AP Physics 1 Fluids Practice Problems with Answers

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Density of a rectangular slab

   A solid rectangular slab measures 0.30 m by 0.15 m by 0.020 m and has a mass of 1.26 kg. (a) Find its density. (b) Express that in grams per cubic centimetre. (c) Will it float or sink in fresh water, of density $1000\ \text{kg/m}^3$? (d) A second slab of the same material has twice the volume. What are its mass and its density?

   Show the worked solution

   1. (a) Volume first, since the density formula needs it: $V = (0.30)(0.15)(0.020) = 9.0 \times 10^{-4}\ \text{m}^3$.

   2. Then $\rho = m/V = (1.26\ \text{kg})/(9.0 \times 10^{-4}\ \text{m}^3) = 1400\ \text{kg/m}^3$.

   3. (b) There are $100\ \text{cm}$ in a metre, so $1\ \text{m}^3 = 10^6\ \text{cm}^3$, and $1000\ \text{kg/m}^3$ is exactly $1\ \text{g/cm}^3$. So $1400\ \text{kg/m}^3 = 1.4\ \text{g/cm}^3$. Water at $1.0\ \text{g/cm}^3$ makes that conversion a useful mental anchor.

   4. (c) It sinks. $1400\ \text{kg/m}^3$ is greater than the $1000\ \text{kg/m}^3$ of fresh water, so the weight of the slab exceeds the weight of the water it could displace, and no floating position exists. That is a one-line comparison of two densities, and it needs no forces.

   5. (d) The mass doubles to $2.52\ \text{kg}$ and the density does not change at all: it is still $1400\ \text{kg/m}^3$.

   6. Check that claim rather than asserting it: $\rho = (2.52\ \text{kg})/(1.8 \times 10^{-3}\ \text{m}^3) = 1400\ \text{kg/m}^3$. Density is a ratio of two quantities that scale together, so it describes the material and not the lump. Cut the slab in half and each half still measures $1400\ \text{kg/m}^3$.

   7. A dimensional check on part (a) worth doing every time: $\text{kg}/\text{m}^3$ came out of $\text{kg}$ divided by $\text{m}^3$, so the fraction was the right way up. Inverting it gives $7.1 \times 10^{-4}\ \text{m}^3/\text{kg}$, which is a real quantity but is not a density.

   (a) $\rho = 1400\ \text{kg/m}^3$. (b) $1.4\ \text{g/cm}^3$. (c) It sinks, because 1400 is greater than the $1000\ \text{kg/m}^3$ of fresh water. (d) The mass doubles to 2.52 kg and the density is unchanged at $1400\ \text{kg/m}^3$.

2. 2. The same block, two different pressures

   A rectangular block of weight 84 N rests on a flat table. Its base measures 0.30 m by 0.14 m. (a) Find the pressure it exerts on the table. (b) The block is tipped onto a face measuring 0.14 m by 0.060 m. Find the new pressure. (c) By what factor did the pressure change, and why did the weight not? (d) Express the larger pressure as a fraction of one atmosphere.

   Show the worked solution

   1. Pressure is $P = F_{\perp}/A$, the perpendicular force component divided by the area it is spread over. The block's weight acts straight down and the table's surface is horizontal, so the whole 84 N is perpendicular to the surface and no components are needed.

   2. (a) $A_1 = (0.30)(0.14) = 0.042\ \text{m}^2$, so $P_1 = (84\ \text{N})/(0.042\ \text{m}^2) = 2000\ \text{Pa}$.

   3. (b) $A_2 = (0.14)(0.060) = 0.0084\ \text{m}^2$, so $P_2 = (84\ \text{N})/(0.0084\ \text{m}^2) = 10000\ \text{Pa}$.

   4. (c) The factor is $10000/2000 = 5.0$. That is exactly the area ratio upside down: $0.042/0.0084 = 5.0$. At fixed force, pressure and area are inversely proportional, so standing the block on a fifth of the area quintuples the pressure.

   5. The weight did not change because tipping the block moved no matter into or out of it. Weight is a force the Earth exerts on the block's mass; pressure is a statement about how that force is distributed over a contact area. Confusing the two is what makes drawing pins and knife edges seem paradoxical: a small force over a tiny area is a very large pressure.

