Answer key · Unit 7

Pressure and fluids.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 25 minutes

Pressure

Pressure is force spread over an area: a gas that pushes because its particles strike the walls, Boyle's law, particles that speed up as the square root of temperature, the gas law, and pressure that grows with depth under water.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Hits on the walls
Practice · Choose
Question
What makes the pressure the gas in the cylinder puts on the piston?

Correct answer
- Correct Countless hits by particles that fly about and strike the piston and the walls
Setup, choices and feedback
The Scene at this step
Volume
70 %
Temperature
300 K
Show the wall hits
On
Choices
Shown in a random order.
- Correct Countless hits by particles that fly about and strike the piston and the walls
- Wrong The particles swell and push each other outward
  Misconception: Thinks a gas presses harder because its particles swell when heated, or shrink when it is squeezed.
  Feedback from the catalogue
  A gas’s particles don’t change size. Squeeze the gas and the same particles just have less room, so they strike the walls more often. Heat it and they move faster, striking harder and more often. The pressure is the sum of those hits, not the size of the particles.
- Wrong The weight of the gas pressing down
  Feedback
  The gas weighs almost nothing, and it pushes just as hard upward, on the piston, and sideways. Weight cannot explain an upward push.
Feedback for a right answer
Each hit gives the wall a tiny shove, and billions of billions of them every second add up to a steady push. Pressure is that force divided by the area: $p = \dfrac{F}{A}$ ⁠, in pascals, newtons per square metre.

Worked solution
1. Each particle flies in a straight line until it strikes a wall and bounces off.
2. Each bounce shoves the wall outward a little.
3. The vast number of hits gives a steady force on every part of the wall, and force per unit area is pressure.
Step 2 of 7
Open this step
Squeeze it
Practice · Predict, then look
Question
The Volume is 100%, and the gauge reads about 71 kPa. The piston is pushed down to 50%, without changing the Temperature. Predict: what will the gauge read?

Correct answer
141.8 kPa ± 4 kPa
A number prediction counts as right within ± 4 kPa
Setup, choices and feedback
The Scene at this step
Volume
100 %
Temperature
300 Kfixed
Show the wall hits
On
Once the prediction is locked in, the Scene changes
Volume
50 %
The Scene then shows: 141.8 kPa · Pressure of the gas, as the gauge reads it/nodes/gas/outputs/pressure
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Half the volume, twice the pressure: about 142 kPa. This is Boyle’s law: at a fixed temperature $pV$ ⁠ stays the same.

Worked solution
1. The temperature is fixed, so $p_1V_1 = p_2V_2$ ⁠.
2. $V_2 = \tfrac{1}{2}V_1$ ⁠, so $p_2 = 2p_1$ ⁠.
3. $p_2 = 2 \times 71 \approx 142$ ⁠ kilopascals.
The same particles now strike the walls twice as often, because the trip from wall to wall is shorter.
Step 3 of 7
Open this step
Heat it
Practice · Calculate
Question
The gas is heated from 300 K to 600 K. How many times faster do the particles typically move?

Correct answer
1.414 ± 0.03
Setup, choices and feedback
The Scene at this step
Volume
70 %
Temperature
600 Kfixed
Show the wall hits
On
Accepted within
± 0.03
Computed from the Scene at this step’s setting
Typical particle speed, times its speed at 300 K/nodes/gas/outputs/speedRatio
Wrong answers it recognises
- 2 ± 0.03
  Misconception: Thinks a gas's particles move twice as fast when its absolute temperature doubles.
  Feedback from the catalogue
  The temperature in kelvin measures a particle’s average kinetic energy, $\tfrac{1}{2}mv^2$ ⁠, not its speed. For the energy to double, $v^2$ ⁠ must double, so the speed grows only $\sqrt{2} \approx 1.41$ ⁠ times. For the particles to move twice as fast you need four times the absolute temperature.
Feedback for a right answer
$\sqrt{2} \approx 1.41$ ⁠ times. Each hit is 1.41 times as hard, and the hits come 1.41 times as often, so together the pressure grows twofold, like the temperature.

Hint
The temperature in kelvin grows in step with the average kinetic energy, $\tfrac{1}{2}mv^2$ ⁠.

Worked solution
1. A particle’s average kinetic energy grows in step with $T$ ⁠ in kelvin: twofold.
2. The energy is $\tfrac{1}{2}mv^2$ ⁠, so $v^2$ ⁠ doubles.
3. $v$ ⁠ grows $\sqrt{2} \approx 1.41$ ⁠ times.
If you answered 2, you assumed the speed, not the energy, grows in step with the temperature.
Step 4 of 7
Open this step
The gas law
Practice · Predict, then look
Question
The gas is at 70% Volume and 300 K, and the gauge reads about 101 kPa, one atmosphere. It is squeezed to 35% and heated to 600 K. Predict: what will the gauge read?

Correct answer
405.2 kPa ± 15 kPa
A number prediction counts as right within ± 15 kPa
Setup, choices and feedback
The Scene at this step
Volume
70 %
Temperature
300 K
Show the wall hits
On
Once the prediction is locked in, the Scene changes
Volume
35 %
Temperature
600 K
The Scene then shows: 405.2 kPa · Pressure of the gas, as the gauge reads it/nodes/gas/outputs/pressure
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Half the volume doubles the pressure, and twice the temperature doubles it again: four times, about 405 kPa. This is the gas law: $\dfrac{pV}{T}$ ⁠ stays the same.

Worked solution
1. $\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$ ⁠, so $p_2 = p_1 \times \dfrac{V_1}{V_2} \times \dfrac{T_2}{T_1}$ ⁠.
2. $\dfrac{V_1}{V_2} = \dfrac{70}{35} = 2$ ⁠, and $\dfrac{T_2}{T_1} = \dfrac{600}{300} = 2$ ⁠.
3. $p_2 = 101 \times 2 \times 2 \approx 405$ ⁠ kilopascals, four atmospheres.
Step 5 of 7
Open this step
Two atmospheres
Practice · Reach the target
Question
Change the Volume and the Temperature until the pressure is two atmospheres.

