JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Current Electricity

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✓ Correct
(C)4
(D)1

Answer: (B)

Solution

For a metre bridge at balance

\frac{P}{Q} = \frac{ℓ}{100 - ℓ}

. With

P = 2 Ω

in the left gap and

Q = 3 Ω

in the right gap,

\frac{2}{3} = \frac{ℓ}{100 - ℓ}

gives

ℓ = 40

cm. Connecting the unknown

R

in parallel with the

3 Ω

lowers that arm and moves the null

22.5

cm towards

Y

, to

ℓ^{'} = 62.5

cm. Then

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

, so

\frac{2 ( 3 + R )}{3 R} = \frac{5}{3}

, i.e.

6 (3 + R) = 15 R

and

18 = 9 R

. The unknown resistance is

2 Ω

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Current Electricity

Solution

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