   6. (d) One atmosphere is $1.0 \times 10^5\ \text{Pa}$, printed among the constants on the equation sheet, so $10000/(1.0 \times 10^5) = 0.10$ atm.

   7. That last number is worth pausing on. Even standing on its smallest face, the block adds only a tenth of an atmosphere to what the table was already carrying. Atmospheric pressure is large, and most everyday pressures are corrections to it rather than replacements for it, which is exactly why the gauge and absolute distinction in problem 3 exists.

   8. One more feature of the calculation: pressure came out as a plain number with no direction. Essential knowledge 8.2.A.2 says pressure is a scalar quantity, even though the force that produces it is a vector.

   (a) $P_1 = 2000\ \text{Pa}$. (b) $P_2 = 10000\ \text{Pa}$. (c) Five times larger, the exact inverse of the area ratio; the weight is unchanged because tipping the block changes how the force is spread, not how much of it there is. (d) $0.10$ atm.

3. 3. Gauge and absolute pressure in an open tank

   A storage tank with vertical sides and a flat bottom of area $0.80\ \text{m}^2$ is open to the atmosphere and holds a liquid of density $1250\ \text{kg/m}^3$ to a depth of 3.20 m. Take atmospheric pressure as $1.0 \times 10^5\ \text{Pa}$ and $g = 9.8\ \text{m/s}^2$. Find (a) the gauge pressure at the bottom, (b) the absolute pressure at the bottom, (c) the gauge pressure 1.20 m below the surface, and (d) the total downward force on the bottom, separated into the atmosphere's share and the liquid's share.

   Show the worked solution

   1. (a) The gauge pressure of a vertical column of fluid is $P_{\text{gauge}} = \rho g h$, where $h$ is depth below the free surface. Compute $\rho g$ once and reuse it: $(1250)(9.8) = 12250\ \text{Pa/m}$.

   2. At the bottom, $h = 3.20\ \text{m}$: $P_{\text{gauge}} = (12250)(3.20) = 39200\ \text{Pa}$, which is $3.92 \times 10^4\ \text{Pa}$.

   3. (b) Absolute pressure adds the reference pressure pushing down on the free surface, which here is the atmosphere: $P = P_0 + \rho g h = 1.0 \times 10^5 + 3.92 \times 10^4 = 1.392 \times 10^5\ \text{Pa}$. Essential knowledge 8.2.B.2 defines this as the sum of a reference pressure and the gauge pressure.

   4. Note the size of the two pieces. The 3.2 m of liquid contributes 39.2 kPa, and the air above contributes 100 kPa. Most of the pressure at the bottom of this tank has nothing to do with the liquid at all.

   5. (c) At $h = 1.20\ \text{m}$: $P_{\text{gauge}} = (12250)(1.20) = 14700\ \text{Pa}$. Gauge pressure is linear in depth, so this had to be $(1.20/3.20) = 0.375$ of the bottom value, and $(0.375)(39200) = 14700\ \text{Pa}$. It agrees.

   6. (d) The bottom is horizontal, so the whole force is perpendicular to it and $F = PA$. Using the absolute pressure, $F = (1.392 \times 10^5)(0.80) = 1.11 \times 10^5\ \text{N}$.

   7. Split that. The atmosphere's share is $(1.0 \times 10^5)(0.80) = 8.0 \times 10^4\ \text{N}$, transmitted down through the liquid. The liquid's own share is the gauge part, $(3.92 \times 10^4)(0.80) = 3.136 \times 10^4\ \text{N}$.

   8. Now check the liquid's share against something it must equal. The tank has vertical sides, so the liquid sitting directly above the bottom is all of it: volume $(0.80)(3.20) = 2.56\ \text{m}^3$, mass $(1250)(2.56) = 3200\ \text{kg}$, weight $(3200)(9.8) = 31360\ \text{N}$. That is the gauge force to the newton, which is the sense in which the pressure at depth really is the weight of what is stacked above you.

   9. Using absolute pressure where the question wanted gauge, or the reverse, is the error this problem exists to train out. The two differ here by $8.0 \times 10^4\ \text{N}$, which is more than twice the liquid's entire contribution.

   (a) $P_{\text{gauge}} = 3.92 \times 10^4\ \text{Pa}$. (b) $P_{\text{abs}} = 1.392 \times 10^5\ \text{Pa}$. (c) $1.47 \times 10^4\ \text{Pa}$ gauge at 1.20 m. (d) $1.11 \times 10^5\ \text{N}$ in total: $8.0 \times 10^4\ \text{N}$ from the atmosphere and $3.136 \times 10^4\ \text{N}$ from the liquid, which is exactly the liquid's own weight.

