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A wire of resistance R = 100.0 Omega and...

A wire of resistance `R = 100.0 Omega` and length `l = 50.0cm` is put between the jaws of screw gauge Its reading is shown in Pitch of the scregauge is `0.5mm` and there are 50 division on circular scale Find its resistivity in correct significant and maximum permissible error in p (resistivity) .

Text Solution

Verified by Experts

`R = (rhol)/(pid^(2)//4)`
`rho=(Rpid^(2))/(4l)= ((1000.0)(3.14)(8.42xx10^(-3)))/(4(50.0xx10^-2))= 1.32Omega//m` Object thickness `= 8mm 42((1//2mm)/(50))`
`=8.42mm`
`(drho)/(rho)=(dR)/(R)+(2d(D))/(D)+(dl)/(l)=(0.1)/(100.0)+2xx(0.01)/(8.42)+(0.1)/(5.0)=0.00537(~~0.52%)`
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