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(a) Circuit diagram requirements:
- Battery of 12 V shown with correct long-short line cell symbols.
- Fixed resistor and LDR connected in series with the battery.
- Voltmeter connected in parallel across the LDR.

(b) (i) As light intensity increases, the resistance of the LDR decreases.
(ii) Since the LDR resistance decreases, the total resistance of the series circuit decreases, causing the overall current in the circuit to increase. However, because the resistance of the LDR is now a smaller fraction of the total resistance, it takes a smaller share of the battery's voltage (potential divider rule). Therefore, the reading on the voltmeter decreases.

(c) Total resistance \(R_{\text{total}} = 250\ \Omega + 750\ \Omega = 1000\ \Omega\).
Using Ohm's Law: \(I = V / R_{\text{total}} = 12\text{ V} / 1000\ \Omega = 0.012\text{ A}\) (or \(12\text{ mA}\)).
Voltmeter reading across LDR: \(V_{\text{LDR}} = I \times R_{\text{LDR}} = 0.012\text{ A} \times 750\ \Omega = 9.0\text{ V}\).

- (a) [3 marks]: 1 mark for correct battery and series loop (resistor and LDR); 1 mark for correct symbols for resistor and LDR; 1 mark for voltmeter connected in parallel across the LDR.
- (b)(i) [1 mark]: stating that LDR resistance decreases.
- (b)(ii) [3 marks]: 1 mark for stating total resistance decreases / circuit current increases; 1 mark for stating that the LDR takes a smaller fraction of the total voltage; 1 mark for concluding the voltmeter reading decreases.
- (c) [3 marks]: 1 mark for calculating total resistance (1000 ohms) and finding current \(I = 0.012\text{ A}\); 1 mark for formula \(V = I \times R\) used with LDR value; 1 mark for final voltmeter reading of \(9.0\text{ V}\) (allow ecf).

(a) A parallel connection ensures both the heating element and the fan receive the full mains voltage of \(230\text{ V}\). It also allows them to be operated independently with separate switches, and if one component fails, the other can still run.

(b) Using the power formula: \(P = I \times V\)
\(I = P / V = 46\text{ W} / 230\text{ V} = 0.2\text{ A}\).

(c) (i) Using Ohm's Law: \(I = V / R\)
\(I = 230\text{ V} / 23\ \Omega = 10\text{ A}\).

(ii) Total current in the hair dryer: \(I_{\text{total}} = 0.2\text{ A} + 10\text{ A} = 10.2\text{ A}\).
Total power: \(P_{\text{total}} = V \times I_{\text{total}} = 230\text{ V} \times 10.2\text{ A} = 2346\text{ W}\).
Alternatively: \(P_{\text{total}} = P_{\text{fan}} + P_{\text{heater}} = 46\text{ W} + (230\text{ V} \times 10\text{ A}) = 46\text{ W} + 2300\text{ W} = 2346\text{ W}\).

- (a) [2 marks]: 1 mark for stating that both receive full voltage (230 V); 1 mark for stating they can be controlled independently / if one breaks the other still works.
- (b) [2 marks]: 1 mark for rearranging formula \(I = P / V\) and substituting; 1 mark for correct calculation of \(0.2\text{ A}\).
- (c)(i) [2 marks]: 1 mark for formula \(I = V / R\) and substituting; 1 mark for correct current of \(10\text{ A}\).
- (c)(ii) [4 marks]: 1 mark for calculating power of heating element as \(2300\text{ W}\) (or summing currents to get \(10.2\text{ A}\)); 1 mark for method to find total power (adding powers or multiplying total current by 230 V); 1 mark for correct calculation of \(2346\text{ W}\) (or \(2.35\text{ kW}\)); 1 mark for correct unit (W or kW).

(a) Description of experiment:
- Set up a clamp stand and hang the spring vertically from the clamp.
- Position a meter ruler vertically next to the spring.
- Measure the original length of the spring using the ruler at eye level.
- Add a mass hanger of known weight (e.g., 1.0 N) to the spring and measure the new length.
- Calculate extension as: \(\text{extension} = \text{loaded length} - \text{original length}\).
- Repeat this process by adding additional weights one at a time, recording the extension for each total load.
- To ensure accuracy, view the ruler at eye level to avoid parallax error, make sure the ruler is vertical, and use a fiducial marker to pinpoint the bottom of the spring.