4. 4. Oil floating on water in a U-tube

   A U-shaped tube open at both ends holds water of density $1000\ \text{kg/m}^3$. Oil of density $800\ \text{kg/m}^3$ is poured slowly into the left arm, where it floats on the water without mixing, until the oil column is 0.150 m tall. Find (a) the height of the water in the right arm above the oil-water interface in the left arm, and (b) how far the top of the oil stands above the top of the water in the right arm.

   Show the worked solution

   1. The key idea is that within one connected body of the same fluid, two points at the same height are at the same pressure. Pick the level of the oil-water interface in the left arm as your reference height, because below that line the tube contains nothing but water all the way round.

   2. (a) Work out the pressure at that level from each side. On the left, you have the atmosphere on top of 0.150 m of oil: $P_{\text{left}} = P_0 + \rho_{\text{oil}} g (0.150)$. On the right, you have the atmosphere on top of an unknown height $h$ of water: $P_{\text{right}} = P_0 + \rho_{\text{water}} g h$.

   3. Set them equal. $P_0$ appears on both sides and cancels, and so does $g$:

      $$\rho_{\text{oil}} (0.150) = \rho_{\text{water}}\, h$$

   4. So $h = (800)(0.150)/(1000) = 0.120\ \text{m}$. The water in the right arm stands 0.120 m above the interface.

   5. Check it as pressures, with the numbers restored. Oil column: $(800)(9.8)(0.150) = 1176\ \text{Pa}$. Water column: $(1000)(9.8)(0.120) = 1176\ \text{Pa}$. Equal, as required.

   6. (b) The oil's top surface is 0.150 m above the interface and the water's top surface in the right arm is 0.120 m above it, so the oil stands $0.150 - 0.120 = 0.030\ \text{m}$ higher.

   7. The result is a ratio, not a coincidence. The two columns must weigh the same per unit area, so the less dense fluid needs the taller column, by exactly the inverse density ratio: $h_{\text{oil}}/h_{\text{water}} = \rho_{\text{water}}/\rho_{\text{oil}} = 1000/800 = 1.25$, and $(0.120)(1.25) = 0.150\ \text{m}$.

   8. Notice that $g$ cancelled. This U-tube would give the same 0.030 m step on any planet, which is exactly why a U-tube is a good way to measure a density ratio and a bad way to measure gravity.

   (a) The water in the right arm stands $0.120\ \text{m}$ above the oil-water interface. (b) The top of the oil is $0.030\ \text{m}$ higher than the top of the water. The oil column has to be taller by the inverse density ratio, $1000/800 = 1.25$, because equal pressures at the interface level mean equal weight per unit area.

5. 5. A submerged block, and the density that gets grabbed by mistake

   A solid block of volume $3.0 \times 10^{-3}\ \text{m}^3$ and mass 8.4 kg is held completely under fresh water of density $1000\ \text{kg/m}^3$. Use $g = 9.8\ \text{m/s}^2$. Find (a) the buoyant force on it, (b) its weight, (c) the reading on a spring scale holding it while it is submerged, (d) whether it floats or sinks when released, and (e) its acceleration at the instant of release.

   Show the worked solution

   1. Take up as positive. Write the two densities out separately before anything else: $\rho_{\text{fluid}} = 1000\ \text{kg/m}^3$ and $\rho_{\text{object}} = m/V = (8.4)/(3.0 \times 10^{-3}) = 2800\ \text{kg/m}^3$. Both appear below, in different roles.

   2. (a) $F_b = \rho_{\text{fluid}} V_{\text{displaced}}\, g$. The block is fully submerged, so the displaced volume is the block's whole volume: $F_b = (1000)(3.0 \times 10^{-3})(9.8) = 29.4\ \text{N}$, upward.

   3. (b) $W = mg = (8.4)(9.8) = 82.32\ \text{N}$, downward.

   4. (c) With the block hanging still, the three forces balance: $T + F_b - W = 0$, so the scale reads $T = 82.32 - 29.4 = 52.92\ \text{N}$. The block has lost $29.4/82.32 = 35.7$ percent of its apparent weight, and it lost exactly the weight of the water it pushed aside.

