A potentiometer wire of length 4 m and resistance 5 Ω is connected in series with a resistance of 992 Ω and a cell of e.m.f. 4 V with internal resistance 3 Ω. The length of 0.75 m on potentiometer wire balances the e.m.f. of (a) 4.00 mV (b) 3.75 mV (c) 3.00 mV (d) 2.50 mV

Answer: (b). I = (4)/(5+992+3) = 4 mA k = 4×10⁻³× 54 = 5×10⁻³ V/m E = 5×10⁻³× 0.75 = 3.75 mV So the answer is (b) 3.75 mV.

A potentiometer wire of length 4 m and resistance 5 Ω is connected in… | MHT CET 2023 (12 May, Shift 2)

Q43MHT CET 2023 · 12 MAY · SHIFT 2Medium

Potentiometer

A potentiometer wire of length $4\text{ m}$ and resistance $5\ \Omega$ is connected in series with a resistance of $992\ \Omega$ and a cell of e.m.f. $4\text{ V}$ with internal resistance $3\ \Omega$ . The length of $0.75\text{ m}$ on potentiometer wire balances the e.m.f. of

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Answer: (b)

$I = \dfrac{4}{5+992+3} = 4$ mA
$k = \dfrac{4\times10^{-3}\times 5}{4} = 5\times10^{-3}$ V/m
$E = 5\times10^{-3}\times 0.75 = 3.75$ mV
So the answer is (b) 3.75 mV.

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Potentiometer question from MHT CET 2023 (12 May, Shift 2)

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