MHT CET · Physics · Current Electricity
A potentiometer wire of length 1 m is connected in series with \(495 \Omega\) resistance and 2 V battery. If \(0.2 \mathrm{mV} / \mathrm{cm}\) is the potential gradient, then the resistance of the potentiometer wire is
- A \(8 \Omega\)
- B \(7 \Omega\)
- C \(6 \Omega\)
- D \(5 \Omega\)
Answer & Solution
Correct Answer
(D) \(5 \Omega\)
Step-by-step Solution
Detailed explanation
\(I=\frac{V}{R}=\frac{2}{R+495}\)
Potential gradient given as,
\(\begin{array}{ll}
& \phi=\frac{V}{L}=\frac{\mathrm{I} \times \mathrm{R}}{\mathrm{~L}}=\frac{2 \mathrm{R}}{(\mathrm{R}+495)} \quad \ldots .(\because \mathrm{L}=1 \mathrm{~m}) \\
\therefore \quad & 0.02=\frac{2 \mathrm{R}}{(\mathrm{R}+495)} \ldots(\text { given } \phi=0.2 \mathrm{mV} / \mathrm{cm}) \\
\therefore \quad & 0.02 \times(\mathrm{R}+495)=2 \mathrm{R} \\
\therefore \quad & 1.98 \mathrm{R}=9.9 \\
\therefore \quad & \mathrm{R}=\frac{9.9}{1.98}=5 \Omega
\end{array}\)
Potential gradient given as,
\(\begin{array}{ll}
& \phi=\frac{V}{L}=\frac{\mathrm{I} \times \mathrm{R}}{\mathrm{~L}}=\frac{2 \mathrm{R}}{(\mathrm{R}+495)} \quad \ldots .(\because \mathrm{L}=1 \mathrm{~m}) \\
\therefore \quad & 0.02=\frac{2 \mathrm{R}}{(\mathrm{R}+495)} \ldots(\text { given } \phi=0.2 \mathrm{mV} / \mathrm{cm}) \\
\therefore \quad & 0.02 \times(\mathrm{R}+495)=2 \mathrm{R} \\
\therefore \quad & 1.98 \mathrm{R}=9.9 \\
\therefore \quad & \mathrm{R}=\frac{9.9}{1.98}=5 \Omega
\end{array}\)
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