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 the wire becomes $\frac{3}{2}$ times 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 $\frac{1}{8}$

Let us choose the length of the resistance wire initially as $l$ and cross sectional area as $A$.$$R_{in}=\frac{\rho l}{\text{A}}$$


Let us choose a part whose length is $x$.

So, the length of remaining portion is $l-x$.

Final length of the wire is $\frac{3}{2}l$.

After we stretch the chosen part whose initial length was $x$.

As volume of this chosen part remains same.


${\text{A}}^{\prime }=\frac{x\text{A}}{\frac{l}{2}+x}$

We finally have two resistances in series as same amount of current will flow through them. So, the final resistance of the wire,

$R_f=R_1+R_2$

$\Rightarrow R_f=\frac{\rho (l-x)}{\text{A}}+\frac{\rho (\frac{l}{2}+x)}{{\text{A}}^{\prime }}$

On substituting the value of $A^{\prime }$

$\Rightarrow R_f=\frac{\rho (l-x)}{\text{A}}+\frac{\rho (\frac{l}{2}+x{)}^2}{x\text{A}}$


Given, $R_f=4R_{in}$

$\Rightarrow \frac{\rho (l-x)}{\text{A}}+\frac{\rho (\frac{l}{2}+x{)}^2}{x\text{A}}=4\times \frac{\rho l}{A}$

$\Rightarrow xl-x^2+\frac{l^2}{4}+lx+x^2=4lx$

$\Rightarrow \frac{l^2}{4}=2lx$

$\Rightarrow x=\frac{l}{8}$

$\therefore$ value of fraction $\frac{x}{l}=\frac{1}{8}$

Hence, option $\left(\text{B}\right)$ is the correct answer.