College Physics Quiz: Pressure
20 questions · exam conditions
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PressureQuestion 1 of 20

A submarine experiences increasing outside pressure with depth; which quantity primarily sets the rate of pressure increase in seawater?

Seawater density and gravitational acceleration.
Submarine mass and its cross-sectional area.
Ocean surface area and Earth's radius.
Water viscosity and submarine speed.
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College Physics Quiz

College Physics Quiz: Pressure

Practice Pressure in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Pressure, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A submarine experiences increasing outside pressure with depth; which quantity primarily sets the rate of pressure increase in seawater?

  1. Seawater density and gravitational acceleration. (correct answer)
  2. Submarine mass and its cross-sectional area.
  3. Ocean surface area and Earth's radius.
  4. Water viscosity and submarine speed.
Explanation: This question tests college-level understanding of pressure in fluids and hydrostatics, focusing on the application of fundamental principles. In physics, pressure is defined as a force exerted per unit area, and its understanding is crucial in analyzing fluid behavior. The question's context, such as a submarine at depth, highlights specific applications of pressure increase rate. The correct answer accurately applies the principle involving density and gravity in ΔP = ρgΔh, demonstrating comprehension of the factors. A common distractor might include irrelevant quantities like mass, reflecting typical student misconceptions. To aid student understanding, teaching strategies should emphasize hands-on experiments with fluids to visualize pressure effects and reinforce the concepts of buoyancy and pressure measurement techniques.

Question 2

An object floats with 40%40\% of its volume submerged in water; what does this imply about its average density relative to water?

  1. Its density is 0.400.40 times water's density. (correct answer)
  2. Its density is 2.52.5 times water's density.
  3. Its density equals water's density because it floats.
  4. Its density is unrelated to submergence fraction.
Explanation: This question tests college-level understanding of pressure in fluids and hydrostatics, focusing on the application of fundamental principles. In physics, pressure is defined as a force exerted per unit area, and its understanding is crucial in analyzing fluid behavior. The question's context, involving partial submergence, highlights specific applications of density inference from floating. The correct answer accurately applies the principle that density is the submerged fraction times fluid density, demonstrating comprehension of buoyancy balance. A common distractor might invert the ratio, reflecting typical student misconceptions. To aid student understanding, teaching strategies should emphasize hands-on experiments with fluids to visualize pressure effects and reinforce the concepts of buoyancy and pressure measurement techniques.

Question 3

At the same depth in a connected, static fluid of uniform density, what can be said about the pressure at two points?

  1. Pressures are equal because hydrostatic pressure depends only on depth. (correct answer)
  2. Pressures differ because pressure depends on horizontal position.
  3. Pressures differ because pressure depends on container shape.
  4. Pressures are zero gauge at all points of equal depth.
Explanation: This question tests college-level understanding of pressure in fluids and hydrostatics, focusing on the application of fundamental principles. In physics, pressure is defined as a force exerted per unit area, and its understanding is crucial in analyzing fluid behavior. The question's context, involving points at same depth, highlights specific applications of pressure equality. The correct answer accurately applies the principle that pressure depends only on depth in uniform static fluid, demonstrating comprehension of hydrostatics. A common distractor might invoke shape dependence, reflecting typical student misconceptions. To aid student understanding, teaching strategies should emphasize hands-on experiments with fluids to visualize pressure effects and reinforce the concepts of buoyancy and pressure measurement techniques.

Question 4

A U-shaped tube is partially filled with mercury. When water is poured into the left arm of the tube, the mercury level in the right arm rises by 2.0 cm while the mercury level in the left arm drops by 1.5 cm. What is the height of the water column above the mercury surface in the left arm? (Density of mercury = 13.6 g/cm³, density of water = 1.0 g/cm³)

