if the radius of cylinder is increased by 10%,how much will the length be decreased so that volume remains the same?
$\displaystyle V = \pi r^2 L$
So if we increase r by 10% $\displaystyle r \to \frac{11r}{10}$
So what is our new L'?
$\displaystyle V = \pi r^2 L = \pi \left ( \frac{11r}{10} \right )^2 L'$
$\displaystyle \pi r^2 L = \pi \frac{121}{100} r^2 L'$
Cancelling the common $\displaystyle \pi r^2$:
$\displaystyle L = \frac{121}{100} L'$
$\displaystyle L' = \frac{100}{121} L$
$\displaystyle L' = 0.826 L$
or we have reduced the length to about 82.6% of it's original value.
-Dan