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A potentiometer has a uniform wire of length $5 \mathrm{~m}$. A battery of emf $10 \mathrm{~V}$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 \mathrm{~cm}$. The emf of the secondary cell is
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$4 \mathrm{~V}$
Given, length of potentiometer wire
$l=5 \mathrm{~m}$
emf of battery, $E=10 \mathrm{~V}$
Potential gradient across the potentiometer wire,
$K=\frac{E}{l}=\frac{10}{5}=2 \mathrm{Vm}^{-1}$
When secondary cell is connected to the circuit of potentiometer, then balancing length,
$l_{1}=200 \mathrm{~cm}=2 \mathrm{~m}$
$\therefore$ emf of the secondary cell, $E_{s}=K l_{1}=2 \times 2=4 \mathrm{~V}$
$l=5 \mathrm{~m}$
emf of battery, $E=10 \mathrm{~V}$
Potential gradient across the potentiometer wire,
$K=\frac{E}{l}=\frac{10}{5}=2 \mathrm{Vm}^{-1}$
When secondary cell is connected to the circuit of potentiometer, then balancing length,
$l_{1}=200 \mathrm{~cm}=2 \mathrm{~m}$
$\therefore$ emf of the secondary cell, $E_{s}=K l_{1}=2 \times 2=4 \mathrm{~V}$
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