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A potentiometer balances at $44 \mathrm{~cm}$ when a cell of internal resistance $1 \Omega$ is in the secondary circuit. To obtain the balancing point at $40 \mathrm{~cm}$, the resistance to be connected parallel to cell is
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The correct answer is:
$10 \Omega$
(b) Internal Resistance, $\mathrm{r}=1 \Omega$
Balancing Point, $\ell_1=44 \mathrm{~cm}$.
Obtain Balancing Point $\ell_2=40 \mathrm{~cm}$.
Internal Resistance is given by
$$
\mathrm{r}=\mathrm{R}\left[\frac{\ell_1-\ell_2}{\ell_2}\right]
$$
$\begin{aligned} & \mathrm{R}=\frac{\mathrm{r} \ell_2}{\ell_1-\ell_2}=\frac{1 \times 40}{44-40} \\ & \Rightarrow \mathrm{R}=\frac{40}{4}=10 \\ & \Rightarrow \mathrm{R}=10 \Omega\end{aligned}$
Balancing Point, $\ell_1=44 \mathrm{~cm}$.
Obtain Balancing Point $\ell_2=40 \mathrm{~cm}$.
Internal Resistance is given by
$$
\mathrm{r}=\mathrm{R}\left[\frac{\ell_1-\ell_2}{\ell_2}\right]
$$
$\begin{aligned} & \mathrm{R}=\frac{\mathrm{r} \ell_2}{\ell_1-\ell_2}=\frac{1 \times 40}{44-40} \\ & \Rightarrow \mathrm{R}=\frac{40}{4}=10 \\ & \Rightarrow \mathrm{R}=10 \Omega\end{aligned}$
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