A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of 5 Ω. But when it is shunted by a resistance of 10 Ω, then balancing length increases by 25 cm. The balancing length when the cell is in an open circuit is (a) 200 cm (b) 210 cm (c) 225 cm (d) 250 cm

Answer: (b). l = l₀R/R+r: 150 = l₀5/5+r, 175 = l₀(10)/(10+r). Dividing: 7/6 = (2(5+r))/(10+r) ⇒ 70+7r = 60+12r ⇒ r = 2 Ω l₀ = 150×5+2/5 = 210 cm So the answer is (b) 210 cm.

A cell balances against a length of 150 cm on a potentiometer wire… | MHT CET 2024 (15 May, Shift 1)

Q35MHT CET 2024 · 15 MAY · SHIFT 1Medium

Potentiometer

A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of $5\ \Omega$ . But when it is shunted by a resistance of $10\ \Omega$ , then balancing length increases by 25 cm. The balancing length when the cell is in an open circuit is

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Answer: (b)

$l = l_0\dfrac{R}{R+r}$ : $150 = l_0\dfrac{5}{5+r}$ , $175 = l_0\dfrac{10}{10+r}$ .
Dividing: $\dfrac{7}{6} = \dfrac{2(5+r)}{10+r} \Rightarrow 70+7r = 60+12r \Rightarrow r = 2\ \Omega$
$l_0 = 150\times\dfrac{5+2}{5} = 210$ cm
So the answer is (b) 210 cm.

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Potentiometer question from MHT CET 2024 (15 May, Shift 1)

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