A wire of length $ 25 \, \text{m} $ and cross-sectional area $ 5 \, \text{mm}^2 $ having resistivity $ 2 \times 10^{-6} \, \Omega \cdot \text{m} $ is bent into a complete circle. The resistance between diametrically opposite points will be:
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- 12.5 \( \Omega \)
- 50 \( \Omega \)
- 100 \( \Omega \)
- 2.5 \( \Omega \)
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to calculate the resistance between diametrically opposite points on a circular wire. Given are the following:
- Length of the wire, \(L = 25 \, \text{m}\)
- Cross-sectional area, \(A = 5 \, \text{mm}^2 = 5 \times 10^{-6} \, \text{m}^2\)
- Resistivity, \(\rho = 2 \times 10^{-6} \, \Omega \cdot \text{m}\)
Let's calculate the resistance using the formula for resistance of a wire:
\(R = \frac{\rho \cdot L}{A}\)
Substituting the values,
\(R = \frac{2 \times 10^{-6} \cdot 25}{5 \times 10^{-6}}\)
Simplifying this gives:
\(R = \frac{50 \times 10^{-6}}{5 \times 10^{-6}} = 10 \, \Omega\)
This resistance is for the full length of the wire. Since the wire is bent into a circle, we treat it as two equal halves. Therefore, the resistance between the two diametrically opposite points is equivalent to two halves of \(10 \, \Omega\) in parallel:
The resistance of each half-circle is \(R_{\text{half}} = \frac{R}{2} = \frac{10}{2} = 5 \, \Omega\).
The total resistance \(R_{\text{total}}\) across the diameter is then given by the parallel resistance formula:
\(\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{half}}} + \frac{1}{R_{\text{half}}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\)
This simplifies to:
\(R_{\text{total}} = \frac{5}{2} = 2.5 \, \Omega\)
Thus, the resistance between the diametrically opposite points is 2.5 \( \Omega \), which matches the given correct answer.
Approach Solution -2
The wire is bent into a circle. The resistance between diametrically opposite points will be the resistance of two semicircles in parallel.
\( L = 25 \) m, \( A = 5 mm^2 = 5 \times 10^{-6} m^2 \)
\( \rho = 2 \times 10^{-6} \Omega m \) \( R_{wire} = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10 \Omega \)
The resistance of each semicircle is \( \frac{R_{wire}}{2} = \frac{10}{2} = 5 \Omega \)
The equivalent resistance of two semicircles in parallel is: \( R_{eq} = \frac{R/2}{2} = \frac{10}{4} = 2.5 \Omega \)
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