Physics (9-1) 0972 | Paper 2 Mock 01 | Multiple Choice (Extended)

Trapped air is a poor conductor of thermal energy, so the many small pockets of air held within the foam reduce heat loss by conduction across the cavity. Just as importantly, because the air is broken up into many small, separate pockets, it cannot circulate freely as a single body of air, which prevents convection currents from forming across the cavity.

The foam reduces thermal energy loss mainly by trapping air in small pockets, which are poor conductors of heat and cannot set up convection currents. Without the foam, a large open cavity of air could circulate freely, carrying warm air away by convection much faster.

The table shows that for every pair of volume and pressure readings, the product \(pV\) comes out to the same value, 20 000 kPa cm³, even though the individual pressure and volume readings change considerably (from 100 cm³/200 kPa up to 25 cm³/800 kPa).

This constant product is the defining feature of Boyle's law: for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume, so that \(pV\) remains constant as one increases while the other decreases. The conclusion best supported by the table is that pressure and volume are inversely proportional at constant temperature.

The potential difference across the fixed resistor is:

For \( V_{\text{fixed}} = 3.0 \) V:

From the table, the thermistor has a resistance of 600 \(\Omega\) at 30 \(^\circ\)C. When the thermistor and fixed resistor are equal, the supply voltage is shared equally.

A generator with slip rings produces alternating current (a.c.). The slip rings provide a continuous connection between the rotating coil and the external circuit without reversing the connections. As the coil rotates, the direction of the induced e.m.f. reverses every half turn, producing a current that alternates in direction.

If the generator used a split-ring commutator instead of slip rings, the output would be direct current (d.c.), because the commutator reverses the external connections every half turn, keeping the current flowing in one direction.

The key to this problem is recognising that the person's mass stays constant regardless of location. Since weight = mass × gravitational field strength (W = mg), the mass can be found from the Earth data: m = W / g = 700 / 10 = 70 kg.

With the mass known, the weight on each body is calculated using that body's value of g. On Jupiter: W = 70 × 25 = 1750 N. On Mars: W = 70 × 3.7 = 259 N. The difference is 1750 − 259 = 1491 N.

A common error is to subtract the two values of g and then multiply by the Earth weight, which gives a meaningless result. Another pitfall is forgetting to find the mass first and instead trying to scale the Earth weight directly by a ratio of g values for the wrong calculation. Always extract the mass first, then apply each planet's g separately before finding the difference.

Mercury freezes at approximately -39 °C, so a mercury thermometer cannot measure temperatures below that. In very cold climates, temperatures can drop well below -39 °C.

Alcohol (ethanol) has a freezing point of approximately -114 °C, which is far lower than mercury's. This means an alcohol thermometer remains liquid and functional at temperatures where mercury would solidify and become useless.

While alcohol is indeed less toxic than mercury, that is a safety advantage rather than the reason it is chosen for cold-climate thermometers. Alcohol actually boils at a lower temperature than mercury (about 78 °C vs 357 °C), which limits its upper range. Both mercury and alcohol expand fairly evenly, so even expansion is not the distinguishing factor for cold climates.

Pressure is defined as force per unit area: \(P = \frac{F}{A}\). The block weighs 50 N in all three orientations, so the force on the surface is the same each time. What changes is the contact area.

The greatest pressure occurs when the same force is applied over the smallest area. From the diagram, position 3 has the block standing on its smallest face, giving the smallest contact area and therefore the greatest pressure.

Position 1 (wide face down) gives the largest contact area and therefore the lowest pressure. The option stating all positions give the same pressure is incorrect because, although the weight is the same, the contact area differs. Pressure depends on both force and area, not force alone.

When the coil plane is parallel to the magnetic field lines (the coil is horizontal), the sides of the coil are cutting through the field lines at the maximum rate. This is the position where the rate of change of flux linkage is greatest, so the induced e.m.f. is at its maximum value.

By contrast, when the coil plane is perpendicular to the field (coil vertical), the sides move parallel to the field lines and do not cut through them, so the induced e.m.f. is zero at that instant. The e.m.f. varies sinusoidally as the coil rotates.

The wavefront spacing represents the wavelength. In medium 1, the spacing is 3 cm, and in medium 2, the spacing is 4 cm. Since the frequency of a wave does not change when crossing a boundary between media, and \(v = f\lambda\):

The wave speed in medium 2 is greater than in medium 1 (by a factor of 4/3). A larger wavelength at the same frequency means a higher speed.