  1. 47.6 cm (correct answer)
  2. 3.5 cm
  3. 27.2 cm
  4. 34.0 cm
  5. 20.4 cm
Explanation: When you encounter a U-tube problem with different fluids, you're dealing with hydrostatic pressure equilibrium. The key insight is that pressure at any horizontal level must be equal on both sides of the tube. Let's set up the problem systematically. When water is added to the left arm, it creates pressure that pushes the mercury down on the left and up on the right. The mercury moves a total distance of 2.0 cm + 1.5 cm = 3.5 cm from left to right. For pressure equilibrium at the mercury surface level in the left arm, the pressure from the water column must equal the pressure from the extra mercury height on the right side: ρwater⋅g⋅hwater=ρmercury⋅g⋅hmercury\rho_{water} \cdot g \cdot h_{water} = \rho_{mercury} \cdot g \cdot h_{mercury} The height difference in mercury between the two arms is 3.5 cm, so: 1.0 g/cm³×hwater=13.6 g/cm³×3.5 cm1.0 \text{ g/cm³} \times h_{water} = 13.6 \text{ g/cm³} \times 3.5 \text{ cm} hwater=13.6×3.5=47.6 cmh_{water} = 13.6 \times 3.5 = 47.6 \text{ cm} This confirms answer A is correct. Answer B (3.5 cm) represents the total mercury displacement, not the water height. Answer C (27.2 cm) likely comes from using only the 2.0 cm mercury rise instead of the full 3.5 cm difference. Answer D (34.0 cm) might result from incorrectly using 2.5 cm as the mercury height difference. Study tip: In U-tube problems, always identify the total height difference between fluid levels on both sides, then apply the principle that pressure differences must balance using the density ratio.

Question 5

A cylindrical container with cross-sectional area 0.50 m² contains water to a depth of 2.0 m. A wooden block with density 600 kg/m³ and volume 0.20 m³ is placed in the water and floats. By how much does the water level in the container rise?

  1. 0.24 m (correct answer)
  2. 0.40 m
  3. 0.12 m
  4. 0.30 m
  5. 0.20 m
Explanation: When you encounter floating object problems, you need to apply Archimedes' principle: a floating object displaces a volume of water equal to its weight divided by water's density. The key insight is that the water level rise equals the volume of water displaced by the floating block, not the block's total volume. Since the wooden block floats, it displaces only enough water to support its weight. First, find the block's weight: Weight=ρwood×Vblock×g=600 kg/m3×0.20 m3×g\text{Weight} = \rho_{wood} \times V_{block} \times g = 600 \text{ kg/m}^3 \times 0.20 \text{ m}^3 \times g Next, find the volume of water displaced: Vdisplaced=Weight of blockρwater×g=600×0.201000=0.12 m3V_{displaced} = \frac{\text{Weight of block}}{\rho_{water} \times g} = \frac{600 \times 0.20}{1000} = 0.12 \text{ m}^3 Finally, calculate the water level rise: Rise=VdisplacedAcontainer=0.12 m30.50 m2=0.24 m\text{Rise} = \frac{V_{displaced}}{A_{container}} = \frac{0.12 \text{ m}^3}{0.50 \text{ m}^2} = 0.24 \text{ m} Answer A (0.24 m) is correct. Answer B (0.40 m) represents the mistake of dividing the block's total volume by the container area, ignoring that only part of the block is submerged. Answer C (0.12 m) is the volume displaced but fails to account for the container's cross-sectional area. Answer D (0.30 m) appears to be an arbitrary intermediate value with no clear physical basis. Remember: for floating objects, always calculate the displaced volume using the object's weight and water's density, then divide by the container's area to find the level change.

Question 6

A pressure gauge reads 250 kPa when connected to a tank. If atmospheric pressure is 101 kPa, what is the absolute pressure inside the tank?

  1. 351 kPa (correct answer)
  2. 250 kPa
  3. 149 kPa
  4. 202 kPa
  5. 125 kPa
Explanation: When you encounter pressure gauge problems, you need to distinguish between gauge pressure (what the instrument reads) and absolute pressure (the total pressure including atmospheric pressure). Most pressure gauges are designed to read gauge pressure, which measures pressure relative to atmospheric pressure. The relationship is: Absolute Pressure = Gauge Pressure + Atmospheric Pressure In this problem, the gauge reads 250 kPa (gauge pressure) and atmospheric pressure is 101 kPa. Therefore: Absolute Pressure = 250 kPa + 101 kPa = 351 kPa Looking at the answer choices: A) 351 kPa correctly applies the absolute pressure formula. This represents the total pressure inside the tank relative to a perfect vacuum. B) 250 kPa is simply the gauge pressure reading. This would only be correct if you were asked for gauge pressure, not absolute pressure. C) 149 kPa results from incorrectly subtracting atmospheric pressure from gauge pressure (250 - 101 = 149). This misconception treats atmospheric pressure as something to remove rather than add. D) 202 kPa appears to come from doubling atmospheric pressure (101 × 2 = 202), which has no physical basis in pressure relationships. Remember this key distinction: gauge pressure is what most instruments display because we typically care about pressure differences relative to our atmosphere. However, for thermodynamic calculations and theoretical work, you often need absolute pressure. When a problem mentions "absolute pressure," always add atmospheric pressure to the gauge reading.

