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Question

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 uncertainly of ±0.05mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4mm with an uncertainly of ±0.01mm. Take g=9.8m/s2(exact). The Young's modulus obtained from the reading is?

A
(2.0±0.3)×1011N/m2
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B
(2.0±0.2)×1011N/m2
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C
(2.0±0.1)×1011N/m2
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D
(2.0±0.05)×1011N/m2
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Solution

The correct option is B (2.0±0.2)×1011N/m2
Given, length of wire =2 m extension in wire =0.8±0.05 mm mass =1 kg diameter of wire =0.4±0.01 mm

FA=y(ΔLL) Here F=mgA=πd24\

y=(9180)(2)(4)(3114)(4×104)2(8×104)=2×1011 N/m2

y=FLA(ΔL) Taking log on both sides : logy=logF+logLlogAlog(ΔL)

Here F and L are fixed, so. ΔF=0,ΔL=0

Δyy=ΔAA+Δ(ΔL)ΔL
A=πd24dA=πd2(Δd)dAA=2Δdd

Δyy=2(0.01)0.40+0.050.80=0.05+0.06=0.11coΔy=0.22×1011 N/m2

The expression of young's medulus is :(2+0.2)×1011 N/m2


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