Saved Bookmarks
| 1. |
A metal plate of area `0.10 m^(2)` is connected to a `0.04 kg` mass via a string that passes over an ideal pulley (considered massless and frictionless) A liquid with a film of thickness of `0.3mm` is placed between the plate and the table. When system is released, the plate moves up with constant speed of `0.085 m//s`. Find the approximate value of coefficient of viscosity of the liquid. (mass of metal plate is `0.02 kg`, `g=9.8 m//s^(2)`) A. `3.46xx10^(-3)Pa sec`B. `3.46xx10^(-4)Pa sec`C. `6.90xx10^(-3)Pa sec`D. `6.90xx10^(-4)Pa sec` |
|
Answer» `T=m_(2)gsin30^(@)`….(`i`) `T=m_(1)gsin30^(@)+(etaAV)/(l)`….(`ii`) From equation (`i`) and (`ii`) we can write `(etaAV)/(l)=(m_(2)-m_(1))gsin30^(@)=mgsin30^(@)` `eta=(mglsin30^(@))/(vA)=(0.02xx10xx0.3xx10^(-3))/(2xx0.085xx0.1)3.46xx10^(-3)PaS` |
|