Question 7

A submarine is at a depth where the absolute pressure is 4.0 times atmospheric pressure. If atmospheric pressure is 1.01 × 10⁵ Pa and the density of seawater is 1025 kg/m³, at what depth is the submarine located?

  1. 297 m
  2. 396 m
  3. 302 m (correct answer)
  4. 405 m
  5. 248 m
Explanation: This question tests your understanding of hydrostatic pressure in fluids. When dealing with pressure at depth in a fluid, you need to consider that the total (absolute) pressure equals atmospheric pressure plus the pressure from the fluid column above. The key relationship is: Pabsolute=Patmospheric+ρghP_{absolute} = P_{atmospheric} + \rho gh, where ρ is fluid density, g is gravitational acceleration, and h is depth. Since the absolute pressure is 4.0 times atmospheric pressure: Pabsolute=4.0×1.01×105 Pa=4.04×105 PaP_{absolute} = 4.0 \times 1.01 \times 10^5 \text{ Pa} = 4.04 \times 10^5 \text{ Pa} The pressure from the water column alone is: Pwater=Pabsolute−Patmospheric=4.04×105−1.01×105=3.03×105 PaP_{water} = P_{absolute} - P_{atmospheric} = 4.04 \times 10^5 - 1.01 \times 10^5 = 3.03 \times 10^5 \text{ Pa} Now solve for depth: h=Pwaterρg=3.03×1051025×9.8=3.03×10510,045=302 mh = \frac{P_{water}}{\rho g} = \frac{3.03 \times 10^5}{1025 \times 9.8} = \frac{3.03 \times 10^5}{10,045} = 302 \text{ m} Choice A (297 m) is too small and might result from calculation errors or rounding mistakes. Choice B (396 m) could come from incorrectly using the total absolute pressure instead of just the water pressure contribution. Choice D (405 m) is significantly too large and might result from using incorrect values for density or gravitational acceleration. Remember: always subtract atmospheric pressure first to find the pressure contribution from the fluid alone, then use ρgh\rho gh to find depth. Don't forget that "absolute pressure" includes atmospheric pressure.

Question 8

A hydraulic press has a small piston with area 10 cm² and a large piston with area 200 cm². If a force of 50 N is applied to the small piston, what force can be exerted by the large piston?

  1. 1000 N (correct answer)
  2. 250 N
  3. 500 N
  4. 100 N
  5. 2000 N
Explanation: When you encounter a hydraulic press problem, you're dealing with Pascal's principle: pressure applied to a confined fluid transmits equally throughout the system. This means the pressure on both pistons must be equal. Since pressure equals force divided by area (P=F/AP = F/A), you can set up the equation: F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}, where subscript 1 represents the small piston and subscript 2 represents the large piston. Substituting the given values: 50 N10 cm2=F2200 cm2\frac{50 \text{ N}}{10 \text{ cm}^2} = \frac{F_2}{200 \text{ cm}^2} Solving for F2F_2: F2=50×20010=1000 NF_2 = \frac{50 \times 200}{10} = 1000 \text{ N} The force is amplified by the ratio of the areas: 20010=20\frac{200}{10} = 20, so the output force is 20 times the input force. Looking at the wrong answers: B) 250 N results from incorrectly using a 5:1 ratio instead of 20:1, possibly confusing linear dimensions with areas. C) 500 N comes from using a 10:1 ratio, which might happen if you mistakenly used the area of the small piston as the multiplier instead of the area ratio. D) 100 N represents only a 2:1 force multiplication, far too small for this area ratio. Remember this key pattern: in hydraulic systems, the force multiplication equals the ratio of the piston areas. Always identify which piston is larger, calculate the area ratio, then multiply the input force by this ratio. The mechanical advantage comes directly from the geometry.

Question 9

A rectangular gate 2.0 m wide and 3.0 m tall is installed vertically in the wall of a water tank. The top edge of the gate is 1.0 m below the water surface. What is the total hydrostatic force acting on the gate?

