In a metre bridge experiment the balance point is obtained if the gaps are closed by 2 Ω and 3 Ω. A shunt of X Ω is added to 3 Ω resistor to shift the balancing point by 22.5 cm. The value of X is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b). Initially 2/3 = (l)/(100-l) ⇒ l = 40 cm. Shunting 3 Ω lowers the right resistance, so the null point moves right to 40 + 22.5 = 62.5 cm. 2/R' = (62.5)/(37.5) = 5/3 ⇒ R' = 1.2 Ω. 3x/3+x = 1.2 ⇒ 1.8x = 3.6 ⇒ x = 2 Ω. So the answer is (b) 2.

In a metre bridge experiment the balance point is obtained if the… | MHT CET 2026 (11 Apr, Shift 1)

Q19MHT CET 2026 · 11 APR · SHIFT 1Difficult

Wheatstone Bridge & Metre Bridge

In a metre bridge experiment the balance point is obtained if the gaps are closed by $2\ \Omega$ and $3\ \Omega$ . A shunt of $X\ \Omega$ is added to $3\ \Omega$ resistor to shift the balancing point by $22.5$ cm. The value of $X$ is

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Answer: (b)

Initially $\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}$ .
Shunting $3\ \Omega$ lowers the right resistance, so the null point moves right to $40 + 22.5 = 62.5\text{ cm}$ .
$\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega$ .
$\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega$ .
So the answer is (b) 2.

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Wheatstone Bridge & Metre Bridge question from MHT CET 2026 (11 Apr, Shift 1)

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