Question

A steel sphere with a diameter of \(0.250 \mathrm{~m}\) is submerged in the ocean to a depth of \(500.0 \mathrm{~m}\). What is the percentage change in the volume of the sphere? The bulk modulus of steel is \(160 \cdot 10^{9}\) Pa. a) 0.003196 c) \(0.33 \%\) e) 1.596 b) 0.04596 d) 0.559

Short answer

Answer: The percentage change in the volume of the steel sphere is approximately 0.032%.

Step-by-step solution

Calculate the initial volume of the steel sphere

Given, diameter of the sphere, \(d = 0.250\:m\). We need to calculate the radius of the sphere. The radius is half of the diameter, so \(r = \frac{d}{2} = \frac{0.250}{2} = 0.125\:m\) Now, we can find the initial volume of the steel sphere using the formula \(V = \frac{4}{3}\pi r^3\): \(V = \frac{4}{3}\pi (0.125)^3 = 0.00818 m^3\)

Calculate the pressure at a depth of 500 meters

We need to find the pressure acting on the sphere when it is submerged at a depth of 500 meters in the ocean. The pressure is calculated using the formula \(P=\rho g h\), where \(\rho\) is the density of the water, \(g\) is the acceleration due to gravity, and \(h\) is the depth. For seawater, the density is approximately \(\rho = 1025\: kg/m^3\) and the acceleration due to gravity is \(g = 9.81\: m/s^2\). So, the pressure at a depth of 500 meters is: \(P = 1025\: kg/m^3 \times 9.81\: m/s^2 \times 500\: m = 5.03 \times 10^6\: Pa\)

Use the bulk modulus to find the change in volume

The bulk modulus of steel is defined as \(B = -V \frac{\Delta P}{\Delta V}\), where \(\Delta P\) is the change in pressure, and \(\Delta V\) is the change in volume. Rearranging the formula to solve for \(\Delta V\), we get: \(\Delta V = -\frac{V \Delta P}{B}\) So, the change in volume is: \(\Delta V = -\frac{(0.00818\:m^3) (5.03 \times 10^6\: Pa)}{160 \times 10^9\: Pa} = -2.61 \times 10^{-6}\: m^3\) (The negative sign indicates that the volume decreases)

Calculate the final volume

The final volume is the difference between the initial volume and the change in volume: \(V_f = V + \Delta V = 0.00818 m^3 - 2.61 \times 10^{-6} m^3 = 0.00818\: m^3\)

Calculate the percentage change in volume

The percentage change in volume is: \(Percentage\: Change\: in\: Volume = \frac{|\Delta V|}{V} \times 100\%\) \(Percentage\: Change\: in\: Volume = \frac{2.61 \times 10^{-6} m^3}{0.00818 m^3} \times 100\%= 0.032\%\) By comparing the calculated percentage change in volume with the given options, we find that the percentage change in volume is closest to option (a), 0.003196. So, the correct answer is (a) 0.003196.

Key concepts

Pressure Calculation

Understanding pressure calculation is fundamental in physics, particularly when evaluating the effects of fluids at different depths. Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. Generally expressed in Pascals (Pa), the formula used to compute pressure in a fluid at a specific depth is given by:
\[ P = \rho g h \]
Here, \(\rho\) represents the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth of the fluid. When dealing with water, particularly seawater, you must consider its density (~1025 kg/m³), which is greater than freshwater due to the dissolved salts.When submerged objects are involved, calculating the pressure at the depth of submersion provides insight into the forces they will experience. For example, a steel sphere submerged to a depth of 500 meters in seawater will encounter considerable pressure due to the weight of the water above it. Recognizing how to compute this pressure is crucial for applications in engineering, oceanography, and physics.

Volume Change Under Pressure

Materials can deform when subjected to external forces, and when these forces are uniform and result in compression, the volume of the material can decrease. The bulk modulus (B) quantifies a material's resistance to uniform compression and is defined as the ratio of pressure increase to relative volume decrease.
The formula to calculate the change in volume (\(\Delta V\)) under a change in pressure (\(\Delta P\)) using the bulk modulus is:\[ \Delta V = -\frac{V \Delta P}{B} \]
The negative sign in the formula signifies that an increase in pressure leads to a reduction in volume. For incompressible fluids, the bulk modulus is very high, signifying that the fluid doesn't compress much under pressure. On the contrary, gases or highly compressible fluids have a much lower bulk modulus.In the context of the steel sphere submerged in seawater, knowing the bulk modulus of steel enables us to calculate precisely how much the steel's volume would decrease when subjected to the immense oceanic pressure. This computation is crucial for the design and analysis of structures intended for environments involving high pressure, like deep-sea submarines or oil extraction equipment.

Physics of Submersion

The physics of submersion deals with the behavior of objects when they are placed in a fluid. A submerged object in a fluid like water experiences a force due to the fluid's pressure, which increases with depth. This force acts on all sides of the object and can lead to deformation, a concept integral to the study of materials science and engineering.
Moreover, when an object is submerged, it also experiences an upward buoyant force equal to the weight of the fluid displaced by the object, as stated by Archimedes' principle. The balance between this buoyant force and the object's weight determines if the object will sink, float, or remain neutral in the fluid. For example, the steel sphere in the exercise will experience a buoyant force attempting to push it upward, while the pressure at that depth will push inwards, deforming the material.The depth of submersion significantly affects the forces acting on the object due to pressure, as illustrated in our steel sphere example. This is critical for designing equipment and structures for use under water, such as submarines, diving suits, and underwater habitat modules. Understanding how pressure and volume change interact under submersion is key to ensuring the safety and functionality of such technology in the challenging environment found beneath the ocean's surface.