Goal
- 2 atm ± 0.02961 atm
  At the start of the step: 0.9998 atm
Setup, choices and feedback
The Scene at this step
Volume
70 %
Temperature
300 K
Show the wall hits
On
Read from the Scene at
Pressure of the gas, as the gauge reads it/nodes/gas/outputs/pressure
One setting that reaches it
Volume
20 %
Temperature
169 K
Feedback for a right answer
Any setting where $\dfrac{T}{V}$ ⁠ has doubled gets there: 35% at 300 K, or 70% at 600 K, or 50% at 429 K.

Hint
It is one atmosphere now. Half the volume, or twice the temperature, doubles it.

Worked solution
1. At the start, at 70% and 300 K, the pressure is one atmosphere.
2. $p \propto \dfrac{T}{V}$ ⁠, so $\dfrac{T}{V}$ ⁠ must double.
3. For example: keep the temperature and squeeze to 35%, or keep the volume and heat to 600 K.
Step 6 of 7
Open this step
Pressure with depth
Practice · Choose
Question
Every ten metres of water adds a pressure of about one atmosphere. What is the absolute pressure on a diver 20 m deep in the sea?

Correct answer
- Correct About three atmospheres
Setup, choices and feedback
The Scene at this step
Volume
70 %
Temperature
300 K
Show the wall hits
On
Choices
Shown in a random order.
- Correct About three atmospheres
- Wrong About two atmospheres
  Misconception: Counts only the water above a diver and forgets the atmosphere pressing on the water's surface.
  Feedback from the catalogue
  The air presses on the water’s surface with one atmosphere, and the water passes that pressure on downward. So the absolute pressure at a depth is the air’s pressure plus that of the water above: two atmospheres at 10 m, not one.
- Wrong About twenty atmospheres
  Feedback
  One atmosphere for every ten metres, not every metre: twenty metres are two atmospheres of water.
Feedback for a right answer
One atmosphere of air above the sea, and two of water: three. So the air in the diver’s lungs is squeezed to a third of its volume, and the diver must rise slowly.

Worked solution
1. The water’s pressure at that depth: $\rho g h = 1025 \times 9.8 \times 20 \approx 2 \times 10^5$ ⁠ pascals, about two atmospheres.
2. Above the sea the air presses down with one more atmosphere.
3. The absolute pressure: $1 + 2 = 3$ ⁠ atmospheres.
Step 7 of 7
Open this step
Hits in a hot gas
Practice · Calculate
Question
The gas is at 70% Volume, and it is heated from 300 K to 600 K. How many times as many hits does each small patch of wall get each second?

Correct answer
1.414 ± 3%
Setup, choices and feedback
The Scene at this step
Volume
70 %fixed
Temperature
600 Kfixed
Show the wall hits
On
Accepted within
± 3%
Computed from the Scene at this step’s setting
Hits on each patch of wall per second, times the number at 70 % and 300 K/nodes/gas/outputs/hitRatio
Wrong answers it recognises
- 2 ± 3%
  Misconception: Thinks a gas's particles move twice as fast when its absolute temperature doubles.
  Feedback from the catalogue
  The temperature in kelvin measures a particle’s average kinetic energy, $\tfrac{1}{2}mv^2$ ⁠, not its speed. For the energy to double, $v^2$ ⁠ must double, so the speed grows only $\sqrt{2} \approx 1.41$ ⁠ times. For the particles to move twice as fast you need four times the absolute temperature.
Feedback for a right answer
About 1.41 times, $\sqrt{2}$ ⁠ times. The pressure doubles because each hit is also $\sqrt{2}$ ⁠ times harder.

Hint
The volume has not changed, so the particles hit the wall more often only because they are faster. How many times faster?

Worked solution
1. The particles’ typical speed goes with the square root of the absolute temperature: $\sqrt{\dfrac{600}{300}} = \sqrt{2} \approx 1.41$ ⁠.
2. The volume is fixed, so each particle comes back to the wall sooner in the same ratio: 1.41 times as many hits a second.
3. Each hit is also 1.41 times harder, because the particle is faster. Together: $1.41 \times 1.41 = 2$ ⁠, and the pressure has doubled.
If you answered $2$ ⁠, you put the whole doubling of the pressure down to the number of hits. Half the rise comes from each hit being harder.

Lesson 2 of 4 · About 35 minutes

Floating and buoyancy

Why things float or sink: density, the buoyant force and Archimedes' principle, what share of a floating cube is under water and why it doesn't depend on size, what holds up a cube that has sunk, and steel that floats on mercury.

Open the lesson

The 10 steps and their answers

Step 1 of 10
Open this step
Density
Practice · Choose
Question
A log of hundreds of kilograms floats in a river, and a pebble of a few grams sinks in it. What explains this?

Correct answer
- Correct The wood is less dense than water, and the stone denser
Setup, choices and feedback
The Scene at this step
Cube density
500 kg/m³
Liquid
Fresh water
Cube size
8 cm
Show the forces
Off
Choices
Shown in a random order.
- Correct The wood is less dense than water, and the stone denser
- Wrong The pebble is too heavy for the river, and the log big enough to float
  Misconception: Thinks heavy things sink and light things float, whatever they are made of.
  Feedback from the catalogue
  What decides is density, the mass in each unit of volume, not the weight. A huge log floats because each cubic centimetre of it is lighter than a cubic centimetre of water, and a small pebble sinks because each cubic centimetre of it is heavier. A body floats if it is less dense than the liquid.
- Wrong The log is long and flat, and the pebble round
  Feedback
  Shape changes how a body floats, not whether it floats. A round lump of wood floats, and a flat slab of stone sinks.
Feedback for a right answer
Density is the mass in each unit of volume, $\rho = \dfrac{m}{V}$ ⁠. Water has 1000 kg/m³, pine about 500, and stone about 2700. A body less dense than the liquid floats, at any size.

