Physics 0625 | Paper 2 Mock 01 | Multiple Choice (Extended)

For the rule to be in equilibrium, the total clockwise moment about the pivot must equal the total anticlockwise moment. The pivot is at the 50 cm mark.

Objects P and Q are on the left of the pivot, so they produce anticlockwise moments. Object R is on the right.

Anticlockwise moments:

• P: \( 4.0 \times (50 - 10) = 4.0 \times 40 = 160 \text{ N cm} \)
• Q: \( 2.0 \times (50 - 25) = 2.0 \times 25 = 50 \text{ N cm} \)

Total anticlockwise moment = 210 N cm.

Clockwise moment from R: \( W \times (75 - 50) = W \times 25 \)

Setting clockwise equal to anticlockwise:

\[ W \times 25 = 210 \]

\[ W = \frac{210}{25} = 8.4 \text{ N} \]

The electronic balance displays 0.0082 kg. To convert kilograms to grams, multiply by 1000, since 1 kg = 1000 g. Therefore 0.0082 kg × 1000 = 8.2 g.

Unit conversion errors are the most common trap in this type of question. Dividing by 1000 instead of multiplying would give 0.0000082 g, which is far too small. Moving the decimal point the wrong number of places gives values like 0.082 g or 0.82 g. A useful check: grams should always be a larger number than kilograms for the same object, because a gram is a smaller unit.

When converting between metric mass units, remember the chain: 1 kg = 1000 g, 1 g = 1000 mg. Moving from a larger unit to a smaller unit means multiplying; moving from a smaller unit to a larger unit means dividing.

Gravitational potential energy depends on the vertical height gained, not on the distance travelled along the ramp:

\[ \text{GPE} = W \times h \]

where \( W \) is the weight of the box (50 N) and \( h \) is the vertical height (1.5 m):

\[ \text{GPE} = 50 \times 1.5 = 75 \text{ J} \]

Using the ramp length (4.0 m) instead of the height would give \( 50 \times 4.0 = 200 \) J, which is incorrect. The ramp makes it easier to push the box up (less force needed over a longer distance), but the energy stored as GPE depends only on how high the box ends up, not the path taken to get there.

The upper fixed point (100 °C) on the Celsius scale is defined as the temperature of steam above pure water boiling at standard atmospheric pressure (1 atm or 101 325 Pa).

The thermometer bulb is placed in the steam rather than in the boiling water itself, because water can superheat slightly and because dissolved impurities raise the boiling point. Steam from pure water at a fixed pressure provides a reliable, reproducible temperature.

Using salt water would raise the boiling point above 100 °C. Boiling under 'any conditions' does not control pressure, and the boiling point changes with pressure. Heating water to 'its highest possible temperature' is vague and not a defined fixed point.

Electrical power is calculated using the equation \(P = V \times I\), where \(P\) is the power in watts, \(V\) is the potential difference in volts, and \(I\) is the current in amperes. The circuit shows a 6 V battery and an ammeter reading 3 A, with a single lamp as the only component:

\[P = 6 \times 3 = 18 \text{ W}\]

Since the lamp is the only component in the circuit (apart from the battery and ammeter, which has negligible resistance), the full 6 V of the battery acts across the lamp, and the full 3 A flows through it. The power dissipated by the lamp is therefore 18 W. Other options such as 2 W (which would be \(V/I\), an incorrect formula) or 9 W (which would be \(V + I\) or some other erroneous combination) do not result from the correct application of the power equation.

When light reflects off the first mirror, the angle of reflection equals the angle of incidence (40°), both measured from the normal to that mirror. The reflected ray then travels towards the second mirror. To find the angle of incidence on the second mirror, consider the geometry of the triangle formed by the two mirrors and the light path between them.

The angle between the two mirrors is 60°. Inside the triangle formed by the first mirror's normal, the reflected ray, and the second mirror, the angles must sum to 180°. The angle of reflection at the first mirror is 40°, so the angle between the reflected ray and the first mirror surface is \( 90° - 40° = 50° \). The interior angle at the point where the mirrors meet is \( 180° - 60° = 120° \) (or, more directly, the angle in the triangle at the junction is \( 60° \)). In the triangle: \( 50° + 60° + \theta_{\text{surface}} = 180° \), giving \( \theta_{\text{surface}} = 70° \). The angle of incidence at the second mirror is \( 90° - 70° = 20° \).

