In searle's experiment, which is used to find Young's modulus of elasticity, the diameter of experimental wire is D= 0.05cm(measursd by a scale of least count 0.001 cm) and length is L =110cm( measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of l= 0.125 cm( measured by a micrometer of least count 0.001 cm). Find maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free from error.

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)` .