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Test: Fluids at Rest - MCAT MCQ


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10 Questions MCQ Test Physics for MCAT - Test: Fluids at Rest

Test: Fluids at Rest for MCAT 2024 is part of Physics for MCAT preparation. The Test: Fluids at Rest questions and answers have been prepared according to the MCAT exam syllabus.The Test: Fluids at Rest MCQs are made for MCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Fluids at Rest below.
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Test: Fluids at Rest - Question 1

On which of the following does the buoyant force on an object floating on the surface of a liquid most directly depend?

Detailed Solution for Test: Fluids at Rest - Question 1
  • The buoyant force is the upward force by the fluid that opposes the downward force of gravity, and in the case of a floating object, the two forces are equal and opposite.
  • The formula for buoyant force is FB = pVg, where V is the volume of the object submerged. So the buoyant force does not depend on the volume of the object.
  • The buoyant force does not depend on the viscosity of the liquid since viscosity refers to fluids in motion and is the resistance to shearing flows.
  • In the formula for buoyant force, the density in question is that of the liquid, so to reiterate:
Test: Fluids at Rest - Question 2

There is a 1-liter container filled with 500 milliliters of water placed on top of a scale displaying 80 pounds. When a metal cube is submerged tied to a string end into the container, which of the following statements accurately describes the result?

Detailed Solution for Test: Fluids at Rest - Question 2
  • When the cube is submerged, the object floats in equilibrium.
  • There must be a buoyant force keeping the object afloat pushing upward.
  • According to Newton’s Third Law, there should be an equal and opposite force downwards.
  • That additional force will cause the scale to register a weight greater than 80 pounds.
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Test: Fluids at Rest - Question 3

A gold crown recently acquired is suspected to be brass covered in gold. The crown weighs 25.0 N in air and has an apparent weight of 22.5 N in fresh water. Based on these results, which of the following conclusions is correct?
gold = 19,300 kg/m3, ρbrass = 8300-8700 kg/m3)

Detailed Solution for Test: Fluids at Rest - Question 3
  • Look at the forces in the y-axis to figure out the buoyant force: Fy = may becomes Wa + Fb - mg = may 
  • 22.5 + Fb - 25.0 = 0, so the buoyant force Fb is equal to 2.5 N. 
  • To find the volume of the crown, approximate g as 10 and use 1000 for ρ of water. So ρg = 10000. and Fb = ρgV, so V = 2.5 / 10000 = 2.5 x 10-4 m3  
  • To find the mass of the crown. divide the weight by 10. so mass of crown is 2.5 kg. 
  • To find the density, divide mass by volume: 2.5 kg/2.5 x 10-4 m3 = 10.000 kg/m3
  • Based on the density value. it is likely that the crown is made of brass and onjly covered in gold. 
Test: Fluids at Rest - Question 4

When the gauge pressure is doubled, what happens to the absolute pressure?

Detailed Solution for Test: Fluids at Rest - Question 4
  • Absolute pressure is equal to atmospheric pressure plus gauge pressure.
  • Atmospheric pressure is 101,300 Pa, while a typical gauge pressure for a tire is 30 psi or 207 Pa.
  • Since atmospheric pressure is relatively large compared to most gauge pressures, doubling the gauge pressure would not end up doubling the absolute pressure.
  • To use the tire example, if the gauge is 414 Pa, then absolute pressure would be 101,314 Pa. Absolute pressure increase by a factor less than 2.
Test: Fluids at Rest - Question 5

A patient is scheduled to receive an intravenous injection of medication that must be at least 109.3 kilopascals at the injection point before going into surgery. If the patient is lying on a bed 0.9 meters high, what is the minimum height above the ground that the nurse must suspend the bag to achieve that pressure at the injection point?
fluid = 1020 kg/m3)

Detailed Solution for Test: Fluids at Rest - Question 5
  • The formula for hydrostatic pressure is P = P0 + ρgh, where ρ is the density of the fluid.
  • Plug in 109.3 kPa for absolute pressure, P, and use 101.3 kPa for the atmospheric pressure (P0)
  • Solve for h: h = ΔP/ρg and plug in the values.
  • 109.3 kPa - 101.3 kPa / (9.8 m/s2 * 1020 kg/m3) = 8000 Pa / 10000 kg/m2s2 = 0.8 m
  • This will give us the height above the patient’s arm to place the bag, so we must add 0.9 meters to figure out the height above the ground.
    The answer is 1.7 meters.
Test: Fluids at Rest - Question 6

Which of the following graphs represents the hydrostatic pressure for an incompressible fluid in a container open to the atmosphere on the surface of the Moon? (P0 = atmospheric pressure on Earth’s surface, h = height above Moon’s surface)

Detailed Solution for Test: Fluids at Rest - Question 6
  • Recall the hydrostatic pressure equation, P = P+ gh. Recognize that it is a manifestation of y = mx +b, which is the formula for a linear function.
  • y is P, and x is h. Po​ start subscript, o, end subscript would be b, and ρg together represents m. Eliminate any choices that are not showing a linear or straight line function.
  • On the moon, there is less atmospheric pressure due to its much smaller mass and would be less than Po, specifically defined as atmospheric pressure on Earth. The y-intercept would start below Po.  
  • This is the correct graph:
Test: Fluids at Rest - Question 7

Which of the following scenarios most accurately represents Pascal’s law?

Detailed Solution for Test: Fluids at Rest - Question 7
  • Pascal’s Law states that pressure is transmitted undiminished to every portion of the fluid and to the walls of the container.
  • This allows us to write the equation P1 = P2, and  then F1/A1 = F2/A2, as well as F1d1 = F2d2.
  • As a result, we can explorate that a small force applied on a small area over a long distance will produce a large force applied on a large area over a short distance.
  • We also can deduce that W1/V1 = W2/V2. Since work is the same on both sides, the volume displaced by one side will be equal to the volume displaced on the other side in a hydraulic lift.
Test: Fluids at Rest - Question 8

A 30 kilogram object is dropped into acetic acid, which has a specific gravity of 1.06. Once it is immersed in acetic acid, the object has an apparent weight of 240 N. What is the object’s specific gravity?

Detailed Solution for Test: Fluids at Rest - Question 8
  • Convert 30 kg to Newtons, which would be 300 N. With an apparent weight of 240 N, the buoyant force gives it an upward push of 60 N.
  • The buoyant force of 60 N accounts for the displaced volume of acetic acid, and we know the object weighs 300 N.
  • Setup a ratio of FB(aceticacid)/Wobject = paceticacid/pobject
  • ​60/300 = 1.06/x, so 60·5 = 300 and 1.06·5 must equal x, which is 5.3.
Test: Fluids at Rest - Question 9

Which of the following statements about fluids at rest is true?

Detailed Solution for Test: Fluids at Rest - Question 9

According to Pascal's law, the pressure in a fluid increases with depth. This is because the weight of the fluid above exerts a force on the fluid below, resulting in an increase in pressure as we move deeper into the fluid.

Test: Fluids at Rest - Question 10

What property of a fluid allows it to exert pressure equally in all directions?

Detailed Solution for Test: Fluids at Rest - Question 10

Incompressibility is the property of a fluid that allows it to exert pressure equally in all directions. Since the fluid cannot be compressed, any force applied to it is transmitted equally in all directions, resulting in equal pressure distribution.

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