   5. (d) It sinks. The weight of 82.32 N beats the buoyant force of 29.4 N, and the same verdict comes out of one density comparison: $2800 > 1000$, so the block is heavier than the water it could displace.

   6. (e) Released, the net force is $F_{\text{net}} = F_b - W = 29.4 - 82.32 = -52.92\ \text{N}$, that is 52.92 N downward, and $a = F_{\text{net}}/m = (-52.92)/(8.4) = -6.3\ \text{m/s}^2$: $6.3\ \text{m/s}^2$ downward.

   7. Check that against the general form, which is worth knowing because the volume drops out of it: $a = g\left(1 - \rho_{\text{fluid}}/\rho_{\text{object}}\right) = 9.8\left(1 - 1000/2800\right) = 9.8(0.6429) = 6.3\ \text{m/s}^2$. The block's size never mattered; only the density ratio did.

   8. Now the error this problem was built for. Substitute the object's density into the buoyancy line instead of the fluid's: $F_b = (2800)(3.0 \times 10^{-3})(9.8) = 82.32\ \text{N}$. That is the block's weight, exactly. It has to be, because $\rho_{\text{object}} V g$ is $mg$ rewritten. So the mistake predicts a buoyant force that cancels the weight perfectly, a scale reading of zero, and a block that hovers motionless at any depth in any liquid.

   9. That is what makes this particular slip so easy to catch once you know to look. If your buoyant force comes out equal to the weight, you have not discovered neutral buoyancy, you have written $mg$ twice. Essential knowledge 8.3.B.3 is the fix: the buoyant force is the weight of the fluid displaced, so the density belongs to the fluid and the volume to the hole the object makes in it.

   (a) $F_b = 29.4\ \text{N}$ up. (b) $W = 82.32\ \text{N}$ down. (c) The scale reads $52.92\ \text{N}$. (d) It sinks: $\rho_{\text{object}} = 2800\ \text{kg/m}^3$ against the water's 1000. (e) $6.3\ \text{m/s}^2$ downward. Using the object's density in $F_b = \rho V g$ would have given 82.32 N, exactly the weight, and predicted the block hovering.

6. 6. A floating block, and how much of it is under

   A block of wood floats in fresh water ($1000\ \text{kg/m}^3$) with 0.72 of its volume below the surface. (a) Find the wood's density. (b) The same block is moved to a liquid of density $1300\ \text{kg/m}^3$. What fraction is submerged now? (c) A block of this wood has a mass of 0.90 kg. What volume of fresh water does it displace while floating, and what is the block's own volume?

   Show the worked solution

   1. Take up as positive. A floating object is in equilibrium, so the buoyant force exactly equals its weight, and that single statement carries the whole problem: $\rho_{\text{fluid}} V_{\text{sub}}\, g = \rho_{\text{object}} V_{\text{object}}\, g$.

   2. Cancel $g$ and divide through by $V_{\text{object}}$ to get the submerged fraction:

      $$\frac{V_{\text{sub}}}{V_{\text{object}}} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$$

   3. (a) Read that backwards. The fraction is 0.72 and the fluid is water, so $\rho_{\text{object}} = (0.72)(1000) = 720\ \text{kg/m}^3$.

   4. (b) Same wood, denser liquid: fraction $= 720/1300 = 0.554$. Just over half is under, against 72 percent in water. A denser liquid supports the same weight with less of the block below the surface, so the block rides higher.

   5. Sanity check the direction before trusting the number: heavier liquid, more buoyant force per cubic metre displaced, so less volume needs displacing. The fraction fell, which is the right way.

   6. (c) Floating means the weight of displaced water equals the block's weight, so the displaced water has the same mass as the block, 0.90 kg. Its volume is $V_{\text{sub}} = m/\rho_{\text{water}} = (0.90)/(1000) = 9.0 \times 10^{-4}\ \text{m}^3$, which is 0.90 litres.

   7. The block's own volume comes from its density: $V_{\text{object}} = m/\rho_{\text{object}} = (0.90)/(720) = 1.25 \times 10^{-3}\ \text{m}^3$.

   8. Cross-check the two against part (a): $(9.0 \times 10^{-4})/(1.25 \times 10^{-3}) = 0.72$. The fraction comes back out, as it must.

