A cell in secondary circuit gives null deflection for $2.5 \mathrm{~m}$ length of potentiometer having $10…
A cell in secondary circuit gives null deflection for $2.5 \mathrm{~m}$ length of potentiometer having $10 \mathrm{~m}$ length of wire. If the length of the potentiometer wire is increased by $1 \mathrm{~m}$ without changing the cell in the primary, the position of the null point now is
3.5 m
3 m
2.75 m
2.0 m
Solution
Resistance of potentiometer wire
$R=\rho \times \frac{l}{A}$
or $\quad R=\left(\rho \times \frac{10}{A}\right)$
The value of $2.5 \mathrm{~m}$ length wire
$R^{\prime}=\frac{\rho \times 10}{A \times 10} \times 2.5$
or $\quad R^{\prime}=\left(\frac{2.5 \rho}{A \times 10}\right)$
Potential
$\begin{aligned}
V^{\prime} & =I \times R^{\prime} \\
& =I\left(\frac{2.5 \rho}{A \times 10}\right)
\end{aligned}$
Now, again the length of potentiometer wire is increased by $1 \mathrm{~m}$, then resistance of null position wire.
$\begin{aligned}
R^{\prime \prime} & =\left(\frac{\rho \times l}{11 \times A}\right) \\
V^{\prime \prime} & =I R^{\prime \prime} \\
and \quad V & =V^{\prime}
\end{aligned}$
$\begin{aligned} \frac{I \times 2.5 \rho}{A \times 10} & =\frac{\rho \times l}{11 \times A} \times I \\ \text { or } \quad \frac{2.5 \times 11}{10} & =l=2.75 \mathrm{~m}\end{aligned}$