Worked solution
1. A body in a liquid pushes aside its own volume of liquid, and gets a buoyant force equal to that liquid’s weight.
2. If each cubic centimetre of it is lighter than a cubic centimetre of water, the buoyancy beats its weight before it is fully under, and it floats.
3. It is a comparison of densities, not of weights: the log is heavy, but less dense than water.
Step 2 of 10
Open this step
Archimedes' principle
Practice · Calculate
Question
A Pine cube, 500 kg/m³, with 8 cm sides, floats in Fresh water. What is the buoyant force on it? Take g = 9.81 m/s².

Correct answer
2.511 N ± 2%
Setup, choices and feedback
The Scene at this step
Cube density
500 kg/m³fixed
Liquid
Fresh waterfixed
Cube size
8 cmfixed
Show the forces
Offfixed
Accepted within
± 2%
Computed from the Scene at this step’s setting
Buoyant force: the weight of the liquid the cube pushes aside/nodes/tank/outputs/buoyancy
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 5.023 N ± 2%
  Misconception: Works out the buoyancy of a floating body from its whole volume, not the part under the surface.
  Feedback from the catalogue
  The buoyant force equals the weight of the liquid pushed aside, and a floating body pushes aside only what is below the surface. A cube floating half under gets the buoyancy of half its volume of water, and that is exactly its weight: which is why it floats, neither rising nor sinking.
Feedback for a right answer
The cube floats, so the buoyancy equals its weight: $0.256 \times 9.81 \approx 2.5$ ⁠ newtons. It sinks to half its height, where the water pushed aside weighs exactly as much as it does.

Hint
A floating body: the forces on it balance.

Worked solution
1. The volume: $(0.08)^3 = 5.12 \times 10^{-4}$ ⁠ cubic metres.
2. The mass: $500 \times 5.12 \times 10^{-4} = 0.256$ ⁠ kilograms.
3. The weight: $0.256 \times 9.81 \approx 2.51$ ⁠ newtons.
4. A floating body is in balance, so the buoyancy equals the weight: about 2.5 N.
If you got about 5.0 N, you worked out the weight of water in the whole cube’s volume. Only the half below the surface pushes water aside.
Step 3 of 10
Open this step
What share is under
Practice · Calculate
Question
An Ice cube, 917 kg/m³, floats in Sea water, 1025 kg/m³. What share of its height is below the surface? Write it as a decimal.

Correct answer
0.8946 ± 0.005
Setup, choices and feedback
The Scene at this step
Cube density
917 kg/m³fixed
Liquid
Sea waterfixed
Cube size
8 cm
Show the forces
On
Accepted within
± 0.005
Computed from the Scene at this step’s setting
Share of the cube below the surface, 0..1/nodes/tank/outputs/submergedShare
Wrong answers it recognises
- 0.917 ± 0.005
  Misconception: Thinks a floating body sits equally deep in any liquid.
  Feedback from the catalogue
  A floating body sinks in until the liquid it pushes aside weighs as much as it does. In a denser liquid less volume weighs the same, so the body floats higher: the share below the surface is $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠. A ship sits a little deeper in a river than at sea.
Feedback for a right answer
$\dfrac{917}{1025} \approx 0.89$ ⁠: almost nine tenths under. In fresh water it would be 0.917, so ice floats a little higher at sea.

Hint
The share under is $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠.

Worked solution
1. A floating body sinks until the liquid it pushes aside weighs as much as it does: $\rho_{\text{liquid}} V_{\text{under}} g = \rho_{\text{obj}} V g$ ⁠.
2. $\dfrac{V_{\text{under}}}{V} = \dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}} = \dfrac{917}{1025} \approx 0.895$ ⁠.
3. About 0.89 of the cube is below the surface.
If you answered 0.917, you used the density of fresh water. Sea water is denser, and less of it is needed.
Step 4 of 10
Open this step
A bigger cube
Practice · Predict, then look
Question
A Pine cube with 8 cm sides floats half under. Cube size is raised to 12 cm. Predict: what share of its height will be under?

Correct answer
- Correct Still half
Setup, choices and feedback
The Scene at this step
Cube density
500 kg/m³fixed
Liquid
Fresh waterfixed
Cube size
8 cm
Show the forces
On
Choices
- Correct Still half
- Wrong More than half, because it is heavier
  Misconception: Thinks a bigger cube of the same material floats lower, or higher, as a share of its height.
  Feedback from the catalogue
  A bigger cube is heavier, but it also pushes aside more liquid for every centimetre it sinks, in exactly the same proportion. So the share below the surface, $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠, does not depend on size: a big cube sinks more centimetres, but the same share of its height.
- Wrong Less than half, because a big cube has more buoyancy
  Misconception: Thinks a bigger cube of the same material floats lower, or higher, as a share of its height.
  Feedback from the catalogue
  A bigger cube is heavier, but it also pushes aside more liquid for every centimetre it sinks, in exactly the same proportion. So the share below the surface, $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠, does not depend on size: a big cube sinks more centimetres, but the same share of its height.
Once the prediction is locked in, the Scene changes
Cube size
12 cm
The Scene then shows: 0.5 · Share of the cube below the surface, 0..1/nodes/tank/outputs/submergedShare
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Still half: the cube sinks 6 cm instead of 4 cm, but the same share of its height. The level in the tank rose more, because it pushes more water aside.

Worked solution
1. Made 1.5 times as wide, the cube has $1.5^3 \approx 3.4$ ⁠ times the volume and weight.
2. But now each centimetre it sinks pushes aside water over a base $1.5^2$ ⁠ times as big, and it sinks 1.5 times as far: the liquid pushed aside also grows 3.4 times.
3. The ratio is kept: the share under is $\dfrac{500}{1000} = 0.5$ ⁠ at any size.
Step 5 of 10
Open this step
A cube that has sunk
Practice · Calculate
Question
A Steel cube, 7850 kg/m³, with 8 cm sides, sinks to the floor of a tank of Fresh water. How hard does the floor push up on it? Take g = 9.81 m/s².

