Question

A wire of resistance R is stretched so that its length increases by 10%. The resistance of the wire increases by:

• A. 11%
• B. 15%
• C. 21%
• D. 28%

Answer 1

Answer: (C)

21%

Resistance of the wire $R=\frac{\rho L}{A}$

New length of the wire after stretching $L^{\prime }=1.1L$

As volume of the wire remains constant i.e. $A^{\prime }{L}^{\prime }=AL$

$\therefore$ $A^{\prime }\left(1.1L\right)=AL$ $⟹A^{\prime }=\frac{A}{1.1}$

Thus new resistance $R^{\prime }=\frac{\rho \left(1.1L\right)}{\frac{A}{1.1}}=1.21R$

$\therefore$ Percentage increase in resistance $\frac{R^{\prime }-R}{R}\times 100=\frac{1.21R-R}{R}\times 100=21$%