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

512000 Pa

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51200 Pa

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5120 Pa

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1 atm

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Solution

The correct option is A
512000 Pa

The expression of pressure at a depth of ‘h’ inside a liquid with density $ρ$ is given by $ρ g h$ .
The given values are: $ρ = 1024 k g / m^{3}, h = 50 m a n d g = 10 m / s^{2}$
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.

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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 1024 kg/m3.

The correct option is A 512000 Pa The expression of pressure at a depth of ‘h’ inside a liquid with density ρ is given by ρgh. The given values are: ρ=1024 kg/m3, h=50 m and g=10 m/s2 Thus, pressure =ρgh=1024×50×10=512000 Pa

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

The correct option is A
512000 Pa

The expression of pressure at a depth of ‘h’ inside a liquid with density $ρ$ is given by $ρ g h$ .
The given values are: $ρ = 1024 k g / m^{3}, h = 50 m a n d g = 10 m / s^{2}$
Thus, pressure $= ρ g h = 1024 \times 50 \times 10 = 512000 P a$