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 Pa

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51200 Pa

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5120 Pa

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None of these

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Solution

The correct option is A
512000 Pa

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: $d = 1024 k g / m^{3}$ , h = 50 m and g = 10 $m / s^{2}$
Thus, pressure $= ρ g h = 1024 \times 50 \times 10 = 512000 P a$ .

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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.

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

The correct option is A 512000 Pa 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=1024kg/m3, h = 50 m and g = 10 m/s2 Thus, pressure =ρgh=1024×50×10=512000Pa.

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

The correct option is A
512000 Pa

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: $d = 1024 k g / m^{3}$ , h = 50 m and g = 10 $m / s^{2}$
Thus, pressure $= ρ g h = 1024 \times 50 \times 10 = 512000 P a$ .