Question

Calculate the pressure experienced by a driver at the depth of 40m of sea water. Density of sea water is 1.3×10^3 kg/m^3
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Answer 1

GIVEN :-

• The driver is in the sea at the depth of 40m.

• Density of sea water is 1.3×10³kg/m³.

TO FIND :-

• Pressure experienced by the driver.

FORMULA USED :-

$\\ \bigstar \boxed{\sf \: P = P_{0} + \rho \: gh}$

Here ,

• P → Pressure experienced by driver

• P₀ → Pressure at surface of sea

• ρ → density

• g → Gravity of earth

• h → depth

CONCEPT :-

• As we go deeper , that means towards the core of earth , pressure increses by ρgh. Final pressure becomes initial pressure + increased pressure. Initially , Pressure is 10⁵Pa(1atm) exerted by the air.

So, Pressure exerted = P₀ + ρgh

SOLUTION :-

Pressure exerted on driver ,

★ P = P₀ + ρgh

We have ,

• P = ?

• P₀ = 10⁵ Pa (pressure exerted by the air at sea level)

• ρ= 1.3×10³kg/m³

• g = 9.8m/sec²

• h = 40m

Substituting values , we get...

➟ P = 10⁵ + (1.3×10³)(9.8)(40)

➟ P = 10⁵ + (509.6×10³)

➟ P = 10⁵ + 5.09×10⁵

➟ P = 10⁵ ( 1 + 5.09 )

➟ P = 10⁵ ( 6.09 )

➟ P = 6.09 × 10⁵ Pa

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P(atm) = P(Pa)/10⁵

ㅤㅤㅤㅤㅤㅤㅤㅤ[1atm=10⁵Pa ]

P(atm) = 6.09

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Hence ,

Pressure exerted on driver is 6.09×10⁵ Pa or 6.09atm at depth of 40m in sea water of 1.3×10³ kg/m³ density.