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(B) Let $\lambda$ be the resistance per unit length of the potentiometer wire $AB$.When the key $k$ is open,the cell $E_1$ is in an open circuit,so th

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$6$

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(B) Let $\lambda$ be the resistance per unit length of the potentiometer wire $AB$.When the key $k$ is open,the cell $E_1$ is in an open circuit,so the balance length $x_1 = 75\,cm$ corresponds to the $EMF$ of the cell:$E_1 = \lambda x_1 \ldots (1)$When the key $k$ is closed,the cell $E_1$ sends a current $i$ through the external resistor $R = 24\,\Omega$. The terminal voltage $V$ is balanced at $x_2 = 60\,cm$:$V = E_1 - ir = \lambda x_2 \ldots (2)$From equations $(1)$ and $(2)$,we have:$\frac{E_1}{V} = \frac{x_1}{x_2} = \frac{75}{60} = 1.25$Since $V = E_1 - ir$ and $i = \frac{E_1}{R+r}$,we have $V = E_1 - \left(\frac{E_1}{R+r}ight)r = E_1 \left(1 - \frac{r}{R+r}ight) = E_1 \left(\frac{R}{R+r}ight)$.Therefore,$\frac{E_1}{V} = \frac{R+r}{R} = 1 + \frac{r}{R}$.Substituting the values: $1.25 = 1 + \frac{r}{24} \Rightarrow 0.25 = \frac{r}{24}$.$r = 0.25 	imes 24 = 6\,\Omega$.

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

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