Question

Equal amounts of a metal are converted into cylindrical wires of different lengths (\(L\)) and cross-sectional area (\(A\)). The wire with the maximum resistance is the one, which has

• A. length = \(L\) and area = \(A\)
• B. length = \(\frac{L}{2}\) and area = \(2A\)
• C. length = \(2L\) and area = \(\frac{A}{2}\)
• D. all have the same resistance, as the amount of metal is the same

Hint:

Resistance increases when length increases and area decreases because \(R = \rho \frac{L}{A}\).

Answer 1

The Correct Option is C

Step 1: Recall resistance of a wire.
Resistance of a cylindrical wire is:
\[ R = \rho \frac{L}{A} \]
Step 2: Condition of equal amount of metal.
Equal amount of metal means same volume:
\[ V = LA = \text{constant} \]
Step 3: Compare resistance for each option.
Option (A):
\[ R_A = \rho \frac{L}{A} \]
Option (B):
\[ R_B = \rho \frac{L/2}{2A} = \rho \frac{L}{4A} = \frac{R_A}{4} \]
Option (C):
\[ R_C = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} = 4R_A \]
Step 4: Identify maximum resistance.
Clearly, \(R_C\) is maximum because it becomes \(4\) times \(R_A\).
Final Answer:
\[ \boxed{\text{(C) length = \(2L\) and area = \(\frac{A}{2}\)}} \]