To find the centre of gravity of an irregular lamina, it is suspended from two different points (A and B). Each time, a vertical plumb line is drawn on the lamina. The centre of gravity lies at the intersection of these two lines.

The two dashed lines from pins A and B intersect at point P. Therefore, the centre of gravity of the lamina is at point P. Points Q, R, and S are not at the intersection and so cannot be the centre of gravity.

In a potential divider, the output voltage across \( R_2 \) is:

The output voltage is 6.0 V. The larger resistor (\( R_2 = 6 \) k\(\Omega\)) takes a larger share of the total voltage. The remaining 3.0 V is dropped across \( R_1 \). A common error is to reverse the resistors in the formula, which would give 3.0 V instead.

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.

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.

To melt a given mass of a metal, the energy needed is found from \(E = mL\), so with the same mass (2.0 kg) used for every metal, the energy required is directly proportional to each metal's specific latent heat of fusion.

Since aluminium has the largest specific latent heat of fusion (397 kJ/kg) among the four metals, it requires the most energy (794 kJ) to melt 2.0 kg at its melting point.

So aluminium requires the most energy. A common mistake is to assume the metal with the highest melting point requires the most energy to melt; the energy needed to melt a substance depends on its specific latent heat of fusion, not directly on how hot its melting point is.

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.

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.

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.

With the earth wire disconnected, when the live wire touches the metal case, the case becomes live at 230 V. There is no safe path for the fault current to flow to ground.

When a person touches the case, current flows through their body to the ground, causing a potentially fatal electric shock. The fuse does not blow because the current through a human body (typically a few tens of milliamps) is much less than the fuse rating. Without the earth wire providing a low-resistance alternative path, the person's body is the only available path to earth.

Using \(R = \frac{V}{I}\), calculate the resistance of each resistor:

• A: \(R = \frac{4.5}{0.30} = 15\;\Omega\)
• B: \(R = \frac{7.2}{0.60} = 12\;\Omega\)
• C: \(R = \frac{3.6}{0.15} = 24\;\Omega\)
• D: \(R = \frac{9.0}{0.45} = 20\;\Omega\)
• E: \(R = \frac{5.4}{0.36} = 15\;\Omega\)

Resistor C has a resistance of 24 Ω.

From the CRO trace, each complete wave (one full cycle) occupies 2 divisions on the screen.

The period \(T\) of the wave is:

In the first 8 divisions, the number of complete cycles is:

The frequency is:

The correct answer is 4 complete cycles with a frequency of 125 Hz.

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.

Frequency and amplitude are the two wave properties that determine the perceived pitch and loudness of a sound respectively: frequency determines pitch, and amplitude determines loudness.

Since sound N is stated to have the same frequency and the same amplitude as sound M, both of the properties that determine how a sound is perceived by ear are identical between the two.

So sound N has the same pitch and the same loudness as sound M. Although the two waveforms could in principle still differ in more subtle ways (such as the mixture of overtones present, which affects tone quality or timbre), for the properties directly linked to frequency and amplitude at this level, pitch and loudness are the two quantities that must match.

Since the springs are identical, they have the same spring constant. From spring P:

For spring Q with a 6.0 N load:

The extension of spring Q is 12 cm. Since the load is doubled compared to P, and the spring obeys Hooke's law, the extension also doubles from 6.0 cm to 12 cm.

Taking moments about the hinge, the system is in equilibrium when the clockwise moment equals the anticlockwise moment.

The weight (120 N) acts at the centre of the trapdoor, 0.40 m from the hinge. The tension T acts at the opposite edge, 0.80 m from the hinge.

The tension in the rope is 60 N, which is half the weight. The rope acts at twice the distance from the hinge compared to the weight, so only half the force is needed.

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.

Impulse (equal to the change in momentum) depends on how much the velocity of an object changes, \(\Delta p = m\Delta v\), not just on whether it stops. Both balls have the same mass and hit the wall at the same speed \(u\), but they behave differently afterwards.

The clay ball sticks to the wall, so its final velocity is zero: its change in momentum has magnitude \(mu\). The rubber ball bounces back at (approximately) the same speed, reversing its direction: its final velocity is \(-u\), so its change in momentum has magnitude \(m(u-(-u)) = 2mu\), twice as large as the clay ball's.

