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

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.

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.

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.

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 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.

A beam balance works by comparing the unknown object against standard masses on opposite sides of a pivot. When the beam is level, the turning effects (moments) on each side are equal, which happens when the masses are equal. Crucially, both the object and the standard masses experience the same gravitational field strength, so any change in g (for example, moving to the Moon) affects both sides equally and the balance point does not change.

This means a beam balance measures mass, not weight. A spring balance, by contrast, measures weight (the gravitational force on the object) and would give a lower reading on the Moon because g is smaller there. Since the beam balance gives the same reading regardless of location, it is not measuring weight, gravitational field strength, or density.

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.

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.

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 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 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.

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.

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.

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.

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.

This question tests identifying visible light by its wavelength. The visible light spectrum spans approximately \( 4 \times 10^{-7} \) to \( 7 \times 10^{-7} \) m (400-700 nm).

• Wave P: \( 3.0 \times 10^{-10} \) m is in the X-ray range (much shorter than visible).
• Wave Q: \( 5.0 \times 10^{-7} \) m = 500 nm, which is green visible light. This is in the visible range.
• Wave R: \( 2.0 \times 10^{-2} \) m is in the microwave range (much longer than visible).

Wave Q is the only one with a wavelength in the visible range. All three travel at the same speed because they are all electromagnetic waves in a vacuum.

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.

This question tests applying alpha and beta decay rules to a decay series.

Starting with X: nucleon number 232, proton number 90.

Alpha decay (X then Y): nucleon number decreases by 4, proton number decreases by 2. So Y has nucleon number 232 - 4 = 228 and proton number 90 - 2 = 88.

Beta decay (Y then Z): nucleon number stays the same, proton number increases by 1 (a neutron converts to a proton). So Z has nucleon number 228 and proton number 88 + 1 = 89.

Z has nucleon number 228 and proton number 89.

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 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.

In a mercury barometer, the height of the mercury column is determined by the balance between atmospheric pressure pushing down on the mercury reservoir and the weight of the mercury column. This gives \(h = \dfrac{P_{\text{atm}}}{\rho g}\).

The tube diameter does not appear in this equation. A wider tube holds more mercury by volume, but the extra weight is spread over a proportionally larger cross-sectional area, so the height remains the same. Both barometers measure the same atmospheric pressure using the same liquid, so \(h_1\) equals \(h_2\).

This is a fundamental property of fluid pressure: it depends on depth (height), density, and gravitational field strength, not on the shape or width of the container.

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.

Weight is calculated using W = m g. First find the total mass of the two blocks: 1.5 + 3.5 = 5.0 kg. Then multiply by the gravitational field strength: W = 5.0 x 9.8 = 49 N.

Alternatively, you can calculate each weight separately and add them: block X weighs 1.5 x 9.8 = 14.7 N and block Y weighs 3.5 x 9.8 = 34.3 N, giving a total of 14.7 + 34.3 = 49.0 N. Both methods give the same result. A common error is to use g = 10 N/kg instead of the stated 9.8 N/kg, which would give 50 N. Always use the value of g provided in the question rather than a rounded approximation.

Bubbles forming throughout the body of the liquid and rising to the surface is the defining characteristic of boiling. During boiling, the liquid has reached its boiling point and is changing state from liquid to gas (steam) at locations throughout the liquid, not just at the surface.

Evaporation, by contrast, occurs only at the surface of a liquid and happens at temperatures below the boiling point. The bubbles described in this question form deep within the liquid, which rules out surface-only evaporation. They also cannot be air bubbles: dissolved air produces tiny bubbles when a liquid is first warmed, but the large bubbles of steam described here indicate a full change of state.

Condensation is the reverse process (gas turning to liquid) and does not produce bubbles rising through a liquid. The correct description is that the water is boiling and changing to steam throughout the liquid.

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.

The pressure exerted by an object resting on a horizontal surface is:

\[ p = \frac{F}{A} = \frac{W}{A} \]

Substituting the values:

\[ p = \frac{180}{0.060} = 3000 \, \text{Pa} \]

The value 10 800 Pa would result from multiplying weight by area (\( 180 \times 0.060 \times 1000 \)), which is not the correct operation. Always divide force by area to find pressure.