The age of the universe can be estimated from the reciprocal of the Hubble constant:

\[ t = \frac{1}{H} = \frac{1}{2.2 \times 10^{-18} \;\text{s}^{-1}} \]

Taking the reciprocal of \( 2.2 \times 10^{-18} \): the reciprocal of \( 2.2 \) is approximately \( 0.4545 \), and the reciprocal of \( 10^{-18} \) is \( 10^{18} \). Combining:

\[ t = 0.4545 \times 10^{18} = 4.5 \times 10^{17} \;\text{s} \]

A common mistake is to simply invert the sign of the exponent without inverting the coefficient, which would give \( 2.2 \times 10^{18} \) instead. Remember that \( 1/(a \times 10^n) = (1/a) \times 10^{-n} \), so both the number and the power of ten must be inverted.

The newton meter reads 45 N when the lift is stationary, which means the weight of the parcel is 45 N (no acceleration, so the reading equals the true weight). To find mass from weight, rearrange W = m g to give m = W / g. Substituting: m = 45 / 9.0 = 5.0 kg.

A frequent error is to divide by 10 instead of 9.0, which would give 4.5 kg. Another mistake is to multiply weight by g, producing 405 kg, which is clearly unreasonable for a parcel. When a lift is stationary, there is no net acceleration, so the apparent weight shown on the meter equals the actual gravitational weight. If the lift were accelerating, the reading would differ from the true weight, but that is not the case here.

Lateral inversion is a specific property of plane mirror images. It means that the image appears to have its left and right sides swapped compared to the object. If you raise your right hand in front of a plane mirror, the image appears to raise what looks like its left hand.

The image in a plane mirror is not smaller than the object (it is the same size), not upside down (it is upright), and not further away (the image distance equals the object distance). These are all separate, incorrect descriptions that do not relate to lateral inversion.

The fundamental principle of thermal energy transfer is that energy flows spontaneously from a region of higher temperature to a region of lower temperature. The hand is at body temperature (approximately 37 °C), while the ice is at 0 °C. Because the hand is warmer than the ice, thermal energy transfers from the hand to the ice.

This energy transfer is what causes the ice to melt: it receives the latent heat of fusion needed to change from solid to liquid. The transfer is not in both directions equally - net transfer is always from hot to cold. It does not bypass the hand through the air, and it certainly does not flow from the cold ice to the warm hand. Energy always flows down the temperature gradient until thermal equilibrium is reached.

The graph shows a straight line through the origin, meaning extension is directly proportional to weight. Reading the plotted points: at 1 N the extension is 2 cm, at 2 N it is 4 cm, at 3 N it is 6 cm, and at 4 N it is 8 cm. This gives a constant ratio of 2 cm per newton.

For a weight of 2.5 N, the extension is 2.5 × 2 = 5.0 cm. This can also be found by reading directly from the straight-line graph at the midpoint between the 2 N and 3 N data points, which lies at 5.0 cm on the vertical axis.

Selecting 2.5 cm would result from dividing rather than multiplying, while 25 cm might arise from misreading the scale. The proportional relationship (Hooke's law region) means any intermediate value can be found by interpolation along the straight line.

Precision describes how close repeated measurements are to each other. Accuracy describes how close measurements are to the true value.

All five readings (52.0 to 52.4 g) are tightly clustered within a range of only 0.4 g, so they are precise. However, every reading is about 2 g above the true mass of 50.0 g, so they are not accurate. This pattern suggests a systematic error, such as a zero error on the balance (the scale reading is consistently too high), rather than random fluctuations.

If the readings were scattered widely around 50.0 g, they would be accurate on average but not precise. If they were both clustered tightly and centred on 50.0 g, they would be both precise and accurate.

When two objects at different temperatures are in thermal contact, energy transfers from the hotter object to the cooler one. This continues until both objects reach the same temperature. Once the temperatures are equal, there is no net transfer of energy between them. This condition is called thermal equilibrium.

