Question

A wire of stretched 50%, calculate percentage change in its resistance.

• A. 0.50%
• B. 25%
• C. 100%
• D. 125%

Answer 1

The correct option is D 125%

Let initial length of the wire be $l$.

Resistance of the wire $R=\frac{\rho l^2}{V}$

New length of the wire $l^{\prime }=l+0.5l=1.5l$

New resistance of the wire $R^{\prime }=\frac{\rho (1.5l{)}^2}{V}=2.25R$

Percentage change in resistance $\frac{R^{\prime }-R}{R}\times 100=\frac{2.25R-R}{R}\times 100=125$ %