Question

A uniform metal wire of length \(l\) has 10 Ω resistance. Now this wire is stretched to a length \(2l \) and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is :

• A. 10 Ω
• B. 5 Ω
• C. 40 Ω
• D. 20 Ω

Answer 1

The Correct Option is A

To solve this problem, we need to consider the properties of resistors, particularly how resistance changes with the length of the conductor.

1. The initial resistance of the wire of length \( l \) is 10 Ω. 
2. When the wire is stretched to double its length, \( 2l \), the resistance changes due to the formula for resistance: \(R = \rho \frac{L}{A}\), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire.
3. Since the volume of the wire remains constant when stretched, \((l \cdot A_1) = (2l \cdot A_2)\), where \( A_1 \) and \( A_2 \) are the original and new cross-sectional areas, respectively. So, \(A_2 = \frac{A_1}{2}\).
4. Therefore, the new resistance is given by \(R' = \rho \frac{2l}{A_2} = \rho \frac{2l}{A_1/2} = 4 \cdot \rho \frac{l}{A_1} = 4 \cdot 10 \, \Omega = 40\, \Omega\).
5. Now, this wire is bent to form a circle. In a circle, the resistance across any diameter consists of two resistors, each of resistance equal to half of the total resistance of the circular wire: \(\frac{40}{2} = 20 \, \Omega\) per semicircle.
6. These two semicircular resistors are in parallel when measured across a diameter. The formula for equivalent resistance \( R \) of two parallel resistors is: \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\) 
   Here, both \( R_1 \) and \( R_2 \) are 20 Ω, so: \(\frac{1}{R} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}\).
7. Therefore, the equivalent resistance, \( R \), across a diameter is 10 Ω.

Hence, the correct answer is 10 Ω.