Question

If the diameter of cross-section of a wire is decreased by $5$% how much present will the length be increased so that the volume remains the same

Answer 1

Let $r$ be the radius of cross-section of wire and $h$ be its length.

Then volume $=\pi r^{2}h....\left(i\right)$
$5$% of diameter of cross-seciton $=\frac{5}{100}\times 2r=\frac{r}{10}$
New diameter $=2r-\frac{r}{10}=\frac{19r}{10}$
New radius $=\frac{19r}{20}$

Let new length be $h_1$.

Volume $=\pi {\left(\frac{19r}{20}\right)}^2{h}_1...\left(ii\right)$

From (i) and (ii) we have

$\pi r^{2}h={\left(\frac{19r}{20}\right)}^2{h}_1$

$\Rightarrow h=\frac{361}{400}{h}_1\Rightarrow h_1=\frac{400}{361}h$

Increase in length $=h_1-h=\frac{400}{361}-h=\frac{39h}{361}$

Percentage increase in length $=\frac{h_1-h}{h}\times 100=\frac{39h}{h}\times 100$

$=\frac{3900}{361}$

$=10.8$%