(b) (i) The limit of proportionality is the point beyond which force and extension are no longer directly proportional to each other (Hooke's law is no longer obeyed, and the graph ceases to be a straight line).

(ii) Extension \(x = 17.4\text{ cm} - 12.0\text{ cm} = 5.4\text{ cm} = 0.054\text{ m}\).
Using Hooke's Law: \(F = k \times x\)
\(k = F / x = 4.5\text{ N} / 0.054\text{ m} = 83.3\text{ N/m}\) (or \(83\text{ N/m}\)).

- (a) [5 marks]: 1 mark for apparatus setup (clamp stand, spring, ruler); 1 mark for measuring original and loaded lengths; 1 mark for calculating extension; 1 mark for repeating with different masses; 1 mark for accuracy detail (e.g., eye level to avoid parallax, use of fiducial marker, ensuring ruler is vertical).
- (b)(i) [2 marks]: 1 mark for stating that up to this point extension is proportional to force; 1 mark for stating that beyond this point they are no longer proportional.
- (b)(ii) [3 marks]: 1 mark for converting extension to meters (\(0.054\text{ m}\)); 1 mark for formula \(k = F/x\); 1 mark for correct calculation of \(83.3\text{ N/m}\) (accept \(83\) or \(83.3\), reject if unit is missing or incorrect).

(a) (i) Using Snell's law: \(n = \frac{\sin(i)}{\sin(r)}\)
\(n = \frac{\sin(42^\circ)}{\sin(26^\circ)} = \frac{0.6691}{0.4384} = 1.53\).

(ii) The frequency of the light remains unchanged. When entering the optically denser glass block, the speed of the light decreases. Since \(v = f \lambda\), the wavelength of the light also decreases.

(b) The critical angle is the angle of incidence in the more dense medium that results in an angle of refraction of \(90^\circ\) in the less dense medium.
Using the formula: \(\sin(c) = \frac{1}{n}\)
\(\sin(c) = \frac{1}{1.53} = 0.6536\)
\(c = \sin^{-1}(0.6536) = 40.8^\circ\) (accept \(41^\circ\)).

(c) The two conditions for total internal reflection are:
1. The light must travel from a medium of higher refractive index to a medium of lower refractive index (more dense to less dense medium).
2. The angle of incidence must be greater than the critical angle.

- (a)(i) [2 marks]: 1 mark for formula \(n = \sin(i)/\sin(r)\) and substitution; 1 mark for correct calculation of \(1.53\) (allow 1.5).
- (a)(ii) [3 marks]: 1 mark for stating frequency remains constant; 1 mark for stating wave speed decreases; 1 mark for concluding wavelength decreases.
- (b) [3 marks]: 1 mark for definition of critical angle; 1 mark for using formula \(\sin(c) = 1/n\); 1 mark for correct calculation of \(40.8^\circ\) or \(41^\circ\) (allow ecf from a(i)).
- (c) [2 marks]: 1 mark for light going from high to low refractive index medium; 1 mark for angle of incidence being greater than critical angle.

(a) Principle of operation:
- An alternating current (AC) is supplied to the primary coil.
- This alternating current creates a continuously changing magnetic field around the primary coil.
- This changing magnetic field is guided by/carried through the soft iron core to the secondary coil.
- The changing magnetic field cuts through the secondary coil, inducing an alternating voltage (and current if there is a complete circuit) across its ends due to electromagnetic induction.

(b) (i) Using the transformer turns ratio equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{240}{V_s} = \frac{1200}{90}\)
\(V_s = 240 \times \frac{90}{1200} = 18\text{ V}\).

(ii) For a 100% efficient transformer, input electrical power equals output electrical power:
\(V_p \times I_p = V_s \times I_s\)
\(240 \times I_p = 18 \times 4.0\)
\(240 \times I_p = 72\)
\(I_p = 72 / 240 = 0.30\text{ A}\).

