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Current Electricity question

2021 · 18 Mar · Shift 1 · Q56
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  5. /2021 · 18 Mar · Shift 1 · Q56

Current Electricity question

2021 · 18 Mar · Shift 1 · Q56

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :
  1. A
    3.9
  2. B
    8.4
  3. C
    7.5
  4. D
    3.0
View written solutionFree

Correct answer: A

  1. Formula for resistivity

For a conductor,

R=ρLA⇒ρ=RALR=\rho\frac{L}{A} \quad\Rightarrow\quad \rho = R\frac{A}{L}R=ρAL​⇒ρ=RLA​

Using Ohm’s law, R=VIR=\dfrac{V}{I}R=IV​, so

ρ=VI⋅AL\rho = \frac{V}{I}\cdot \frac{A}{L}ρ=IV​⋅LA​

For a wire of diameter ddd,

A=πd24A = \frac{\pi d^2}{4}A=4πd2​

Hence,

ρ∝V d2 I−1 L−1\rho \propto V\, d^2\, I^{-1}\, L^{-1}ρ∝Vd2I−1L−1
  1. Maximum permissible percentage error

The maximum fractional error is

Δρρ=ΔVV+2Δdd+ΔII+ΔLL\frac{\Delta \rho}{\rho} = \frac{\Delta V}{V} + 2\frac{\Delta d}{d} + \frac{\Delta I}{I} + \frac{\Delta L}{L}ρΔρ​=VΔV​+2dΔd​+IΔI​+LΔL​
  1. Least count / absolute errors from the given measured values

From the data:

  • V=5.0 VV = 5.0\,\text{V}V=5.0V, so maximum absolute error is ±0.1 V\pm 0.1\,\text{V}±0.1V
  • L=10.0 cmL = 10.0\,\text{cm}L=10.0cm, so maximum absolute error is ±0.1 cm\pm 0.1\,\text{cm}±0.1cm
  • d=5.00 mmd = 5.00\,\text{mm}d=5.00mm, so maximum absolute error is ±0.01 mm\pm 0.01\,\text{mm}±0.01mm
  • I=2.00 AI = 2.00\,\text{A}I=2.00A, so maximum absolute error is ±0.01 A\pm 0.01\,\text{A}±0.01A

Thus,

ΔVV=0.15.0=0.02=2%\frac{\Delta V}{V} = \frac{0.1}{5.0} = 0.02 = 2\%VΔV​=5.00.1​=0.02=2% Δdd=0.015.00=0.002=0.2%\frac\Delta d d = \frac{0.01}{5.00} = 0.002 = 0.2\%dΔ​d=5.000.01​=0.002=0.2%

So,

2Δdd=0.4%2\frac{\Delta d}{d} = 0.4\%2dΔd​=0.4% ΔII=0.012.00=0.005=0.5%\frac{\Delta I}{I} = \frac{0.01}{2.00} = 0.005 = 0.5\%IΔI​=2.000.01​=0.005=0.5% ΔLL=0.110.0=0.01=1%\frac{\Delta L}{L} = \frac{0.1}{10.0} = 0.01 = 1\%LΔL​=10.00.1​=0.01=1%
  1. Total percentage error
Δρρ×100=2+0.4+0.5+1=3.9%\frac{\Delta \rho}{\rho}\times 100 = 2 + 0.4 + 0.5 + 1 = 3.9\%ρΔρ​×100=2+0.4+0.5+1=3.9%
  1. Correct option

Therefore, the maximum permissible percentage error in resistivity is

3.9%\boxed{3.9\%}3.9%​

So the correct option is A.

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