The radius of an iron rod is decreased to one fourth of its original rad

Question

The radius of an iron rod is decreased to one fourth of its original radius. If its volume remains constant, then the length will become.

Options:

A . 2 times

B . 12 times

C . 8 times

D . 16 times

Answer: Option D
:
D

Solution:- Volume of original rod = $π r^{2}$ h
Changed volume of changed rod = $π {(\frac{r}{4})}^{2} h_{1}$
But, Volume remains constant.
$π r^{2}$ h = $π {(\frac{r}{4})}^{2} h_{1} \Rightarrow h_{1}$ = 16h

Option(d).

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