   9. Worth carrying away: a floating object always displaces its own mass of fluid, while a fully submerged one always displaces its own volume. Which of those is fixed is the difference between a floating problem and a submerged one.

   (a) $\rho_{\text{wood}} = 720\ \text{kg/m}^3$. (b) $0.554$, just over half, so the block floats higher in the denser liquid. (c) It displaces $9.0 \times 10^{-4}\ \text{m}^3$ of water, and the block's own volume is $1.25 \times 10^{-3}\ \text{m}^3$; the ratio is 0.72, as part (a) requires.

7. 7. How much load can the raft carry?

   A foam raft has a total volume of $0.240\ \text{m}^3$ and a mass of 36 kg. It floats in fresh water ($1000\ \text{kg/m}^3$). Use $g = 9.8\ \text{m/s}^2$. Find (a) the fraction of the raft that is submerged when it floats empty, (b) the largest load it can carry before water reaches its top surface, and (c) the fraction submerged when it carries a 150 kg load.

   Show the worked solution

   1. Take up as positive. The raft floats, so the buoyant force equals the total weight it is supporting at every stage of this problem. What changes is how much of the raft has to go under to supply it.

   2. (a) The raft's average density is $\rho_{\text{raft}} = (36\ \text{kg})/(0.240\ \text{m}^3) = 150\ \text{kg/m}^3$, so the submerged fraction is $150/1000 = 0.15$. Empty, 15 percent of it is under water and 85 percent is above.

   3. (b) The most buoyancy the raft can ever supply is the weight of the water displaced when it is submerged to the brim, that is when $V_{\text{sub}}$ equals the whole $0.240\ \text{m}^3$: $F_{b,\text{max}} = (1000)(0.240)(9.8) = 2352\ \text{N}$.

   4. That force can support a total weight of 2352 N, which is a total mass of $(2352)/(9.8) = 240\ \text{kg}$. The raft is 36 kg of that, so the load can be at most $240 - 36 = 204\ \text{kg}$.

   5. Notice the shortcut hiding in those numbers: the maximum supported mass, 240 kg, is the mass of water the raft's full volume would displace, $(1000)(0.240) = 240\ \text{kg}$. You can skip $g$ entirely, because it appears on both sides of the equilibrium.

   6. (c) With a 150 kg load the total mass is $36 + 150 = 186\ \text{kg}$, so the displaced water must also have mass 186 kg, giving $V_{\text{sub}} = (186)/(1000) = 0.186\ \text{m}^3$. The fraction is $0.186/0.240 = 0.775$.

   7. Check that against the limit: 77.5 percent submerged, so there is still 22.5 percent of the raft's depth above the water and the load is comfortably inside the 204 kg maximum. Check the endpoints too, since they are free: 0 kg of load gives $36/240 = 0.15$, matching part (a), and 204 kg gives $240/240 = 1.00$, matching part (b).

   8. The submerged fraction is linear in the total mass here, which is only true because the raft has vertical sides over the range that matters. A hull that widens as it rises would need less extra depth for each extra kilogram.

   (a) 0.15, so 15 percent of the raft is under water when empty. (b) $204\ \text{kg}$ of load, which brings the total to the 240 kg of water the raft's full volume can displace. (c) 0.775, so 77.5 percent submerged.

8. 8. Continuity: the same water through a narrower pipe

   Water flows steadily through a horizontal pipe that is completely full. In the wide section the inside diameter is 8.0 cm and the water moves at 1.20 m/s. Further along, the pipe narrows to an inside diameter of 3.0 cm. Find (a) the volume flow rate, (b) the water's speed in the narrow section, and (c) how long it takes to fill a 25 L container from the open end.

   Show the worked solution

   1. Work in metres from the start. The wide section has radius $0.040\ \text{m}$ and the narrow section $0.015\ \text{m}$. Halving a diameter to get a radius, then squaring, is where most of the errors in this problem live.

   2. (a) $A_1 = \pi r_1^2 = \pi (0.040)^2 = 5.03 \times 10^{-3}\ \text{m}^2$, so the volume flow rate is $Q = A_1 v_1 = (5.03 \times 10^{-3})(1.20) = 6.03 \times 10^{-3}\ \text{m}^3/\text{s}$. Since $1\ \text{m}^3 = 1000\ \text{L}$, that is $6.03\ \text{L/s}$.

