Question

In case of four wires of same material, the resistance will be minimum if the diameter and length of the wire respectively are

• A. D/2 and L/4
• B. D/4 and 4L
• C. 2D and L
• D. 4D and 2L

Answer 1

The Correct Option is D

Step 1: Understanding the Factors Affecting Resistance:
The resistance \( R \) of a wire is given by the formula: \[ R = \rho \times \frac{L}{A} \] where:
- \( \rho \) is the resistivity of the material (constant for wires of the same material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
The cross-sectional area \( A \) of a wire with a circular cross-section is related to the diameter \( D \) by the formula: \[ A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]
Step 2: Minimizing the Resistance:
- For the resistance to be minimized, the wire should have the maximum possible cross-sectional area \( A \) (which means a larger diameter \( D \)) and the smallest possible length \( L \). - Since the area of the wire increases with the square of the diameter, a larger diameter results in lower resistance. A smaller length \( L \) also reduces resistance because resistance is directly proportional to the length.

Step 3: Answer Explanation:
To achieve the minimum resistance with four wires of the same material:
- The diameter should be as large as possible, and it should be 4 times greater than the original diameter \( D \) (i.e., 4D).
- The length should be reduced to 2 times the original length \( L \) (i.e., 2L).

Step 4: Conclusion:
The resistance will be minimum if the diameter is \( 4D \) and the length is \( 2L \), as this combination minimizes resistance by increasing the cross-sectional area and reducing the length of the wire.