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 determined by the formula:
\[R = \rho \times \frac{L}{A}\]
Here:
- \( \rho \) represents the material's resistivity, which is constant for wires made of the same material.
- \( L \) denotes the wire's length.
- \( A \) signifies the wire's cross-sectional area.
For a wire with a circular cross-section, the area \( A \) is related to the diameter \( D \) by:
\[A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}\]

Step 2: Minimizing the Resistance:
- To achieve minimum resistance, the wire must have the largest possible cross-sectional area \( A \) (implying a larger diameter \( D \)) and the shortest possible length \( L \).
- Resistance decreases as the cross-sectional area increases, which is proportional to the square of the diameter. A shorter length \( L \) also directly reduces resistance.

Step 3: Answer Explanation:
When considering four wires of the same material to achieve minimum resistance:
- The diameter should be maximized, specifically 4 times 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:
Minimum resistance is achieved when the diameter is \( 4D \) and the length is \( 2L \). This combination minimizes resistance by maximizing the cross-sectional area and minimizing the wire's length.