In this scenario, the water starts at 60 °C and the room is at 20 °C. Energy transfers from the warmer water to the cooler surroundings until the water cools to 20 °C. At that point, water and room are at the same temperature and no further net energy flow occurs.

Absolute zero is the lowest possible temperature (-273 °C), not a condition of energy balance. Boiling point refers to a specific phase-change temperature. Thermal insulation is a property of a material that slows heat transfer - it does not describe the state of equal temperatures.

The relationship between weight and mass is W = m g, which rearranges to m = W / g. The lamp has a weight of 18 N and the gravitational field strength is given as g = 9.0 N/kg. Substituting: m = 18 / 9.0 = 2.0 kg.

The two strings and the ceiling are distractors in this question. Regardless of how the lamp is supported, its weight is still 18 N and the relationship W = m g still applies. A common error is to multiply weight by g instead of dividing, which would give 162 kg. Another mistake is to simply divide by 10 instead of 9.0, or to confuse the weight value with the mass value. Always check that your answer has sensible units and magnitude: a household lamp with a mass of 2.0 kg is reasonable, whereas 162 kg is not.

Neutrons are found inside the nucleus of an atom, along with protons. In the diagram, the nucleus is the central solid circle. Label W points directly to this central region, making it the correct indicator for where neutrons are located.

The small filled circles on the dashed orbital rings represent electrons, which orbit the nucleus in energy levels (shells). Labels pointing to these electrons or to the shells themselves indicate parts of the atom's electron structure, not the nucleus. Neutrons never exist in the electron shells; they are bound within the dense central nucleus by the strong nuclear force.

When interpreting atomic structure diagrams, the nucleus is always the central body containing protons and neutrons, while the concentric dashed rings represent electron shells. Protons and neutrons together make up the mass of the atom, while the electrons determine the atom's chemical behaviour. Being able to identify each part of these diagrams is essential for questions on atomic structure.

Gravitational field strength g relates weight W to mass m through the equation W = m × g. Rearranging gives g = W / m. Since the toolbox has a mass of 5.0 kg on every planet (mass does not change with location), the gravitational field strength on each planet equals the bar-chart weight divided by 5.0 kg.

Reading the bar chart: Earth shows a weight of about 50 N, giving g = 50 / 5.0 = 10 N/kg. Mars shows approximately 30 N, giving g = 30 / 5.0 = 6 N/kg. Neptune shows roughly 47 N (g ≈ 9.4 N/kg) and Uranus about 40 N (g ≈ 8 N/kg). Only Mars yields a gravitational field strength of approximately 6 N/kg.

A common mistake is to confuse mass and weight or to misread the bar chart scale. Remember that the mass of an object is the same everywhere; what changes from planet to planet is the weight, because each planet has a different gravitational field strength.

The table shows that when a fixed mass of gas is heated at constant pressure, the ratio \( V/T \) (volume divided by kelvin temperature) remains at 0.20 for every data point. A constant ratio \( V/T = k \) means \( V = kT \), which is the mathematical definition of direct proportionality between volume and kelvin temperature.

This is Charles's law: at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. When temperature doubles (from 200 K to 400 K, for example), volume doubles (from 40 cm\(^3\) to 80 cm\(^3\)).

The data cannot confirm a relationship about constant volume (since volume is changing) or about pressure being proportional to volume (since pressure is held constant, not varying). An inverse proportionality would require \( V \times T = \text{constant} \), but \( 250 \times 50 = 12500 \) while \( 400 \times 80 = 32000 \), so that does not hold.