Correct answer
34.41 N ± 2%
Setup, choices and feedback
The Scene at this step
Cube density
7850 kg/m³fixed
Liquid
Fresh waterfixed
Cube size
8 cmfixed
Show the forces
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
How hard the floor pushes up on a sunken cube, weight − buoyancy; 0 when it floats/nodes/tank/outputs/floorPush
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 39.43 N ± 2%
  Misconception: Thinks a body resting on the bottom gets no buoyancy, so the floor carries its whole weight.
  Feedback from the catalogue
  Buoyancy acts on every body in a liquid, one that has sunk too. It is just smaller than the weight, so the body doesn’t float, and the floor pushes the difference: $N = mg - F_b$ ⁠. That is why a stone feels lighter when you lift it under water.
Feedback for a right answer
The weight is about 39.4 N, and the buoyancy, the weight of 512 cm³ of water, about 5.0 N. The floor pushes the difference: about 34.4 N.

Hint
Three forces balance: the weight down, the buoyancy and the floor’s push up.

Worked solution
1. The volume: $(0.08)^3 = 5.12 \times 10^{-4}$ ⁠ cubic metres.
2. The weight: $7850 \times 5.12 \times 10^{-4} \times 9.81 \approx 39.4$ ⁠ newtons.
3. The whole cube is under water, so the buoyancy is $1000 \times 5.12 \times 10^{-4} \times 9.81 \approx 5.0$ ⁠ newtons.
4. The forces balance: $N = 39.4 - 5.0 \approx 34.4$ ⁠ newtons.
If you answered about 39.4 N, you forgot that a sunken body gets buoyancy too.
Step 6 of 10
Open this step
Steel that floats
Practice · Reach the target
Question
Cube density is fixed at Steel. Make it float.

Goal
- 1 ± 0.1
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Cube density
7850 kg/m³fixed
Liquid
Fresh water
Cube size
8 cm
Show the forces
On
Read from the Scene at
1 when the cube floats, 0 when it rests on the floor/nodes/tank/outputs/floats
One setting that reaches it
Liquid
Mercury
Cube size
3 cm
Feedback for a right answer
In Mercury, 13,534 kg/m³, steel floats with about 0.58 of its height under. Only gold and platinum are dense enough to sink in it.

Hint
You can’t change the cube. What else is in the tank?

Worked solution
1. A body floats if it is less dense than the liquid.
2. Steel, 7850 kg/m³, is denser than water, sea water and oil.
3. Only Mercury is denser, so choose it under Liquid.
4. The share under: $\dfrac{7850}{13\,534} \approx 0.58$ ⁠.
Step 7 of 10
Open this step
Weighing with water
Practice · Calculate
Question
Cube density is hidden. The cube, with 8 cm sides, floats in Fresh water with 4 cm of its height below the surface. What is its mass?

Correct answer
256 g ± 3%
Setup, choices and feedback
The Scene at this step
Cube density
500 kg/m³hidden
Liquid
Fresh waterfixed
Cube size
8 cmfixed
Show the forces
On
Accepted within
± 3%
Computed from the Scene at this step’s setting
Mass of the cube/nodes/tank/outputs/mass
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 512 g ± 3%
  Misconception: Works out the buoyancy of a floating body from its whole volume, not the part under the surface.
  Feedback from the catalogue
  The buoyant force equals the weight of the liquid pushed aside, and a floating body pushes aside only what is below the surface. A cube floating half under gets the buoyancy of half its volume of water, and that is exactly its weight: which is why it floats, neither rising nor sinking.
Feedback for a right answer
256 g: the weight of the water pushed aside, 256 cm³. That is how a ship is weighed: by how deep it sits.

Hint
A floating cube pushes aside water that weighs as much as it does. How much water does it push aside?

Worked solution
1. The volume under water: $8 \times 8 \times 4 = 256$ ⁠ cubic centimetres.
2. A cubic centimetre of fresh water is one gram, so the cube pushes aside 256 g of water.
3. It floats, so the buoyancy equals its weight: its mass is 256 g.
4. Check: $\dfrac{256}{512} = 0.5$ ⁠ grams per cubic centimetre, 500 kg/m³, the density of pine.
If you answered 512 g, you worked out the water the whole cube would push aside, not just the part under water.
Step 8 of 10
Open this step
Pine in oil
Practice · Calculate
Question
A Pine cube, 500 kg/m³, with 8 cm sides, floats in Cooking oil, whose density is 920 kg/m³. How much of its height is below the surface of the oil?

Correct answer
4.348 cm ± 3%
Setup, choices and feedback
The Scene at this step
Cube density
500 kg/m³fixed
Liquid
Cooking oilfixed
Cube size
8 cmfixed
Show the forces
On
Accepted within
± 3%
Computed from the Scene at this step’s setting
How much of the cube’s height is under the surface: its draught when it floats, its whole height once it sinks/nodes/tank/outputs/submergedDepth
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4 cm ± 3%
  Misconception: Thinks a floating body sits equally deep in any liquid.
  Feedback from the catalogue
  A floating body sinks in until the liquid it pushes aside weighs as much as it does. In a denser liquid less volume weighs the same, so the body floats higher: the share below the surface is $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠. A ship sits a little deeper in a river than at sea.
Feedback for a right answer
About 4.35 cm, more than the 4 cm in water: in a lighter liquid the cube sits deeper.

Hint
The share of the cube’s height under the surface is the ratio of the densities.

Worked solution
1. The share under the surface: $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}} = \dfrac{500}{920} \approx 0.54$ ⁠.
2. Of a height of 8 cm: $0.54 \times 8 \approx 4.35$ ⁠ centimetres.
If you answered 4 cm, you used the density of water. Oil is lighter than water, so the cube has to push aside more of it to carry the same weight.
Step 9 of 10
Open this step
The level rises
Practice · Calculate
Question
The tank is 30 cm wide and 20 cm deep, filled with Fresh water. A Pine cube with 8 cm sides goes in, and floats with half its height in the water. By how much does the water level rise?

