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Units and Measurements question

2012 · Shift 0 · Q69
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Units and Measurements question

2012 · Shift 0 · Q69

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is
  1. A
    6 %
  2. B
    zero
  3. C
    1 %
  4. D
    3 %
View written solutionFree

Correct answer: A

  1. Relation used

    Resistance is given by Ohm’s law: R=VIR = \frac{V}{I}R=IV​ where VVV is voltage and III is current.

  2. Error propagation rule

    For a quantity involving division, ΔRR=ΔVV+ΔII\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}RΔR​=VΔV​+IΔI​ when we are dealing with maximum percentage error.

  3. Given percentage errors

    ΔVV×100=3%\frac{\Delta V}{V} \times 100 = 3\%VΔV​×100=3% ΔII×100=3%\frac{\Delta I}{I} \times 100 = 3\%IΔI​×100=3%

  4. Calculate percentage error in resistance

    ΔRR×100=3%+3%=6%\frac{\Delta R}{R} \times 100 = 3\% + 3\% = 6\%RΔR​×100=3%+3%=6%

  5. Match with options

    So, the error in resistance is: 6%\boxed{6\%}6%​

    Hence, the correct option is A.

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