- (a) [4 marks]: 1 mark for AC in primary; 1 mark for creating a changing magnetic field; 1 mark for core linking/carrying magnetic field to secondary; 1 mark for inducing voltage in secondary due to electromagnetic induction.
- (b)(i) [3 marks]: 1 mark for formula \(V_p/V_s = N_p/N_s\); 1 mark for substitution; 1 mark for correct calculation of \(18\text{ V}\).
- (b)(ii) [3 marks]: 1 mark for using power equation \(V_p I_p = V_s I_s\); 1 mark for substitution of values; 1 mark for correct primary current of \(0.30\text{ A}\) (or \(0.3\text{ A}\)).

(a) Electrostatic spray painting process:
- The nozzle of the spray gun is connected to a high voltage source, giving all the exiting paint droplets the same type of charge (e.g., positive).
- Since they all have the same charge, the droplets repel each other, creating a very fine, uniform mist of paint.
- The metal bicycle frame is given an opposite charge (e.g., negative) or is earthed.
- The charged paint droplets are electrostatically attracted to the oppositely charged frame.
- Advantages: The paint wraps around the frame, ensuring that hard-to-reach areas and the backs of tubes get painted. This results in an even finish and significantly reduces paint waste.

(b) (i) Static charge builds up due to friction between the fuel rubbing against the inside of the delivery pipes during transfer (or friction between the tires of the tanker and the road/air resistance).
(ii) The build-up of static charge can create a potential difference, resulting in a spark that could ignite highly flammable fuel vapors, causing a fire or explosion. This is prevented by connecting an earthing wire (bonding strap) to the tanker before transferring fuel, allowing the charge to flow safely to the ground instead of building up.

- (a) [5 marks]: 1 mark for explaining paint droplets get the same charge; 1 mark for explaining that droplets repel each other to form an even mist; 1 mark for explaining bicycle frame is oppositely charged/earthed; 1 mark for explaining attraction of paint to frame; 1 mark for stating an advantage (less waste / even coating / wraps around objects).
- (b)(i) [2 marks]: 1 mark for friction/rubbing mentioned; 1 mark for transfer of electrons leading to charge build-up.
- (b)(ii) [3 marks]: 1 mark for identifying the risk of a spark causing fire/explosion; 1 mark for using an earthing wire/bonding strap; 1 mark for explaining this discharges the tanker safely to the ground.

(a) (i) Volume of the block \(V = 5.0\text{ cm} \times 8.0\text{ cm} \times 15.0\text{ cm} = 600\text{ cm}^3\).
Mass \(m = 1.35\text{ kg} = 1350\text{ g}\).
Density \(\rho = m / V = 1350\text{ g} / 600\text{ cm}^3 = 2.25\text{ g/cm}^3\).

(ii) To convert from \(\text{g/cm}^3\) to \(\text{kg/m}^3\), multiply by 1000:
\(2.25 \times 1000 = 2250\text{ kg/m}^3\).

(b) (i) \(p = h \times \rho \times g\) (or \(\Delta p = \Delta h \times \rho \times g\)).

(ii) Height difference \(\Delta h = 35\text{ cm} - 20\text{ cm} = 15\text{ cm} = 0.15\text{ m}\).
Pressure difference \(\Delta p = \Delta h \times \rho \times g\)
\(\Delta p = 0.15\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg} = 1500\text{ Pa}\) (or \(1500\text{ N/m}^2\)).

(iii) Pressure increases with depth, which means that the water pressure acting on the bottom surface of the block is greater than the water pressure acting on the top surface. Since \(\text{Force} = \text{Pressure} \times \text{Area}\), this pressure difference creates a larger upward force on the bottom surface than the downward force on the top surface, resulting in a net upward force called upthrust.

- (a)(i) [3 marks]: 1 mark for volume calculation (600 cm³); 1 mark for density formula \(\rho = m/V\); 1 mark for correct density of \(2.25\text{ g/cm}^3\).
- (a)(ii) [1 mark]: correct conversion to \(2250\text{ kg/m}^3\).
- (b)(i) [1 mark]: correct formula \(p = h \rho g\).
- (b)(ii) [3 marks]: 1 mark for identifying \(\Delta h = 0.15\text{ m}\); 1 mark for substitution of values; 1 mark for correct pressure difference of \(1500\text{ Pa}\) (or \(\text{N/m}^2\)).
- (b)(iii) [2 marks]: 1 mark for noting pressure is higher at the bottom than the top; 1 mark for linking pressure difference to a net upward force (force = pressure × area).

