Current Electricity

JEE Main level, numerical answer, medium

277 of 380

Question 277 of 380

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.

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Answer: 6

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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