This circuit is a potential divider with the LDR on top and the fixed 470 \(\Omega\) resistor on the bottom. The output voltage \(V_{\text{out}}\) is taken across the 470 \(\Omega\) resistor:

\[ V_{\text{out}} = V_{\text{battery}} \times \frac{R_{\text{fixed}}}{R_{\text{LDR}} + R_{\text{fixed}}} \]

In bright light, the LDR resistance is low (200 \(\Omega\)), so the fixed resistor takes a large share of the total voltage:

\[ V_{\text{out}} = 6.0 \times \frac{470}{200 + 470} = 6.0 \times \frac{470}{670} \approx 4.2 \;\text{V} \]

In darkness, the LDR resistance is very high (15 000 \(\Omega\)), so almost all the voltage drops across the LDR and very little across the fixed resistor:

\[ V_{\text{out}} = 6.0 \times \frac{470}{15000 + 470} \approx 0.18 \;\text{V} \]

Therefore \(V_{\text{out}}\) is higher in bright light. The general principle is that in this arrangement, as the LDR resistance decreases (brighter light), a larger fraction of the supply voltage appears across the fixed resistor.

The Sun releases energy through nuclear fusion, a process in which small atomic nuclei are combined to form larger nuclei. In the Sun's core, hydrogen nuclei (protons) fuse together under extreme temperature and pressure to form helium nuclei, releasing enormous amounts of energy in the process.

Nuclear fission is a different process in which large, heavy nuclei (such as uranium-235) are split into smaller fragments. Fission is used in nuclear power stations on Earth but is not the process that powers the Sun. Radioactive decay of helium does not occur - helium is the stable product of the Sun's fusion reactions, not a radioactive source.

This question tests magnetic induction. When a magnet is brought near an unmagnetised magnetic material, the material becomes temporarily magnetised by induction. The end nearest the magnet acquires the opposite polarity to the magnet's approaching pole.

The N pole approaches end X. By induction, X becomes a S pole (opposite to the approaching N), and Y becomes a N pole. This is why the iron bar is attracted to the magnet: the induced S pole at X is attracted to the N pole of the magnet.

If X became N (same as the approaching pole), the bar would be repelled, which contradicts the observation that unmagnetised iron is always attracted to a magnet.

Day and night are caused by the rotation of the Earth on its own axis. The Earth completes one full rotation approximately every 24 hours. At any moment, the half of the Earth facing the Sun is lit (daytime) while the half facing away is in shadow (night-time). As the Earth rotates, different parts of its surface move into and out of sunlight, producing the daily cycle of day and night.

The Earth's orbit around the Sun takes one year, not one day, so orbital motion does not explain the 24-hour day-night cycle. The Moon's orbit around the Earth is responsible for the phases of the Moon and for eclipses, but it plays no role in causing day and night. The tilt of the Earth's axis (23.5° from the vertical) explains why seasons occur and why day length varies throughout the year, but it is not the cause of the basic day-night alternation. Without any tilt at all, a rotating Earth would still have day and night; every location would simply experience exactly 12 hours of each.

During the flat section from 0 to \( t_1 \), the ice is melting at a constant temperature of 0 degrees C. All the energy supplied by the heater goes into breaking the bonds between water molecules in the solid lattice, not into raising the temperature. This energy is the latent heat of fusion.

The energy supplied during this time is:

\[ E = Pt = 200 \times 83.5 = 16\,700 \text{ J} \]

The specific latent heat of fusion is defined as the energy required per kilogram to change a substance from solid to liquid at constant temperature:

\[ L_f = \frac{E}{m} = \frac{16\,700}{0.050} = 334\,000 \text{ J/kg} = 3.34 \times 10^5 \text{ J/kg} \]

If you accidentally divide by the mass in grams (50) instead of kilograms (0.050), you would get 334 J/kg, which is far too small. If you forget to multiply by time and just use the power (200 J), you would get 4000 J/kg. The key steps are: calculate total energy first (power multiplied by time), then divide by the mass in kilograms.

In the kinetic particle model, the three states of matter are distinguished by particle arrangement and motion. Solids have particles in a regular, closely packed arrangement, vibrating in fixed positions. Liquids have particles that are irregularly arranged, close together, and able to slide over each other. Gases have particles that are randomly distributed, widely spaced, and moving rapidly in all directions.

Looking at the table, materials P and T both have regular, closely packed particles vibrating in fixed positions - these are solids. Material R has randomly spaced particles moving fast in all directions - this is a gas. Materials Q and S both have irregular, close-together particles that slide over each other - this matches the liquid state exactly.

