___ KPa the pressure at a depth of 200 m below the surface of the sea. The density of seawater is 1024 kg/m³.

Open in App

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 = 200 m and g = $10 \frac{m}{{s e c}^{2}}$ .
Thus, pressure $= ρ g h = 1024 \times 200 \times 10$
= 2048,000 Pa = 2048 KPa .

Suggest Corrections

Similar questions

Calculate the pressure at a depth of 50 m below the surface of the sea. The density of seawater is 1024 kg/m³. Take $g = 10 m / s^{2}$

Q. Calculate the pressure at a depth of 50m below the surface of sea. The density of sea water is

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

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 = $10 m / s^{2}$ )

Calculate the pressure at a depth of $50 m$ below the surface of the sea. The density of sea water is $1024 k g / m^{3}$ . Take acceleration due to gravity $g = 10 m / s^{2}$

Join BYJU'S Learning Program

___ KPa the pressure at a depth of 200 m below the surface of the sea. The density of seawater 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 = 200 m and g = 10msec2. Thus, pressure =ρgh=1024×200×10 = 2048,000 Pa = 2048 KPa .

___ KPa the pressure at a depth of 200 m below the surface of the sea. The density of seawater 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 = 200 m and g = $10 \frac{m}{{s e c}^{2}}$ .
Thus, pressure $= ρ g h = 1024 \times 200 \times 10$
= 2048,000 Pa = 2048 KPa .