A metallic cylindrical rod is melted and recast as a rod which has one-fourth of the radius of the original rod. Its length will be how many times the original length?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 16 times
Melting and recasting does not change the volume of metal, so the volume \(\pi r^2h\) of the new rod equals that of the old one.
Old rod: radius \(r\), length \(h\); new rod: radius \(\frac{r}{4}\), length \(H\)
\(\displaystyle \pi\left(\frac{r}{4}\right)^2H=\pi r^2h\)
\(\displaystyle \frac{H}{16}=h\), so \(H=16h\)