Question

A metal rod is melted and reshaped such that the radius of the rod is reduced to one – fourth.
The new length of the rod is how many times the original length?

• A. Remains the same
• B. 4 times
• C. times
• D. 16 times
• E. 8 times

Answer 1

The correct option is D

16 times

Clearly, the volume of the rod remains the same.
Let the original radius = $r$ and original length = $h$
$\therefore$ the new radius = $\frac{r}{4}$ and let new length = $H$
Then, $\pi r^2\times h=\pi {\left(\frac{r}{4}\right)}^2\times H\to H=16h$