Wheatstone bridge principle is used to measure the specific resistance $$S_1$$ of given wire, having length $$L$$, radius $$r$$. If X is the resistance of wire, then specific resistance is : $$S_1 = X\frac{\pi r^2}{L}$$. If the length of the wire gets doubled then the value of specific resistance will be :

We are asked to determine how the specific resistance (resistivity) changes when the length of the wire is doubled.

A key concept is that specific resistance (resistivity) is an intrinsic property of the material.

The relationship between resistance and resistivity is given by $$R = \frac{\rho L}{A} = \frac{S_1 L}{\pi r^2}$$, where $$R$$ is the resistance, $$\rho = S_1$$ is the specific resistance (resistivity), $$L$$ is the length, and $$A = \pi r^2$$ is the cross-sectional area.

Rewriting this expression gives $$S_1 = \frac{R \cdot \pi r^2}{L} = \frac{X \cdot \pi r^2}{L}$$.

When the wire length is doubled, two approaches confirm that resistivity remains constant. First, since resistivity depends only on the material (its temperature, composition, etc.) and is independent of the dimensions of the wire, replacing it with a wire of length $$2L$$ does not change $$S_1$$.

Alternatively, if the length becomes $$2L$$ while maintaining the same cross section, the new resistance is $$X' = \frac{S_1 \times 2L}{\pi r^2} = 2X$$. Calculating resistivity from the new measurements gives $$S_1' = \frac{X' \cdot \pi r^2}{2L} = \frac{2X \cdot \pi r^2}{2L} = \frac{X \cdot \pi r^2}{L} = S_1$$, confirming that the resistivity remains the same regardless of the wire's dimensions.

The correct answer is Option 4: $$S_1$$.