A student performs an experiment to determine the Young’s modulus of a wire, exactly 2m  long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8mm  with an uncertainty of  ±0.05 mm at a load of exactly  1.0kg. The student also measures the diameter of the wire to be  0.4mm with an uncertainty of  ±0.01 mm. Take g=9.8 m/s2 (exact). The Young’s modulus obtained from the reading is[ Take π=3.14 ]

A student performs an experiment to determine the Young’s modulus of a wire, exactly 2m  long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8mm  with an uncertainty of  ±0.05 mm at a load of exactly  1.0kg. The student also measures the diameter of the wire to be  0.4mm with an uncertainty of  ±0.01 mm. Take g=9.8 m/s2 (exact). The Young’s modulus obtained from the reading is[ Take π=3.14 ]

  1. A

    2.0±0.3×1011 N/m2

  2. B

    2.0±0.2×1011 N/m2

  3. C

    2.0±0.1×1011 N/m2

  4. D

    2.0±0.05×1011 N/m2

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    Solution:

    Given Length of wire l=2 m
    Extension= 0.8 mm
    Uncertainty =0.05 mm
    Diameter of the wire   d=0.4 mm
    Mass  m=1.0 kg
    Acceleration due to gravity  
    Using Hooke’s Law
    Y=FLAl=4FLπd2l.....1
    Also   Y=FLAl=mgLπr2lF=mgA=πr2(or)πd24
    Where  L= length of the wire
    l= elongation of the wire
    d= diameter of the wire
     Y=1.0×9.8×23.14×0.22×0.8Y=194.96×109     =1.95×1011 N/m2
    From equation (1)
    Y=FLAl=4FLπd2l
    Taking log on both sides
    logY=log4FLlogπd2l
    Now partially differentiating
    ΔYY=2Δdd+Δll=20.010.4+0.050.8=980ΔY=980×Y=980×1.95×1011=2.19×1011N/m2Y=(2±0.219)×1011N/m2Y(2±0.2)×1011N/m2
    Therefore, the correct answer is (B).
     

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