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.

Answer

A, B

Explanation

`Y = (W)/((piD^(2))/(4)) xx(L)/(X)`
KEY CONCEPT : Maximum error in `Y` is given by
`((DeltaY)/(Y))_(max) = ((DeltaD)/(D)) + (DeltaX)/(X)) + (DeltaL)/(L))`
`=2((0.001)/(0.05)) + ((0.001)/(0.125)) + ((0.1)/(110)) = 0.0489`
It is given that `W = 50 N`, D = 0.05cm = 0.05 xx 10^(-2)m`,
`X = 0.125cm = 0.125 xx 10^(-2)m`,
`L = 110 = 110 xx 10^(-2)m`
:. `Y = (50xx4xx110xx10^(-2))/(3.14(0.05xx10^(-2))xx(0.125xx10^(-2))) = 2.24xx10^(11) N//m^(2)`
:. Maximum possible error in the value of
`Y = DeltaY = 0.0489 xx 2.24 xx 10^(11)`
`=1.09 xx 10^(10)N//m^(2)`