  1. 147 kN (correct answer)
  2. 73.5 kN
  3. 294 kN
  4. 98 kN
  5. 220 kN
Explanation: When you encounter hydrostatic pressure problems involving submerged surfaces, you need to account for how pressure varies with depth. Unlike atmospheric pressure, water pressure increases linearly with depth, so the force calculation requires integration or using the average pressure. For this rectangular gate, the pressure varies from the top edge (1.0 m deep) to the bottom edge (4.0 m deep). The hydrostatic pressure at any depth h is P=ρghP = \rho gh, where ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3 for water and g=9.8 m/s2g = 9.8 \text{ m/s}^2. The total force equals the average pressure times the gate's area. The average depth of the gate is 1.0+4.02=2.5 m\frac{1.0 + 4.0}{2} = 2.5 \text{ m}. The average pressure is Pavg=(1000)(9.8)(2.5)=24,500 PaP_{avg} = (1000)(9.8)(2.5) = 24,500 \text{ Pa}. The gate's area is (2.0)(3.0)=6.0 m2(2.0)(3.0) = 6.0 \text{ m}^2. Therefore, the total force is F=Pavg×A=24,500×6.0=147,000 N=147 kNF = P_{avg} \times A = 24,500 \times 6.0 = 147,000 \text{ N} = 147 \text{ kN}. Answer A (147 kN) is correct. Answer B (73.5 kN) represents half the correct value—likely from using only the pressure at the top edge instead of the average pressure. Answer C (294 kN) is double the correct answer, possibly from incorrectly adding pressures instead of averaging them. Answer D (98 kN) might result from calculation errors or using an incorrect average depth. Remember: for submerged surfaces, always use the pressure at the centroid (average depth for rectangles) multiplied by the total area, not the pressure at just one edge.

Question 10

A barometer uses mercury with density 13.6 g/cm³. If atmospheric pressure supports a mercury column of height 76.0 cm, what height of water column would be supported by the same atmospheric pressure?

  1. 10.3 m (correct answer)
  2. 5.6 m
  3. 15.2 m
  4. 7.6 m
  5. 20.6 m
Explanation: This problem tests your understanding of hydrostatic pressure and how different fluids relate to atmospheric pressure measurements. When you see barometer problems involving different liquids, remember that atmospheric pressure remains constant—only the height of the supporting column changes based on the fluid's density. The key principle is that atmospheric pressure equals the pressure exerted by any fluid column: Patm=ρghP_{atm} = \rho gh. Since atmospheric pressure is the same in both cases, you can set up the equation: ρHgghHg=ρwaterghwater\rho_{Hg} g h_{Hg} = \rho_{water} g h_{water}. The gravitational acceleration cancels out, giving you: ρHghHg=ρwaterhwater\rho_{Hg} h_{Hg} = \rho_{water} h_{water}. Solving for the water column height: hwater=ρHghHgρwater=13.6×76.01.0=1034 cm=10.3 mh_{water} = \frac{\rho_{Hg} h_{Hg}}{\rho_{water}} = \frac{13.6 \times 76.0}{1.0} = 1034 \text{ cm} = 10.3 \text{ m} Choice A (10.3 m) is correct—water, being much less dense than mercury, requires a much taller column to exert the same pressure. Choice B (5.6 m) appears to use an incorrect density ratio, possibly confusing the relationship between the fluids. Choice C (15.2 m) might result from incorrectly adding rather than using proportional relationships. Choice D (7.6 m) seems to simply convert the mercury height to meters without accounting for density differences. Remember this pattern: when comparing barometric columns of different fluids, the height is inversely proportional to density. Denser fluids need shorter columns; less dense fluids need taller columns to match atmospheric pressure.

Question 11

A cube with edge length 0.20 m and density 800 kg/m³ floats in water. What fraction of the cube's volume is above the water surface?

  1. 0.20 (correct answer)
  2. 0.80
  3. 0.50
  4. 0.60
  5. 0.40
Explanation: When you encounter floating object problems, you're dealing with Archimedes' principle and the concept of buoyant equilibrium. A floating object displaces a volume of fluid whose weight equals the object's total weight. To find what fraction is above water, start with the equilibrium condition. The cube's weight equals the buoyant force from the displaced water: ρcube×Vcube×g=ρwater×Vsubmerged×g\rho_{cube} \times V_{cube} \times g = \rho_{water} \times V_{submerged} \times g The gravitational acceleration cancels out, leaving: ρcube×Vcube=ρwater×Vsubmerged\rho_{cube} \times V_{cube} = \rho_{water} \times V_{submerged} Solving for the submerged fraction: VsubmergedVcube=ρcubeρwater=8001000=0.80\frac{V_{submerged}}{V_{cube}} = \frac{\rho_{cube}}{\rho_{water}} = \frac{800}{1000} = 0.80 Since 80% of the cube is submerged, 20% must be above water. Looking at the wrong answers: Choice B (0.80) represents the submerged fraction, not the fraction above water—a common mistake when students calculate correctly but answer the wrong question. Choice C (0.50) would apply to an object with the same density as water, which isn't the case here. Choice D (0.60) has no physical basis in this problem and likely results from calculation errors. The correct answer is A (0.20). Key strategy: In buoyancy problems, always identify whether the question asks for the submerged or exposed fraction. Calculate the density ratio first—this gives you the submerged fraction directly, then subtract from 1 to get the exposed fraction.

