Question

A wire of length L and radius R is stretched to get the radius of cross-section halved. What is the new resistance?

• A. \( 4R \)
• B. \( 5R \)
• C. \( 8R \)
• D. \( 16R \)

Hint:

When a wire is stretched, its length increases, and the cross-sectional area decreases. The new resistance can be calculated by considering the change in both length and area.

Answer 1

The Correct Option is D

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire. The area of the cross-section of the wire is: \[ A = \pi R^2 \] When the wire is stretched, the length \( L \) increases, and the radius \( R \) is halved. Let the new length be \( L' \) and the new radius be \( R' = \frac{R}{2} \). Since the volume of the wire remains constant (because the wire is just stretched, not material added or removed), the volume before and after stretching is the same: \[ \pi R^2 L = \pi (R/2)^2 L' \] Simplifying: \[ R^2 L = \frac{R^2}{4} L' \] \[ L' = 4L \] So, the new resistance \( R' \) is: \[ R' = \rho \frac{L'}{A'} = \rho \frac{4L}{\pi (R/2)^2} = \rho \frac{4L}{\pi \frac{R^2}{4}} = \rho \frac{16L}{\pi R^2} = 16R \]