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) 2.50 mV (b) 3 mV (c) 3.75 mV (d) 4 mV

Answer: (c). I = (4)/(5+992+3) = 4×10⁻³ A Vwire = 4×10⁻³× 5 = 0.02 V, so k = 0.02/4 = 5×10⁻³ V/m. E = k l = 5×10⁻³× 0.75 = 3.75×10⁻³ V So the answer is (c) 3.75 mV.

A potentiometer wire of length 4 m and resistance 5 Ω is connected in… | MHT CET 2026 (18 Apr, Shift 1)

Q18MHT CET 2026 · 18 APR · SHIFT 1Medium

Potentiometer

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

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Answer: (c)

$I = \dfrac{4}{5+992+3} = 4\times10^{-3}$ A
$V_{wire} = 4\times10^{-3}\times 5 = 0.02$ V, so $k = \dfrac{0.02}{4} = 5\times10^{-3}$ V/m.
$E = k l = 5\times10^{-3}\times 0.75 = 3.75\times10^{-3}$ V
So the answer is (c) 3.75 mV.

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Potentiometer question from MHT CET 2026 (18 Apr, Shift 1)

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