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Units and Measurements question

2018 · 16 Apr · Shift 1 · Q69
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Units and Measurements question

2018 · 16 Apr · Shift 1 · Q69

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity A = P3Q2RS.{{{P^3}{Q^2}} \over {\sqrt R S}}.R​SP3Q2​. The maximum percentage error in the value of A will be :
  1. A
    6.0%
  2. B
    7.5%
  3. C
    8.5%
  4. D
    6.5%
View written solutionFree

Correct answer: D

  1. The quantity is
A=P3Q2R S=P3Q2R−1/2S−1A=\frac{P^3Q^2}{\sqrt{R}\,S}=P^3Q^2R^{-1/2}S^{-1}A=R​SP3Q2​=P3Q2R−1/2S−1
  1. For maximum percentage error in a product/power expression, the fractional errors add with the absolute values of the powers:
ΔAA×100=3(ΔPP×100)+2(ΔQQ×100)+12(ΔRR×100)+1(ΔSS×100)\frac{\Delta A}{A}\times 100 = 3\left(\frac{\Delta P}{P}\times 100\right)+2\left(\frac{\Delta Q}{Q}\times 100\right)+\frac{1}{2}\left(\frac{\Delta R}{R}\times 100\right)+1\left(\frac{\Delta S}{S}\times 100\right)AΔA​×100=3(PΔP​×100)+2(QΔQ​×100)+21​(RΔR​×100)+1(SΔS​×100)
  1. Substitute the given percentage errors:
  • In PPP: 0.5%0.5\%0.5%
  • In QQQ: 1%1\%1%
  • In RRR: 3%3\%3%
  • In SSS: 1.5%1.5\%1.5%

So,

Maximum % error in A=3(0.5)+2(1)+12(3)+1(1.5)\text{Maximum \% error in }A = 3(0.5)+2(1)+\frac{1}{2}(3)+1(1.5)Maximum % error in A=3(0.5)+2(1)+21​(3)+1(1.5)
  1. Calculate:
3(0.5)=1.53(0.5)=1.53(0.5)=1.5 2(1)=22(1)=22(1)=2 12(3)=1.5\frac{1}{2}(3)=1.521​(3)=1.5 1(1.5)=1.51(1.5)=1.51(1.5)=1.5

Adding,

1.5+2+1.5+1.5=6.51.5+2+1.5+1.5=6.51.5+2+1.5+1.5=6.5
  1. Therefore,
Maximum percentage error in A=6.5%\boxed{\text{Maximum percentage error in }A=6.5\%}Maximum percentage error in A=6.5%​
  1. Checking options:
  • A: 6.0%6.0\%6.0% ❌
  • B: 7.5%7.5\%7.5% ❌
  • C: 8.5%8.5\%8.5% ❌
  • D: 6.5%6.5\%6.5% ✅

Hence the correct option is D.

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