JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

Two resistors $2Ω$ and $3Ω$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with $3Ω$ resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is____ $Ω$ .

A
$3$
B
$2$
C
$4$
D
$1$

Solution

Meter-bridge balance: $\frac{R _{left}}{R _{right}} = \frac{x}{100 - x}$ where $x$ is the balance length from X.

Initially: $\frac{2}{3} = \frac{x}{100 - x} \Rightarrow 2 (100 - x) = 3 x \Rightarrow x = 40$ cm.

After connecting unknown $R$ in parallel with $3 Ω$ : new right-gap resistance is $\frac{3 R}{3 + R}$ ; null point shifts 22.5 cm toward Y, so new $x^{'} = 62.5$ cm.

$\frac{2}{3 R / ( 3 + R )} = \frac{62.5}{37.5} = \frac{5}{3}$

$\frac{2 ( 3 + R )}{3 R} = \frac{5}{3} \Rightarrow 6 (3 + R) = 15 R \Rightarrow 18 = 9 R \Rightarrow R = 2 Ω$ .

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