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(A) In a meter bridge,the null point condition is given by the Wheatstone bridge principle: $\frac{R_1}{R_2} = \frac{P}{Q}$,where $P$ and $Q$ are the

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$x$

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(A) In a meter bridge,the null point condition is given by the Wheatstone bridge principle: $\frac{R_1}{R_2} = \frac{P}{Q}$,where $P$ and $Q$ are the resistances of the two segments of the wire $AB$ divided by the jockey.Let the resistance per unit length of the wire be $\sigma$. If the length of the wire is $L = 100 \ cm$,then $P = \sigma x$ and $Q = \sigma (L - x)$.The condition becomes $\frac{R_1}{R_2} = \frac{\sigma x}{\sigma (L - x)} = \frac{x}{L - x}$.Since the resistance per unit length $\sigma = \frac{\rho}{A}$ (where $\rho$ is resistivity and $A$ is the cross-sectional area),$\sigma$ cancels out from the ratio.Therefore,the null point position $x$ is independent of the radius (and thus the cross-sectional area) of the wire $AB$.Hence,if the radius of the wire is doubled,the null point position remains unchanged at $x$.

The experimental setup of a meter bridge is shown in the figure. If the null deflection point of the galvanometer is at distance $x$ from end $A$,what will be the new null point if the radius of the wire $AB$ is doubled?

Similar Questions

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