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?

### Concept & Volume Conservation A wire is treated as a cylinder. The volume of a cylinder depends linearly on its length (height) and quadratically on its radius. If the volume remains constant, any change in the radius squared must be inversely compensated by the length. $$V = \pi r^2 h$$ ### Step-by-Step Solution 1. **Define the original parameters:** - Let original radius $= r$ - Let original length (height) $= h$ - Original Volume $V_1 = \pi r^2 h$ 2. **Define the new parameters:** - New radius $r_2 = \frac{r}{3}$ - Let new length $= h_2$ - New Volume $V_2 = \pi \left(\frac{r}{3}\right)^2 h_2 = \pi \frac{r^2}{9} h_2$ 3. **Equate the volumes:** - $V_1 = V_2$ - $\pi r^2 h = \pi \frac{r^2}{9} h_2$ 4. **Solve for the new length:** - Divide both sides by $\pi r^2$: - $h = \frac{1}{9} h_2$ - $h_2 = 9h$ ### Exam Strategy & Shortcut Since volume is directly proportional to the square of the radius ($V \propto r^2 h$), if the radius becomes $\frac{1}{3}$, the $r^2$ factor becomes $\frac{1}{9}$. To keep the volume constant, the height must be multiplied by the reciprocal, which is $9$. ### Common Pitfall A common mistake is forgetting to square the change in radius, leading students to incorrectly guess that the length should increase by only $3$ times. ### Final Answer Therefore, the correct answer is **9 times**.