(a) Life cycle of a low-mass star (like our Sun):
- A nebula (cloud of hydrogen gas and dust) collapses under its own gravity.
- As it contracts, the gravitational potential energy turns into thermal energy, heating up to form a protostar.
- When the temperature and pressure are high enough, hydrogen fusion starts, making it a main sequence star.
- When the hydrogen in the core runs out, the core contracts, heating up, while the outer layers expand and cool to form a red giant.
- Eventually, the outer layers are ejected as a planetary nebula, leaving behind a hot, dense core called a white dwarf, which eventually cools into a black dwarf.

(b) Differences for a massive star:
- After the main sequence, a massive star expands into a red supergiant (instead of a red giant).
- When nuclear fusion in the core stops, it undergoes a sudden, violent collapse and explodes as a supernova (instead of slowly ejecting a planetary nebula).
- The remaining core collapses into either an incredibly dense neutron star or a black hole (where gravity is so strong that not even light can escape) instead of a white dwarf.

(c) A main sequence star is in a stable state where the inward gravitational force pulling the matter of the star together is perfectly balanced by the outward thermal/radiative pressure produced by nuclear fusion in its core.

- (a) [5 marks]: 1 mark for nebula collapsing under gravity to form a protostar; 1 mark for start of hydrogen fusion to enter main sequence; 1 mark for expansion into red giant when hydrogen runs out; 1 mark for ejection of outer layers as planetary nebula; 1 mark for leaving behind a white dwarf.
- (b) [3 marks]: 1 mark for expansion into red supergiant; 1 mark for supernova explosion; 1 mark for forming a neutron star or black hole.
- (c) [2 marks]: 1 mark for outward pressure from fusion; 1 mark for inward force of gravity balancing it.

(a) The principle of conservation of momentum states that the total momentum of a closed system remains constant, provided no external forces act. (b) Total initial momentum is zero because both the skater and the ball are stationary. Let the direction to the right be positive. Momentum of the ball after the throw: \(p_{\text{ball}} = 5.0 \text{ kg} \times 4.8 \text{ m/s} = 24 \text{ kg m/s}\). According to the conservation of momentum, total final momentum must also be zero: \(p_{\text{skater}} + p_{\text{ball}} = 0 \implies (60 \text{ kg} \times v) + 24 \text{ kg m/s} = 0\). Solving for \(v\): \(v = \frac{-24}{60} = -0.40 \text{ m/s}\). Therefore, the magnitude of the velocity is 0.40 m/s, and the direction is to the left. (c) Newton's third law states that when the skater exerts a force on the ball to the right, the ball simultaneously exerts an equal and opposite force on the skater to the left. Since the skater experiences an unbalanced force to the left, this force causes them to accelerate and move to the left according to \(F = ma\).

(a) 1 mark for stating total momentum before equals total momentum after; 1 mark for specifying this occurs in a closed system (or no external forces acting). (b) 1 mark for calculating the momentum of the ball as 24 kg m/s; 1 mark for equating the skater's momentum to the negative of the ball's momentum; 1 mark for calculating the correct magnitude of velocity as 0.40 m/s; 1 mark for stating the correct direction (to the left / opposite to the ball). (c) 1 mark for stating that Newton's third law involves equal and opposite forces; 1 mark for applying it directly to the skater and the ball; 1 mark for identifying that the force on the skater is directed to the left; 1 mark for explaining that this unbalanced force causes acceleration/movement of the skater using F = ma.

