In a meter bridge experiment, the balance point is obtained at length ₁ cm from left end when resistances in the left gap and right gap are 5 Ω and R Ω respectively. When the resistance R is shunted with equal resistance, the new balance point is at 1.6₁. The resitance R in ohm is (a) 25 (b) 15 (c) 10 (d) 20

Answer: (b). Metre bridge balance: 5/R = (l₁)/(100-l₁). Shunting R with an equal resistance makes it R/2, so 10/R = (1.6l₁)/(100-1.6l₁). Dividing: 2 = (1.6(100-l₁))/(100-1.6l₁) ⇒ 200-3.2l₁ = 160-1.6l₁ ⇒ l₁ = 25 cm. 5/R = 25/75 ⇒ R = 15 Ω. So the answer is (b) 15.

In a meter bridge experiment, the balance point is obtained at length… | MHT CET 2021 (21 Sep, Shift 1)

Q57MHT CET 2021 · 21 SEP · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge

In a meter bridge experiment, the balance point is obtained at length $\ell_1\text{ cm}$ from left end when resistances in the left gap and right gap are $5\ \Omega$ and $R\ \Omega$ respectively. When the resistance $R$ is shunted with equal resistance, the new balance point is at $1.6\ell_1$ . The resitance $R$ in ohm is

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Answer: (b)

Metre bridge balance: $\dfrac{5}{R} = \dfrac{l_1}{100-l_1}$ .
Shunting $R$ with an equal resistance makes it $\dfrac{R}{2}$ , so $\dfrac{10}{R} = \dfrac{1.6l_1}{100-1.6l_1}$ .
Dividing: $2 = \dfrac{1.6(100-l_1)}{100-1.6l_1} \Rightarrow 200-3.2l_1 = 160-1.6l_1 \Rightarrow l_1 = 25\text{ cm}$ .
$\dfrac{5}{R} = \dfrac{25}{75} \Rightarrow R = 15\ \Omega$ .
So the answer is (b) 15.

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Wheatstone Bridge & Metre Bridge question from MHT CET 2021 (21 Sep, Shift 1)

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