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 pm 0.3) xx10^(11)N//m^(2)`
B. `(2.0pm 0.2)xx10^(11)N//m^(2)`
C. `(2.0 pm 0.1)xx10^(11)N//m^(2)`
D. `(2.0 pm 0.05)xx10^(11)N//m^(2)`

Answer

B

Explanation

`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)=2xx0.025+0.0625`
`[because" the value of m, g and L are exact"]`
`=0.05+0.0625=0.1125`
`rArr" "DeltaY=2xx10^(11)xx0.1125=0.2xx10^(11)N//m^(2)`