Saved Bookmarks
| 1. |
The depth of a lake at which density of water is 2% greater than on surface. [Compressibility of water =50xx10^(-6)/atm , 1 atm = 10^5N//m^2] |
|
Answer» 2 km `because Deltap=hrhog` and `1/B=C` so r'=r(1+Ch`rho`g) `rArr (rho'-rho)/rho=Chrhog` `rArr H = (Deltarho)/rhoxx1/(Crhog)` `=2/100xx1/((50xx10^(-6))/105 xx10^3xx10)` `=2/5xx10^4 =20/5xx10^3`m = 4 km OR `rho=M/V rArr drho=-M/V^2 dV` `rArr (drho)/rho=(-M/V^2dV)/(M/V)` `rArr (drho)/rhoxx100% =-(dV)/Vxx100` `rArr (dV)/V xx 100=2%[ "as" (drho)/rhoxx100%= 2%]` Now `B=(DeltaP)/((DELTAV)/V)` and compressibility `C=1/B` and `DeltaP=hrho g ` `1/C=(hrhog)/(2/100)` `rArr h=2/100xx1/(V rhog) = 2/100xx1/((50xx10^(-6))/10^5 xx10^3xx10)m` = 4 km |
|