(a) As the magnet enters the coil, the magnetic field lines from the magnet cut across the copper wire of the coil. This creates a continuous change in magnetic flux linkage, which induces an electromotive force (voltage) across the ends of the coil. (b) To increase the maximum induced voltage, the student can: 1. Use a stronger magnet (which increases the magnetic field strength and flux density); 2. Increase the number of turns in the coil (increasing the total flux linkage); 3. Drop the magnet from a greater height (which increases the speed of the magnet as it passes through, causing a higher rate of change of magnetic flux). (c) According to Lenz's law, the induced voltage and current oppose the change in magnetic flux that creates them. As the magnet enters the coil, the magnetic flux linkage is increasing, creating an induced voltage with a positive polarity. As the magnet leaves the coil, the magnetic flux linkage is decreasing and the relative direction of motion is reversed, which reverses the direction of the induced voltage, resulting in a negative reading on the voltmeter.

(a) 1 mark for mentioning the magnetic field or magnetic flux of the magnet; 1 mark for stating that the field lines cut across the wire/turns of the coil; 1 mark for linking this cutting of flux to the induction of a voltage. (b) 1 mark for each valid modification (maximum 3 marks): use a stronger magnet, increase the number of turns in the coil, drop from a higher position/increase the speed of the magnet. (c) 1 mark for stating that the polarity of the induced voltage depends on the direction of change in magnetic flux; 1 mark for noting that the flux increases as the magnet enters but decreases as it exits; 1 mark for mentioning Lenz's law / that the induced voltage opposes the change; 1 mark for linking this reversal of change to the opposite voltmeter readings.

(a) The formula is pressure difference = height \(\times\) density \(\times\) gravitational field strength (\(p = h \rho g\)). (b) First, calculate the pressure difference due to the depth of the seawater: \(p = 25 \text{ m} \times 1030 \text{ kg/m}^3 \times 10 \text{ N/kg} = 257500 \text{ Pa}\). Convert this pressure difference to kilopascals (kPa): \(257500 \text{ Pa} = 257.5 \text{ kPa}\). The total pressure is the sum of atmospheric pressure and the liquid pressure: \(p_{\text{total}} = 101 \text{ kPa} + 257.5 \text{ kPa} = 358.5 \text{ kPa}\). (c) When the temperature of the gas increases, the average kinetic energy of the gas molecules increases, which means they move at higher average speeds. Since the volume of the tank is constant, these faster-moving molecules collide with the internal walls of the tank more frequently and with greater force (or change in momentum per collision). Since pressure is defined as the total force per unit area, this increased rate and intensity of collisions results in an increase in the pressure inside the tank.

(a) 1 mark for writing the correct formula: \(p = h \rho g\) (or equivalent symbols defined). (b) 1 mark for substituting correct values into the pressure difference equation (\(25 \times 1030 \times 10\)); 1 mark for calculating the correct pressure difference (\(257500 \text{ Pa}\) or \(257.5 \text{ kPa}\)); 1 mark for adding the atmospheric pressure (\(101 \text{ kPa}\)); 1 mark for the correct final answer of \(358.5 \text{ kPa}\) (accept \(359 \text{ kPa}\)). (c) 1 mark for stating that a temperature rise increases the kinetic energy/speed of the molecules; 1 mark for identifying that molecules collide with the container walls; 1 mark for stating that collisions occur more frequently (more collisions per second); 1 mark for stating that collisions occur with greater force / change in momentum; 1 mark for linking the increased average force on the walls to a higher pressure.

(a) The circuit diagram must include: a series circuit containing a cell or power supply, an ammeter, the thermistor, and a variable resistor (or variable voltage source). A voltmeter must be connected in parallel across the thermistor.

(b) Place the thermistor inside a waterproof container or boiling tube, then submerge it in a water bath filled with crushed ice. Record the temperature and the current/voltage readings. Slowly heat the water bath using a Bunsen burner (or electric heater), stirring regularly, and take readings of temperature, current, and voltage at regular intervals (e.g., every \(10\ ^\circ\text{C}\)) up to \(100\ ^\circ\text{C}\).

(c) Use the equation: \(R = \frac{V}{I}\). Convert current to amperes: \(4.5\text{ mA} = 4.5 \times 10^{-3}\text{ A}\) (or \(0.0045\text{ A}\)). Calculate resistance: \(R = \frac{3.6}{0.0045} = 800\ \Omega\).