Question 12

A rectangular tank 2.0 m wide, 3.0 m long, and 1.5 m deep is completely filled with water. What is the total force exerted by the water on the bottom of the tank?

  1. 88.2 kN (correct answer)
  2. 44.1 kN
  3. 132.3 kN
  4. 176.4 kN
  5. 66.2 kN
Explanation: When you encounter problems about fluid pressure on submerged surfaces, think about hydrostatic pressure. The key insight is that pressure in a fluid increases linearly with depth due to the weight of the fluid above. For the force on the bottom of the tank, you need to find the pressure at the bottom and multiply by the surface area. The pressure at depth hh in a fluid is P=ρghP = \rho gh, where ρ\rho is the fluid density, gg is gravitational acceleration, and hh is the depth. At the bottom of this tank: P=(1000 kg/m3)(9.8 m/s2)(1.5 m)=14,700 PaP = (1000 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(1.5 \text{ m}) = 14,700 \text{ Pa} The bottom area is: A=2.0 m×3.0 m=6.0 m2A = 2.0 \text{ m} \times 3.0 \text{ m} = 6.0 \text{ m}^2 Total force: F=PA=(14,700 Pa)(6.0 m2)=88,200 N=88.2 kNF = PA = (14,700 \text{ Pa})(6.0 \text{ m}^2) = 88,200 \text{ N} = 88.2 \text{ kN} This confirms answer A is correct. Answer B (44.1 kN) likely comes from using only half the depth or half the area—a common calculation error. Answer C (132.3 kN) might result from incorrectly using 1.5 times the correct answer, possibly from misapplying pressure formulas. Answer D (176.4 kN) is exactly double the correct answer, suggesting someone might have incorrectly added atmospheric pressure or doubled the calculation somewhere. Remember: for hydrostatic pressure problems, always use P=ρghP = \rho gh for pressure at depth, then multiply by the surface area. Don't overthink it—atmospheric pressure cancels out when calculating the force due to the water alone.

Question 13

An object with density ρ_obj is completely submerged in a fluid with density ρ_fluid. If ρ_obj = 1.2ρ_fluid, what is the ratio of the object's apparent weight in the fluid to its weight in air?

  1. 0.167 (correct answer)
  2. 0.833
  3. 1.200
  4. 0.200
  5. 0.600
Explanation: When an object is submerged in a fluid, you need to understand buoyancy and apparent weight. The apparent weight is the actual weight minus the buoyant force, which equals the weight of the displaced fluid. Let's work through this systematically. If the object has volume V, its weight in air is Wair=ρobj⋅V⋅gW_{air} = \rho_{obj} \cdot V \cdot g. When submerged, the buoyant force equals the weight of displaced fluid: Fb=ρfluid⋅V⋅gF_b = \rho_{fluid} \cdot V \cdot g. The apparent weight becomes: Wapparent=Wair−Fb=ρobj⋅V⋅g−ρfluid⋅V⋅gW_{apparent} = W_{air} - F_b = \rho_{obj} \cdot V \cdot g - \rho_{fluid} \cdot V \cdot g Taking the ratio: WapparentWair=ρobj−ρfluidρobj\frac{W_{apparent}}{W_{air}} = \frac{\rho_{obj} - \rho_{fluid}}{\rho_{obj}} Since ρobj=1.2ρfluid\rho_{obj} = 1.2\rho_{fluid}, substituting gives: WapparentWair=1.2ρfluid−ρfluid1.2ρfluid=0.2ρfluid1.2ρfluid=0.21.2=16=0.167\frac{W_{apparent}}{W_{air}} = \frac{1.2\rho_{fluid} - \rho_{fluid}}{1.2\rho_{fluid}} = \frac{0.2\rho_{fluid}}{1.2\rho_{fluid}} = \frac{0.2}{1.2} = \frac{1}{6} = 0.167 This confirms answer A is correct. Answer B (0.833) represents the buoyant force ratio, not apparent weight ratio. Answer C (1.200) is the density ratio itself, which students might mistakenly think applies directly to weight. Answer D (0.200) comes from incorrectly using just the numerator 0.2ρfluidρfluid\frac{0.2\rho_{fluid}}{\rho_{fluid}} instead of the proper denominator. Remember: apparent weight problems always follow the pattern "original weight minus buoyant force." Set up the ratio algebraically before plugging in numbers to avoid arithmetic errors.

