The radius of a wire is decreased to one-third. If volume remains the same, the length will become

(d) 9 times

Let the new radius be $\frac{1}{3} r$.

 Suppose that the new height is H.
The volume remains the same.

i. e., $\pi r^{2} h=\pi \times\left(\frac{1}{3} r\right)^{2} \times H$

$\Rightarrow h=\frac{1}{9} H$

$\Rightarrow H=9 h$

$\therefore$ The new height becomes nine times the original height.