A resistance of 5 ohm is further drawn so that its length becomes double. Its resistance now is
- A. 5 ohms
- B. 7.5 ohms
- C. 10 ohms
- D. 20 ohms
Answer: D. 20 ohms
Explanation: When a wire is drawn to double its length, its new length $l' = 2l$. Because the volume stays constant, its new cross-sectional area becomes $A' = A/2$. New resistance $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R$. Therefore, $R' = 4 \times 5 = 20~\Omega$.