A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by 5 Ω. When a resistance of 15 Ω is used for shunting, null point moves to 300 cm. The internal resistance of the cell is (a) 3 Ω (b) 4 Ω (c) 5 Ω (d) 6 Ω

Answer: (c). l = l₀R/R+r: 200 = l₀5/5+r, 300 = l₀(15)/(15+r). Dividing: 1.5 = (3(5+r))/(15+r) ⇒ 22.5+1.5r = 15+3r r = 5 Ω So the answer is (c) 5 Ω.

A null point is obtained at 200 cm on potentiometer wire when cell in… | MHT CET 2025 (20 Apr, Shift 1)

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Potentiometer

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by $5\ \Omega$ . When a resistance of $15\ \Omega$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is

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Answer: (c)

$l = l_0\dfrac{R}{R+r}$ : $200 = l_0\dfrac{5}{5+r}$ , $300 = l_0\dfrac{15}{15+r}$ .
Dividing: $1.5 = \dfrac{3(5+r)}{15+r} \Rightarrow 22.5+1.5r = 15+3r$
$r = 5\ \Omega$
So the answer is (c) $5\ \Omega$ .

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Potentiometer question from MHT CET 2025 (20 Apr, Shift 1)

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