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

Explanation

Least count of screw gauge `= 0.001 cm`
Diameter of the wire `= 0.05 cm`
Least count of micrometer `= 0.001 cm`
Extension `= 0.125 cm`
Least count of scale `= 0.01 cm`
Length `= 110 cm`
`Y = (F//A)/(Delta L // L), A = (pi d^(2))/(4)`
`= (4 FL)/((pi d^(2)) (Delta L))`
`(Delta Y)/(Y) = 2 (Delta d)/(d) + (Delta (Delta L))/(Delta L) + (Delta L)/(L)`
` = 2 xx (0.001)/(0.05) + (0.001)/(0.125) + (0.1)/(110)`
`= 0.0489`
`Y = (4FL)/(pi d^(2) (Delta L))`
`= (4 xx 50 xx 1.1)/(pi (0.05 xx 10^(-2))^(2) xx 0.125 xx 10^(-2))`
`= 2.24 xx 10^(11) N//m^(2)`
`Delta Y = 0.0489 xx 2.24 xx 10^(11)`
`= 0.109 xx 10^(11) N//m^(2)`