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(A) In a metre bridge experiment,the null point condition is given by the Wheatstone bridge principle:$\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$Where $

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$x$

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(A) In a metre bridge experiment,the null point condition is given by the Wheatstone bridge principle:$\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$Where $R_{AC}$ is the resistance of the wire segment $AC$ and $R_{CB}$ is the resistance of the wire segment $CB$.Since 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 $(\pi r^2)$,$\frac{R_{AC}}{R_{CB}} = \frac{\rho (x) / A}{\rho (100-x) / A} = \frac{x}{100-x}$Thus,the condition for null deflection is $\frac{R_1}{R_2} = \frac{x}{100-x}$.This equation shows that the null point position $x$ depends only on the values of the resistors $R_1$ and $R_2$ and the total length of the wire. It is independent of the cross-sectional area (and thus the radius) of the wire $AB$,provided the wire is uniform.Therefore,if the radius of the wire $AB$ is doubled,the null position $x$ will remain unchanged.

In the shown arrangement of the experiment of the metre bridge,if $AC$ corresponding to 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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