The rubber ball exerts the greater impulse on the wall, because reversing an object's momentum requires twice the impulse needed to simply bring the same momentum to zero. It is a common misconception that the ball which stops (the clay ball) must experience the bigger change, but stopping is only "half" of the momentum reversal that the bouncing ball undergoes.

In a series circuit, there is only one path for the current to flow through. This means the current is the same at every point in the circuit. The ammeter near the battery reads 0.5 A, so the current flowing through every component, and through every section of wire, is also 0.5 A.

The current between L2 and L3 is therefore 0.5 A, identical to the current measured at the ammeter. The current does not split, divide, or reduce as it passes through successive lamps in a series arrangement. Each lamp may have a voltage drop across it, but the current remains constant throughout. This is a direct consequence of conservation of charge: charge cannot accumulate at any point in the circuit, so the same amount of charge per second must pass through every cross-section of the series loop.

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 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.

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.

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.

In the unfilled cavity wall, the air gap allows convection currents to form: warm air near the inner wall rises and cool air near the outer wall sinks, transferring energy across the gap. Injecting foam into the cavity traps the air in small pockets, preventing convection. The foam itself and the trapped air are poor conductors, so conduction through the cavity is also reduced. Together these effects significantly lower the rate of energy transfer through the wall.

The diagram shows all galaxies moving away from the observer in every direction. This observation tells us that the universe is expanding.

The fact that galaxies in all directions are receding means this is not local motion but a universal expansion of space itself. Every observer, regardless of position, would see the same pattern. There is no centre of the expansion; space expands uniformly, carrying all galaxies apart.

The bar chart shows the weight of a 4.0 kg metal block on three bodies. To find the gravitational field strength on Venus, read the weight from the chart and use g = W / m.

From the chart, the Venus bar reaches approximately 30 N. Dividing by the mass: g = 30 N / 4.0 kg = 7.5 N/kg. For comparison, the Earth bar reads about 40 N (g = 40 / 4.0 = 10 N/kg, as expected) and the Moon bar reads roughly 14 N (g = 14 / 4.0 = 3.5 N/kg).

The calculation requires recognizing that mass stays constant at 4.0 kg regardless of location, and that weight varies because each body has a different gravitational field strength. A common mistake is to read the weight value directly as the gravitational field strength without dividing by the mass.

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.

The photographic film at position Y, just beyond the violet end of the visible spectrum, is exposed by ultraviolet radiation. When white light is dispersed by a prism, the visible spectrum extends from red to violet. Beyond the violet end lies ultraviolet, which is invisible to the eye but has enough energy to cause chemical changes in photographic film.

Although ultraviolet is not visible, it is still present in the white light from the source and is refracted more than violet light by the prism, directing it to position Y beyond the visible violet.

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.

For each situation, the centripetal force must be identified:

• Satellite orbiting Earth: the centripetal force is provided by gravitational attraction, not rocket thrust. Satellites in orbit need no engines.
• Car turning on a flat road: the centripetal force is provided by friction between the tyres and the road. This is correct.
• Electron orbiting a nucleus: the centripetal force is provided by electrostatic (Coulomb) attraction between the negative electron and the positive nucleus, not gravity.
• Ball on a string in a horizontal circle: the centripetal force is provided by the tension in the string, not the weight of the ball.

The row identifying friction for a car on a flat road is correct.

A beam balance compares masses by balancing moments on each side. Both blocks have the same mass (1.5 kg each). On the Moon, both blocks experience the same reduced gravitational field strength, so both sides are equally affected. The beam remains level. The reading does not change because a beam balance measures mass, not weight, and mass is independent of location.

Useful output power equals the input power multiplied by the efficiency (expressed as a decimal):

\[ P_{\text{useful}} = P_{\text{input}} \times \text{efficiency} \]

• Motor P: \( 200 \times 0.90 = 180 \text{ W} \)
• Motor Q: \( 500 \times 0.40 = 200 \text{ W} \)
• Motor R: \( 300 \times 0.60 = 180 \text{ W} \)
• Motor S: \( 150 \times 0.80 = 120 \text{ W} \)

Motor Q provides the greatest useful output power at 200 W, despite having the lowest efficiency. A high input power can compensate for low efficiency. This illustrates why both the power rating and the efficiency must be considered together when determining useful output.