A potentiometer wire of length $4 \mathrm{~m}$ and resistance $5 \Omega$ is connected in series with a…
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
$4.00 \mathrm{mV}$
$3.75 \mathrm{mV}$
$3.00 \mathrm{mV}$
$2.50 \mathrm{mV}$
Solution
$\therefore \quad$ 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:
$\begin{aligned}
0.004 \times 0.75 & =0.00375 \\
& =3.75 \mathrm{mV}
\end{aligned}$