A student performs an experiment to determine the Young's modulus of a wire, exactly`2m` long, by Searle's method. In a partcular reading, the student measures the extension in the length of the wire to be `0.8mm with an uncertainty of `+- 0.05mm` at a load of exactly `1.0kg`, the student also measures the diameter of the wire to be `04mm` with an uncertainty of `+-0.01mm`. Take `g=9.8m//s^(2)` (exact). the Young's modulus obtained from the reading is

Options

A. `(2.0 +- 0.3)xx10^(11) N//m^2`
B. `(2.0 +- 0.2)xx10^(11) N//m^2)`
C. `(2.0 +- 0.2 xx 10^(11) N//m^2)`
D. `(2.0 +- 0.05) xx10^(11) N//m^2`

Answer

(b)

Explanation

Here, `F = 1` kg wt =9.8 N, L =2m,
`Deltad = 0.01 mm, DeltaI = 0.05 mm`
Y = `("Normal stress")/("longitudinal strain") = (FI)/(Axx DeltaI)`
`Y = (F L)/(pi r^2 I) = (4 F L)/(pi d^2 I)`
`Y = (4xx9.8xx2)/((22)/(7)xx(0.4xx10^(-3))^2xx0.8xx10^("-3))`
`=(548.8)/(2.816xx10^(-9))`
` ~= 2xx10^(11) N//m^2`
`:. (DeltaY)/(Y) = (2 Delta d)/(d) + (DeltaI)/(I)`
`[I "is the extension in wire". As F, i.e., "load and L are exactly known", :. Delta F and DeltaL are zero]`
`(DeltaY)/(Y)= (2xx0.01)/(0.4) + (0.05)/(0.8) = 0.1125 = 0.1`
`Delta Y =0.1xx Y = 0.1xx2xx10^(11) = 0.2xx10^(11) N//m^2`
Young modulus obtained from reading
`=Y +- DeltaY = (2.0 +- 0.2)xx10^(11) N//m^2`