In the shown arrangement of the meter bridge | vedclass

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(A) In a meter bridge,the balancing condition is given by the ratio of resistances: $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$.Here,$R_{AC}$ is the res

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$x$

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(A) In a meter bridge,the balancing condition is given by the ratio of resistances: $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$.Here,$R_{AC}$ is the resistance of the wire segment $AC$ and $R_{CB}$ is the resistance of the wire segment $CB$.The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.Substituting this into the balancing condition: $\frac{R_1}{R_2} = \frac{\rho (x) / A}{\rho (100-x) / A} = \frac{x}{100-x}$.Since the cross-sectional area $A$ cancels out from the numerator and the denominator,the balancing length $x$ depends only on the ratio of the resistors $R_1$ and $R_2$.Therefore,changing the radius of the wire $AB$ does not affect the balancing length $x$.

In the shown arrangement of the meter bridge experiment,if the length $AC$ corresponding to the null deflection of the galvanometer is $x$,what would be its value if the radius of the wire $AB$ is doubled?

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