Two spheres of the same diameter (in air), one of a lead alloy the other one made of steel are submerged at a depth h = 2932 m below the surface of the ocean. The ratio of the volumes of the two spheres at this depth is VSteel(h)/VLead(h) = 1.001206.Knowing the density of ocean water ñ = 1024 kg/m3 and the bulk modulus of steel, 2. 1011 N/m2, calculate the bulk modulus of the lead alloy used in the sphere.
To solve this problem, we can start by using the definition of the bulk modulus formula, which states that the ratio of volume change to the original volume of an object submerged in a fluid is equal to the ratio of the pressure change to the bulk modulus.
Let's denote the bulk modulus of the lead alloy as B.
We can set up the equation for the steel sphere as follows:
VSteel(h) / VSteel(0) = (PSteel(h) - P0) / BSteel
Where:
VSteel(h) represents the volume of the steel sphere at depth h.
VSteel(0) represents the volume of the steel sphere in air (atmospheric pressure).
PSteel(h) represents the pressure acting on the steel sphere at depth h.
P0 represents atmospheric pressure.
BSteel represents the bulk modulus of steel.
Since the spheres have the same diameter and are submerged at the same depth, we can assume that their volumes without compression are the same. Therefore, we can rewrite the equation as:
1 = (PSteel(h) - P0) / BSteel
Similarly, for the lead alloy sphere, the equation can be written as:
1.001206 = (PLead(h) - P0) / B
Where:
PLead(h) represents the pressure acting on the lead alloy sphere at depth h.
B represents the bulk modulus of the lead alloy.
Now, we can set the two equations equal to each other and solve for B.
(PSteel(h) - P0) / BSteel = (PLead(h) - P0) / B
Since the spheres are submerged at the same depth, their pressures at that depth are equal:
(PSteel(h) - P0) = (PLead(h) - P0)
We can simplify the equation to:
1 / BSteel = 1.001206 / B
Now, we can solve for B, the bulk modulus of the lead alloy:
B = BSteel / 1.001206
Substituting the known value of BSteel (2.1011 × 10^11 N/m^2):
B = (2.1011 × 10^11 N/m^2) / 1.001206
B ≈ 2.0989 × 10^11 N/m^2
Therefore, the approximate bulk modulus of the lead alloy used in the sphere is 2.0989 × 10^11 N/m^2.