Question

The radius of a wire is decreased to one-third, and its volume remains the same. The new length is how many times the original length?

• A. 2 times
• B. 4 times
• C. 5 times
• D. 9 times

Hint:

Volume depends on r²h; shrinking r to 1/3 shrinks r² to 1/9, so h must grow 9 times to keep volume constant.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept.
A wire is shaped like a cylinder, with volume \(=\pi r^2h\).

Step 2: Set up the equal-volume equation.
Let the original radius be r and length h, and the new radius be \(\dfrac{r}{3}\) with new length \(h'\). Since volume stays the same: \(\pi r^2h=\pi\left(\dfrac{r}{3}\right)^2h'\).

Step 3: Simplify.
\(\pi r^2h=\pi\dfrac{r^2}{9}h' \Rightarrow h=\dfrac{h'}{9} \Rightarrow h'=9h\).

Step 4: Final Answer.
The new length is 9 times the original length, so option D is correct.