Pipe Radius Reduction and Length Calculation

This problem involves understanding the relationship between the dimensions (radius and length) of a cylinder (pipe) and its volume. We are given that the pipe's radius changes, and we need to find the resulting change in its length while keeping the volume constant.

Volume Calculation for a Pipe

The volume ($V$) of a pipe, modeled as a cylinder, is calculated using the formula:

$ V = \pi r^2 h $

where $r$ is the radius and $h$ is the length.

Analyzing the Radius Change

Let the original radius be $r_1$ and the original length be $h_1$. The original volume is $V_1 = \pi r_1^2 h_1$.

The problem states the radius is "reduced by one fourth". Based on the provided options and correct answer, we interpret this to mean the new radius ($r_2$) is one-fourth of the original radius ($r_1$).

Therefore, the relationship is:

$ r_2 = \frac{1}{4} r_1 $

Calculating the New Length

Let the new length be $h_2$. The new volume is $V_2 = \pi r_2^2 h_2$.

We are given that the volume remains unchanged, so $V_1 = V_2$.

1. Set the volume expressions equal: $ \pi r_1^2 h_1 = \pi r_2^2 h_2 $
2. Substitute the expression for $r_2$: $ \pi r_1^2 h_1 = \pi \left(\frac{1}{4} r_1\right)^2 h_2 $
3. Simplify the equation: $ \pi r_1^2 h_1 = \pi \left(\frac{1}{16} r_1^2\right) h_2 $
4. Cancel out $\pi$ and $r_1^2$ from both sides (since $r_1$ is not zero): $ h_1 = \frac{1}{16} h_2 $
5. Solve for the new length $h_2$: $ h_2 = 16 h_1 $

This shows that the new length ($h_2$) is 16 times the original length ($h_1$).