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.

A student performing Searle's experiment for finding the Young's modul

1.	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.
Answer» Solution :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%`