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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51200 P a

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5120 P a

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512 P a

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Solution

The correct option is A
$512000 P a$

The expression of pressure at the bottom of a liquid of height $h$ and density $d$ is given as:
$P = ρ g h$
Given, density, $ρ = 1024 k g m^{- 3}$ , depth, $h = 50 m$ and acceleration due to gravity, $g = 10 m s^{- 2}$ .
Hence, pressure, P = $ρ g h = 1024 \times 50 \times 10 = 512000 k g m^{- 1} s^{- 2}$ .
We know, $1 k g m^{- 1} s^{- 2} = 1 P a s c a l (P a)$
$∴$ The pressure is $512000 P a$ .

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Similar questions

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

1024 \frac{k g}{m^{3}}

___ Pa is the pressure at a depth of 50 m below the surface of the sea. The density of sea water is 1024 kg/m³.

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

( Accelaration due to gravity = $10 m / s^{2}$ )

50 m

ρ

1024

k g m^{- 3}

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