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Units and Measurements question

2021 · 16 Mar · Shift 1 · Q66
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Units and Measurements question

2021 · 16 Mar · Shift 1 · Q66

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
The resistance R = VI{V \over I}IV​, where V = (50 ±\pm± 2)V and I = (20 ±\pm± 0.2)A. The percentage error in R is 'x'%. The value of 'x' to the nearest integer is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 5

  1. We are given: R=VIR=\frac{V}{I}R=IV​ with V=(50±2) V,I=(20±0.2) AV=(50\pm 2)\,\text{V}, \qquad I=(20\pm 0.2)\,\text{A}V=(50±2)V,I=(20±0.2)A

  2. For division, the fractional errors add: ΔRR=ΔVV+ΔII\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}RΔR​=VΔV​+IΔI​

  3. Compute the percentage error in VVV: ΔVV×100=250×100=4%\frac{\Delta V}{V}\times 100=\frac{2}{50}\times 100=4\%VΔV​×100=502​×100=4%

  4. Compute the percentage error in III: 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 RRR is: x=4%+1%=5%x=4\%+1\%=5\%x=4%+1%=5%

  6. To the nearest integer: x=5x=5x=5

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