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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
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The correct answer is:
$250 \mathrm{~mA}$
$$
l=300 \mathrm{~cm}=3 \mathrm{~m}
$$
Total resistance of wire,
$$
\mathrm{R}=3 \times 2=6 \Omega
$$
Since, the potentiometer is balanced. Voltage across wire segment $=1.5 \mathrm{~V}$
$$
\begin{array}{ll}
\therefore & \mathrm{IR}=1.5 \mathrm{~V} \\
\therefore & \mathrm{I}=\frac{1.5}{6}=250 \mathrm{~mA}
\end{array}
$$
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}
$$
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