Therefore the two materials that are both liquids are Q and S.

Hubble's law is expressed as \( v = H_0 \, d \), where \( v \) is the recession velocity of a galaxy, \( d \) is its distance from Earth, and \( H_0 \) is the Hubble constant.

The graph plots recession velocity (\( v \)) on the vertical axis against distance (\( d \)) on the horizontal axis. Rearranging Hubble's law gives:

\[ H_0 = \frac{v}{d} \]

This is exactly the definition of the gradient of a straight line passing through the origin on a \( v \) versus \( d \) graph. The gradient of the best-fit line therefore represents the Hubble constant.

The reciprocal \( 1/H_0 \) gives an estimate of the age of the universe, but it is the Hubble constant itself that is read directly from the gradient.

Weight on Earth: W_E = m × g_E = 4500 kg × 9.8 N/kg = 44 100 N. Weight on the Moon: W_M = m × g_M = 4500 kg × 1.6 N/kg = 7200 N. The difference in weight is 44 100 N - 7200 N = 36 900 N.

Mass remains 4500 kg in both locations because mass is a measure of the quantity of matter in an object and does not depend on gravity. Weight, however, is the gravitational force on that mass and changes with the local gravitational field strength. The Moon's much weaker gravity (1.6 N/kg compared to Earth's 9.8 N/kg) means the truck weighs far less there.

A common mistake is to subtract the two values of g first and then multiply by the mass: (9.8 - 1.6) × 4500 = 8.2 × 4500 = 36 900 N. This shortcut gives the same result because multiplication distributes over subtraction, but students should understand both approaches.

This is a head-on collision where two vehicles travelling in opposite directions lock together. The key is to assign a positive direction and apply conservation of momentum.

Taking east as positive:

\[ p_{\text{total}} = m_{\text{car}} \times v_{\text{car}} + m_{\text{van}} \times v_{\text{van}} \]

The car's momentum is \( 1200 \times 15 = 18\,000 \text{ kg m/s} \) (east), and the van's momentum is \( 900 \times (-20) \) (west, so negative). Since the vehicles lock together, the combined mass is \( 1200 + 900 = 2100 \text{ kg} \).

\[ v = \frac{p_{\text{total}}}{m_{\text{total}}} \]

Substituting gives a small positive velocity, meaning the locked vehicles move east. The result is approximately 0.71 m/s, confirming the direction is east. When two objects collide head-on with nearly equal momenta, the combined object moves slowly in the direction of whichever had the slightly greater momentum.

Boyle's law applies to a fixed mass of gas at constant temperature: \( P_1 V_1 = P_2 V_2 \). As the balloon rises, the atmospheric pressure decreases, allowing the gas inside to expand.

\[ 100 \times 0.50 = 40 \times V_2 \]

\[ V_2 = \frac{100 \times 0.50}{40} = \frac{50}{40} = 1.25 \text{ m}^3 \]

The pressure drops to \( \frac{40}{100} = \frac{2}{5} \) of its original value, so the volume increases to \( \frac{5}{2} = 2.5 \) times the original. \( 0.50 \times 2.5 = 1.25 \text{ m}^3 \). Selecting 0.20 m\(^3\) would result from multiplying pressure ratio by volume instead of dividing. Selecting 2.50 m\(^3\) would come from applying the factor of 2.5 to 1.0 rather than 0.50, or confusing the multiplier.

This question uses a stroboscopic photograph to determine speed. The equally spaced images show the ball moves at constant speed. There are 7 images, creating 6 intervals. Each interval is 0.10 s.

Total time = 6 × 0.10 = 0.60 s. Total distance = 30.0 cm = 0.300 m.

As with ticker tape, the number of intervals is one less than the number of images. Using 7 intervals (0.70 s) would give 0.43 m/s, which is incorrect.

The centre of gravity of a composite object is the single point where the entire weight can be considered to act. For a T-shape made of two identical strips, each strip has the same mass. The horizontal strip has its centre of gravity at its geometric centre (the midpoint of the top bar), and the vertical strip has its centre of gravity at its geometric centre (the midpoint of the vertical bar, well below the junction).

