Saved Bookmarks
| 1. |
A rectangular metal plate has dimensions of `10cmxx20cm`. A thin film of oil separates the plate from a fixed horizontal surface. The separation between the rectangular plate and the horizontal surface is `0.2mm`. An ideal string is attached to the plate and passes over an ideal pulley to a mass `m`. When `m=125gm`, the metal plate moves at constant speed of `5(cm)/(s)`, across the horizontal surface. Then the coefficient of viscosity of oil in `("dyne"-s)/(cm^2)` is (Use `g=1000(cm)/(s^2)`) |
|
Answer» The coefficient of viscosity is the ratio of tangential stress on top surface of film (exerted by block) to that of velocity gradient (vertically downwards) of film. Since mass m moves with constant velocity, the string exerts a force equal to mg on plate towards right. hence oil shall exert tangential force mg on plate towards left. `:. eta=(F//A)/((v-0)//Deltax)=(125xx1000//10xx20)/((5-0)//0.2) =2.5 ` dyne-s/`cm^(2)` |
|