To determine the internal resistance of a cell by using a potentiometer, the null point is at $1…

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \mathrm{~m}$, when shunted by $3 \Omega$ resistance and at a length $1.5 \mathrm{~m}$, when cell is shunted by $6 \Omega$ resistance. The internal resistance of the cell is
  1. $1 \Omega$
  2. $4 \Omega$
  3. $2 \Omega$
  4. $6 \Omega$

Solution

$\begin{aligned} & \mathrm{r}=\mathrm{R}\left(\frac{\ell_1}{\ell_2}-1\right)=\mathrm{R}^{\prime}\left(\frac{\ell_1}{\ell_2^{\prime}}-1\right) \\ & \ell_2=1 \mathrm{~m}, \ell_2^{\prime}=1.5 \mathrm{~m} \mathrm{R}=3 \Omega, \mathrm{R}^{\prime}=6 \Omega \\ & \therefore 3\left(\frac{\ell_1}{1}-1\right)=6\left(\frac{\ell_1}{1.5}-1\right) \end{aligned}$ Solving we get $\ell_1=3 \mathrm{~m}$ $\therefore \mathrm{r}=3\left(\frac{3}{1}-1\right)=3 \times 2=6 \Omega$ :

Asked in: MHT CET 2021 (24 Sep Shift 1)

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