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Units and Measurements question

2024 · 29 Jan · Shift 1 · Q61
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Units and Measurements question

2024 · 29 Jan · Shift 1 · Q61

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The resistance R=VIR=\frac{V}{I}R=IV​ where V=(200±5)V\mathrm{V}=(200 \pm 5) \mathrm{V}V=(200±5)V and I=(20±0.2)AI=(20 \pm 0.2) \mathrm{A}I=(20±0.2)A, the percentage error in the measurement of R\mathrm{R}R is :
  1. A
    5.5%
  2. B
    3%
  3. C
    7%
  4. D
    3.5%
View written solutionFree

Correct answer: D

  1. The resistance is given by R=VIR=\frac{V}{I}R=IV​ where V=(200±5) V,I=(20±0.2) A.V=(200\pm 5)\,\text{V},\qquad I=(20\pm 0.2)\,\text{A}.V=(200±5)V,I=(20±0.2)A.

  2. For a quantity involving division, ΔRR=ΔVV+ΔII.\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}.RΔR​=VΔV​+IΔI​. So the percentage error in RRR is the sum of percentage errors in VVV and III.

  3. Percentage error in voltage: ΔVV×100=5200×100=2.5%\frac{\Delta V}{V}\times 100=\frac{5}{200}\times 100=2.5\%VΔV​×100=2005​×100=2.5%

  4. Percentage error in current: ΔII×100=0.220×100=1%\frac{\Delta I}{I}\times 100=\frac{0.2}{20}\times 100=1\%IΔI​×100=200.2​×100=1%

  5. Therefore, percentage error in resistance: 2.5%+1%=3.5%2.5\%+1\%=3.5\%2.5%+1%=3.5%

  6. Hence the correct option is D: 3.5%\boxed{\text{D: }3.5\%}D: 3.5%​

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