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MHT CET 2026 (20 Apr, Shift 2) · Potentiometer

Potentiometer question from MHT CET 2026 (20 Apr, Shift 2)

Q17MHT CET 2026 · 20 APR · SHIFT 2Medium

Potentiometer

Figure shows a potentiometer wire AB having resistance of 5 Ω5\ \Omega and length 10 m. An e.m.f. is 0.40.4 V of battery, the balancing length AP is (internal resistance is negligible)

Figure

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Answer: (a)

  1. From the figure, driver: 5 V cell, 45 Ω45\ \Omega in series with wire of 5 Ω5\ \Omega.
  2. I=545+5=0.1I = \dfrac{5}{45+5} = 0.1 A, so VAB=0.1×5=0.5V_{AB} = 0.1\times 5 = 0.5 V.
  3. k=0.510=0.05k = \dfrac{0.5}{10} = 0.05 V/m
  4. AP=0.40.05=8AP = \dfrac{0.4}{0.05} = 8 m
  5. So the answer is (a) 8 m.
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