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JEE Main 2026 (24 Jan, Shift 1) · Wheatstone Bridge & Metre Bridge

Wheatstone Bridge & Metre Bridge question from JEE Main 2026 (24 Jan, Shift 1)

Q15JEE Main 2026 · 24 JAN · SHIFT 1Medium

Wheatstone Bridge & Metre Bridge

Two resistors 2 Ω2\ \Omega and 3 Ω3\ \Omega 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 XYXY. When an unknown resistor is connected in parallel with 3 Ω3\ \Omega resistor, the null point is shifted by 22.5 cm toward YY. The resistance of unknown resistor is ‾ Ω\underline{\qquad}\ \Omega.

Figure

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Answer: (d)

  1. Initially 23=l100−l⇒l=40 cm\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\text{ cm}.
  2. Shunting 3 Ω3\ \Omega lowers the right resistance, so the null point moves right to 40+22.5=62.5 cm40 + 22.5 = 62.5\text{ cm}.
  3. 2R′=62.537.5=53⇒R′=1.2 Ω\dfrac{2}{R'} = \dfrac{62.5}{37.5} = \dfrac{5}{3} \Rightarrow R' = 1.2\ \Omega.
  4. 3x3+x=1.2⇒1.8x=3.6⇒x=2 Ω\dfrac{3x}{3+x} = 1.2 \Rightarrow 1.8x = 3.6 \Rightarrow x = 2\ \Omega.
  5. So the answer is (d) 2.
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