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MHT CET 2025 (19 Apr, Shift 2) · Potentiometer

Potentiometer question from MHT CET 2025 (19 Apr, Shift 2)

Q28MHT CET 2025 · 19 APR · SHIFT 2Medium

Potentiometer

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of 5 Ω5\ \Omega the balancing length is 250 cm. When the cell is shunted by 20 Ω20\ \Omega, the balancing length of potentiometer wire is 400 cm. The internal resistance, of the cell is

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Answer: (c)

  1. l=l0RR+rl = l_0\dfrac{R}{R+r}: 250=l055+r250 = l_0\dfrac{5}{5+r}, 400=l02020+r400 = l_0\dfrac{20}{20+r}.
  2. Dividing: 1.6=4(5+r)20+r⇒32+1.6r=20+4r1.6 = \dfrac{4(5+r)}{20+r} \Rightarrow 32+1.6r = 20+4r
  3. 2.4r=12⇒r=5 Ω2.4r = 12 \Rightarrow r = 5\ \Omega
  4. So the answer is (c) 5 Ω5\ \Omega.
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