Test: Fluids in Motion - MCAT MCQ

The absolute and gauge pressures depend only on the density of the fluid, not that of the object. When the pressure at the surface is equal to atmospheric pressure, the gauge pressure is given by Pgauge = ρgz, where ρ represents the density of the fluid, not the object. These objects are also at the same depth, so they must have the same gauge pressure.

Using Newton's second law, F_net = ma, we obtain the following equation:
F_buoy ? mg = ma
Thus,

Both balls experience the same buoyant force because they are in the same liquid and have the same volume (F_buoy = ρVg). Thus, the ball with the smaller mass experiences the greater acceleration. Because both balls have the same volume, the ball with the smaller density has the smaller mass (m = ρV), which is ball A.

This question tests our understanding of Pascal's principle, which states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. We are told that the work required to lift the bed with the patient is double the work needed to lift just the bed. In other words, the force required doubles when both the bed and the patient have to be lifted. To maintain the same pressure, we must double the cross-sectional area of the platform of the hydraulic lever on which the patient and the bed are lifted.

The continuity equation states that the flow rate of a fluid must remain constant from one cross-section to another. In other words, when an ideal fluid flows from a pipe with a large cross-sectional area to one that is narrower, its speed decreases. This can be illustrated through the equation A₁ν₁ = A₂ν₂. If blood flows much more slowly through the capillaries, we can infer that the cross-sectional area is larger. This might seem surprising at first glance, but given that each blood vessel divides into thousands of little capillaries, it is not hard to imagine that adding the cross-sectional areas of each capillary from an entire capillary bed results in an area that is larger than the cross-sectional area of the aorta.

The first step in answering this question is defining the different types of pressures. Atmospheric pressure is the pressure at the top of the first fluid exerted by air (at sea level, it is equal to 1 atm). Gauge pressure is the pressure inside the balloon above and beyond atmospheric pressure; gauge pressure is the total (absolute or hydrostatic) pressure inside the balloon minus the atmospheric pressure. Gauge pressure depends on the density of the fluid, the constant of gravity, and the depth at which the object is submerged. Hydrostatic or absolute pressure is the total pressure in the balloon (that is, the gauge pressure and the atmospheric pressure together). Because we are given the gauge pressure at the bottom of the first fluid as 3 atm, our task now is to calculate the gauge pressure accounted for by the second fluid. The hydrostatic pressure at the bottom of the cylinder is 8 atm. One of these atmospheres is atmospheric pressure pushing on the fluids. Another 3 atmospheres are accounted for by the first fluid that is pushing on the second fluid. Thus, the gauge pressure due to the second fluid is 8 ? 1 ? 3 = 4 atm. The ratio of the gauge pressures is therefore 3:4.

The buoyant force (Fbuoy) is equal to the weight of water displaced, which is quantitatively expressed as

F_buoy = m_{fluid displaced}g = ρ_fluidV_{fluid displaced}g
The volume of displaced fluid is equal to the volume of the ball. The density of the fluid remains constant. Therefore, because ball A has a larger volume, it will displace more water and experience a larger buoyant force.

This question is a simple application of the definition of pressure, which is force per area. If pressure decreases 1 percent and area does not change, the force will be decreased by 1 percent. Note that the other measurements given do not play a role in our calculations.

This is a basic interpretation of Bernoulli's equation that states, at equal heights, speed and pressure of a fluid are inversely related (the Venturi effect). Decreasing the speed of the water will therefore increase its pressure. An increase in pressure over a given area will result in increased force being transmitted to the piston.

It is not necessary to do any calculations to answer this question. The open vertical pipes are exposed to the same atmospheric pressure; therefore, differences in the heights of the columns of water in the vertical pipes is dependent only on the differences in hydrostatic pressures in the horizontal pipe. Because the horizontal pipe has variable cross-sectional area, water will flow the fastest and the hydrostatic pressure will have its lowest value where the horizontal pipe is narrowest; this is called the Venturi effect. As a result, pipe 2 will have the lowest water level.

The data given in choice (C) are sufficient to determine the flow rate through Poiseuille's law, which can then be used to determine the linear speed by dividing by the cross-sectional area (which could be determined from the radius, as well). Choice (A) would be sufficient if we also knew the flow rate in the other segment of pipe; one could use the continuity equation to determine the linear speed. The data in choice (B) could be used to determine the critical speed at which turbulent flow begins, but there is no indication that there is turbulent flow. The data in choice (D) could be used to determine the depth of an object in a fluid.