Question 14

A laboratory experiment uses a piston-cylinder apparatus to study gas pressure. The cylinder has a cross-sectional area of 0.012 m² and contains gas at atmospheric pressure (101 kPa). A 50 kg mass is placed on top of the frictionless piston.

What is the new gas pressure inside the cylinder after the mass is added?

  1. 142 kPa (correct answer)
  2. 101 kPa
  3. 151 kPa
  4. 192 kPa
  5. 242 kPa
Explanation: When you encounter piston-cylinder problems, you're dealing with pressure equilibrium. The gas pressure must balance all the forces pressing down on the piston from above. Initially, the gas is at atmospheric pressure (101 kPa), which balances the atmospheric pressure pushing down on the piston from outside. When you add the 50 kg mass, this creates additional downward force that the gas must support. The additional pressure from the mass equals the weight divided by the piston area: Padded=mgA=(50 kg)(9.8 m/s2)0.012 m2=490 N0.012 m2=40,833 Pa=40.8 kPaP_{added} = \frac{mg}{A} = \frac{(50 \text{ kg})(9.8 \text{ m/s}^2)}{0.012 \text{ m}^2} = \frac{490 \text{ N}}{0.012 \text{ m}^2} = 40,833 \text{ Pa} = 40.8 \text{ kPa} The new gas pressure becomes: Pnew=Patmospheric+Padded=101+40.8=141.8 kPaP_{new} = P_{atmospheric} + P_{added} = 101 + 40.8 = 141.8 \text{ kPa} This rounds to 142 kPa, making (A) correct. (B) 101 kPa assumes the gas pressure doesn't change, ignoring the added mass entirely. (C) 151 kPa likely comes from rounding errors or using 10 m/s² for gravity instead of 9.8 m/s². (D) 192 kPa suggests doubling something incorrectly, perhaps confusing this with a different type of pressure relationship. Study tip: In piston problems, always remember that gas pressure equals atmospheric pressure plus the pressure contribution from any additional masses on top. The key formula is Pgas=Patm+mgAP_{gas} = P_{atm} + \frac{mg}{A} for each added mass.

Question 15

Two identical containers are filled with different fluids to the same height h. Container A contains a fluid with density ρ₁ and container B contains a fluid with density ρ₂ = 2ρ₁. What is the ratio of the pressure at the bottom of container B to the pressure at the bottom of container A?

  1. 2.0 (correct answer)
  2. 1.5
  3. 4.0
  4. 1.0
  5. 0.5
Explanation: When you encounter fluid pressure problems, remember that pressure at any depth depends on the weight of the fluid column above that point. The fundamental relationship is P=P0+ρghP = P_0 + \rho gh, where P0P_0 is atmospheric pressure, ρ\rho is fluid density, gg is gravitational acceleration, and hh is depth. Since both containers are open to atmosphere and filled to the same height hh, the atmospheric pressure P0P_0 affects both equally. The key difference is the hydrostatic pressure contribution ρgh\rho gh from each fluid. For container A: PA=P0+ρ1ghP_A = P_0 + \rho_1 gh For container B: PB=P0+ρ2gh=P0+2ρ1ghP_B = P_0 + \rho_2 gh = P_0 + 2\rho_1 gh To find the ratio PBPA\frac{P_B}{P_A}: PBPA=P0+2ρ1ghP0+ρ1gh\frac{P_B}{P_A} = \frac{P_0 + 2\rho_1 gh}{P_0 + \rho_1 gh} In typical physics problems, the hydrostatic pressure ρgh\rho gh dominates over atmospheric pressure, so we can approximate: PBPA≈2ρ1ghρ1gh=2.0\frac{P_B}{P_A} \approx \frac{2\rho_1 gh}{\rho_1 gh} = 2.0 This confirms answer (A) 2.0 is correct. (B) 1.5 might result from incorrectly averaging the densities. (C) 4.0 could come from squaring the density ratio instead of using it directly. (D) 1.0 would imply pressure is independent of density, which contradicts the fundamental pressure equation. Study tip: In hydrostatic pressure problems, pressure scales linearly with fluid density when all other factors (height, gravity) remain constant. The denser fluid always produces proportionally higher pressure.

