A null point is found at 200 cm in potentiometer 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 ___________ Ω. Numerical answer type. Work it out, then open the answer.

Answer: 5. 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

A null point is found at 200 cm in potentiometer when cell in… | JEE Main 2023 (29 Jan, Shift 2)

Q45JEE Main 2023 · 29 JAN · SHIFT 2Medium

Potentiometer

A null point is found at 200 cm in potentiometer 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 ___________ $\Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 5

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

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Potentiometer question from JEE Main 2023 (29 Jan, Shift 2)

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