Question

In order to increase the resistance of a given wire of uniform cross-section to four times its value, a fraction of its length is stretched uniformly till the full length of wire becomes $3/2$ times of the original length. What is the value of this fraction?

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B $\frac{1}{8}$

During streching the volume of wire remains constant.



The resistance of the original wire of length $l$ is

$R=\frac{\rho l}{A}$

Now after stretching,

$R^{\prime }=4R......\left(1\right)$

From volume conservation on the stretched portion:

$Ax=A^{\prime }(0.5l+x)$

$\Rightarrow A^{\prime }=\frac{Ax}{0.5l+x}......\left(2\right)$

Since, $R^{\prime }=4R$

$\Rightarrow \frac{\rho (l-x)}{A}+\frac{\rho (0.5l+x)}{A^{\prime }}=\frac{4\rho l}{A}....\left(3\right)$

[Applying formula for resistance for individual parts of wire]

From eqs. (2) & (3)

$\frac{4\rho l}{A}=\frac{\rho (l-x)}{A}+\frac{\rho (0.5l+x{)}^2}{Ax}$

$\Rightarrow 4lx=(lx-x^2)+(0.5l{)}^2+x^2+(2\times 0.5lx)$

$\Rightarrow 4lx=lx-x^2+(0.5l{)}^2+x^2+xl$

$\Rightarrow 2lx=\left(0.5\right)\left(0.5\right)l^2$

$\Rightarrow x=\frac{25l^2}{200\times l}=\frac{l}{8}$

Hence, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: In such problems, think to apply}\\ \text{volume conservation and modified}\\ \text{relation for resistance in terms of resistivity,}\\ \text{length and cross-sectional area.}\end{array}$