Correct answer
4.267 mm ± 3%
Setup, choices and feedback
The Scene at this step
Cube density
500 kg/m³fixed
Liquid
Fresh waterfixed
Cube size
8 cmfixed
Show the forces
On
Accepted within
± 3%
Computed from the Scene at this step’s setting
How much the liquid level rose when the cube went in/nodes/tank/outputs/levelRise
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 8.533 mm ± 3%
  Misconception: Works out the buoyancy of a floating body from its whole volume, not the part under the surface.
  Feedback from the catalogue
  The buoyant force equals the weight of the liquid pushed aside, and a floating body pushes aside only what is below the surface. A cube floating half under gets the buoyancy of half its volume of water, and that is exactly its weight: which is why it floats, neither rising nor sinking.
Feedback for a right answer
About 4.3 mm: 256 cm³ of water pushed aside, spread over 600 cm².

Hint
The water the cube pushes aside spreads out in a thin layer over the whole tank.

Worked solution
1. The volume under water: $8 \times 8 \times 4 = 256$ ⁠ cubic centimetres.
2. The water’s surface area: $30 \times 20 = 600$ ⁠ square centimetres.
3. The rise: $\dfrac{256}{600} \approx 0.43$ ⁠ centimetres, about 4.3 mm.
If you answered about 8.5 mm, you divided the volume of the whole cube. Only the part under water pushes water aside.
Step 10 of 10
Open this step
Which liquid?
Practice · Calculate
Question
The Liquid is hidden. An Ice cube, 917 kg/m³, with 8 cm sides, floats in it with 7.16 cm of its height below the surface. What is the liquid’s density?

Correct answer
1025 ± 1.5%
Setup, choices and feedback
The Scene at this step
Cube density
917 kg/m³fixed
Liquid
Sea waterhidden
Cube size
8 cmfixed
Show the forces
On
Accepted within
± 1.5%
Computed from the Scene at this step’s setting
Density of the liquid, kg/m³/nodes/tank/outputs/liquidDensity
Wrong answers it recognises
- 917 ± 0.5%
  Misconception: Thinks a floating body sits equally deep in any liquid.
  Feedback from the catalogue
  A floating body sinks in until the liquid it pushes aside weighs as much as it does. In a denser liquid less volume weighs the same, so the body floats higher: the share below the surface is $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠. A ship sits a little deeper in a river than at sea.
Feedback for a right answer
About 1025 kg/m³: Sea water. In fresh water the ice would sit deeper.

Hint
Write the answer in kg/m³. The share under the surface is $\dfrac{\rho_{\text{obj}}}{\rho_{\text{liquid}}}$ ⁠: turn the ratio round.

Worked solution
1. The share under the surface: $\dfrac{7.16}{8} \approx 0.895$ ⁠.
2. $0.895 = \dfrac{917}{\rho_{\text{liquid}}}$ ⁠, so $\rho_{\text{liquid}} = \dfrac{917}{0.895} \approx 1025$ ⁠ kilograms per cubic metre.
3. That is denser than fresh water: 1025 kg/m³ is Sea water.
In fresh water the same cube would sit 7.34 cm deep. The salt adds mass to every litre, and the ice floats a little higher.

Lesson 3 of 4 · About 25 minutes

Flow and Bernoulli's principle

Water in a pipe that narrows: continuity and water that speeds up in the throat, pressure that falls where the water is fast, Bernoulli's equation, a pressure drop that grows as the square of the flow, air drawn in, and a perfume atomiser.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Faster where it is narrow
Practice · Calculate
Question
The pipe is 80 mm wide, and two litres a second flow through it at about 0.40 m/s. The Throat width is 40 mm. How fast does the water flow in the throat?

Correct answer
1.592 m/s ± 3%
Setup, choices and feedback
The Scene at this step
Flow rate
2 L/sfixed
Throat width
40 mmfixed
Show pressure and speed
Flowing water
Accepted within
± 3%
Computed from the Scene at this step’s setting
Mean speed in the throat/nodes/meter/outputs/throatSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.7958 m/s ± 3%
  Misconception: Assumes a quantity that goes with the square of another changes in step with it: half the width, twice the speed; twice the flow, twice the drop.
  Feedback from the catalogue
  Two links here are squares. A round section’s area grows as the square of its width, so a throat half as wide has a quarter of the area, and the water in it is four times as fast. And the energy of motion, $\tfrac{1}{2}\rho v^2$ ⁠, grows as the square of the speed, so twice the flow gives four times the pressure drop.
Feedback for a right answer
Half the width is a quarter of the area, so four times the speed: about 1.6 m/s.

Hint
The same flow passes every section: $Q = Av$ ⁠. A round section’s area grows as the square of its width.

Worked solution
1. Continuity: water hardly compresses, so $A_1v_1 = A_2v_2$ ⁠.
2. $A \propto d^2$ ⁠, so $\dfrac{A_1}{A_2} = \left(\dfrac{80}{40}\right)^2 = 4$ ⁠.
3. $v_2 = 4 \times 0.40 \approx 1.6$ ⁠ metres a second.
If you answered 0.8 m/s, you multiplied only by the ratio of the widths. The area is four times smaller, not two.
Step 2 of 6
Open this step
The columns
Practice · Predict, then look
Question
In a Straight pipe every column stands at the same height. It is changed to a Venturi meter, whose throat is 40 mm wide. Predict: what will happen to the column over the throat?

Correct answer
- Correct It will drop: the pressure in the throat is lower
Setup, choices and feedback
The Scene at this step
Flow rate
2 L/sfixed
Throat width
80 mm
Show pressure and speed
Flowing water
Choices
- Correct It will drop: the pressure in the throat is lower
- Wrong It will rise: the water is squeezed in the throat
  Misconception: Thinks the pressure is highest where a pipe narrows, because the water is squeezed there.
  Feedback from the catalogue
  For water to speed up on its way into the throat, something has to push it forward: the pressure behind it must be greater than the pressure ahead. So the pressure falls where the water speeds up, and is lowest in the throat, where it is fastest. That is Bernoulli’s principle: $p + \tfrac{1}{2}\rho v^2$ ⁠ stays the same along the flow.
- Wrong It will stay level: the same water flows
  Feedback
  The same flow, yes, but not the same speed: in the throat the water is four times as fast, and that energy comes out of the pressure.
Once the prediction is locked in, the Scene changes
Throat width
40 mm
The Scene then shows: 12.1 cm · How much lower the throat column stands than the inlet column/nodes/meter/outputs/headDrop
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The column over the throat drops by about 12 cm. The water speeds up into the throat only if the pressure behind it is higher than the pressure ahead, so the lowest pressure is where it is fastest.

