Calculate the pressure at a depth of $50 m$ below the surface of the sea. (The density of sea water is $1024 k g m^{- 3}$ ).

512000 P a

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512 P a

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5120 P a

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612000 P a

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Solution

The correct option is D $612000 P a$
Given:
Density, $ρ = 1024 k g m^{- 3}$
Depth, $h = 50 m$

The expression of pressure at the bottom of a liquid of depth $h$ and density $d$ is given as:
$P = P_{o} + ρ g h$
where, the constants have the values:
Acceleration due to gravity, $g = 10 m s^{- 2}$
Atmospheric pressure, $P_{o} \approx 10^{5} P a$

Substituting values, we get:
$P = 10^{5} + 1024 \times 10 \times 50$
$P = 10^{5} + 5.12 \times 10^{5}$
$P = 612000 P a$

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