A potentiometer wire of length 1 m is connected in series with $495 \Omega$ resistance and 2 V battery. If…

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
  1. $8 \Omega$
  2. $7 \Omega$
  3. $6 \Omega$
  4. $5 \Omega$

Solution

$I=\frac{V}{R}=\frac{2}{R+495}$
Potential gradient given as, $\begin{array}{ll} & \phi=\frac{V}{L}=\frac{\mathrm{I} \times \mathrm{R}}{\mathrm{~L}}=\frac{2 \mathrm{R}}{(\mathrm{R}+495)} \quad \ldots .(\because \mathrm{L}=1 \mathrm{~m}) \\ \therefore \quad & 0.02=\frac{2 \mathrm{R}}{(\mathrm{R}+495)} \ldots(\text { given } \phi=0.2 \mathrm{mV} / \mathrm{cm}) \\ \therefore \quad & 0.02 \times(\mathrm{R}+495)=2 \mathrm{R} \\ \therefore \quad & 1.98 \mathrm{R}=9.9 \\ \therefore \quad & \mathrm{R}=\frac{9.9}{1.98}=5 \Omega \end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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