A SCUBA diver produces a spherical air bubble with bulk modulus = 1.2 x 105 Pa and radius = 2.9 mm while exploring a shipwreck below the surface of the ocean. The air bubble rises to the surface of the ocean and expands to a radius = 3.8 mm. What is the difference in pressure between the surface and at the SCUBA diver's depth, ( )?


Answer 1


ΔP = 0.497 10⁵ Pa


All materials are elastic, so they deform for the pressures applied, in bulk module is the deformation spreads for a given pressure, it is defined by

    B = - P /(ΔV/V)

Where the negative sign is for the module to be positive, P is the pressure and ΔV/V is the unit volume reduction

Let's apply that equation to our case, let's look at the volumes of the bubbles

In the water

     r₁ = 2.9 mm = 2.9 10⁻³ m

    V = 4/3 π R3

    V₁ = 4/3 π (2.9 10⁻³)³

    V₁ = 102.2 10⁻⁹ m³

On the surface

    r₂ = 3.8mm = 3.8 10⁻³m

    V₂ = 4/3 π (3.8 10⁻³)³

    V₂ = 229.8 10⁻⁹ m³

    Δv = V₁-V₂

    ΔV = (102.2 - 229.8) 10⁻⁹ m³

    ΔV = -127.6 10⁻⁹  m³

Let's calculate the pressure

    B = - P / (ΔV / V)

    P = - B ΔV / V

    P = - 1.2 10⁵ (-127.6 10⁻⁹/102.2 10⁻⁹)

    P = 1,498 10⁵ Pa

The pressure difference between this depth and the surface is the difference with the atmospheric pressure (1.01 105 Pa)

   ΔP = P2 - Patm

   ΔP = (1,498 -1.01) 10⁵

   ΔP = 0.497 10⁵ Pa

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Distance covered in snow, d = 1.1 m  


Work done (W) = change in kinetic energy


W =

in substituting the values, we get

W =


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