___ 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.

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Solution

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by $ρ g h$ .
The given values are: d = $1024 \frac{k g}{m^{3}}$ , h = 50 m and g = $10 \frac{m}{{s e c}^{2}}$ .
Thus, pressure $= ρ g h = 1024 \times 50 \times 10 = 512000 P a$ .

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___ 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.

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by ρgh. The given values are: d = 1024kgm3, h = 50 m and g = 10msec2. Thus, pressure =ρgh=1024×50×10=512000 Pa.

___ 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.

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by $ρ g h$ .
The given values are: d = $1024 \frac{k g}{m^{3}}$ , h = 50 m and g = $10 \frac{m}{{s e c}^{2}}$ .
Thus, pressure $= ρ g h = 1024 \times 50 \times 10 = 512000 P a$ .