Question

The radius of a pipe is reduced by one fourth. If its volume remains unchanged, what will be its length now?

• A. \(3\) times the original length
• B. Will remain unchanged
• C. \(9\) times the original length
• D. \(16\) times the original length

Hint:

For a cylinder with constant volume, length is inversely proportional to the square of radius.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A pipe is mathematically a cylinder. If the radius is modified while keeping the total volume constant, we need to calculate the ratio of change in its length.

Step 2: Key Formula or Approach:
The volume $V$ of a cylinder is: \[ V = \pi r^2 L \] where $r$ is the radius and $L$ is the length. Since the volume remains constant: \[ V_1 = V_2 \] \[ \pi r_1^2 L_1 = \pi r_2^2 L_2 \implies L_2 = L_1 \left( \frac{r_1}{r_2} \right)^2 \]

Step 3: Detailed Explanation:

• Let the initial radius be $r_1$ and initial length be $L_1$.

• The question uses the phrase "reduced by one fourth", which can sometimes be translated colloquially or from local papers as "reduced to one-fourth". Let us evaluate both cases to match the standard multiple-choice answer:

• Case I: Radius reduced to one-fourth
If $r_2 = \frac{1}{4} r_1$: \[ L_2 = L_1 \left( \frac{r_1}{\frac{1}{4} r_1} \right)^2 \] \[ L_2 = L_1 (4)^2 = 16 L_1 \] This matches Option (D) perfectly.

• Case II: Radius reduced by one-fourth
If $r_2 = r_1 - \frac{1}{4} r_1 = \frac{3}{4} r_1$: \[ L_2 = L_1 \left( \frac{r_1}{\frac{3}{4} r_1} \right)^2 = \frac{16}{9} L_1 \] Since $\frac{16}{9}$ is not among the given options, the question setter intended Case I.

Step 4: Final Answer:
The length of the pipe becomes 16 times the original length. Thus, option (D) is the correct answer.