Question

A wire of length \(L\) and resistance \(R\) is stretched to twice its length; what is the new resistance?

• A. \(R\)
• B. \(2R\)
• C. \(3R\)
• D. \(4R\)

Hint:

When a wire is stretched without changing volume: \[ A \propto \frac{1}{L} \] Thus resistance varies as \(R \propto L^2\).

Answer 1

The Correct Option is D

Concept: Resistance of a wire is given by \[ R=\rho\frac{L}{A} \] where \(\rho\) = resistivity, \(L\) = length of wire, \(A\) = cross-sectional area. When a wire is stretched, its volume remains constant. \[ A_1L_1=A_2L_2 \]

Step 1: Apply the change in length. Initial length: \[ L_1=L \] Final length: \[ L_2=2L \] Using constant volume: \[ A_1L_1=A_2L_2 \] \[ A_1L=A_2(2L) \] \[ A_2=\frac{A_1}{2} \]

Step 2: Calculate the new resistance. Initial resistance: \[ R_1=\rho\frac{L}{A_1} \] Final resistance: \[ R_2=\rho\frac{2L}{A_2} \] Substitute \(A_2=\frac{A_1}{2}\): \[ R_2=\rho\frac{2L}{A_1/2} \] \[ R_2=4\rho\frac{L}{A_1} \] \[ R_2=4R \]

Step 3: Final result. \[ \boxed{R_2=4R} \]