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.

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.

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:

\[ W \times 25 = 210 \]

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

Total internal reflection occurs when light inside a material strikes the boundary at an angle greater than the critical angle for that material. The question states that the ray strikes at 45° in each material.

Checking each material against 45°:

• Diamond: critical angle = 24°. Since 45° > 24°, total internal reflection occurs.
• Glass: critical angle = 42°. Since 45° > 42°, total internal reflection occurs.
• Perspex: critical angle = 43°. Since 45° > 43°, total internal reflection occurs.
• Water: critical angle = 49°. Since 45° < 49°, total internal reflection does not occur. The light partially refracts out of the water.

Total internal reflection therefore occurs in diamond, glass, and perspex, but not in water.

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.

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.

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.

When light passes from air into an optically denser medium (such as glass or water), two things happen: the speed decreases and the wavelength decreases.

The key relationship is \( v = f\lambda \). The frequency of a wave is determined by the source and does not change when the wave enters a new medium. Since frequency stays constant and speed decreases, the wavelength must also decrease to keep the equation balanced:

\[ \lambda = \frac{v}{f} \]

If \( v \) decreases and \( f \) stays the same, \( \lambda \) must decrease. Therefore the correct description is that both wavelength and speed decrease.

This question tests the principle of flotation and density calculation. When an object floats, the weight of water displaced equals the weight of the object. The cube has sides of 5 cm, and 3 cm is submerged.

The volume of the whole cube is \( 5 \times 5 \times 5 = 125 \, \text{cm}^3 \). The submerged volume is \( 5 \times 5 \times 3 = 75 \, \text{cm}^3 \). Since the cube floats, the mass of the cube equals the mass of water displaced:

The density of the wood is:

Alternatively, for a floating object, the fraction submerged equals the ratio of its density to the liquid's density: \( 3/5 = 0.6 \), giving 0.6 g/cm³ directly.

When the gymnast lands, her momentum changes from a downward value to zero. This change in momentum is the same regardless of whether she lands on a mat or a hard floor, because her mass and landing speed are the same in both cases.

The relationship between force, time, and momentum change is:

\[F = \frac{\Delta p}{\Delta t}\]

The thick soft mat compresses on impact, and this compression takes time. By increasing the time \(\Delta t\) over which the gymnast decelerates, the mat reduces the average force on her legs. The mat does not absorb all the kinetic energy before landing (the gymnast still makes contact), nor does it reduce her mass or her change in momentum.

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.

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.

In beta decay, a neutron inside the nucleus converts into a proton, and the nucleus emits a high-speed electron (the beta particle). The mass number does not change because the total number of nucleons (protons + neutrons) stays the same: one neutron is lost but one proton is gained. However, the proton number increases by 1.

Carbon-14 has 6 protons and 8 neutrons. After beta decay, one neutron becomes a proton, so the daughter nucleus has 7 protons and 7 neutrons. The mass number remains 14. The element with proton number 7 is nitrogen, so the daughter is nitrogen-14.

A proton number of 5 would mean a proton was lost, which does not occur in beta decay. A proton number of 8 would mean two protons were gained. A proton number of 6 would mean no change at all. In beta decay, exactly one neutron transforms into one proton, so the proton number always increases by exactly one.

During melting, a substance changes from solid to liquid. The particles in a solid are held in fixed positions by intermolecular forces. To convert the solid into a liquid, these forces must be partially overcome so that particles can move more freely.

The energy being supplied during melting is used entirely to break or weaken these intermolecular bonds. This energy is called the latent heat of fusion. Because the energy goes into changing the arrangement of particles rather than increasing their speed, the average kinetic energy of the particles does not increase, and the temperature remains constant.

The substance has not reached any 'maximum temperature' - once melting is complete, further heating will raise the temperature again. The explanation is not about energy losses balancing gains either; it is specifically about the energy being redirected to overcome intermolecular forces.