In searle's experiment, the diameter of the wire, as measured by a screw gauge of least count 0.001 cm is 0.500 cm. The length, measured by a scale of least count 0.1cm is 110.0cm. When a weight of 40N is suspended from the wire, its extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the Young's modulus of the meterial of the wire from this data.

Explanation

`E=40 N=40xx10^(5)` dyme, `L=110.0 cm l=0.125 cm, D=0.500 cm`
the Young's modulus is given by
`Y=(F//A)/(l//L)=(W//pir^(2))/(l//L)=(4WL)/(pi D^(2)l)` ...(i)
`Y=(4xx40xx10^(5)xx110xx7)/(22xx(0.500)^(2)xx0.125)`
`Y=17.92xx10^(9)" dyne"//cm^(2)`
`Y=17.92xx10^(10) N//m^(2)`
`Y=1.8xx10^(11) N//m^(2)`
Taking log of eqn. (i) we have
`log Y=log 4+log W+log L-log pi - 2log D - log l`
Differentiating,
`(Delta Y)/(Y)=(Delta L)/(L)-(2 Delta D)/(D)-(Delta l)/l`
The maximum permissible error in Y is given by
`(Delta Y)/(Y)=(Delta L)/L+(2 Delta D)/(D)+(Delta l)/(l)`
`=0.1/110.0+2(0.001/0.500)+0.001/0.125`
`=0.009+0.004+0.008=0.021`
`DeltaY=0.021 Y=0.021xx1.792xx10^(11)`
`=0.037362xx10^(11)`
`=3.8xx10^(9) N//m^(2)`
`:.` The Young's modulud of the wire is
`Y=(1.8xx10^(11) pm 3.8xx10^(9))N//m^(2)`