Two resistors 2 and 3 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 X Y . When an unknown resistor is connected in parallel with 3 resistor, the null point is shifted by 22.5 cm toward Y . The resistance of unknown resistor is _ _ _ _ .

Step-by-step solution

In a Wheatstone bridge with 2Ω and 3Ω, the null point is found when R₁ R₂ = l L-l . Initial balance: 2 3 = l₁ L-l₁ gives l₁ = 2L 5 . For standard wire length L = 100 cm, l₁ = 40 cm. When unknown resistance R is connected in parallel with 3Ω, equivalent resistance becomes 3R 3+R . New null point: 2 3R 3+R = l₂ L-l₂ , which simplifies to 2(3+R) 3R = l₂ L-l₂ . Null point shifts 22.5 cm toward Y: l₂ = 40 + 22.5 = 62.5 cm. Substituting: 2(3+R) 3R = 62.5 37.5 = 5 3 6(3+R) = 15R 18 = 9R R = 2 Ω