions of the cylinder:
Let the radius of the cylindrical iron rod be $r$.
According to the problem, the height ($h$) of the cylinder is twelve times its radius:

$h=12r$

2. Calculate the volume of the cylindrical rod:
The formula for the volume of a cylinder ($V_{\text{cylinder}}$) is:

$V_{\text{cylinder}}=\pi r^{2}h$

Substituting $h=12r$ into the formula:

$V_{\text{cylinder}}=\pi r^2\times 12r=12\pi r^3$

3. Define the dimensions and volume of one spherical ball:
Let the radius of each spherical ball be $r_{\text{ball}}$.
It is given that the radius of each ball is one-third the radius of the cylinder:

$r_{\text{ball}}=\frac{r}{3}$

The formula for the volume of a single sphere ($V_{\text{ball}}$) is:

$V_{\text{ball}}=\frac{4}{3}\pi r_{\text{ball}}^3$

Substituting $r_{\text{ball}}=\frac{r}{3}$:

$V_{\text{ball}}=\frac{4}{3}\pi {\left(\frac{r}{3}\right)}^3=\frac{4}{3}\pi \times \frac{r^3}{27}=\frac{4\pi r^3}{81}$

4. Calculate the number of spherical balls:
Since the cylinder is melted and recast into spherical balls, the total volume of all the spherical balls is equal to the volume of the cylindrical rod.
Let $n$ be the number of spherical balls.

$n\times V_{\text{ball}}=V_{\text{cylinder}}$

$n\times \frac{4\pi r^3}{81}=12\pi r^3$

Solving for $n$:

$n=\frac{12\pi r^3\times 81}{4\pi r^3}$

Canceling $\pi r^3$ from both numerator and denominator:

$n=3\times 81=243$

Thus, the number of spherical balls formed is 243.