In a meter bridge experiment,resistances are | vedclass

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(D) In the balanced position of the meter bridge,the condition is given by $\frac{P}{Q} = \frac{l}{100-l}$,where $l$ is the distance of the null point

Vedclass · Accepted Answer

$\frac{20}{3}\,\Omega$

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(D) In the balanced position of the meter bridge,the condition is given by $\frac{P}{Q} = \frac{l}{100-l}$,where $l$ is the distance of the null point from end $A$.Initially,$P = 4\,\Omega$ and $l = 60\,cm$. Therefore,$100-l = 40\,cm$.Substituting these values into the balance condition:$\frac{4}{Q} = \frac{60}{40}$$\frac{4}{Q} = \frac{3}{2}$$Q = \frac{8}{3}\,\Omega$.Now,an unknown resistance $R$ is connected in series with $P$,so the new resistance is $P' = P + R = 4 + R$. The new null point is at $l' = 80\,cm$,so $100-l' = 20\,cm$.Using the balance condition again:$\frac{4+R}{Q} = \frac{80}{20}$$\frac{4+R}{8/3} = 4$$4+R = 4 	imes \frac{8}{3}$$4+R = \frac{32}{3}$$R = \frac{32}{3} - 4 = \frac{32-12}{3} = \frac{20}{3}\,\Omega$.Thus,the value of the unknown resistance $R$ is $\frac{20}{3}\,\Omega$.

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