   3. (b) $A_2 = \pi (0.015)^2 = 7.07 \times 10^{-4}\ \text{m}^2$. The continuity equation $A_1 v_1 = A_2 v_2$ says the same volume per second passes every cross-section, so $v_2 = Q/A_2 = (6.03 \times 10^{-3})/(7.07 \times 10^{-4}) = 8.53\ \text{m/s}$.

   4. Check without ever computing an area. The $\pi$ and the factor of four cancel between the two sections, leaving $v_2 = v_1 (d_1/d_2)^2 = (1.20)(8.0/3.0)^2 = (1.20)(7.111) = 8.53\ \text{m/s}$. Same answer, and it makes the scaling obvious.

   5. The scaling is the point. The diameter fell by a factor of 2.67, and the speed rose by the square of that, 7.11. Speed in a pipe responds to the square of the diameter, so a modest narrowing produces a dramatic jet.

   6. (c) $25\ \text{L} = 0.025\ \text{m}^3$, so $t = V/Q = (0.025)/(6.03 \times 10^{-3}) = 4.1\ \text{s}$.

   7. Note that part (c) never needed the narrow section. The flow rate is the same everywhere in a full pipe with no branches, which is what essential knowledge 8.4.A.1.i is saying: the rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which it exits.

   8. Two assumptions are doing quiet work here. The water is treated as incompressible, so a cubic metre entering means a cubic metre leaving. And the Topic 8.4 boundary statement says all pipes are assumed to be completely filled by the fluid unless otherwise stated, which is what lets the pipe's cross-section stand in for the flow's cross-section.

   (a) $Q = 6.03 \times 10^{-3}\ \text{m}^3/\text{s}$, about $6.0\ \text{L/s}$. (b) $v_2 = 8.53\ \text{m/s}$, faster by the square of the diameter ratio. (c) About $4.1\ \text{s}$ to fill 25 L.

9. 9. Bernoulli across a constriction

   Water of density $1000\ \text{kg/m}^3$ flows steadily through a horizontal pipe. At point 1 the cross-sectional area is $1.20 \times 10^{-3}\ \text{m}^2$, the speed is 1.50 m/s and the absolute pressure is $1.80 \times 10^5\ \text{Pa}$. At point 2 the pipe has narrowed to $4.0 \times 10^{-4}\ \text{m}^2$. Find (a) the speed at point 2, (b) the absolute pressure at point 2, and (c) both pressures expressed as gauge pressures, taking atmospheric pressure as $1.0 \times 10^5\ \text{Pa}$.

   Show the worked solution

   1. (a) Continuity first, always, because Bernoulli needs both speeds: $v_2 = A_1 v_1/A_2 = (1.20 \times 10^{-3})(1.50)/(4.0 \times 10^{-4}) = (1.80 \times 10^{-3})/(4.0 \times 10^{-4}) = 4.50\ \text{m/s}$. The area fell to a third, so the speed tripled.

   2. (b) Bernoulli's equation as printed is $P_1 + \rho g y_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g y_2 + \frac{1}{2}\rho v_2^2$. The pipe is horizontal, so $y_1 = y_2$ and the two gravitational terms are identical and cancel. What is left is a trade between pressure and speed alone.

   3. Rearrange for the unknown pressure: $P_2 = P_1 + \frac{1}{2}\rho\left(v_1^2 - v_2^2\right)$.

   4. Evaluate the two kinetic terms separately, since that is where sign errors happen. $\frac{1}{2}\rho v_1^2 = (500)(1.50)^2 = (500)(2.25) = 1125\ \text{Pa}$, and $\frac{1}{2}\rho v_2^2 = (500)(4.50)^2 = (500)(20.25) = 10125\ \text{Pa}$.

   5. So $P_2 = 1.80 \times 10^5 + 1125 - 10125 = 1.80 \times 10^5 - 9000 = 1.71 \times 10^5\ \text{Pa}$.

   6. Check by putting both sides back together. Left: $180000 + 1125 = 181125\ \text{Pa}$. Right: $171000 + 10125 = 181125\ \text{Pa}$. Equal, so the total is conserved along the pipe, which is what essential knowledge 8.4.B.2 means by Bernoulli's equation describing the conservation of mechanical energy in fluid flow.

