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. (a) `(2.0+-0.3)xx10^(11)N//m^(2)`
B. (b) `(2.0+-0.2)xx10^(11)N//m^(2)`
C. ( c ) `(2.0+-0.1)xx10^(11)N//m^(2)`
D. (d) `(2.0+-0.05)xx10^(11)N//m^(2)`

Answer

B

Explanation

(b) `Y= (4mgL)/(piD^(2)l) = (4xx1xx9.8xx2)/(pi(0.4xx10^(-3))^(2) xx (0.8xx10^(-3))`
`=2.0xx10^(11)N//m^(2)`
Now `(DeltaY)/(Y) = (2DeltaD)/(D) + (Deltal)/(l)`
[because the value of `m,g and L` are exact]
`=2xx(0.01)/(0.4) + (0.05)/(0.8) = 2xx0.025 + 0.0625`
`= 0.05 + 0.0625 = 0.1125`
rArr `DeltaY = 2xx10^(11)xx0.1125 = 0.225xx10^(11)`
`=0.2xx10^(11) N//m^(2)`