The resistance per centimeter of a meter brid | vedclass

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(D) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{R_1}{R_2}$,where $P$ and $Q$ are the resistances in the gaps,and $R_1$

Vedclass · Accepted Answer

$40$

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(D) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{R_1}{R_2}$,where $P$ and $Q$ are the resistances in the gaps,and $R_1$ and $R_2$ are the resistances of the wire segments.Let the resistance per unit length be $r$. For a balancing length $\ell_1 = 40 \ cm$,the other segment length is $\ell_2 = 100 - 40 = 60 \ cm$.The resistance of the segments are $r \ell_1$ and $r \ell_2$.The balance condition is $\frac{X}{25} = \frac{r \ell_1}{r \ell_2} = \frac{\ell_1}{\ell_2} = \frac{40}{60} = \frac{2}{3}$.When the wire is replaced by one with resistance $2r$ per unit length,the new resistances of the segments become $(2r) \ell_1'$ and $(2r) \ell_2'$.The new balance condition is $\frac{X}{25} = \frac{(2r) \ell_1'}{(2r) \ell_2'} = \frac{\ell_1'}{\ell_2'}$.Since the ratio $\frac{X}{25}$ remains constant,$\frac{\ell_1'}{\ell_2'} = \frac{2}{3}$.Given $\ell_1' + \ell_2' = 100 \ cm$,we have $\ell_1' = 40 \ cm$ and $\ell_2' = 60 \ cm$.Thus,the balancing length remains unchanged at $40 \ cm$.

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