In a potentiometer experiment, the balancing length for a cell is 240 cm. On shunting the cell with a resistance of 2 Ω, the balancing length becomes half the initial balancing length. The internal resistance of the cell is (a) 1.5 Ω (b) 1 Ω (c) 0.5 Ω (d) 2 Ω

Answer: (d). r = R(l₀/l-1) l = 240/2 = 120 cm, so r = 2((240)/(120)-1) r = 2 Ω So the answer is (d) 2 Ω.

In a potentiometer experiment, the balancing length for a cell is 240… | MHT CET 2021 (22 Sep, Shift 2)

Q60MHT CET 2021 · 22 SEP · SHIFT 2Medium

Potentiometer

In a potentiometer experiment, the balancing length for a cell is $240\text{ cm}$ . On shunting the cell with a resistance of $2\ \Omega$ , the balancing length becomes half the initial balancing length. The internal resistance of the cell is

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Answer: (d)

$r = R\left(\dfrac{l_0}{l}-1\right)$
$l = \dfrac{240}{2} = 120$ cm, so $r = 2\left(\dfrac{240}{120}-1\right)$
$r = 2\ \Omega$
So the answer is (d) $2\ \Omega$ .

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Potentiometer question from MHT CET 2021 (22 Sep, Shift 2)

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