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(C) Let $E$ be the $EMF$ of the cell and $r$ be its internal resistance. The balance length $l_1 = 60 \ cm$ corresponds to the $EMF$ of the cell: $E =

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$1.2$

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(C) Let $E$ be the $EMF$ of the cell and $r$ be its internal resistance. The balance length $l_1 = 60 \ cm$ corresponds to the $EMF$ of the cell: $E = k l_1$,where $k$ is the potential gradient.When the cell is shunted by a resistance $R' = 6 \ \Omega$,the terminal potential difference $V$ is given by $V = E - Ir = E - \frac{E}{R' + r} r = E \left( \frac{R'}{R' + r} ight)$.The new balance length $l_2 = 50 \ cm$ corresponds to this terminal potential difference: $V = k l_2$.Dividing the two equations: $\frac{E}{V} = \frac{l_1}{l_2} = \frac{R' + r}{R'}$.Substituting the values: $\frac{60}{50} = \frac{6 + r}{6}$.$1.2 = 1 + \frac{r}{6} \implies 0.2 = \frac{r}{6}$.$r = 0.2 	imes 6 = 1.2 \ \Omega$.

Similar Questions

Resistance of $100 \, cm$ long potentiometer wire is $10 \, \Omega$. It is connected to a battery of $2 \, V$ and a resistance $R$ in series. $A$ source of $10 \, mV$ gives a null point at $40 \, cm$ length. The value of external resistance $R$ is ........... $\Omega$.

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Explain the method to measure the internal resistance of a cell using a potentiometer.

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