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A $2 \mathrm{~V}$ battery, a $990 \Omega$ resistor and $\mathrm{a}$ potentiometer of $2 \mathrm{~m}$ length, all are connected in series of the resistance of potentiometer wire is $10 \Omega$, then the potential gradient of the potentiometer wire is
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The correct answer is:
$0.01 \mathrm{Vm}^{-1}$
Potential gradient $x=\frac{e}{\left(R+R_{h}+r\right)} \cdot \frac{R}{L}$
$$
\Rightarrow \quad x=\frac{2}{(990+10)} \times \frac{10}{2}=0.01 \mathrm{Vm}^{-1}
$$
$$
\Rightarrow \quad x=\frac{2}{(990+10)} \times \frac{10}{2}=0.01 \mathrm{Vm}^{-1}
$$
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