In meter bridge experiment, two resistances X and Y in the two gaps give a null point dividing the wire in the ratio 2:3. When each resistance is increased by 30 Ω, the null point divides the wire in the ratio 5:6. The resistance X and Y are respectively (a) 20 Ω, 30 Ω (b) 22 Ω, 33 Ω (c) 32 Ω, 48 Ω (d) 40 Ω, 60 Ω

Answer: (a). X/Y = 2/3, so X = 2k, Y = 3k. (2k+30)/(3k+30) = 5/6 ⇒ 12k + 180 = 15k + 150 ⇒ k = 10. X = 20 Ω, Y = 30 Ω. So the answer is (a).

In meter bridge experiment, two resistances X and Y in the two gaps… | MHT CET 2026 (20 Apr, Shift 1)

Q18MHT CET 2026 · 20 APR · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge

In meter bridge experiment, two resistances X and Y in the two gaps give a null point dividing the wire in the ratio $2:3$ . When each resistance is increased by $30\ \Omega$ , the null point divides the wire in the ratio $5:6$ . The resistance X and Y are respectively

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Answer: (a)

$\dfrac{X}{Y} = \dfrac{2}{3}$ , so $X = 2k$ , $Y = 3k$ .
$\dfrac{2k+30}{3k+30} = \dfrac{5}{6} \Rightarrow 12k + 180 = 15k + 150 \Rightarrow k = 10$ .
$X = 20\ \Omega$ , $Y = 30\ \Omega$ . So the answer is (a).

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