Fluids: Unit Diagnostic (Physics 1 Unit 8)

This diagnostic is for the algebra-based Physics 1 course, Unit 8 (Fluids). It shows you which topics to revisit. Allow about 30 minutes, then use the table at the end.

These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator in degree mode, g = 9.8 m/s², water density 1000 kg/m³ and 1 atm = 1.0 × 10⁵ Pa. Treat every fluid as ideal (incompressible, no viscosity), every pipe as full, and ignore drag.

Question 1 (multiple choice · 8.1)

A flask holds 150 cm³ of water (1.00 g/cm³) with 100 cm³ of oil (0.80 g/cm³) floating on top. The liquids do not mix. What is the average density of the flask’s contents?

(A) 900 kg/m³
(B) 920 kg/m³
(C) 880 kg/m³
(D) 0.92 kg/m³

Answer and explanation

Answer: (B). Average density = total mass ÷ total volume. Mass = 150 × 1.00 + 100 × 0.80 = 230 g. Volume = 250 cm³. So ρ_avg = 0.92 g/cm³ = 920 kg/m³.

(A) averages the two densities without weighting them by volume. There is more water than oil.
(C) swaps the weights, as if there were more oil than water.
(D) gets 0.92 g/cm³ but forgets to multiply by 1000 to convert to kg/m³.

If you missed this: read “Average density of an object made of parts” in the Topic 8.1 study guide.

Question 2 (multiple choice · 8.1)

A student models honey running slowly off a spoon as an ideal fluid. Which part of that model fits honey worst?

(A) Incompressible, because honey shrinks a lot when squeezed
(B) Being a fluid at all, because honey keeps a fixed shape
(C) Having a density, because a thick liquid has no definite mass per volume
(D) No viscosity, because honey has a lot of internal friction as its layers slide

Answer and explanation

Answer: (D). An ideal fluid is incompressible and has no viscosity. Honey flows slowly because its layers drag on each other: high viscosity. It is still very hard to compress.

(A) is false. Liquid particles are already close together, so honey barely compresses.
(B) is false. Honey takes the shape of its container, so it is a fluid.
(C) is false. Any sample of honey has a mass and a volume, so it has a density.

If you missed this: read “The ideal-fluid model” in the Topic 8.1 study guide.

Question 3 (multiple choice · 8.2)

A straight tube, closed at its lower end, slopes down at 30° below the horizontal. It is full of water and open to the air at its top. Point X is 1.2 m along the tube from the water surface. What is the gauge pressure at X?

(A) 1.2 × 10⁴ Pa
(B) 1.0 × 10⁴ Pa
(C) 5.9 × 10³ Pa
(D) 1.1 × 10⁵ Pa

Answer and explanation

Answer: (C). Only the vertical depth counts: h = 1.2 sin 30° = 0.60 m. P_gauge = ρgh = 1000 × 9.8 × 0.60 = 5880 Pa ≈ 5.9 × 10³ Pa.

(A) uses the 1.2 m measured along the tube as the depth.
(B) uses 1.2 cos 30° = 1.04 m, the horizontal distance.
(D) is the absolute pressure (5880 Pa + 1.0 × 10⁵ Pa). The question asks for gauge pressure.

If you missed this: read “Pressure under a column of fluid” in the Topic 8.2 study guide.

Question 4 (multiple choice · 8.2)

A solid concrete block measures 0.10 m × 0.20 m × 0.30 m and has density 2500 kg/m³. It stands upright on a table on its smallest face. What pressure does it exert on the table?

(A) 7.4 × 10³ Pa
(B) 4.9 × 10³ Pa
(C) 2.5 × 10³ Pa
(D) 7.5 × 10² Pa

Answer and explanation

Answer: (A). Mass = 2500 × 0.0060 = 15 kg, so the weight is 147 N. The smallest face is 0.10 m × 0.20 m = 0.020 m². P = 147 ÷ 0.020 = 7350 Pa. Shortcut: for an upright uniform block, P = ρgh = 2500 × 9.8 × 0.30, the same answer.

