Question

If the diameter of wire is decreased by $10\%$, by how much percent approximately will the length be increased to keep the volume constant?

• A. $5\%$
• B. $17\%$
• C. $20\%$
• D. $23\%$

Answer 1

Answer: (D)

$23\%$

Since the wire is cylindrical in shape, its $volume=\frac{1}{4}\pi d^{2}l$

Where, $d$ is the diameter, $l$ is the length

Also, $r^2={\left(\frac{d}{2}\right)}^2=\frac{d^2}{4}$

Decreased diameter $=d-10$ of $d$ $=d(1-\frac{10}{100})=\frac{9d}{10}$

New length $=l_n$. Then

New volume $=$ original volume ...given

$\Rightarrow \frac{1}{4}\pi {\left(\frac{9d}{10}\right)}^2{l}_n=\frac{1}{4}\pi d^{2}l$

$\Rightarrow \frac{l_n}{l}=\frac{{10}^2}{9^2}=\frac{100}{8l}\Rightarrow \frac{100}{81}l$

$\therefore$ % increase in length $=(\frac{\frac{100}{81}l-l}{l}\times 100)$

$=\frac{19}{81}\times 100=23\%$ ( approx,)