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Current Electricity question

2019 · 10 Apr · Shift 1 · Q60
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Current Electricity question

2019 · 10 Apr · Shift 1 · Q60

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 248 English
SI. No. R(Ω\OmegaΩ) l(cm)
1. 1000 60
2. 100 13
3. 10 1.5
4. 1 1.0
Which of the readings is inconsistent?
  1. A
    4
  2. B
    3
  3. C
    2
  4. D
    1
View written solutionFree

Correct answer: A

  1. Principle of meter bridge

At balance point of a meter bridge,

RS=l100−l\frac{R}{S}=\frac{l}{100-l}SR​=100−ll​

where:

  • RRR = resistance in one gap,
  • SSS = unknown/fixed resistance in the other gap,
  • lll = balancing length from one end.

If the same unknown resistance SSS is used in all observations, then

S=R⋅100−llS=R\cdot \frac{100-l}{l}S=R⋅l100−l​

For all consistent readings, the value of SSS should be nearly the same.

  1. Check each observation

Reading 1:

R=1000 ΩR=1000\,\OmegaR=1000Ω, l=60 cml=60\,\text{cm}l=60cm

S=1000⋅100−6060S=1000\cdot \frac{100-60}{60}S=1000⋅60100−60​ S=1000⋅4060=1000⋅23=666.7 ΩS=1000\cdot \frac{40}{60}=1000\cdot \frac{2}{3}=666.7\,\OmegaS=1000⋅6040​=1000⋅32​=666.7Ω

Reading 2:

R=100 ΩR=100\,\OmegaR=100Ω, l=13 cml=13\,\text{cm}l=13cm

S=100⋅8713=669.2 ΩS=100\cdot \frac{87}{13}=669.2\,\OmegaS=100⋅1387​=669.2Ω

Reading 3:

R=10 ΩR=10\,\OmegaR=10Ω, l=1.5 cml=1.5\,\text{cm}l=1.5cm

S=10⋅98.51.5=656.7 ΩS=10\cdot \frac{98.5}{1.5}=656.7\,\OmegaS=10⋅1.598.5​=656.7Ω

Reading 4:

R=1 ΩR=1\,\OmegaR=1Ω, l=1.0 cml=1.0\,\text{cm}l=1.0cm

S=1⋅991=99 ΩS=1\cdot \frac{99}{1}=99\,\OmegaS=1⋅199​=99Ω

  1. Compare the values

From readings 1, 2, and 3, the value of SSS is approximately:

∼660 to 670 Ω\sim 660\text{ to }670\,\Omega∼660 to 670Ω

But from reading 4,

S=99 ΩS=99\,\OmegaS=99Ω

which is clearly inconsistent.

  1. Conclusion

The inconsistent reading is 4.

So, the correct option is:

A\boxed{\text{A}}A​

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