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

Since volume is fixed, shrinking the radius must be compensated by stretching the length, and the compensation follows the square of the radius change because volume depends on radius squared.

1. Volume of a cylindrical wire is $\pi r^2 h$, so for a fixed volume, $r^2 h$ stays constant.
2. If the radius shrinks to $\frac{1}{3}$ of its original value, $r^2$ shrinks to $\frac{1}{9}$ of its original value.
3. To keep $r^2 h$ constant, $h$ must grow by exactly the factor that cancels this, which is $9$.

So the correct answer is option D, 9 times.