CBSE
2020
Class Class 12 · Physics
2 Marks · Short
✅ Verified
A wire of resistance 4 $\Omega$ is stretched to twice its original length. What will be the resistance of the stretched wire?
✅ Answer & Solution
New length $L' = 2L$. Volume is conserved: $A'L' = AL \Rightarrow A' = A/2$.
$$R' = \frac{\rho L'}{A'} = \frac{\rho \cdot 2L}{A/2} = 4 \cdot \frac{\rho L}{A} = 4R = 4 \times 4 = 16\ \Omega$$
✅ Verified by Super Admin