Question 16

A manometer consists of a U-shaped tube partially filled with oil (density 850 kg/m³). One end is connected to a gas tank, and the other end is open to the atmosphere. If the oil level in the open end is 25 cm higher than in the closed end, what is the gauge pressure of the gas in the tank?

  1. 2.08 kPa (correct answer)
  2. 4.16 kPa
  3. 1.04 kPa
  4. 2.45 kPa
  5. 3.12 kPa
Explanation: When you encounter a manometer problem, you're dealing with pressure differences in connected fluids. The key principle is that pressure must balance at any horizontal level in the connected system. In this U-tube manometer, the gas pressure pushes down on the oil in the closed end, while atmospheric pressure pushes down on the oil in the open end. Since the oil level is 25 cm (0.25 m) higher in the open end, the gas pressure must be greater than atmospheric pressure by exactly the weight of that extra oil column. The gauge pressure equals the pressure from the height difference: Pgauge=ρgh=(850 kg/m3)(9.8 m/s2)(0.25 m)=2,082.5 Pa=2.08 kPaP_{gauge} = \rho gh = (850 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(0.25 \text{ m}) = 2,082.5 \text{ Pa} = 2.08 \text{ kPa} This confirms answer A is correct. Looking at the wrong answers: B (4.16 kPa) is exactly double the correct answer, likely from mistakenly doubling the height or density. C (1.04 kPa) is half the correct value, possibly from using half the height or forgetting to account for the full column difference. D (2.45 kPa) might result from rounding errors or using an incorrect value for gravitational acceleration. Study tip: In manometer problems, always identify which side is higher and remember that gauge pressure equals the pressure from the height difference of the fluid column. The fluid that's pushed down more (lower level) is experiencing higher pressure. Practice converting units early—many errors come from mixing centimeters and meters.

Question 17

A spherical balloon filled with helium (density 0.18 kg/m³) has a radius of 1.5 m. If the balloon material and basket have a combined mass of 2.0 kg, what is the net upward force on the balloon in air (density 1.29 kg/m³)?

  1. 156 N
  2. 176 N
  3. 136 N (correct answer)
  4. 196 N
  5. 116 N
Explanation: When you encounter buoyancy problems involving balloons, you're dealing with Archimedes' principle: the buoyant force equals the weight of displaced fluid. The net force determines whether the balloon rises, falls, or hovers. To find the net upward force, calculate the buoyant force from displaced air, then subtract the total weight of the balloon system. First, find the balloon's volume: V=43πr3=43π(1.5)3=14.14 m3V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1.5)^3 = 14.14 \text{ m}^3 The buoyant force equals the weight of displaced air: Fb=ρair⋅V⋅g=1.29×14.14×9.8=179.1 NF_b = \rho_{air} \cdot V \cdot g = 1.29 \times 14.14 \times 9.8 = 179.1 \text{ N} The total downward weight includes both the helium and the balloon material/basket:
  • Weight of helium: WHe=0.18×14.14×9.8=24.9 NW_{He} = 0.18 \times 14.14 \times 9.8 = 24.9 \text{ N}
  • Weight of material/basket: Wmaterial=2.0×9.8=19.6 NW_{material} = 2.0 \times 9.8 = 19.6 \text{ N}
  • Total weight: Wtotal=24.9+19.6=44.5 NW_{total} = 24.9 + 19.6 = 44.5 \text{ N}
Net upward force: Fnet=179.1−44.5=134.6 N≈136 NF_{net} = 179.1 - 44.5 = 134.6 \text{ N} \approx 136 \text{ N} Answer C (136 N) is correct. Answer A (156 N) likely forgot to include the helium's weight. Answer B (176 N) probably calculated only the buoyant force minus the balloon material, ignoring the helium entirely. Answer D (196 N) appears to have calculation errors, possibly in the volume determination. Remember: in buoyancy problems, always account for ALL masses in the system—both the lifting gas and the structural components contribute to the total weight.

Question 18

Water at rest exerts pressure on a submerged wall; in hydrostatics, how does pressure depend on direction at a point?

  1. It acts only upward because gravity points downward.
  2. It is the same in all directions at a given depth in a fluid at rest. (correct answer)
  3. It is larger horizontally than vertically due to buoyancy.
  4. It depends on container shape, not on depth.
Explanation: This question tests college-level understanding of pressure in fluids and hydrostatics, focusing on the application of fundamental principles. In physics, pressure is defined as a force exerted per unit area, and its understanding is crucial in analyzing fluid behavior. The question's context, involving water on a submerged wall, highlights specific applications of pressure directionality. The correct answer accurately applies the principle that pressure is isotropic at a point in a static fluid, demonstrating comprehension of direction independence. A common distractor might assume directional dependence, reflecting typical student misconceptions. To aid student understanding, teaching strategies should emphasize hands-on experiments with fluids to visualize pressure effects and reinforce the concepts of buoyancy and pressure measurement techniques.

