MHT CET · Physics · Current Electricity
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
- A \(2.5 \mathrm{~mA}\)
- B \(7.5 \mathrm{~mA}\)
- C \(250 \mathrm{~mA}\)
- D \(750 \mathrm{~mA}\)
Answer & Solution
Correct Answer
(C) \(250 \mathrm{~mA}\)
Step-by-step Solution
Detailed explanation
\(
l=300 \mathrm{~cm}=3 \mathrm{~m}
\)
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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