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

Maximum percentage error in `y` is given by
`y = (W)/((pi D^(2))/(4)) xx (L)/(x) implies ((Delta Y)/(Y)) = 2 ((Delta D)/(D)) + (Delta x)/(x) + (Delta L)/(L)`
`= 2((0.001)/(0.005)) + ((0.001)/(0.125)) + ((0.1)/(110)) = 0.0489`