JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

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 ____ $Ω$ .

Your answer

Solution

Meter bridge balance: $\frac{R _{left}}{R _{right}} = \frac{ℓ}{100 - ℓ}$ .

Case I (gaps have 2 Ω and 3 Ω): $\frac{2}{3} = \frac{ℓ}{100 - ℓ} \Rightarrow ℓ = 40 cm$ .

Case II (right gap now $R = 3 ∥ x = \frac{3 x}{3 + x}$ ; null shifts to 50 cm):

$\frac{2}{R} = \frac{50}{50} = 1 \Rightarrow R = 2 Ω$ .

$\frac{3 x}{3 + x} = 2 \Rightarrow 3 x = 6 + 2 x \Rightarrow x = 6 Ω$ .

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