The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20 Ω, the null point on it is found to be a 1000 cm. The resistance of whole wire is: (a) 80 Ω (b) 60 Ω (c) 120 Ω (d) 100 Ω

Answer: (d). At null point no current flows in the cell, so its emf 5 V is balanced by 1000 cm. k = 5/1000 = 5×10⁻³ V/cm, so Vwire = 5×10⁻³×1200 = 6 V. R = V/I = 6/0.06 = 100 Ω So the answer is (d) 100 Ω.

The length of a potentiometer wire is 1200 cm and it carries a… | JEE Main 2020 (8 Jan, Shift 1)

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Potentiometer

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20 $\Omega$ , the null point on it is found to be a 1000 cm. The resistance of whole wire is:

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Answer: (d)

At null point no current flows in the cell, so its emf 5 V is balanced by 1000 cm.
$k = \dfrac{5}{1000} = 5\times10^{-3}$ V/cm, so $V_{wire} = 5\times10^{-3}\times1200 = 6$ V.
$R = \dfrac{V}{I} = \dfrac{6}{0.06} = 100\ \Omega$
So the answer is (d) $100\ \Omega$ .

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