Question

The resistance of a wire of length \(l\) and cross sectional area \(A\) is \(R\). The resistance of another wire of the same material of length \(3l\) and cross sectional area \(\frac{A}{3}\) is

• A. 3R
• B. R
• C. 9R
• D. R/3
• E. R/9

Hint:

Resistance is directly proportional to length and inversely proportional to cross-sectional area.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The resistance of a wire depends on its intrinsic properties (resistivity) and its physical dimensions (length and cross-sectional area). The question asks how the resistance changes when these dimensions are altered.
Step 2: Key Formula or Approach:
The formula for the resistance (\(R\)) of a wire is:
\[ R = \rho \frac{l}{A} \] where \(\rho\) (rho) is the resistivity of the material, \(l\) is the length of the wire, and \(A\) is its cross-sectional area.
Step 3: Detailed Explanation:
Initial Wire:
Length = \(l\)
Cross-sectional area = \(A\)
The resistance is given as \(R\). So, we have:
\[ R = \rho \frac{l}{A} \quad \text{--- (1)} \] (The resistivity \(\rho\) is the same for both wires because they are made of the "same material".)
New Wire:
New length, \(l' = 3l\)
New cross-sectional area, \(A' = \frac{A}{3}\)
Let the new resistance be \(R'\). Using the resistance formula for the new wire:
\[ R' = \rho \frac{l'}{A'} \] Substitute the new dimensions into this formula:
\[ R' = \rho \frac{3l}{\frac{A}{3}} \] \[ R' = \rho \frac{3l \times 3}{A} \] \[ R' = \rho \frac{9l}{A} \] \[ R' = 9 \left( \rho \frac{l}{A} \right) \quad \text{--- (2)} \] Now, substitute the expression for \(R\) from equation (1) into equation (2):
\[ R' = 9R \] Step 4: Final Answer:
The resistance of the new wire is 9R, which corresponds to option (C).