Answer:
Option C
Explanation:
R1 = $\frac{\rho l_{1}}{A_{1}}$ , now l2 = 2l1 r2 = 2r1
A1 = πr12
A2 = π(r2)2 = π(2r1)2 = 4πr1 2= 4A1
R2 = $\frac{\rho \left(2l_{1}\right)}{4A_{1}} = \frac{\rho l_{1}}{2A_{1}}$
= $\frac{R_{1}}{2}$
.'. Resistance is halved, but specific resistance remains the same.