In a metre-bridge when a resistance in the left gap is 2 Ω and unknown resistance in the right gap, the balance length is found to be 40 cm. On shunting the unknown resistance with 2 Ω, the balance length changes by: (a) 62.5 cm (b) 22.5 cm (c) 20 cm (d) 65 cm

Answer: (b). 2/X = 40/60 ⇒ X = 3 Ω. Shunted: X' = (3 × 2)/(3+2) = 1.2 Ω. 2/1.2 = (l)/(100-l) ⇒ l = 62.5 cm. Change = 62.5 - 40 = 22.5 cm. So the answer is (b).

In a metre-bridge when a resistance in the left gap is 2 Ω and… | JEE Main 2024 (1 Feb, Shift 2)

Q34JEE Main 2024 · 1 FEB · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

In a metre-bridge when a resistance in the left gap is $2\ \Omega$ and unknown resistance in the right gap, the balance length is found to be $40\text{ cm}$ . On shunting the unknown resistance with $2\ \Omega$ , the balance length changes by:

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Answer: (b)

$\dfrac{2}{X} = \dfrac{40}{60} \Rightarrow X = 3\ \Omega$ .
Shunted: $X' = \dfrac{3 \times 2}{3+2} = 1.2\ \Omega$ .
$\dfrac{2}{1.2} = \dfrac{l}{100-l} \Rightarrow l = 62.5\text{ cm}$ .
Change $= 62.5 - 40 = 22.5\text{ cm}$ . So the answer is (b).

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