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Q.

Find gauge pressure at a depth of 1000m in an ocean. The density of sea water is $1.03 \times 10^{3} k g / m^{3}$ $(in 10^{5} P a) (g = 10 m / s^{2})$

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a

101

b

102

c

104

d

103

answer is C.

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Detailed Solution

Gauge pressure $P_{g} = P - P_{0} = P_{0} + ρ g h - P_{0}$

$P_{g} = ρ g h = {1.0}^{3} \times 10^{3} \times 10 \times 10^{3} = 10.3 \times 10^{6} P a = 103 \times 10^{5} P a$

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