(a)
- 1 mark: Voltmeter in parallel across the thermistor.
- 1 mark: Ammeter in series with the thermistor.
- 1 mark: Method of varying voltage/current (variable resistor or variable power supply in series) and cell/power source complete.

(b)
- 1 mark: Use ice to reach \(0\ ^\circ\text{C}\) and heat/boil to reach \(100\ ^\circ\text{C}\).
- 1 mark: Stir the water to ensure uniform temperature, and measure temperature using a thermometer.

(c)
- 1 mark: Recall formula \(V = IR\) or \(R = V/I\).
- 1 mark: Convert milliamperes to amperes (\(4.5\text{ mA} = 0.0045\text{ A}\)).
- 1 mark: Correct substitution: \(R = 3.6 / 0.0045\).
- 0.75 mark: Correct calculation: \(800\ \Omega\).

(a) \text{momentum} = \text{mass} \times \text{velocity}\) (or \(p = m \times v\)).

(b) Total initial momentum = \((0.25\text{ kg} \times 3.0\text{ m/s}) + (0.50\text{ kg} \times 0\text{ m/s}) = 0.75\text{ kg m/s}\). Taking initial direction as positive, final velocity of car is \(-0.60\text{ m/s}\). Total final momentum = \((0.25\text{ kg} \times -0.60\text{ m/s}) + (0.50\text{ kg} \times v_{\text{truck}}) = -0.15 + 0.50 v_{\text{truck}}\). By conservation of momentum: \(0.75 = -0.15 + 0.50 v_{\text{truck}}\). Rearranging: \(0.90 = 0.50 v_{\text{truck}}\), so \(v_{\text{truck}} = 1.8\text{ m/s}\).

(c) According to Newton's third law, when the car exerts a force on the truck, the truck exerts an equal and opposite force on the car. These two forces are equal in magnitude, opposite in direction, and act on different bodies (car and truck) for the exact same duration of time.

(a)
- 1 mark: \text{momentum} = \text{mass} \times \text{velocity}

(b)
- 1 mark: Calculation of initial momentum: \(0.75\text{ kg m/s}\).
- 1 mark: Realisation that final velocity of car is negative (direction change): \(-0.15\text{ kg m/s}\).
- 1 mark: Conservation of momentum equation stated or used (Total Initial = Total Final).
- 1 mark: Rearranging equation correctly: \(0.50 v = 0.90\).
- 0.75 mark: Correct evaluation of final velocity of truck: \(1.8\text{ m/s}\).

(c)
- 1 mark: The car exerts a force on the truck, and the truck exerts a force on the car.
- 1 mark: These forces are equal in magnitude.
- 1 mark: These forces are opposite in direction (or act for the same time duration).

(a) \sin(c) = \frac{1}{n}\).

(b) Substitute \(n = 1.52\): \sin(c) = \frac{1}{1.52} \approx 0.6579. Taking the inverse sine: \(c = \sin^{-1}(0.6579) \approx 41.1^\circ\).

(c) (i) The angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)). Therefore, the light does not pass into the air but is entirely reflected back inside the glass block. The angle of reflection will be equal to the angle of incidence, i.e., \(45^\circ\).
(ii) The phenomenon is Total Internal Reflection (TIR). A practical application in telecommunications is in optical fibres (or fibre optic cables).

(a)
- 1 mark: \sin(c) = 1/n

(b)
- 1 mark: Correct substitution: \sin(c) = 1 / 1.52.
- 1 mark: \sin(c) = 0.658 (or 0.66).
- 0.75 mark: Critical angle \(c = 41.1^\circ\) (accept range 41.0 to 41.2).

(c)(i)
- 1 mark: Light is totally internally reflected.
- 1 mark: Because the angle of incidence is greater than the critical angle.
- 1 mark: Angle of reflection is equal to the angle of incidence (\(45^\circ\)).

(c)(ii)
- 1 mark: Name: Total Internal Reflection.
- 1 mark: Application: Optical fibres / broadband cables.

