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?

Answer 1

Let, the length of wire $=lunit$

Fraction required $=a$

Then, the lenght of wire which is stretched $=al$

$R_l=R_{\left(wire\right)}=\frac{\rho l}{A}$

$R_{l-al}=\frac{\rho (l-al)}{A}=R_{l(1-a)}$

$R_{al}=\frac{\rho al}{A}=a{R}_l$

On stretching ${}^{\prime }a{l}^{\prime }$ length of wire, new length of ${}^{\prime }a{l}^{\prime }$ part:

$l-al+l^{\prime }=\frac{3l}{2}$

$l^{\prime }=l(\frac{1}{2}+a)$

Volume of wire ${}^{\prime }a{l}^{\prime }$ is same:

$l^{\prime }{A}^{\prime }=\left(al\right)A$

$l(\frac{1}{2}+a)A^{\prime }=\left(al\right)A$

$A^{\prime }=\frac{aA}{(\frac{1}{2}+a)}$

Now, resistance of stretched parts,

$R^{\prime }=\frac{\rho \times l^{\prime }}{A^{\prime }}=\frac{\rho \times l(\frac{1}{2}+a)}{\frac{aA}{\frac{1}{2}+a}}=\frac{\rho l}{A}\frac{(\frac{1}{2}+a{)}^2}{a}$

$\Rightarrow R^{\prime }=\frac{R_l(\frac{1}{2}+a{)}^2}{a}$

According to the question,

$R_{l-al}+R^{\prime }=4R_l$

$\Rightarrow R_{l(1-a)}+R_l\frac{(\frac{1}{2}+a{)}^2}{a}=4R_l$

$\Rightarrow a-a^2+\frac{1}{4}+a^2+a=4a$

$\Rightarrow \frac{1}{4}+2a=4a$

$\Rightarrow \frac{1}{4}=4a-2a$

$\Rightarrow \frac{1}{4}=2a$

$\Rightarrow a=\frac{1}{8}$