JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

A meter bridge with two resistances $R_{1}$ and $R_{2}$ as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 $Ω$ resistance is connected in parallel to $R_{2}$ . The values of resistances $R_{1}$ and $R_{2}$ are _______.

A
$R_{2} = 16 Ω, R_{1} = \frac{16}{3} Ω$
B
$R_{2} = 4 Ω, R_{1} = \frac{4}{3} Ω$
C
$R_{2} = 8 Ω, R_{1} = \frac{16}{3} Ω$
D
$R_{2} = 12 Ω, R_{1} = \frac{12}{3} Ω$

Solution

Meter bridge balance: $\frac{R _{1}}{R _{2}} = \frac{ℓ}{100 - ℓ}$ .

First balance (40 cm from P): $\frac{R _{1}}{R _{2}} = \frac{40}{60} = \frac{2}{3}$ … (i)

After putting 16 $Ω$ in parallel with $R_{2}$ , the right-gap resistance becomes $\frac{16 R _{2}}{16 + R _{2}}$ . New balance at 50 cm: $\frac{R _{1}}{16 R _{2} / ( 16 + R _{2} )} = \frac{50}{50} = 1$ , so $R_{1} = \frac{16 R _{2}}{16 + R _{2}}$ … (ii)

Substitute (i) into (ii) using $R_{1} = \frac{2}{3} R_{2}$ : $\frac{2}{3} R_{2} = \frac{16 R _{2}}{16 + R _{2}}$ .

Cancel $R_{2}$ (non-zero) and cross-multiply: $\frac{2}{3} (16 + R_{2}) = 16$ , giving $\frac{2 R _{2}}{3} = 16 - \frac{32}{3} = \frac{16}{3}$ , so $R_{2} = 8 Ω$ .

Then $R_{1} = \frac{2}{3} (8) = \frac{16}{3} Ω$ .

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