In Searle's experiment, the diameter of the wire as measured by a screw gauge of least count `0.01cm` is `0.050cm`. The length, measured by a scale of least count `0.1cm`, is `110.0cm`. When a weight of `50N` is suspended from the wire, the extension is measure to be `0.125 cm` by a micrometer of least count `0.01cm`. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data..

Explanation

Young's modulus of elasticity is given by
`Y = (stree)/(strain) = (F//A)/(l//L) = (FL)/(lA) = (FL)/(l((pid^(2))/(4)))`
`Y=(50 xx 1.1xx4)/((1.25xx10^(-3))xxpi(5.0xx10^(-4))^(2))=2.24xx10^(11)N//m^(2)`
Now `(DeltaY)/(Y)= (DeltaL)/(L)+(Deltal)/(l)+2(Deltad)/(d)=((0.1)/(110))+((0.001)/(0.125))+2((0.002)/(0.005))=0.0489`
`DeltaY = (0.0489) Y = (0.0489) xx (2.24 xx 10^(11)) N//m^(2) = 1.09 xx 10^(10) N//m^(2)` .