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Depth of sea is maximum at Mariana Trenc...

Depth of sea is maximum at Mariana Trench in West Pacific Ocean. Trench has a maximum depth of about `11km`. At bottom of trench water column above it exerts `1000 atm` pressure. Percentage change in density of sea water at such depth will be around
(Given, B `= 2xx10^(9) Nm^(-2) and P_(atm) = 1xx10^(5) Nm^(-2))`

A

about `5%`

B

about `10%`

C

about `3 %`

D

about `7 %`

Text Solution

Verified by Experts

The correct Answer is:
A
Change in volume , `DeltaV = (-Delta_(p).V_(i))/B`
Hence , density at depth of about `11 km ` is
`= (Mass)/(Volume) = (rho_(o)xxV_(i))/(V_(i)-(Deltapv_(i))/B) = (rho_(o)B)/((B-Delta_(p)))`
`=(rho_(o))/(1-Delta_(p)/B) = (rho_(o))/(1-(1xx10^(8))/(2xx10^(9)`
`(rho_(o))/(1-(1)/(20))` = `= (rho)/(0.95) rArr rho = rho_(o)/0.95`
` rArr rho_(o) =0.95rho`
`:.` % change in density `= (rho-rho_(o))/rho_(o)xx100`
`= [(1/0.95-1)/1]xx100`
` ~~ 5%`
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