Question

A wire of resistance 100 \(\Omega\) is stretched so that its length increases by 20%. The stretched wire is then bent in the form of a rectangle whose length and breadth are in the ratio 3 : 2. The effective resistance between the ends of any diagonal of the rectangle is

• A. 36 \(\Omega\)
• B. 72 \(\Omega\)
• C. 28.8 \(\Omega\)
• D. 43.2 \(\Omega\)

Hint:

Calculate resistance changes due to stretching and use series-parallel rules to find effective resistance.

Answer 1

The Correct Option is A

Step 1: Calculate new resistance after stretching
Resistance \( R \propto \frac{L}{A} \).
Length increases by 20%, so \[ L' = 1.2 L, \quad A' = \frac{A}{1.2} \quad (\text{assuming volume constant}) \] Resistance after stretching: \[ R' = R \times \frac{L'}{L} \times \frac{A}{A'} = 100 \times 1.2^2 = 100 \times 1.44 = 144 \, \Omega. \] Step 2: Wire bent into rectangle
Let length \( = 3x \), breadth \( = 2x \). Total perimeter = \( 2(3x + 2x) = 10x \).
Original length after stretch is \( L' = 1.2 L \), so \[ 10x = L' \implies x = \frac{L'}{10} \] Step 3: Calculate resistance of each side
Resistance per unit length \( r = \frac{R'}{L'} = \frac{144}{L'} \).
Resistance of length side = \[ R_{length} = r \times 3x = \frac{144}{L'} \times 3 \times \frac{L'}{10} = \frac{432}{10} = 43.2 \Omega \] Resistance of breadth side = \[ R_{breadth} = r \times 2x = \frac{144}{L'} \times 2 \times \frac{L'}{10} = \frac{288}{10} = 28.8 \Omega \] Step 4: Calculate resistance between diagonal ends
Two paths between diagonal ends: one with length and breadth in series \[ R_1 = 43.2 + 28.8 = 72 \, \Omega \] Other path is similar with same resistance \( R_2 = 72 \Omega \).
Equivalent resistance \( R_{eq} \) of two parallel resistors: \[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{72 \times 72}{72 + 72} = \frac{5184}{144} = 36 \, \Omega. \] Step 5: Conclusion
Effective resistance between diagonal ends is 36 \(\Omega\).