(a) After leaving the main sequence, a high-mass star expands and cools to become a red supergiant. In its core, elements heavier than helium are fused. Once fusion reactions in the core can no longer produce energy (when an iron core forms), the outward radiation pressure stops, and the core collapses rapidly. This collapse triggers a massive explosion called a supernova, blowing off the outer layers of the star. The remaining core collapses further to form either an extremely dense neutron star, or, if the initial mass was sufficiently large, a black hole from which not even light can escape.

(b) In a main sequence star, the gravitational forces pulling inward are balanced by the outward radiation and thermal pressure produced by hydrogen fusion in the core. As hydrogen in the core runs out, fusion decreases, causing the outward pressure to drop. Gravity causes the core to contract, heating it up. This extreme heat initiates fusion in a shell surrounding the core, producing high thermal pressure that forces the outer layers of the star to expand dramatically, forming a red supergiant.

(a)
- 1 mark: Star expands and cools to become a red supergiant.
- 1 mark: Core undergoes further fusion of heavier elements.
- 1 mark: Core collapses when fusion stops, resulting in a supernova explosion.
- 1 mark: Outer layers are ejected into space.
- 1 mark: Remnant core becomes a neutron star OR a black hole (must mention both possibilities depending on mass).

(b)
- 1 mark: In main sequence, gravity and outward thermal/radiation pressure are in equilibrium.
- 1 mark: As hydrogen fuel runs out, fusion decreases and gravity causes core contraction.
- 1 mark: Core heating triggers shell fusion, causing outward radiation pressure to increase dramatically in the outer layers.
- 0.75 mark: This net outward force causes the outer layers to expand into a red supergiant.

(a) \text{pressure difference} = \text{height} \times \text{density} \times g\) (or \(p = h \times \rho \times g\)).

(b) Liquid pressure = \(0.80\text{ m} \times 1200\text{ kg/m}^3 \times 10\text{ N/kg} = 9600\text{ Pa}\).

(c) Total pressure at the bottom of the container = \text{atmospheric pressure} + \text{liquid pressure} = 1.01 \times 10^5\text{ Pa} + 9600\text{ Pa} = 101000\text{ Pa} + 9600\text{ Pa} = 110600\text{ Pa}\). Use the formula: \(P = \frac{F}{A}\), which rearranges to \(F = P \times A\). Total force = \(110600\text{ Pa} \times 0.025\text{ m}^2 = 2765\text{ N}\).

(a)
- 1 mark: \text{pressure} = \text{height} \times \text{density} \times g

(b)
- 1 mark: Correct substitution: \(p = 0.80 \times 1200 \times 10\).
- 1 mark: Correct calculation: \(9600\).
- 0.75 mark: Correct unit: \(\text{Pa}\) (or \(\text{N/m}^2\)).

(c)
- 1 mark: State formula \(P = F / A\) or \(F = P \times A\).
- 1 mark: Identify total pressure as the sum of atmospheric and liquid pressure.
- 1 mark: Calculate total pressure: \(110600\text{ Pa}\).
- 1 mark: Correct substitution of total pressure and area: \(F = 110600 \times 0.025\).
- 1 mark: Correct calculation: \(2765\text{ N}\) (accept \(2.8 \times 10^3\text{ N}\)).

(a) As the magnet moves, its magnetic field lines cut through the turns of wire in the coil. This change in magnetic flux linkage induces an electromotive force (or voltage) across the ends of the coil.

(b) To increase the induced voltage, the student can: 1. Move the magnet faster (greater speed of movement). 2. Use a stronger bar magnet. 3. Add more turns of wire to the coil.

(c) (i) When stationary, there is no relative movement, so no magnetic field lines are being cut. The induced voltage is zero, and the reading on the meter is zero.
(ii) When pulled rapidly out, the direction of relative motion is reversed, which reverses the polarity of the induced voltage (deflection in the opposite direction). Since it is pulled rapidly, the field lines are cut at a faster rate, so the magnitude of the peak reading is larger than during the slow insertion.

(a)
- 1 mark: Magnetic field lines are cut by the coil of wire.
- 1 mark: Reference to relative motion between magnet and coil.
- 1 mark: Cutting of field lines induces an electromotive force / voltage.

(b)
- 1 mark: Increase speed of movement of the magnet.
- 1 mark: Use a stronger magnet.
- 1 mark: Increase the number of turns on the coil.

