Question

If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________ %

Answer 1

We know that,

\(R = \frac{\rho l}A\)

Length of new wire 

\(l_1 = l + 0.2l\)

\(l_1 = 1.2l\)

Volume constant when wire stretch then

\(A.l = A_1 \times 1.2l\)

\(A_1 = \frac A{1.2}\)

then 

Resistance of new wire

\(R_1 = \frac{\rho 1.2l}{A_1}\)

\(R_1 = \frac{\rho\times 1.2l}{\frac A{1.2}}\) 

\(R_1 = \frac {\rho l}A 1.44\)

\(R_1 = 1.44R\) 

So change in resistance 

\(\Delta R = R_1 - R\)

\(= 1.44 R - R\)

\(\Delta R = 0.44R\) 

% change = \(\frac{\Delta R}{R} \times 100 \) 

\(=\frac{0.44 R}{R} \times 100 \) 

⇒ 44%