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`]

Options

A. 2 km
B. 4 km
C. 8 km
D. 40 km

Answer

B

Explanation

The variation of density with pressure is given by `rho'=rho(1+(Deltap)/B)`
`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