Worked solution
1. In the throat the water is faster (continuity).
2. To speed up it needs a forward force: the pressure behind is greater than the pressure ahead.
3. So the pressure falls along the way into the throat, and the column above it, which gauges the pressure, stands lower.
4. In the widening after the throat the water slows, and the pressure climbs back.
Step 3 of 6
Open this step
Bernoulli's equation
Practice · Calculate
Question
The water comes in at 0.40 m/s and flows through the throat at 1.59 m/s. How much lower is the pressure in the throat than at the inlet? The water’s density is 1000 kg/m³.

Correct answer
1185 Pa ± 3%
Setup, choices and feedback
The Scene at this step
Flow rate
2 L/sfixed
Throat width
40 mmfixed
Show pressure and speed
Pressure and speed
Accepted within
± 3%
Computed from the Scene at this step’s setting
Inlet pressure minus throat pressure/nodes/meter/outputs/pressureDrop
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$\tfrac{1}{2} \times 1000 \times (1.59^2 - 0.40^2) \approx 1190$ ⁠ pascals, about 1.2 kPa: the column over the throat stands about 12 cm lower.

Hint
$p + \tfrac{1}{2}\rho v^2$ ⁠ stays the same along the flow.

Worked solution
1. Bernoulli: $p_1 + \tfrac{1}{2}\rho v_1^2 = p_2 + \tfrac{1}{2}\rho v_2^2$ ⁠.
2. $p_1 - p_2 = \tfrac{1}{2}\rho (v_2^2 - v_1^2)$ ⁠.
3. $= \tfrac{1}{2} \times 1000 \times (2.53 - 0.16) \approx 1185$ ⁠ pascals.
4. About 1.2 kPa. If you got twice that, you dropped the half in $\tfrac{1}{2}\rho v^2$ ⁠. You can check it on the columns: 10 cm of water is about 1 kPa.
Step 4 of 6
Open this step
Twice the flow
Practice · Predict, then look
Question
At a Flow rate of 1 L/s the column over the throat stands about 3 cm lower than the inlet’s. The flow is doubled to 2 L/s. Predict: how much lower will it stand now?

Correct answer
- Correct About 12 cm, four times as much
Setup, choices and feedback
The Scene at this step
Flow rate
1 L/s
Throat width
40 mmfixed
Show pressure and speed
Pressure and speed
Choices
- Correct About 12 cm, four times as much
- Wrong About 6 cm, twice as much
  Misconception: Assumes a quantity that goes with the square of another changes in step with it: half the width, twice the speed; twice the flow, twice the drop.
  Feedback from the catalogue
  Two links here are squares. A round section’s area grows as the square of its width, so a throat half as wide has a quarter of the area, and the water in it is four times as fast. And the energy of motion, $\tfrac{1}{2}\rho v^2$ ⁠, grows as the square of the speed, so twice the flow gives four times the pressure drop.
- Wrong Still about 3 cm: the throat hasn’t changed
  Feedback
  The throat is the same, but the water everywhere is twice as fast, and the pressure drop depends on the speed.
Once the prediction is locked in, the Scene changes
Flow rate
2 L/s
The Scene then shows: 12.1 cm · How much lower the throat column stands than the inlet column/nodes/meter/outputs/headDrop
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Twice the flow is twice the speed everywhere, and four times $\tfrac{1}{2}\rho v^2$ ⁠. That is why a Venturi meter is sensitive: a small change in flow gives a clear change in pressure.

Worked solution
1. Twice the flow in the same pipe: every speed doubles.
2. The drop $\tfrac{1}{2}\rho (v_2^2 - v_1^2)$ ⁠ grows as the square of the speed: fourfold.
3. $4 \times 3 = 12$ ⁠ centimetres.
Step 5 of 6
Open this step
Below the air's pressure
Practice · Reach the target
Question
Change the Flow rate and the Throat width until the pressure in the throat falls below the air’s, and air is drawn in through the hole.

Goal
- 3 ± 0.1
  At the start of the step: 2
Setup, choices and feedback
The Scene at this step
Flow rate
2 L/s
Throat width
40 mm
Show pressure and speed
Flowing water
Read from the Scene at
0 still, 1 level columns, 2 the throat column lower, 3 air drawn in at the throat/nodes/meter/outputs/regime
One setting that reaches it
Flow rate
2.15 L/s
Throat width
32 mm
Feedback for a right answer
When the drop passes 36 cm of water, the throat’s pressure is below the air’s: the column empties and bubbles of air are swept along the pipe. That is how a laboratory water-jet vacuum pump works.

Hint
The inlet’s pressure is fixed, 36 cm of water above the air’s. The drop in the throat has to beat that.

Worked solution
1. The inlet’s pressure is 36 cm of water above the air’s.
2. For the throat’s pressure to fall below the air’s, $\dfrac{v_2^2 - v_1^2}{2g}$ ⁠ must pass 0.36 m, which takes more than about 2.7 m/s in the throat.
3. A Flow rate of 2.5 L/s with a Tight throat gets there: about 3 m/s.
Step 6 of 6
Open this step
An atomiser
Practice · Choose
Question
In a perfume atomiser air is blown fast across the top of a thin tube whose lower end is in the bottle. Why does the perfume rise up the tube?