(B) uses the middle face (0.030 m²).
(C) uses the largest face (0.060 m²).
(D) divides the mass (15 kg) by the area instead of the weight.

If you missed this: read “Pressure on a surface” and Worked example 1 in the Topic 8.2 study guide.

Question 5 (multiple choice · 8.3)

A flat-bottomed barge floats in fresh water. Its horizontal cross-section at the waterline is 40 m², and its sides are vertical. A 6000 kg load is lowered onto it. How much deeper does the barge sit once it floats at rest again?

(A) 6.0 m
(B) 1.5 m
(C) 0.15 m
(D) 0, because a floating object already has F_b = mg

Answer and explanation

Answer: (C). Floating again, F_b = mg, so the extra buoyant force equals the load’s weight. Extra displaced volume = 6000 kg ÷ 1000 kg/m³ = 6.0 m³. Spread over 40 m², that is 6.0 ÷ 40 = 0.15 m deeper.

(A) treats the extra volume, 6.0 m³, as a depth.
(B) multiplies by g as well. The g in F_b = ρVg and in mg cancels.
(D) is right that F_b = mg, but mg has grown, so F_b must grow too.

If you missed this: read “Floating” and Worked example 2 in the Topic 8.3 study guide.

Question 6 (multiple choice · 8.3)

Take +y up. An ice cube (density 920 kg/m³) is held at the bottom of a glass of water by a finger, then released. What is its acceleration just after release?

(A) −9.8 m/s²
(B) +11 m/s²
(C) +0.78 m/s²
(D) +0.85 m/s²

Answer and explanation

Answer: (D). ma_y = F_b − mg = (ρ_water − ρ_ice)Vg, and m = ρ_ice V. So a_y = g(ρ_water / ρ_ice − 1) = 9.8 × (1000/920 − 1) = +0.85 m/s².

(A) ignores the buoyant force, as if the ice were falling in air.
(B) uses only the buoyant force and leaves out the weight: F_b / m = 9.8 × 1000/920.
(C) divides the net force by the mass of the displaced water (ρ_water V), not the ice’s own mass (ρ_ice V).

If you missed this: read “Sinking and rising: Newton’s second law” in the Topic 8.3 study guide.

Question 7 (multiple choice · 8.4)

Water moves at 2.4 m/s through a main pipe of area A₀. The pipe splits into three identical branches, each of area A₀/2. What is the speed in each branch?

(A) 0.80 m/s
(B) 1.6 m/s
(C) 2.4 m/s
(D) 4.8 m/s

Answer and explanation

Answer: (B). Mass conservation: the flow rate in equals the total flow rate out. A₀ × 2.4 = 3 × (A₀/2) × v, so v = 2.4 ÷ 1.5 = 1.6 m/s. The branches together have more area than the main pipe.

(A) divides the speed by three but ignores each branch’s area.
(C) confuses equal flow rate with equal speed.
(D) uses only one branch, as if all the water went through it.

If you missed this: read “The continuity equation” in the Topic 8.4 study guide.

Question 8 (multiple choice · 8.4)

Water flows steadily up a pipe of uniform cross-section. Point 2 is 3.0 m higher than point 1. The absolute pressure at point 1 is 2.0 × 10⁵ Pa. What is the absolute pressure at point 2?

(A) 1.7 × 10⁵ Pa
(B) 2.0 × 10⁵ Pa
(C) 2.3 × 10⁵ Pa
(D) 2.9 × 10⁴ Pa

Answer and explanation

Answer: (A). Same area, so the same speed (continuity), and the ½ρv² terms cancel in Bernoulli’s equation. P₂ = P₁ − ρg(y₂ − y₁) = 2.0 × 10⁵ − 1000 × 9.8 × 3.0 = 1.7 × 10⁵ Pa.

(B) cancels the speed terms but forgets the height term.
(C) adds ρgΔy. Higher up, the pressure is lower.
(D) is the pressure change, not the pressure at point 2.

If you missed this: read “Conservation of energy: Bernoulli’s equation” in the Topic 8.4 study guide.

