In a meter bridge experiment to determine the value of unknown resistance, first the resistances 2 Ω and 3 Ω are connected in the left and right gaps of the bridge and the null point is obtained at a distance l cm from the left. Now when an unknown resistance x Ω is connected in parallel to 3 Ω resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance x is Ω. Numerical answer type. Work it out, then open the answer.

Answer: 6. Initially 2/3 = (l)/(100-l) ⇒ l = 40 cm. New null point at 50 cm, so the right gap becomes 2 Ω. 3x/3+x = 2 ⇒ x = 6. x = 6 Ω.

In a meter bridge experiment to determine the value of unknown… | JEE Main 2026 (24 Jan, Shift 2)

Q17JEE Main 2026 · 24 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2\ \Omega$ and $3\ \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l\text{ cm}$ from the left. Now when an unknown resistance $x\ \Omega$ is connected in parallel to $3\ \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance $x$ is

$\underline{\qquad}\ \Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Initially $\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}$ .
New null point at 50 cm, so the right gap becomes $2\ \Omega$ .
$\dfrac{3x}{3+x} = 2 \Rightarrow x = 6$ .
$x = 6\ \Omega$ .

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Wheatstone Bridge & Metre Bridge question from JEE Main 2026 (24 Jan, Shift 2)

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