Correct answer
- Correct The pressure is low in the fast air, and the ordinary pressure in the bottle pushes the liquid up
Setup, choices and feedback
The Scene at this step
Flow rate
2.5 L/s
Throat width
32 mm
Show pressure and speed
Flowing water
Choices
Shown in a random order.
- Correct The pressure is low in the fast air, and the ordinary pressure in the bottle pushes the liquid up
- Wrong The fast air squeezes the tube’s mouth, and high pressure there draws the liquid up
  Misconception: Thinks the pressure is highest where a pipe narrows, because the water is squeezed there.
  Feedback from the catalogue
  For water to speed up on its way into the throat, something has to push it forward: the pressure behind it must be greater than the pressure ahead. So the pressure falls where the water speeds up, and is lowest in the throat, where it is fastest. That is Bernoulli’s principle: $p + \tfrac{1}{2}\rho v^2$ ⁠ stays the same along the flow.
- Wrong The air drags the liquid along by friction, like wind moving leaves
  Feedback
  Dragging happens only once the liquid is in the stream. What lifts it up the tube is the pressure difference between its two ends.
Feedback for a right answer
It is the same principle as in the throat: where the fluid is fast, the pressure is low. A liquid isn’t sucked from above; it is pushed from below, by the higher pressure.

Worked solution
1. By Bernoulli, the pressure in the fast air over the tube is below the ordinary air pressure.
2. Air at ordinary pressure presses on the perfume’s surface in the bottle.
3. The difference pushes the liquid up the tube, and at the top the fast stream breaks it into drops.

Lesson 4 of 4 · About 20 minutes

Surface tension and capillarity

A liquid that climbs a thin tube with no pump: surface tension and the glass's pull, why a tube half as wide holds a column twice as tall, Jurin's law, mercury that sinks, and why capillary action alone can't lift water to the top of a tree.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A liquid that climbs
Practice · Choose
Question
The water climbs the thin tubes above the level in the wide vessel, with no pump. What lifts it?

Correct answer
- Correct The glass pulls on the water’s molecules, and the taut surface pulls the column up all round the rim
Setup, choices and feedback
The Scene at this step
Narrowest tube
0.5 mm
Liquid
Water
Zoom to the molecules
Tubes
Choices
Shown in a random order.
- Correct The glass pulls on the water’s molecules, and the taut surface pulls the column up all round the rim
- Wrong The air pressure in the thin tube is lower
  Feedback
  The tubes are open at the top, and the air pressure over every column is the same. The water would climb in a vacuum too.
- Wrong The narrow tube squeezes the water and forces it up
  Feedback
  The wall doesn’t press the water inward; it pulls the water towards itself. That is why the surface curves up next to the glass.
Feedback for a right answer
This is capillary action. Water wets the glass and forms a dish-shaped meniscus, and the surface tension along the rim pulls upward until it holds exactly the weight of the column.

Worked solution
1. A molecule inside the liquid is pulled equally every way; one at the surface is pulled only inward, so the surface behaves like a stretched skin.
2. Glass pulls water molecules nearly as hard as they pull each other, so water spreads over it and the surface turns up at the wall.
3. The skin pulls the column up all round the rim, until the column’s weight equals the pull.
Step 2 of 6
Open this step
The tube's width
Practice · Predict, then look
Question
Narrowest tube is 1 mm. It is changed to 0.5 mm. Predict: what will happen to the height of its column?

Correct answer
- Correct It will double
Setup, choices and feedback
The Scene at this step
Narrowest tube
1 mm
Liquid
Waterfixed
Zoom to the molecules
Tubes
Choices
- Correct It will double
- Wrong It will halve: less rim, less pull
  Misconception: Thinks a wider tube lifts a liquid higher, because it has more surface pulling.
  Feedback from the catalogue
  A wider tube has more rim, so the surface pulls harder, but only in step with the width. The water to be lifted grows as the area, the width squared. So in a tube twice as wide the liquid climbs only half as high: $h \propto \dfrac{1}{r}$ ⁠.
- Wrong It will grow fourfold: the area is four times smaller
  Feedback
  The area is indeed four times smaller, but the rim, which does the pulling, is also half as long. The net effect is twofold.
Once the prediction is locked in, the Scene changes
Narrowest tube
0.5 mm
The Scene then shows: 59.4 mm · Height of the narrowest tube’s column above the reservoir’s level (negative below)/nodes/tubes/outputs/rise
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 30 mm to 59 mm. The pull grows with the rim, $2\pi r$ ⁠, and the weight with the area, $\pi r^2$ ⁠, so the height grows as $\dfrac{1}{r}$ ⁠.

Worked solution
1. The upward pull: $2\pi r \gamma$ ⁠, in proportion to the radius.
2. The column’s weight: $\rho g \pi r^2 h$ ⁠, in proportion to the radius squared.
3. In balance $h = \dfrac{2\gamma}{\rho g r}$ ⁠: half the radius, twice the height.
Step 3 of 6
Open this step
Jurin's law
Practice · Calculate
Question
Use Jurin’s law to work out how high Water climbs in a tube 1 mm across inside. Water’s surface tension is 0.0728 N/m, its density 998 kg/m³, and it wets glass completely, $\theta = 0^\circ$ ⁠. Take g = 9.81 m/s².

Correct answer
29.74 mm ± 2%
Setup, choices and feedback
The Scene at this step
Narrowest tube
1 mmfixed
Liquid
Waterfixed
Zoom to the molecules
Tubes
Accepted within
± 2%
Computed from the Scene at this step’s setting
Jurin’s law for the narrowest tube, 2γ cos θ / (ρ g r)/nodes/tubes/outputs/jurin
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 14.87 mm ± 2%
  Misconception: Puts the tube's diameter where Jurin's law needs its radius.
  Feedback from the catalogue
  In Jurin’s law $r$ ⁠ is the inner radius, half the diameter. A 1 mm tube has a radius of 0.5 mm. Put the diameter in and the height comes out half what it should be.
Feedback for a right answer
About 30 mm. The column in the Scene is a little lower, about 29.6 mm, because some of the water is held in the curved meniscus.

Hint
$h = \dfrac{2\gamma\cos\theta}{\rho g r}$ ⁠, and $r$ ⁠ is the radius.

