In a meter bridge experiment, the balance point is obtained at length ' l₁ ' cm from left hand when resistances in the left gap and right gap are 15 Ω and R Ω respectively. When the resistance R is shunted with equal resistance the new balance point is at (1.6 l₁). The resistance R in ohm is (a) 25 (b) 30 (c) 45 (d) 60

Answer: (c). 15/R = (l₁)/(100 - l₁) and 30/R = (1.6 l₁)/(100 - 1.6 l₁) Divide: 2(100 - 1.6 l₁) = 1.6(100 - l₁) ⇒ l₁ = 25 cm 15/R = 25/75 ⇒ R = 45 Ω So the answer is (c) 45.

In a meter bridge experiment, the balance point is obtained at length… | MHT CET 2026 (19 Apr, Shift 1)

Q12MHT CET 2026 · 19 APR · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge

In a meter bridge experiment, the balance point is obtained at length ' $l_1$ ' cm from left hand when resistances in the left gap and right gap are $15\ \Omega$ and $R\ \Omega$ respectively. When the resistance $R$ is shunted with equal resistance the new balance point is at $(1.6\, l_1)$ . The resistance $R$ in ohm is

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Answer: (c)

$\dfrac{15}{R} = \dfrac{l_1}{100 - l_1}$ and $\dfrac{30}{R} = \dfrac{1.6 l_1}{100 - 1.6 l_1}$
Divide: $2(100 - 1.6 l_1) = 1.6(100 - l_1) \Rightarrow l_1 = 25$ cm
$\dfrac{15}{R} = \dfrac{25}{75} \Rightarrow R = 45\ \Omega$
So the answer is (c) $45$ .

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Wheatstone Bridge & Metre Bridge question from MHT CET 2026 (19 Apr, Shift 1)

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