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

Answer 1

The Correct Option is D

To determine how many times the new length is compared to the original length, given that the radius of the wire is decreased to one-third and its volume remains constant, follow these steps:

1. Start by considering the formula for the volume of a cylindrical wire, \( V = \pi r^2 l \), where \( r \) is the radius and \( l \) is the length of the wire.
2. Let the original radius be \( r \) and the original length be \( l \). Thus, the original volume is \( V = \pi r^2 l \).
3. Now, let the new radius be \(\frac{r}{3}\). Since the volume remains the same, we set up the equation for the new volume: \( V = \pi \left(\frac{r}{3}\right)^2 l_{\text{new}} \).
4. Simplify the new volume equation: \( V = \pi \frac{r^2}{9} l_{\text{new}} \).
5. Set the original volume equal to the new volume: \(\pi r^2 l = \pi \frac{r^2}{9} l_{\text{new}}\).
6. Cancel \(\pi r^2\) from both sides to obtain: \( l = \frac{l_{\text{new}}}{9} \).
7. Solve for \( l_{\text{new}} \): \( l_{\text{new}} = 9l \).
8. This indicates the new length is 9 times the original length.

Therefore, the new length is 9 times the original length.