Saved Bookmarks
| 1. |
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) . |
|
Answer» SOLUTION :`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%)`  .
|
|