The resistances in two arms of the metre bridge are 5 Ω and R Ω respectively, null point is obtained at length l₁. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l₁. The resistance R is (a) 10 Ω (b) 15 Ω (c) 20 Ω (d) 25 Ω

Answer: (b). 5/R = (l₁)/(100 - l₁) and, with R → R/2: 10/R = (1.6 l₁)/(100 - 1.6 l₁) Divide: 2 = (1.6(100 - l₁))/(100 - 1.6 l₁) ⇒ 200 - 3.2 l₁ = 160 - 1.6 l₁ l₁ = 25 cm, so 5/R = 25/75 R = 15 Ω So the answer is (b) 15 Ω.

The resistances in two arms of the metre bridge are 5 Ω and R Ω… | MHT CET 2026 (13 May, Shift 1)

Q10MHT CET 2026 · 13 MAY · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge

The resistances in two arms of the metre bridge are $5\ \Omega$ and $R\ \Omega$ respectively, null point is obtained at length $l_1$ . When the resistance R is shunted with an equal resistance, the new balance point is at $1.6 l_1$ . The resistance R is

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Answer: (b)

$\dfrac{5}{R} = \dfrac{l_1}{100 - l_1}$ and, with $R \to R/2$ : $\dfrac{10}{R} = \dfrac{1.6 l_1}{100 - 1.6 l_1}$
Divide: $2 = \dfrac{1.6(100 - l_1)}{100 - 1.6 l_1} \Rightarrow 200 - 3.2 l_1 = 160 - 1.6 l_1$
$l_1 = 25$ cm, so $\dfrac{5}{R} = \dfrac{25}{75}$
$R = 15\ \Omega$
So the answer is (b) $15\ \Omega$ .

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Wheatstone Bridge & Metre Bridge question from MHT CET 2026 (13 May, Shift 1)

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