With a resistance of ' X ' in the left gap and resistance of 9 Ω in the right gap of a meter bridge, the balance point is obtained at 40 cm from the left end. In what way and to which resistance 3 Ω resistance be connected to obtain the balance at 50 cm from the left end? (a) In series with 9 Ω (b) Parallel to X Ω (c) In series with X Ω (d) Parallel to 9 Ω

Answer: (c). X/9 = 40/60 ⇒ X = 6 Ω. Balance at 50 cm needs equal resistances in both gaps. 6 + 3 = 9 Ω, so 3 Ω must be added in series with X. So the answer is (c) in series with X Ω.

With a resistance of ' X ' in the left gap and resistance of 9 Ω in… | MHT CET 2019 (2 May, Shift 2)

Q67MHT CET 2019 · 2 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

With a resistance of ' $X$ ' in the left gap and resistance of $9\ \Omega$ in the right gap of a meter bridge, the balance point is obtained at 40 cm from the left end. In what way and to which resistance $3\ \Omega$ resistance be connected to obtain the balance at 50 cm from the left end?

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Answer: (c)

$\dfrac{X}{9} = \dfrac{40}{60} \Rightarrow X = 6\ \Omega$ .
Balance at 50 cm needs equal resistances in both gaps.
$6 + 3 = 9\ \Omega$ , so $3\ \Omega$ must be added in series with $X$ .
So the answer is (c) in series with $X\ \Omega$ .

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Wheatstone Bridge & Metre Bridge question from MHT CET 2019 (2 May, Shift 2)

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