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$$=3.14$$ ]

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 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$$=3.14$$ ]

Infinity Learn · Accepted Answer

Given Length of wire $$l=2$$ $$mExtension=$$ $$0.8$$ mmUncertainty $$=0.05$$ mmDiameter of the wire   $$d=0.4$$ mmMass  $$m=1.0$$ kgAcceleration due to gravity  Using Hooke’s $$LawY=FLAl=4FL$$$$π$$$$d2l$$....$$.1Also$$   $$Y=FLAl=mgL$$$$π$$$$r2l$$∵$$F=mg$$∵$$A=$$π$$r2$$ (or) π$$d24Where$$  $$L=$$ length of the $$wirel=$$ elongation of the $$wired=$$ diameter of the wire $$Y=1.0$$×$$9.8$$×$$23.14$$×$$0.22$$×$$0.8Y=194.96$$×$$109$$     $$=1.95$$×$$1011$$ $$N/m2From$$ equation (1)Y=FLAl=4FL$$$$π$$$$d2lTaking$$ log on both sides⇒$$logY=log4FL$$−log$$π$$$$d2lNow$$ partially differentiating⇒Δ$$YY=2$$Δ$$dd+$$Δ$$ll=20.010.4+0.050.8=980$$⇒Δ$$Y=980$$×$$Y=980$$×$$1.95$$×$$1011=2.19$$×$$1011N/m2Y=(2$$±$$0.219)×$$1011N/m2$$∴Y≈(2$$±$$0.2)×$$1011N/m2Therefore$$, the correct answer is (B).

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