Worked solution
1. The radius: $r = 0.5$ ⁠ millimetres, $5 \times 10^{-4}$ ⁠ metres. $\cos 0^\circ = 1$ ⁠.
2. The top: $2 \times 0.0728 = 0.1456$ ⁠.
3. The bottom: $998 \times 9.81 \times 5 \times 10^{-4} \approx 4.895$ ⁠.
4. $h = \dfrac{0.1456}{4.895} \approx 0.0297$ ⁠ metres, about 30 mm.
If you got about 15 mm, you put in the diameter instead of the radius.
Step 4 of 6
Open this step
Forty millimetres
Practice · Reach the target
Question
Change the Narrowest tube until the water climbs 40 mm above the level in it.

Goal
- 40 mm ± 1 mm
  At the start of the step: 29.58 mm
Setup, choices and feedback
The Scene at this step
Narrowest tube
1 mm
Liquid
Waterfixed
Zoom to the molecules
Tubes
Read from the Scene at
Height of the narrowest tube’s column above the reservoir’s level (negative below)/nodes/tubes/outputs/rise
One setting that reaches it
Narrowest tube
0.75 mm
Feedback for a right answer
A tube of about 0.75 mm: $\dfrac{30}{40} = 0.75$ ⁠.

Hint
In a 1 mm tube the water climbs about 30 mm, and the height goes as one over the width.

Worked solution
1. $h \propto \dfrac{1}{d}$ ⁠, and in a 1 mm tube $h \approx 30$ ⁠ millimetres.
2. For $h = 40$ ⁠: $d = 1 \times \dfrac{30}{40} = 0.75$ ⁠ millimetres.
Step 5 of 6
Open this step
Mercury
Practice · Predict, then look
Question
The Liquid is changed to Mercury. Predict: what will happen in the tubes?

Correct answer
- Correct The mercury will stand below the level outside, lowest in the narrowest
Setup, choices and feedback
The Scene at this step
Narrowest tube
1 mm
Liquid
Water
Zoom to the molecules
Tubes
Choices
- Correct The mercury will stand below the level outside, lowest in the narrowest
- Wrong It will climb higher than water: its surface tension is seven times as big
  Misconception: Thinks every liquid climbs up a thin glass tube.
  Feedback from the catalogue
  A liquid climbs only if it wets the glass: if the glass pulls its molecules at least nearly as hard as they pull each other. Mercury pulls on itself far harder than glass pulls on it, so it draws back from the wall, its surface domes upward, and the column sinks below the level outside.
- Wrong It will climb, but less than water, because it is heavy
  Misconception: Thinks every liquid climbs up a thin glass tube.
  Feedback from the catalogue
  A liquid climbs only if it wets the glass: if the glass pulls its molecules at least nearly as hard as they pull each other. Mercury pulls on itself far harder than glass pulls on it, so it draws back from the wall, its surface domes upward, and the column sinks below the level outside.
Once the prediction is locked in, the Scene changes
Liquid
Mercury
The Scene then shows: -11.09 mm · Height of the narrowest tube’s column above the reservoir’s level (negative below)/nodes/tubes/outputs/rise
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Mercury doesn’t wet glass: its contact angle is about $140^\circ$ ⁠, so $\cos\theta$ ⁠ is negative, its surface domes up, and the pull along the rim points down. In a 1 mm tube it sinks about 11 mm.

Worked solution
1. Mercury atoms pull on each other far harder than glass pulls on them.
2. So the mercury draws back from the wall, and its surface meets the glass at about $140^\circ$ ⁠.
3. In Jurin’s law $\cos 140^\circ \approx -0.77$ ⁠, so $h$ ⁠ is negative: the column sinks, deepest in the narrowest tube.
Step 6 of 6
Open this step
Trees
Practice · Choose
Question
The water-carrying vessels in a tree are about 0.04 mm wide. Can capillary action alone lift water to the top of a 30 m tree?

Correct answer
- Correct No: in vessels like that water climbs less than a metre
Setup, choices and feedback
The Scene at this step
Narrowest tube
0.5 mm
Liquid
Water
Zoom to the molecules
Tubes
Choices
Shown in a random order.
- Correct No: in vessels like that water climbs less than a metre
- Wrong Yes: vessels so thin lift water as high as needed
  Feedback
  Even in a very thin tube the height is finite: $h = \dfrac{2\gamma}{\rho g r}$ ⁠. At 0.04 mm it is under a metre.
- Wrong Only if the vessels were wider
  Misconception: Thinks a wider tube lifts a liquid higher, because it has more surface pulling.
  Feedback from the catalogue
  A wider tube has more rim, so the surface pulls harder, but only in step with the width. The water to be lifted grows as the area, the width squared. So in a tube twice as wide the liquid climbs only half as high: $h \propto \dfrac{1}{r}$ ⁠.
Feedback for a right answer
$h = \dfrac{2 \times 0.0728}{1000 \times 9.81 \times 2 \times 10^{-5}} \approx 0.74$ ⁠ metres. Most of the way, the water is pulled by evaporation from the leaves, with the chain of water in the vessels drawn up behind it.

Worked solution
1. The radius: $r = 0.02$ ⁠ millimetres, $2 \times 10^{-5}$ ⁠ metres.
2. $h = \dfrac{2\gamma}{\rho g r} = \dfrac{0.1456}{1000 \times 9.81 \times 2 \times 10^{-5}} \approx 0.74$ ⁠ metres.
3. That is far short of 30 m: capillary action explains only about the first metre.

Across the unit · 13 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Pressure

Hits on the walls

Question

Correct answer

The Scene at this step

Choices

Worked solution

Squeeze it

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

Heat it

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The gas law

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

Two atmospheres

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Pressure with depth

Question

Correct answer

The Scene at this step

Choices

Worked solution

Hits in a hot gas

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Floating and buoyancy

Density

Question

Correct answer

The Scene at this step

Choices

Worked solution

Archimedes' principle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

What share is under

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A bigger cube

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A cube that has sunk

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint