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A potentiometer wire of length $4 \mathrm{~m}$ and resistance $5 \Omega$ is connected in series with a resistance of $992 \Omega$ and a cell of e.m.f. $4 \mathrm{~V}$ with internal resistance $3 \Omega$. The length of $0.75 \mathrm{~m}$ on potentiometer wire balances the e.m.f. of
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Verified Answer
The correct answer is:
$3.75 \mathrm{mV}$
$\therefore \quad$ Total Resistance:
$$
\mathrm{R}=992+5+3=1000 \Omega
$$
Voltage across $4 \mathrm{~m}$ wire:
$$
\frac{5}{995+5} \times 4=0.02 \mathrm{~V}
$$
$\therefore \quad$ For one metre wire:
$$
\frac{0.02}{4}=0.005 \mathrm{~V}
$$
$\therefore \quad$ For $0.75 \mathrm{~m}$ wire:
$$
\begin{aligned}
0.004 \times 0.75 & =0.00375 \\
& =3.75 \mathrm{mV}
\end{aligned}
$$
$$
\mathrm{R}=992+5+3=1000 \Omega
$$
Voltage across $4 \mathrm{~m}$ wire:
$$
\frac{5}{995+5} \times 4=0.02 \mathrm{~V}
$$
$\therefore \quad$ For one metre wire:
$$
\frac{0.02}{4}=0.005 \mathrm{~V}
$$
$\therefore \quad$ For $0.75 \mathrm{~m}$ wire:
$$
\begin{aligned}
0.004 \times 0.75 & =0.00375 \\
& =3.75 \mathrm{mV}
\end{aligned}
$$
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