   7. (c) Gauge pressures subtract the atmosphere: $1.80 \times 10^5 - 1.0 \times 10^5 = 0.80 \times 10^5\ \text{Pa}$ at point 1, and $1.71 \times 10^5 - 1.0 \times 10^5 = 0.71 \times 10^5\ \text{Pa}$ at point 2. The difference between the two points, 9000 Pa, is identical either way, because subtracting the same atmosphere from both changes no difference.

   8. The trap is expecting the pressure to rise where the water speeds up, on the reasoning that fast water is pushing harder. It falls, and by 9000 Pa, which is 5.0 percent of the upstream pressure. Every joule per cubic metre the water gains as kinetic energy has to come out of the pressure term, because a horizontal pipe offers nowhere else to take it from.

   9. One caution on the causation. The narrowing is what forces the speed up, through continuity, and the pressure drop is the consequence. Reading it the other way round, as though a mysterious low pressure were sucking the water faster, will mislead you on the next problem.

   (a) $v_2 = 4.50\ \text{m/s}$, three times faster because the area fell to a third. (b) $P_2 = 1.71 \times 10^5\ \text{Pa}$ absolute, a drop of 9000 Pa. (c) $0.80 \times 10^5\ \text{Pa}$ and $0.71 \times 10^5\ \text{Pa}$ gauge. Pressure is lower where the water moves faster.

10. 10. A balloon in air, where the fluid is a gas

    A balloon of volume $0.85\ \text{m}^3$ is filled with helium of density $0.18\ \text{kg/m}^3$. The balloon skin plus its attached payload have a combined mass of 0.60 kg. The surrounding air has density $1.20\ \text{kg/m}^3$. Use $g = 9.8\ \text{m/s}^2$ and ignore air resistance. Find (a) the buoyant force on the balloon, (b) its total weight, (c) whether it rises or falls when released, and (d) the extra payload that would make it hover.

    Show the worked solution

    1. Take up as positive. A gas is a fluid: essential knowledge 8.1.A.2 defines a fluid as a substance that has no fixed shape, which covers liquids and gases alike, so buoyancy works here exactly as it did under water.

    2. Label the three densities before starting, because this problem has one more than usual. The surrounding fluid is air at $1.20\ \text{kg/m}^3$. The contents are helium at $0.18\ \text{kg/m}^3$. The balloon as a whole has an average density that neither of those numbers is.

    3. (a) The displaced fluid is air, and the displaced volume is the balloon's whole outside volume: $F_b = \rho_{\text{air}} V g = (1.20)(0.85)(9.8) = 9.996\ \text{N}$, which is $10.0\ \text{N}$ upward.

    4. (b) The weight is that of everything the balloon carries, and the helium is part of that: $m_{\text{He}} = \rho_{\text{He}} V = (0.18)(0.85) = 0.153\ \text{kg}$, so the total mass is $0.153 + 0.60 = 0.753\ \text{kg}$ and $W = (0.753)(9.8) = 7.38\ \text{N}$ downward.

    5. (c) It rises. $F_{\text{net}} = 9.996 - 7.379 = 2.62\ \text{N}$ upward. The same verdict from densities: the balloon's average density is $(0.753\ \text{kg})/(0.85\ \text{m}^3) = 0.886\ \text{kg/m}^3$, less than the air's 1.20, so it goes up. Notice that this average, 0.886, is neither the helium's 0.18 nor the air's 1.20; the skin and payload are most of the mass.

    6. (d) Hovering means zero net force, so the extra weight has to swallow the whole 2.6166 N surplus: $\Delta m = (2.6166\ \text{N})/(9.8\ \text{m/s}^2) = 0.267\ \text{kg}$. Confirm it: total mass $0.753 + 0.267 = 1.020\ \text{kg}$, weight $(1.020)(9.8) = 9.996\ \text{N}$, exactly the buoyant force. And the average density becomes $1.020/0.85 = 1.20\ \text{kg/m}^3$, the air's density, which is the density condition for neutral buoyancy.

    7. The trap, in its second costume. Reaching for the contents' density instead of the surrounding fluid's gives $F_b = (0.18)(0.85)(9.8) = 1.50\ \text{N}$, well under the 7.38 N weight, and predicts a helium balloon that drops to the floor. Everyone knows that is wrong, which is what makes this a good place to see the error: the density in $F_b = \rho V g$ belongs to what the object is in, never to the object or to what is inside it.

