Question

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)))

Answer 1

Percentage change in density of sea water at such depth would be around :

• Given: Maximum depth of Trench, H =11km, Pressure at bottom, P =1000atm, B = 2×10^5 N/m^2 and Patm = 1×10^5 N/m^2

• Percentage of change in density of sea water at a such depth is given as ∆P/P×100

• We know that, Pressure at h m in depth is given as

P = Hρg

• P + Patm = Hρg

P = Hρg − Patm .... (1)

• Substitute values of all given data in eq(1)

100×101325 = 11×10^3×∆P/P×100×10 − 1×10^5

11×10^3×∆P/P×1000 = 101325×10^2 +10^5

• (∆P/P)×100 = 101325×10^2 + 10^5 / 11×10^3

(∆P/P)×100 = 930.227

• ∆P/P = 9.3%