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Units and Measurements question

2018 · 15 Apr · Shift 1 · Q65
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Units and Measurements question

2018 · 15 Apr · Shift 1 · Q65

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The relative error in the determination of the surface area of sphere is α\alphaα. Then the relative error in the determination of its volume is :
  1. A
    32α{3 \over 2}\alpha23​α
  2. B
    23α{2 \over 3}\alpha32​α
  3. C
    52α{5 \over 2}\alpha25​α
  4. D
    α\alphaα
View written solutionFree

Correct answer: A

  1. Write the formulas for surface area and volume of a sphere

    S=4πr2S = 4\pi r^2S=4πr2 V=43πr3V = \frac{4}{3}\pi r^3V=34​πr3

  2. Use error propagation for powers

    If a quantity depends on rnr^nrn, then the relative error is:

    ΔQQ=nΔrr\frac{\Delta Q}{Q} = n\frac{\Delta r}{r}QΔQ​=nrΔr​

  3. Given relative error in surface area

    Since S∝r2S \propto r^2S∝r2 we have ΔSS=2Δrr\frac{\Delta S}{S} = 2\frac{\Delta r}{r}SΔS​=2rΔr​

    Given that relative error in surface area is α\alphaα,

    α=2Δrr\alpha = 2\frac{\Delta r}{r}α=2rΔr​

    Therefore,

    Δrr=α2\frac{\Delta r}{r} = \frac{\alpha}{2}rΔr​=2α​

  4. Find relative error in volume

    Since V∝r3V \propto r^3V∝r3 so ΔVV=3Δrr\frac{\Delta V}{V} = 3\frac{\Delta r}{r}VΔV​=3rΔr​

    Substitute Δrr=α2\frac{\Delta r}{r} = \frac{\alpha}{2}rΔr​=2α​:

    ΔVV=3⋅α2=32α\frac{\Delta V}{V} = 3\cdot \frac{\alpha}{2} = \frac{3}{2}\alphaVΔV​=3⋅2α​=23​α

  5. Match with the options

    32α\boxed{\frac{3}{2}\alpha}23​α​

    So the correct option is A.

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