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.

Question

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.