Resistances are connected in a meter bridge c | vedclass

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(D) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l}{100-l}$.Initially,$P = 4\,\Omega$ and $l_1 = 40\,cm$. So,$100 - l_1

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$20$

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(D) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l}{100-l}$.Initially,$P = 4\,\Omega$ and $l_1 = 40\,cm$. So,$100 - l_1 = 60\,cm$.$\frac{P}{Q} = \frac{40}{60} = \frac{2}{3} \implies \frac{4}{Q} = \frac{2}{3} \implies Q = 6\,\Omega$.When an unknown resistance $x$ is connected in series with $P$,the new resistance becomes $P' = P + x = 4 + x$.The new balancing length is $l_2 = 80\,cm$. So,$100 - l_2 = 20\,cm$.Using the balancing condition again:$\frac{P + x}{Q} = \frac{80}{20} = 4$.Substituting $Q = 6\,\Omega$:$\frac{4 + x}{6} = 4$$4 + x = 24$$x = 20\,\Omega$.

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