The overall centre of gravity lies on the axis of symmetry (the vertical line through the middle), at a position between the two individual centres. Since both strips have equal mass, the combined centre of gravity is exactly halfway between them. This point falls near the junction where the two strips meet, slightly below the top of the vertical strip. Of the labelled points, P is nearest to this location.

Points further down the vertical strip (like Q) would only be correct if the vertical strip were much heavier. A point at the centre of the horizontal strip alone (like R) ignores the mass below the junction. The centre of gravity must lie on the axis of symmetry, which eliminates any off-centre point.

An object topples when the vertical line drawn downward from its centre of gravity falls outside its base of support. At Stage 1 the box sits flat on the ground, so the centre of gravity is directly above the middle of the base and the box is stable. At Stage 2 the box is tilted on its bottom-right edge, but the vertical line through the centre of gravity still falls within (or on the edge of) the base, so the box would return to its original position if released.

At Stage 3 the tilt is large enough that the vertical line through the centre of gravity passes beyond the pivot edge. Once this happens, the weight of the box creates a turning moment that rotates it further away from the upright position rather than back towards it, and the box topples over.

The key principle is that stability depends on the horizontal position of the centre of gravity relative to the base. A wider base or a lower centre of gravity makes an object harder to topple. In examination questions involving tilting, always check whether the centre of gravity still lies above the support region.

A concave (converging) barrier acts like a concave mirror for waves. When plane wavefronts strike a concave surface, the shape of the barrier causes different parts of each wavefront to reflect at slightly different angles, all directed inward toward a single point called the focal point.

This is the same principle used in satellite dishes and parabolic reflectors: the curved surface collects parallel incoming waves and concentrates them at the focus. The reflected wavefronts become curved (converging) rather than remaining flat.

The wavefronts do not remain plane after reflection because the barrier is curved, not flat. A flat barrier would produce plane reflected wavefronts. They also do not spread out as circular wavefronts - that would require a convex barrier. And they certainly do not pass through unchanged, since reflection reverses their direction of travel.

From the table, at 200 lux the LDR has a resistance of 5000 \(\Omega\) = 5 k\(\Omega\). The LDR is the lower resistor (\( R_2 \)) and \( R_1 = 5 \text{ k}\Omega \), with a 10 V supply.

Using the potential divider formula:

\[ V_{\text{out}} = V_{\text{supply}} \times \frac{R_2}{R_1 + R_2} = 10 \times \frac{5}{5 + 5} = 10 \times \frac{5}{10} = 5.0 \text{ V} \]

When the two resistors in a potential divider are equal, the supply voltage is split equally between them. Each resistor receives exactly half the supply voltage, giving 5.0 V across the LDR.

Charge is calculated using \( Q = I \times t \). The current is 3.2 A. The time must be converted entirely to seconds: 2 minutes and 30 seconds = \( 2 \times 60 + 30 = 150 \, \text{s} \).

\[ Q = 3.2 \times 150 = 480 \, \text{C} \]

A common error is to forget to convert the time fully to seconds. Using 2 minutes as 120 s (and ignoring the extra 30 s) gives \( 3.2 \times 120 = 384 \, \text{C} \). Using just 30 s gives \( 3.2 \times 30 = 96 \, \text{C} \). Always convert the entire time to seconds before substituting into the equation.

Random errors in timing a pendulum arise mainly from human reaction time when starting and stopping the stopwatch. Each individual measurement of the period includes this uncertainty, which is why the student gets different values each time.

The most effective way to reduce this random error is to time a larger number of oscillations and then divide. If you time 20 complete swings instead of one, the reaction-time error (roughly the same fixed amount) is spread over 20 periods rather than one, reducing the percentage error by a factor of 20.

Changing the string length or bob mass alters the period of the pendulum but does nothing to reduce the timing error itself. Measuring the string length more carefully addresses a different source of error (the length measurement), not the timing uncertainty. Only the averaging approach directly targets the random error described in the question.

When a uranium-235 nucleus undergoes fission, it releases two or three neutrons. In a controlled chain reaction (as needed in a nuclear power station), exactly one neutron from each fission event must, on average, go on to cause another fission.

