Calculate the pressure at a depth of $60 m$ under the sea in $kPa$ , if the density of seawater is $1, 025 {kg m}^{- 3}$ .
( $1 kPa = 1000 Pa$ and $g = 10 {m s}^{- 2}$ )
615

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Solution

The correct option is A 615
Given:
The depth of the sea, $h = 60 m$
The density of seawater, $ρ = 1, 025 {kg m}^{- 3}$
Acceleration due to gravity, $g = 10 {m s}^{- 2}$

Pressure at a depth $h$ in a liquid, is given by $P = ρ gh$

$⟹$ Pressure in the given seawater at a depth of $60 m$ , is $= 1, 025 \times 10 \times 60$
$⟹ 10, 250 \times 60 = 6, 15, 000 Pa$

$1 kPa = 1000 Pa ⟹ P = 615 kPa$

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