Question

A resistor of wire 24 cm length and resistance 8Ω is stretched into a uniform wire of 48 cm length, then the new resistance will be:

• A. 32Ω
• B. 8Ω
• C. 16Ω
• D. 4Ω

Hint:

When a wire is stretched, its resistance increases because the length increases, and the cross-sectional area decreases.

Answer 1

The Correct Option is A

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] where: - \( \rho \) is the resistivity of the material, - \( L \) is the length of the wire, - \( A \) is the cross-sectional area of the wire. When the wire is stretched, its length changes, but the volume of the wire remains constant. Since the volume \( V = A \times L \), if the wire is stretched to double the original length, the new length \( L' \) is 48 cm, and the original length \( L = 24 \) cm. The volume remains constant, so: \[ A_1 L_1 = A_2 L_2 \quad \Rightarrow \quad A_2 = \frac{A_1 L_1}{L_2} \] Thus, the new area \( A_2 \) is half the original are(A) Since resistance is directly proportional to the length and inversely proportional to the area, the new resistance will be: \[ R' = R \times \left( \frac{L'}{L} \right) \times \left( \frac{A}{A'} \right) \] Substitute the values: \[ R' = 8 \times \frac{48}{24} \times 2 = 8 \times 2 \times 2 = 32 \, \Omega \] Thus, the new resistance of the stretched wire is \( 32 \, \Omega \).