10^(-6)m^(3) volume of water is taken from the surface to the bottom of a lake to a depth of 200 m inside helake. If the bulk modulus of elasticity of water is 220 atm., then what will be the change in its volume? (Density of water =10^(3)kgm^(-3))

10^(-6)m^(3) volume of water is taken from the surface to the bottom o

1.	10^(-6)m^(3) volume of water is taken from the surface to the bottom of a lake to a depth of 200 m inside helake. If the bulk modulus of elasticity of water is 220 atm., then what will be the change in its volume? (Density of water =10^(3)kgm^(-3))
Answer» SOLUTION :`V=10^(-6)m h=200m, K=220atm` `P_(e)=h rho g=200xx10^(3)xx9.8=19.6xx10^(5)NM^(-2)` i.e. `P_(e)=(19.6xx10^(5))/(1.013xx10^(5))=19.348atm` `DeltaP=P_(2)-P_(1)=h rho g =19.348atm` But `K=(- DELTA PV)/(DELTAV)` i.e. `DeltaV=(-(DeltaP)V)/K=(-19.348xx10^(-6))/220=-0.0879xx10^(-6)m^(3)` `Delta V=-8.79xx10^(-8)m^(3)` i.e. Decrease in the volume will be `879xx10^(-8)m^(3)`