Question 9 (short answer · 8.3)

Take +y up. A 2.0 kg rock of density 2500 kg/m³ rests on the flat bottom of a pond.

(a) Draw the rock’s free-body diagram. (b) Find the normal force from the pond floor on the rock. (c) A student says: “The rock presses on the floor with its full weight, because the water is above it and only pushes down.” Evaluate this claim.

Worked answer

(a) Three forces: weight mg down; buoyant force F_b up; normal force F_N (floor on rock) up.

(b) V = 2.0 ÷ 2500 = 8.0 × 10⁻⁴ m³. F_b = 1000 × 8.0 × 10⁻⁴ × 9.8 = 7.84 N. At rest: F_N + F_b − mg = 0, so F_N = 19.6 − 7.84 = 12 N (11.8 N).

(c) Incorrect. The water pushes on every face of the rock, and the pressure is greater lower down. Water still reaches the rock’s sides and seeps under its rough base, so there is a net upward push: the buoyant force. By Newton’s third law the rock presses on the floor with 11.8 N, not 19.6 N.

Check yourself: 1 point each for three forces with F_b up, F_b = 7.8 N from the displaced volume, F_N = 12 N, and rejecting the claim because fluid pressure pushes up on lower surfaces.

If you missed this: read “Where the buoyant force comes from” and Figure 2 in the Topic 8.3 study guide.

Question 10 (short answer · 8.4)

A full siphon tube carries water out of a wide tank, open to the air, into a bucket. The tube’s open outlet is 0.30 m below the tank’s water surface. The highest point of the tube is 0.20 m above the surface. The tube has a uniform inside area of 6.0 × 10⁻⁵ m².

(a) Find the speed of the water leaving the outlet. (b) Find how long it takes to collect 2.0 L (1 L = 10⁻³ m³), assuming the tank level barely changes. (c) Find the gauge pressure at the highest point of the tube, and explain what its sign means.

Worked answer

(a) Bernoulli from the surface (atmospheric pressure, v ≈ 0) to the outlet (atmospheric pressure, 0.30 m lower): ρg(0.30) = ½ρv², so v = √(2 × 9.8 × 0.30) = 2.4 m/s (2.42 m/s).

(b) Flow rate Av = 6.0 × 10⁻⁵ × 2.42 = 1.46 × 10⁻⁴ m³/s. Time = 2.0 × 10⁻³ ÷ 1.46 × 10⁻⁴ = 14 s.

(c) The tube is uniform, so the speed at the top is also 2.42 m/s (continuity). Bernoulli from the surface (gauge 0, y = 0, v ≈ 0) to the top (y = 0.20 m): 0 = P + ρg(0.20) + ½ρv² = P + 1960 + 2940, so P = −4.9 × 10³ Pa gauge. The negative sign means the water at the top is below atmospheric pressure. The air pushing on the tank surface is what drives water up the first part of the tube.

Check yourself: 1 point each for 2.4 m/s from the 0.30 m drop, 14 s from Av, the same speed at the top, and −4.9 × 10³ Pa with its meaning.

If you missed this: read “Conservation of energy: Bernoulli’s equation” and “Torricelli’s result” in the Topic 8.4 study guide.

Your next step

Mark each question right or wrong. Count Questions 9 and 10 as right only if you earned every point.

Topic	Question(s)	If you missed it, read
8.1 Internal structure and density	1, 2	Topic 8.1 study guide
8.2 Pressure	3, 4	Topic 8.2 study guide
8.3 Fluids and Newton’s laws	5, 6, 9	Topic 8.3 study guide
8.4 Fluids and conservation laws	7, 8, 10	Topic 8.4 study guide

How to use your result

Missed one question in a topic? Re-read the section named in that answer, then try the topic’s practice set.
Missed two or more questions in a topic? Work through that study guide again, including its worked examples, before mixed practice.
Guessed? Treat the question as missed.
Look for patterns. Errors in Questions 3, 4 and 8 often mean a depth, area or height was taken from the wrong length. Errors in Questions 5, 6 and 9 usually mean a missing free-body diagram.
All correct? Move on to the mixed unit review, which combines the topics.