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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2026 · 24 January · Evening Shift · Physics · Current Electricity

In a meter bridge experiment to determine the value of unknown...

In a meter bridge experiment to determine the value of unknown resistance, first the resistances 2Ω2 \Omega and 3Ω3 \Omega are connected in the left and right gaps of the bridge and the null point is obtained at a distance l cml \mathrm{~cm} from the left. Now when an unknown resistance xΩx \Omega is connected in parallel to 3Ω3 \Omega resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance xx is ____\_\_\_\_ Ω\Omega.

Correct answer

0

Step-by-step solution

Meter bridge null point condition: R1R2=l100−l\frac{R_1}{R_2} = \frac{l}{100-l}
Initial: R1=2ΩR_1 = 2Ω, R2=3ΩR_2 = 3Ω, null at ll cm
23=l100−l\frac{2}{3} = \frac{l}{100-l} gives 2(100−l)=3l2(100-l) = 3l, so l=40l = 40 cm
When xΩxΩ connected in parallel with 3Ω3Ω: R2′=3x3+xΩR_2' = \frac{3x}{3+x}Ω
New null point at (40+10)=50(40+10) = 50 cm from left:
2R2′=5050=1\frac{2}{R_2'} = \frac{50}{50} = 1
2(3+x)3x=1\frac{2(3+x)}{3x} = 1 gives 6+2x=3x6 + 2x = 3x
x=6Ωx = 6Ω
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