In the determination if Young's modulus `((Y = (4MLg)/(pild^(2))` by using searle's method, a wire of length `L=2m` and diameter `d=0.5mm` is used. For a load `M=2.5kg`, an extension `l=0.25mm` in the length of the wire is observed. Quantites `D and l` are measured using a screw gauge and a micrometer, respectively. they have the same pitch of `0.5mm`. the number of divisions on their circular scale is `100`. the contrubution to the maximum probable error of the `Y` measurement

Options

A. due to the errors in the measurement of d and l are the same
B. due to the errors in the measurement of d is twice that due to the error in the measurement of l.
C. due to the error in the measurement of l is twice that due to the error in the measurement of d
D. due to the error in the measurement of d is four times that due to the error in the measurement of l.

Answer

A

Explanation

Here, L=2 m, d= 0.5 mm, M=2 kg. `l=0.25` mm
Least count of screw gauge or micrometer is
`Delta l=Delta d=(0.5 mm)/(100) =0.5xx10^(-2) mm`
`Y=(4M L g)/(pi ld^(2)) =(4M L g)/(pi)l^(-1) d^(-2)`
`Delta Y=(4M Lg)/(pip) [-l^(2) Delta l d^(-2)+(-2)d^(-3)Delta d l^(-1)]`
`(Delta Y)/(Y)=-[(Delta ld^(-2)l^(-2))/(l^(-1)d^(-2))+(2d^(-3)Delta dl^(-1))/(l^(-1)d^(-2))]`
`=-[(Delta l)/(l)+(2Delta d)/(d)]`
Probable error in `Y=(Delta Y)/(Y)=(Delta l)/(l)+(2Delta d)/(d)`
`=[(0.5xx10^(-2))/(0.25)+(2xx0.5xx10^(-2))/(0.5)] =[(1)/(50)+(1)/(50)]`
Thus, the probable error in the measurement of Y is due to the errors in the measurements of d and l which are the same. Thus option (a) is true.