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.

Answer

`4.89 %`

Explanation

`Y=(Fl)/(((piD^(2))/(4))Deltal)=(4Fl)/(piD^(2)Deltal)`
`(DeltaY)/(Y)=(Deltal)/(l)+2(DeltaD)/(D)+(Delta(Deltal))/(110)+2((0.001)/(0.050))+(0.001)/(0.125)`
Maximum percentage error
`(DeltaY)/(Y)xx100=(1)/(11)+4+0.8=4.89%`.