If more than one neutron per fission causes further fission, the reaction rate increases exponentially - this is an uncontrolled chain reaction, which would lead to a dangerous meltdown or explosion. If fewer than one neutron per fission causes another fission, the reaction gradually dies out and power generation stops.

Control rods (made of materials like boron or cadmium that absorb neutrons) are inserted into or withdrawn from the reactor core to maintain this balance, ensuring exactly one neutron per fission sustains the reaction at a constant rate.

The equation \(E = mc\Delta T\) can be rearranged to \(\Delta T = \frac{E}{mc}\). For a given energy input \(E\) and a given mass \(m\), the temperature rise \(\Delta T\) is inversely proportional to the specific heat capacity \(c\).

Water has a very high specific heat capacity (about 4200 J/(kg °C)), while sand has a much lower one (roughly 800 J/(kg °C)). When the Sun delivers roughly the same energy to both the sand and the sea, the sand - with its lower specific heat capacity - undergoes a much larger temperature rise. This is why beach sand feels burning hot underfoot while the sea remains relatively cool on the same sunny day.

Water does not reflect all sunlight, and sand being darker does not give it a higher specific heat capacity (in fact the opposite reasoning would apply). The difference is fundamentally about how much energy each material needs to raise its temperature by one degree.

Examine the data: when the volume halves, the force doubles. Volume goes 40, 20, 10, 5 while force goes 10, 20, 40, 80. The product \( F \times V \) is constant: \( 10 \times 40 = 400 \), \( 20 \times 20 = 400 \), \( 40 \times 10 = 400 \), \( 80 \times 5 = 400 \). A constant product \( FV = k \) means \( F = k/V \), the definition of inverse proportionality.

This follows from Boyle's law. At constant temperature, \( PV = \text{constant} \). Since the plunger has a fixed cross-sectional area \( A \), pressure \( P = F/A \), so \( (F/A) \times V = \text{constant} \), giving \( F \times V = \text{constant} \). Force is therefore inversely proportional to volume.

Direct proportionality would require force and volume to increase together (they do the opposite). Proportionality to volume squared would not produce a constant \( FV \) product. And force is clearly not independent of volume - it changes markedly as volume changes.

The container has a fixed volume, but it also has a pressure-release valve that opens whenever the internal pressure exceeds 101 kPa. As the air is heated, the molecules gain kinetic energy and move faster, creating more frequent and harder collisions with the container walls. This raises the pressure above 101 kPa, causing the valve to open and allow some gas molecules to escape.

Once molecules leave, the total mass of gas inside the container decreases. Since density equals mass divided by volume, and the volume stays constant while the mass falls, the density must decrease. Each time the temperature is raised further, more molecules escape through the valve, reducing the mass and hence the density even more.

The container does not expand (it is described as having constant volume). Individual molecules do not shrink when heated - in fact, molecular size is essentially unchanged by temperature. Molecules speed up at higher temperatures, not slow down.

The two forces act at right angles to each other (one due east, one due north). When forces are perpendicular, the magnitude of the resultant is found using Pythagoras' theorem:

\[R = \sqrt{F_1^2 + F_2^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ N}\]

The resultant is 13 N, directed at an angle between north and east. Simply adding the forces (12 + 5 = 17 N) would only be correct if they acted in the same direction. Multiplying them (12 x 5 = 60 N) or subtracting them (12 - 5 = 7 N) are not valid operations for finding a resultant of perpendicular forces.

For the beam to balance, the principle of moments requires that the clockwise moment about the pivot equals the anticlockwise moment. The 6.0 N weight hangs 0.40 m to the left of the pivot, and weight W hangs 0.60 m to the right.

Taking moments about the pivot:

\[ 6.0 \times 0.40 = W \times 0.60 \]

\[ 2.4 = 0.60\,W \]

\[ W = \frac{2.4}{0.60} = 4.0 \text{ N} \]

The weight on the longer arm must be smaller to produce the same moment. This is the same principle that allows a lighter person to balance a heavier person on a seesaw by sitting further from the pivot.