Resistance of a potentiometer wire is $2 \Omega / \mathrm{m}$. A cell of e.m.f. $1.5 \mathrm{~V}$ balances…
Resistance of a potentiometer wire is $2 \Omega / \mathrm{m}$. A cell of e.m.f. $1.5 \mathrm{~V}$ balances at $300 \mathrm{~cm}$. The current through the wire is
$2.5 \mathrm{~mA}$
$7.5 \mathrm{~mA}$
$250 \mathrm{~mA}$
$750 \mathrm{~mA}$
Solution
$l=300 \mathrm{~cm}=3 \mathrm{~m}$
Total resistance of wire,
$\mathrm{R}=3 \times 2=6 \Omega$
Since, the potentiometer is balanced. Voltage across wire segment $=1.5 \mathrm{~V}$
$\begin{array}{ll}
\therefore & \mathrm{IR}=1.5 \mathrm{~V} \\
\therefore & \mathrm{I}=\frac{1.5}{6}=250 \mathrm{~mA}
\end{array}$