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(D) In a potentiometer,the balancing length $\ell$ is directly proportional to the terminal potential difference $V$ across the cell.For an open circu

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$r=R$

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(D) In a potentiometer,the balancing length $\ell$ is directly proportional to the terminal potential difference $V$ across the cell.For an open circuit,the balancing length $\ell_{1}$ corresponds to the e.m.f. $E$ of the cell: $E \propto \ell_{1}$.When the cell is shunted with an external resistance $R$,the terminal potential difference $V$ is given by $V = E - Ir$,where $I = \frac{E}{R+r}$.Thus,$V = E - \left(\frac{E}{R+r}\right)r = E\left(1 - \frac{r}{R+r}\right) = E\left(\frac{R}{R+r}\right)$.The new balancing length $\ell_{2}$ corresponds to the terminal potential difference $V$: $V \propto \ell_{2}$.Given $\ell_{2} = \frac{\ell_{1}}{2}$,we have $\frac{V}{E} = \frac{\ell_{2}}{\ell_{1}} = \frac{1}{2}$.Substituting the expression for $V/E$: $\frac{R}{R+r} = \frac{1}{2}$.Solving for $r$: $2R = R + r$,which gives $r = R$.

In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

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