Question

The resistance of a 10 m long wire is $10\Omega$. Its length is increased by 25% by streatching the wire uniformly . the resistance of wire will change to (approximately)

• A. 12.5$\Omega$
• B. 14.5 $\Omega$
• C. 15.6$\Omega$
• D. 16.6$\Omega$

Answer 1

The correct option is C 15.6$\Omega$

Given,
Initial resistance $R_1=\rho \frac{L_1}{A_1}=10\Omega$

New length after stretch is $L_2=1.25L_1$

For constant volume

$V=A_1{L}_1=A_2{L}_2$

Hence,

$\frac{A_1}{A_2}=\frac{L_2}{L_1}=1.25$

$\Rightarrow A_2=\frac{A_1}{1.25}$

Where,

$V=Volume$

$A=Area$

$L=Length$

New resistance of stretched wire,

$R_2=\rho \frac{L_2}{A_2}=\rho \frac{1.25L_1}{A_1/1.25}={1.25}^2\rho \frac{L_1}{A_1}$

$R_2={1.25}^2\times 10=15.625\Omega$

the resistance of wire will change to (approximately) $15.6\Omega$