Question

Two spheres of the same diameter (in air), one made of a lead alloy and the other one made of steel, are submerged at a depth \(h=2000\). \(\mathrm{m}\) below the surface of the ocean. The ratio of the volumes of the two spheres at this depth is \(V_{\text {sceel }}(h) / V_{\text {Lead }}(h)=1.001206 .\) Knowing the density of ocean water \(\rho=1024 \mathrm{~kg} / \mathrm{m}^{3}\) and the bulk modulus of steel, \(B_{\text {ged }}=160-10^{9} \mathrm{~N} / \mathrm{m}^{2}\), calculate the bulk modulus of the lead alloy.

Short answer

Answer: The approximate Bulk modulus of the lead alloy sphere is \(1.21 \times 10^{11} \mathrm{~N / m^2}\).

Step-by-step solution

Finding the pressure at depth h

We can find the pressure at a depth \(h\) in a fluid by using the following formula: $$ P =\rho g h $$ where: \(P\) is the pressure at depth h, \(\rho\) is the density of the fluid, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/ s}^2\)). For this problem, we have \(\rho = 1024 \mathrm{~kg / m^3}\) and \(h = 2000 \mathrm{~m}\). Plug in the values to get: $$ P = (1024 \mathrm{~kg / m^3}) (9.81 \mathrm{~m/ s^2}) (2000 \mathrm{~m}) $$

Understand the Bulk modulus and its relationship with pressure

The Bulk modulus, denoted as B, is a measure of the compressibility of a material. It is defined as the ratio of the increase of pressure to the relative decrease in volume: $$ B = \frac{\Delta P}{-\frac{\Delta V}{V}} $$ For this problem, the Bulk modulus of steel is given as \(B_{steel} = 160 \times 10 ^9 \mathrm{~N/ m^2}\), and we want to find the Bulk modulus of the lead alloy. As both spheres are at the same depth, they experience the same pressure, and their volume ratio is given. We can use this information to find the Bulk modulus of the lead sphere.

Set up an equation for the relationship between the volumes

As mentioned, both spheres experience the same pressure. So, $$ \frac{B_{steel}}{B_{lead}} = \frac{\Delta P_{steel}}{\Delta P_{lead}} = \frac{\frac{V_{steel}}{V_{lead}} - 1}{1.001206 - 1} $$ where \(B_{lead}\) is the Bulk modulus of the lead alloy.

Solve for the Bulk modulus of the lead alloy

We now have an equation with one unknown, \(B_{lead}\). We can solve for it by plugging in the given values. $$ \frac{160 \times 10 ^9\mathrm{~N / m^2}}{B_{lead}} = \frac{\frac{V_{steel}}{V_{lead}} - 1}{1.001206 - 1} $$ Solving for \(B_{lead}\), we get: $$ B_{lead} = \frac{160 \times 10 ^9\mathrm{~N / m^2}}{\frac{\frac{V_{steel}}{V_{lead}} - 1}{1.001206 - 1}} $$ $$ B_{lead} \approx 1.21 \times 10^{11} \mathrm{~N/ m^2} $$ So, the Bulk modulus of the lead alloy is approximately \(1.21 \times 10^{11} \mathrm{~N / m^2}\).

Key concepts

Fluid Pressure

Fluid pressure describes the force exerted per unit area by a fluid at rest or in motion. It is critical to understand that at any given depth within a fluid, the pressure is the same in all directions. This pressure increases linearly with depth, which is why divers experience increased pressure as they descend into the ocean.

For the problem we're analyzing, the pressure experienced by the spheres in the ocean is calculated using the formula \( P = \rho gh \), where \( \rho \) is the density of fluid (ocean water in this case), \( g \) is the gravitational acceleration, and \( h \) is the depth below the fluid's surface. Knowing this principle allows us to calculate the exact pressure exerted on the spheres below the ocean at a depth of 2000 meters, which plays a vital role in determining the subsequent compressibility of the materials.

Compressibility of Materials

Compressibility refers to a material's ability to change its volume under external pressure. A material that is highly compressible will deform substantially when pressure is applied, whereas a less compressible material will not change much in volume. In scientific terms, the bulk modulus (B) is the value that quantifies a material's compressibility, defined as \( B = \frac{\Delta P}{-\frac{\Delta V}{V}} \).

The equation indicates that the bulk modulus is the ratio of pressure change to relative volume change, meaning that a higher bulk modulus corresponds to a material that resists compression under pressure. The steel and lead spheres in our given problem demonstrate this concept, with their respective changes in volume under the ocean's pressure linked to their individual bulk moduli.

Volume Ratio

In the discussed exercise, the volume ratio is a critical component in determining the relationship between the compressibilities of two different materials. The volume ratio \( V_{steel}(h) / V_{Lead}(h) \) indicates the proportional difference in volume between the steel and lead spheres under the same pressure conditions.

This ratio can provide insights into how each material is affected by the same amount of pressure—essential information required to solve for the unknown bulk modulus of the lead alloy. The ratio, in this case, was given as 1.001206, a small difference that directly impacts the calculation, as both materials are subjected to identical pressures at the same depth.

Density of Ocean Water

The density of a substance is defined as the mass per unit volume. For ocean water, this value is a crucial part of many calculations regarding subsea objects, buoyancy, and pressure. Ocean water has a higher density than fresh water due to the presence of dissolved salts.

In our scenario, the density of ocean water is given as \( 1024 kg/m^3 \). This density affects the hydrostatic pressure experienced by any submerged object. It is used along with the depth and acceleration due to gravity to calculate the precise pressure at depth, which is then employed in the calculation of the bulk modulus of the submerged spheres. Understanding the exact density of ocean water is essential to ensure the accuracy of such calculations.