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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`V=10^(-6)m h=200m, K=220atm` <br/> `P_(e)=h rho g=200xx10^(3)xx9.8=19.6xx10^(5)<a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a>^(-2)`<br/> i.e. `P_(e)=(19.6xx10^(5))/(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.013xx10^(5))=19.348atm` <br/> `DeltaP=P_(2)-P_(1)=h rho g =19.348atm` <br/> But `K=(- <a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> PV)/(<a href="https://interviewquestions.tuteehub.com/tag/deltav-2053504" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAV">DELTAV</a>)` <br/> i.e. `DeltaV=(-(DeltaP)V)/K=(-19.348xx10^(-6))/220=-0.0879xx10^(-6)m^(3)` <br/> `Delta V=-8.79xx10^(-8)m^(3)` <br/> i.e. Decrease in the volume will be `879xx10^(-8)m^(3)`</body></html>