Current Electricity

JEE Main level, single correct, medium

278 of 380

Question 278 of 380

Two resistors 2Ω2 \Omega and 3Ω3 \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 XYX Y. When an unknown resistor is connected in parallel with 3Ω3 \Omega resistor, the null point is shifted by 22.5 cm toward YY. The resistance of unknown resistor is ____\_\_\_\_ Ω\Omega.

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Answer: (A) 2

In a Wheatstone bridge with 2Ω and 3Ω, the null point is found when R1R2=lL−l\frac{R_1}{R_2} = \frac{l}{L-l}.
Initial balance: 23=l1L−l1\frac{2}{3} = \frac{l_1}{L-l_1} gives l1=2L5l_1 = \frac{2L}{5}.
For standard wire length L=100L = 100 cm, l1=40l_1 = 40 cm.
When unknown resistance RR is connected in parallel with 3Ω, equivalent resistance becomes 3R3+R\frac{3R}{3+R}.
New null point: 23R3+R=l2L−l2\frac{2}{\frac{3R}{3+R}} = \frac{l_2}{L-l_2}, which simplifies to 2(3+R)3R=l2L−l2\frac{2(3+R)}{3R} = \frac{l_2}{L-l_2}.
Null point shifts 22.5 cm toward Y: l2=40+22.5=62.5l_2 = 40 + 22.5 = 62.5 cm.
Substituting: 2(3+R)3R=62.537.5=53\frac{2(3+R)}{3R} = \frac{62.5}{37.5} = \frac{5}{3}
6(3+R)=15R⇒18=9R⇒R=26(3+R) = 15R \Rightarrow 18 = 9R \Rightarrow R = 2 Ω

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