Two resistance X and Y are connected in the two gaps of a meterbridge and the null points is obtained at 20 cm from zero end. When the resistance of 20 Ω is connected in series with the smaller of the two resistance X and Y, the null point shifts to 40 cm from left end. The value of smaller resistance in ohm is (a) 6 (b) 9 (c) 12 (d) 15

Answer: (c). X/Y = 20/80 ⇒ Y = 4X, so X is smaller. With 20 Ω in series with X: (X + 20)/(4X) = 40/60 3X + 60 = 8X ⇒ X = 12 Ω So the answer is (c) 12.

Two resistance X and Y are connected in the two gaps of a meterbridge… | MHT CET 2023 (12 May, Shift 2)

Q42MHT CET 2023 · 12 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

Two resistance $\mathrm{X}$ and $\mathrm{Y}$ are connected in the two gaps of a meterbridge and the null points is obtained at $20\text{ cm}$ from zero end. When the resistance of $20\ \Omega$ is connected in series with the smaller of the two resistance $\mathrm{X}$ and $\mathrm{Y}$ , the null point shifts to $40\text{ cm}$ from left end. The value of smaller resistance in ohm is

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Answer: (c)

$\dfrac{X}{Y} = \dfrac{20}{80} \Rightarrow Y = 4X$ , so $X$ is smaller.
With $20\ \Omega$ in series with $X$ : $\dfrac{X + 20}{4X} = \dfrac{40}{60}$
$3X + 60 = 8X \Rightarrow X = 12\ \Omega$
So the answer is (c) 12.

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Wheatstone Bridge & Metre Bridge question from MHT CET 2023 (12 May, Shift 2)

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