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In the determination of Young's modulus (Y= 4 M L g l d² ) by using Searle's method, a wire of length L=2 ~m and diameter d=0.5 ~mm is used. For a load M=2.5 ~kg , an extension l=0.25 ~mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 ~mm . The number of divisions on their circular scale is 100 . The contributi

Options

  1. Adue to the errors in the measurements of d and l are the same.
  2. Bdue to the error in the measurement of d is twice that due to the error in the measurement of l .
  3. Cdue to the error in the measurement of l is twice that due to the error in the measurement of d .
  4. Ddue to the error in the measurement of d is four times that due to the error in the measurement of l .

Correct answer

A. due to the errors in the measurements of d and l are the same.

Step-by-step solution

The maximum possible error in Y due to l and d Y Y = l l + 2 d d Least count = Pitch No. of division on circular scale = 0.5 100 ~mm =0.005 ~mm Here, d= l=0.005 ~mm Error contribution of l= l l = 0.005 ~mm 0.25 ~mm = 1 50 Error contribution of d= 2 d d = 2 0.005 ~mm 0.5 ~mm = 1 50 Hence contribution to the maximum possible error in the measurement of y due to l and d is the same.

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