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In a Searle's experiment, the diameter of the wire as measured by a screw gauge with least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data. 

y=WA×LX
Where W is the weight suspended from the wire, A is the cross section area, L is the length and X is the extension in the wire.

[IIT JEE 2004]


  1. 15.56 %

  2. 4.89 %

  3. 1.26 %

  4. 9.67 %


Solution

The correct option is B

4.89 %


yy=2DD+xx+ll

(As the Diameter is squared, relative error is added twice)

yy=2(0.0010.05)+(0.0010.125)+(0.1110)=0.0489

percentage error = 4.89%

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