Physics - Current Electricity Question with Solution | TestHub

Given initially a resistance $P=4\Omega$ is connected in the left gap and let $Q$ is connected in the right gap of meter bridge and the null point is obtained at $60cm$, From the balanced condition of the Wheatstone bridge

$\frac{P}{Q}=\frac{R}{S}...\left(1\right)$

Let $\rho$ is the resistance per unit length of meter bridge wire than

For wire $A\text{N}$, $P=\rho l=40\rho \left(\because l=40cm\right)$ And for NB wire

$$\begin{gathered} Q=\rho \left(100-l\right) \\ Q=\rho \left(100-60\right)=40\rho \end{gathered}$$

On substituting the values from equation $\left(1\right)$

$\frac{4}{60}=\frac{Q}{40}$

$\Rightarrow Q=\frac{16}{6}=\frac{8}{3}\Omega$

Now an unknown resistance $R$ is connected in series with the $4\Omega$ resistance and the null point is obtained at $l=80cm$

Now, $P=\rho l=80l$

and $$\begin{gathered} Q=\rho \left(100-l\right) \\ Q=\rho \left(100-80\right)=20\rho \end{gathered}$$

Then from equation $\left(1\right)$

$$\begin{gathered} \frac{4+R}{80}=\frac{Q}{20} \\ \Rightarrow \frac{4+R}{80}=\frac{\frac{8}{3}}{20}\left(\because Q=\frac{8}{3}\right) \\ \Rightarrow \frac{4+R}{80}=\frac{8}{60} \\ \Rightarrow R=\frac{20}{3}\Omega \end{gathered}$$