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In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length $\ell$ is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance $R$. If $R$ is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be
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Verified Answer
The correct answer is:
$\ell / 2$
$\ell / 2$
Balancing length $l$ will give emf of cell $\therefore E=K l$
Here $\mathrm{K}$ is potential gradient.
If the cell is short circuited by resistance 'R'
Let balancing length obtained be $l^{\prime}$ then
$$
\begin{aligned}
& V=k l^{\prime} \\
& r=\left(\frac{E-V}{V}\right) R \\
\Rightarrow & V=E-V \\
\Rightarrow & 2 V=E \\
\text { or, } & 2 K l^{\prime}=K l \\
\therefore \quad & l^{\prime}=\frac{l}{2}
\end{aligned}
$$
Here $\mathrm{K}$ is potential gradient.
If the cell is short circuited by resistance 'R'
Let balancing length obtained be $l^{\prime}$ then
$$
\begin{aligned}
& V=k l^{\prime} \\
& r=\left(\frac{E-V}{V}\right) R \\
\Rightarrow & V=E-V \\
\Rightarrow & 2 V=E \\
\text { or, } & 2 K l^{\prime}=K l \\
\therefore \quad & l^{\prime}=\frac{l}{2}
\end{aligned}
$$
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