    8. Why a helium balloon rises at all, in one line: it is not that helium is light, it is that the balloon plus its contents weigh less than the air they push out of the way. Fill the same skin with air instead and the average density becomes greater than 1.20, because the skin's mass is added to a full load of air, and it falls.

    (a) $F_b = 10.0\ \text{N}$ up. (b) $W = 7.38\ \text{N}$ down, including 0.153 kg of helium. (c) It rises, with a net force of $2.62\ \text{N}$; its average density of $0.886\ \text{kg/m}^3$ is below the air's 1.20. (d) An extra $0.267\ \text{kg}$ makes it hover. Using helium's density in $F_b$ instead of air's gives 1.50 N and predicts the balloon sinking.

11. 11. Three claims about buoyancy, and why each one fails

    A student makes three claims. (a) A steel ship must sink, because steel is about eight times denser than water. (b) A wooden block that floats in fresh water will float lower in salt water, because salt water is heavier. (c) A fully submerged block feels a larger buoyant force the deeper it goes, because the pressure down there is greater. Decide which claims survive, and correct the ones that do not, arguing from $F_b = \rho V g$ rather than from intuition.

    Show the worked solution

    1. All three claims are wrong, and each fails on a different word in $F_b = \rho_{\text{fluid}} V_{\text{displaced}}\, g$. Working out which word is the whole skill.

    2. Claim (a) confuses the material's density with the object's. What floats or sinks is settled by the average density of the whole object, hull and enclosed air together, against the fluid's. A ship's hull encloses a very large volume of air, so although its steel is dense, the ship's average density is well under $1000\ \text{kg/m}^3$. The displaced volume in $F_b = \rho V g$ is the volume of the hull below the waterline, not the volume of the metal.

    3. The test is easy to state and easy to apply: crush the ship into a solid steel cube and it sinks, because the enclosed air is gone and the average density is now the steel's. Nothing about the material changed; only the shape did.

    4. Claim (b) has the physics and the direction pointing opposite ways. Salt water is denser than fresh water, and the submerged fraction of a floating object is $\rho_{\text{object}}/\rho_{\text{fluid}}$. A larger $\rho_{\text{fluid}}$ on the bottom of that fraction makes it smaller, so the block floats higher, not lower.

    5. Problem 6 in this set is the arithmetic version of the correction. Wood of density $720\ \text{kg/m}^3$ sits with 0.72 of its volume under fresh water and only $720/1300 = 0.554$ of it under a liquid of density $1300\ \text{kg/m}^3$. Denser fluid, less of the block submerged, every time. The student's instinct that heavier fluid presses down harder overlooks that it also pushes up harder, and only the pushing up is buoyancy.

    6. Claim (c) reads a real fact and draws the wrong conclusion from it. Pressure does increase with depth, by $\rho g h$, and the deeper block genuinely is squeezed harder on every surface. But the buoyant force is the difference between the upward push on the bottom face and the downward push on the top face, and lowering the block by an extra depth $\Delta h$ raises both of those pressures by the same $\rho g \Delta h$. The difference between them does not move.

    7. The formula says the same thing more bluntly: $F_b = \rho V g$ contains a density, a volume and $g$, and no depth appears anywhere in it. If depth is not in the expression, depth cannot change the answer.

    8. The claim does have a boundary worth naming. It relies on the fluid having one density everywhere, which the Topic 8.4 boundary statement licenses by assuming ideal fluids, and essential knowledge 8.1.A.4 defines an ideal fluid as incompressible with no viscosity. Take a compressible object instead, such as a rubber balloon or a diver's lungs, and its volume really does shrink with depth, so its buoyant force really does fall. That is a change in $V$, not a change in the rule.

    None of the three survives. (a) Ships float because a hull encloses air, so the object's average density, not the steel's, is what competes with the water's; the displaced volume is the hull below the waterline. (b) Backwards: the submerged fraction is $\rho_{\text{object}}/\rho_{\text{fluid}}$, so a denser fluid makes the block float higher. Wood at $720\ \text{kg/m}^3$ sits 0.72 submerged in water and 0.554 submerged in a $1300\ \text{kg/m}^3$ liquid. (c) The pressures on the top and bottom faces both rise by the same amount with depth, so their difference, which is the buoyant force, does not change. No depth appears in $F_b = \rho V g$. The one exception is a compressible object, whose volume shrinks with depth.