Question

If a wire is stretched, so that its length becomes $20$% more than its initial length, the percentage increase in the resistance of the wire is

• A. 40%
• B. 10%
• C. 44%
• D. 25%

Answer 1

The correct option is C 44%

$R=\left(\frac{\rho l}{A}\right)=\frac{\rho l^2}{V}$ ( $\because V=Al$ )

When the wire is stretched, it's volume remains constant.

So, $R_1=\frac{\rho l{{}_1}^2}{V}$

$R_2=\frac{\rho l{{}_2}^2}{V}$

Given, $l_2=1.2l_1$

$R_2=\frac{\rho }{V}(1.2l_1{)}^2$

$R_2=\frac{\rho }{V}\left(1.44l{{}_1}^2\right)$

$R_2=1.44\left(\frac{\rho l{{}_1}^2}{V}\right)$

$R_2=1.44R_1$

$R_2-R_1=0.44R_1$

So, The resistance is increased by $44$%

Hence, option ($c$) is the correct answer.

Alternate method:

Whenever a wire is stretched,

$R\propto l^2$

$\%increase=(\frac{R_2}{R_1}-1)\times 100=(\frac{{l_2}^2}{{l_1}^2}-1)\times 100$

$\%increase=(\left(\frac{36}{25}\right)-1)\times 100=44$