A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter ofthe wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take `g = 9.8 ms^(-2)` (exact). What is the Young's modulus obtained from the reading?

Explanation

Young.s modulus `Y =(MgL)/(pir^(2)DeltaL) = (4MgL)/(piD^(2)DeltaL)`
Substituting the given values, we get
`Y = (4 xx 1 xx 9.8 xx 2)/(pi xx (0.4 xx 10^(-3))^(2) xx 0.8 xx 10^(-3)) = 2 xx 10^(11) Nm^(-2)`
Now, `(DeltaY)/Y = (2DeltaD)/D + (Delta(DeltaL))/(DeltaL) rArr DeltaY = ((2 xx 0.01)/0.4 + 0.05/0.8) xx 2 xx 10^(11) Nm^(-2) = 0.2 xx 10^(11) Nm^(-2)`