Question 19

A closed cylindrical tank contains air above water. The air pressure is 120 kPa (absolute) and the water depth is 2.0 m. What is the absolute pressure at the bottom of the tank?

  1. 139.6 kPa (correct answer)
  2. 120.0 kPa
  3. 220.0 kPa
  4. 19.6 kPa
  5. 101.0 kPa
Explanation: When you encounter fluid pressure problems involving multiple layers or compartments, you need to understand that pressure increases with depth due to the weight of the fluid above. In this tank system, you have air pressure acting on the water surface, plus the additional pressure from the water column itself. To find the absolute pressure at the bottom, you apply the hydrostatic pressure equation: Pbottom=Psurface+ρghP_{bottom} = P_{surface} + \rho gh, where PsurfaceP_{surface} is the pressure at the water surface (120 kPa), ρ\rho is water density (1000 kg/m³), gg is gravitational acceleration (9.8 m/s²), and hh is the water depth (2.0 m). The pressure increase due to the water column is: ρgh=(1000)(9.8)(2.0)=19,600 Pa=19.6 kPa\rho gh = (1000)(9.8)(2.0) = 19,600 \text{ Pa} = 19.6 \text{ kPa} Therefore: Pbottom=120+19.6=139.6 kPaP_{bottom} = 120 + 19.6 = 139.6 \text{ kPa} Answer A (139.6 kPa) correctly accounts for both the air pressure and water column pressure. Answer B (120.0 kPa) incorrectly ignores the water's contribution entirely—this would only be correct at the water surface. Answer C (220.0 kPa) likely results from incorrectly adding atmospheric pressure (101.3 kPa) when it's already included in the given absolute pressure. Answer D (19.6 kPa) gives only the pressure increase from the water column, forgetting the initial air pressure. Remember: absolute pressure at any depth equals the pressure at the surface plus the hydrostatic pressure from the fluid column above that point.

Question 20

A diving bell is lowered into water. When the bottom of the bell is at depth 15 m15 \text{ m}, the air inside is compressed to 60%60\% of its original volume. Assuming isothermal compression and that atmospheric pressure is 1.01×105 Pa1.01 \times 10^5 \text{ Pa}, what is the pressure of the air inside the diving bell at this depth?

  1. 1.48×105 Pa1.48 \times 10^5 \text{ Pa}
  2. 1.01×105 Pa1.01 \times 10^5 \text{ Pa}
  3. 2.48×105 Pa2.48 \times 10^5 \text{ Pa}
  4. 1.68×105 Pa1.68 \times 10^5 \text{ Pa} (correct answer)
Explanation: When you encounter diving bell or submarine problems, you're dealing with gas laws under changing pressure conditions. The key insight is that as depth increases, water pressure increases, compressing the trapped air inside. For isothermal compression, we use Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2. Initially, the air is at atmospheric pressure (P1=1.01×105 PaP_1 = 1.01 \times 10^5 \text{ Pa}) with original volume V1V_1. At 15 m depth, the volume compresses to V2=0.60V1V_2 = 0.60V_1. Applying Boyle's Law: P1V1=P2(0.60V1)P_1V_1 = P_2(0.60V_1) P2=P1V10.60V1=P10.60=1.01×1050.60=1.68×105 PaP_2 = \frac{P_1V_1}{0.60V_1} = \frac{P_1}{0.60} = \frac{1.01 \times 10^5}{0.60} = 1.68 \times 10^5 \text{ Pa} This confirms answer D is correct. Looking at the wrong answers: A (1.48×105 Pa1.48 \times 10^5 \text{ Pa}) might result from incorrectly using 0.68 instead of 0.60 in the denominator. B (1.01×105 Pa1.01 \times 10^5 \text{ Pa}) represents atmospheric pressure only, ignoring the compression entirely. C (2.48×105 Pa2.48 \times 10^5 \text{ Pa}) likely comes from adding hydrostatic pressure to atmospheric pressure rather than using the gas law relationship. Remember: in diving bell problems, don't calculate water pressure separately. Instead, use the volume change to find the new gas pressure directly through Boyle's Law. The compressed volume tells you everything about the pressure increase.