(c)(i)
- 1 mark: Reading returns to zero.
- 0.375 mark: Because there is no cutting of magnetic field lines / no relative motion.

(c)(ii)
- 1 mark: Reading deflects in the opposite direction.
- 0.375 mark: Deflection is larger than before (due to higher rate of cutting of field lines).

(a) (i) The paint drops become positively charged. Since like charges repel, the drops repel each other as they leave the nozzle, causing them to spread out into a very fine, even mist.
(ii) By charging the bicycle frame negatively, the positively charged paint drops are attracted to the frame (unlike charges attract). This ensures paint reaches all sides of the frame, including hard-to-reach recesses, and reduces the amount of paint wasted.

(b) Use the formula: \(Q = I \times t\). Convert time to seconds: \(t = 4.0\text{ minutes} = 4.0 \times 60 = 240\text{ s}\). Calculate charge: \(Q = 1.2 \times 10^{-5}\text{ A} \times 240\text{ s} = 2.88 \times 10^{-3}\text{ C}\) (or \(0.00288\text{ C}\)).

(a)(i)
- 1 mark: Paint drops have a positive charge.
- 1 mark: Like charges repel.
- 0.75 mark: Repulsion causes them to spread out into a fine mist.

(a)(ii)
- 1 mark: Frame is negatively charged so it attracts the positive paint drops.
- 1 mark: This ensures paint wraps around the frame / reaches hard-to-access areas.
- 1 mark: Reduces wasted paint / gives a more even coat.

(b)
- 1 mark: Recall formula \(Q = I \times t\).
- 1 mark: Convert time to seconds: \(4.0 \times 60 = 240\text{ s}\).
- 1 mark: Correct substitution and calculation: \(2.88 \times 10^{-3}\text{ C}\) (or \(2.9 \times 10^{-3}\text{ C}\)).

(a) In ice (solid), particles are in a regular, fixed lattice arrangement, held by strong intermolecular forces, and only vibrate about fixed positions. As temperature rises to \(0\ ^\circ\text{C}\), the kinetic energy of particles increases, causing them to vibrate faster and with greater amplitude. During melting at \(0\ ^\circ\text{C}\), thermal energy is absorbed to break/weaken these intermolecular bonds rather than increasing temperature (particle potential energy increases while kinetic energy remains constant). Once melted (liquid), particles are randomly arranged, closely packed but able to slide past and move relative to each other.

(b) Energy required is the sum of heating the ice (\(Q_1\)) and melting the ice (\(Q_2\)).
1. Heating ice: \(Q_1 = m \times c \times \Delta\theta = 0.50\text{ kg} \times 2100\text{ J/kg}\ ^\circ\text{C} \times (0 - (-10))\ ^\circ\text{C} = 0.50 \times 2100 \times 10 = 10500\text{ J}\).
2. Melting ice: \(Q_2 = m \times L = 0.50\text{ kg} \times 3.3 \times 10^5\text{ J/kg} = 165000\text{ J}\).
Total energy required = \(10500\text{ J} + 165000\text{ J} = 175500\text{ J}\) (or \(1.76 \times 10^5\text{ J}\)).

(a)
- 1 mark: In solid ice, particles are in a regular lattice / vibrate about fixed positions.
- 1 mark: As temperature increases, particles vibrate faster / kinetic energy increases.
- 1 mark: During melting, energy is used to break bonds / increase potential energy (temperature stays constant).
- 1 mark: In liquid water, particles are randomly arranged / can flow or move past each other.

(b)
- 1 mark: Use \(Q = m c \Delta\theta\) to calculate energy to heat ice: \(0.50 \times 2100 \times 10\).
- 1 mark: Correct calculation of \(Q_1 = 10500\text{ J}\).
- 1 mark: Use \(Q = mL\) to calculate energy to melt ice: \(0.50 \times (3.3 \times 10^5)\).
- 1 mark: Correct calculation of \(Q_2 = 165000\text{ J}\).
- 0.75 mark: Correct total energy: \(175500\text{ J}\) (or \(1.76 \times 10^5\text{ J}\)).