The diameter of a right circular cylinder is decreased to one third of its initial value. If the volume of the cylinder remains the same, then the height becomes how many times of the initial height?

9

We know that,

Volume of the cylinder = πr²h

New radius of the cylinder = r × \(\frac{1}{3}\) = \(\frac{r}{3}\) 

Let the new height of the cylinder = H

According to the question, π (\(\frac{r}{3}\))²H = πr²h

(\(\frac{r^2}{9}\) ) × H = r²h

H = 9h

\(\frac{H}{h}\) =\(\frac{9}{1}\) 

So,the height becomes 9 times of the initial height.