A student performing Searle's experiment for finding the Young's modulus Y of the material of a wire takes the following observations:
Length of the wire (L) =2.890 m, diameter of the wire (D) = 0.082cm, mass suspended from the wire (M) =3.00 kg, extension in the length of wire `(l`) = 0.087 cm.
Calculate the maximum permissible error in the value of Y.

Explanation

The Young.s modulus of the material is given by, `Y = (4MgL)/(piD^(2)l)`
Here M = 3.00 kg `" ":. Delta M `=0.01 kg
L = 2.890m `" " :. DeltaL` = 0.001m
D= 0.082 cm `" " :. D ` = 0.001 cm
`l` = 0.087 cm `" " :. Delta l` = 0.001 cm
The maximum permissible percentage error in Y is
`((DeltaY)/(Y))_("max")xx100 = ((DeltaM)/(M)xx100)+((DeltaL)/(L)xx100)+2((DeltaD)/(D)xx100)+((Deltal)/(l)xx100)`
=`(0.01)/(3.00)xx100+((0.001)/(2.890)xx100)+2xx((0.001)/(0.082)xx100)+((0.001)/(0.087)xx100)`
`~~0.33%+0.035%+2.44%+1.15%~~4%`