Wheatstone bridge principle is used to measure the specific resistance \( (S_1) \) of a given wire, having length \( L \), radius \( r \). If \( X \) is the resistance of the wire, then specific resistance is: $$ S_1 = X\left( \frac{\pi r^2}{L} \right). $$ If the length of the wire gets doubled, then the value of specific resistance will be:
- \( \frac{S_1}{4} \)
- \( 2S_1 \)
- \( \frac{S_1}{2} \)
- \( S_1 \)
The Correct Option is D
Solution and Explanation
The specific resistance (or resistivity) of a material is defined as: \(Sl = X (\frac{πr^2}{L})\) ,
where X is the resistance, r is the radius, and L is the length of the wire. Specific resistance is a material property and does not change with changes in dimensions such as length or radius. Doubling the length of the wire affects the resistance X, but the specific resistance Sl remains unchanged.
The Correct answer is: \( S_1 \)
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