The balancing length for a cell is 560 ; cm i | vedclass

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(A) In a potentiometer experiment,the balancing length $l_1$ corresponds to the $EMF$ of the cell,$E = k l_1$,where $k$ is the potential gradient.When

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$12$

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(A) In a potentiometer experiment,the balancing length $l_1$ corresponds to the $EMF$ of the cell,$E = k l_1$,where $k$ is the potential gradient.When an external resistance $R$ is connected in parallel,the terminal voltage $V$ is given by $V = E - Ir = E - \frac{E}{R+r} r = E \left( \frac{R}{R+r} \right) = k l_2$.Thus,$\frac{E}{V} = \frac{l_1}{l_2} = \frac{R+r}{R} = 1 + \frac{r}{R}$.Given $l_1 = 560 \; cm$ and the change in length is $60 \; cm$,the new balancing length $l_2 = 560 - 60 = 500 \; cm$.Substituting the values: $\frac{560}{500} = 1 + \frac{r}{10}$.$1.12 = 1 + \frac{r}{10} \implies 0.12 = \frac{r}{10} \implies r = 1.2 \; \Omega$.Since $r = \frac{N}{10} \; \Omega$,we have $\frac{N}{10} = 1.2$,which gives $N = 12$.

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