Question

A given piece of wire of length l, cross sectional area A and resistance R is stretched uniformly to a wire of length 2l. The new resistance will be

• A. 2 R
• B. 4 R
• C. R/2
• D. Remains unchanged

Answer 1

Answer: (B)

4 R

Volume $=A\times L$

We know,

$R=\frac{\rho \times L}{A}$

Hence,

$\frac{R^{\prime }}{R}=\frac{L^{\prime }}{L}\times \frac{A}{A^{\prime }}=\frac{2L}{L}\times \frac{A}{\frac{A}{2}}=4$

Hence,

$R^{\prime }=4R$