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Q.

In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050cm.  The length, measured by a scale of least count 0.1 cm, is 110.0cm.  When a weight of 50 N (measured with no error) 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.

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a

1.09 × 1010 N/m2

b

1.10 × 1010 N/m2

c

2.09 × 1010N/m2

d

3.09 × 1010 N/m2

answer is A.

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Detailed Solution

y=Fπr2×Le=50π0.050×10-222×1100.124×10-2=2.24×1011---(1) Δy=1.095×1010 N/m2Δyy=2Δrr+Δll+Δee=20.0010.05+0.1110+0.0010.125=0.0489 Δy=0.0489×2.24×1011=1.09×1010  from (1)
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