Physics 0625 | Paper 4 Mock 01 | Theory (Extended)

Marking Scheme

• (a) The energy transferred by a 1 kW device operating for 1 hour [1]
• (b) \(E = 350 \times 1000 \times 3600\) [1]; \(E = 1.26 \times 10^9\) J [1]
• (c) Cost = \(350 \times 0.12 = \$42.00\) [1]
• (d) \(t = 350 / 2.0 = 175\) hours [1]

Explanation

(a) The kilowatt-hour (kWh) is a unit of energy (not power). It is defined as the energy transferred when a device with a power rating of 1 kilowatt (1000 W) operates for 1 hour (3600 s). Despite containing "watt" and "hour" in its name, it is a unit of energy because power multiplied by time gives energy.

(b) To convert kWh to joules, recall that:
1 kWh = 1000 W \(\times\) 3600 s = 3,600,000 J = 3.6 \(\times\) 10⁶ J
Therefore:
\(E = 350 \times 3.6 \times 10^6 = 1.26 \times 10^9\) J
This is 1.26 gigajoules. The conversion requires expressing power in watts and time in seconds because \(1\) J = \(1\) W \(\times\) 1 s.

(c) Electricity bills charge per kWh, so the cost calculation is straightforward:
Cost = energy used \(\times\) price per unit = \(350 \times 0.12 = \$42.00\)

(d) From the definition of energy in kWh: \(E = P \times t\), so \(t = E / P\).
The oven's power must be in kW: 2.0 kW.
\(t = \frac{350 \text{ kWh}}{2.0 \text{ kW}} = 175\) hours
This is about 5.8 hours per day over 30 days, which is a reasonable upper limit for oven usage consuming the entire monthly allocation.

(a) At 20 degrees C, the thermistor has a resistance of 5000 ohms. The total resistance in the sensor circuit is approximately 5000 ohms (the relay coil resistance is comparatively small). Using Ohm's law:

\(I = \dfrac{V}{R} = \dfrac{9.0}{5000} = 0.0018\) A = 1.8 mA

(b) The relay coil requires a minimum current of 25 mA to close the contacts. At 20 degrees C, the current through the relay coil is only 1.8 mA, which is far below the 25 mA threshold. The coil does not generate enough magnetic force to attract the armature and close the contacts, so the alarm circuit remains open and the bell does not sound.

(c) At 60 degrees C, the thermistor resistance drops to 200 ohms. The current is now:

\(I = \dfrac{V}{R} = \dfrac{9.0}{200} = 0.045\) A = 45 mA

This exceeds the 25 mA threshold required to operate the relay.

(d) The 45 mA current flowing through the relay coil creates a magnetic field that magnetises the soft iron core inside the coil. The magnetised core attracts the soft iron armature, which pivots and closes the relay contacts in the alarm circuit. This completes the 240 V alarm circuit, allowing current to flow through the bell, which then sounds the alarm.

(e) There are two key advantages of using a relay:

• The alarm bell requires a 240 V mains supply, which would be dangerous if connected directly to the exposed thermistor wiring in the room. A fire or fault could expose someone to lethal mains voltage through the sensor wires.
• The relay provides electrical isolation between the low-voltage sensor circuit (9 V) and the high-voltage alarm circuit (240 V). The small, safe sensor circuit controls the powerful alarm circuit without any direct electrical connection between them.

(a) This phenomenon is explained in four steps using electromagnetic induction and Lenz's law:

1. When the switch is closed, the current in the solenoid rises rapidly from zero to its maximum value, creating a rapidly increasing magnetic field through and around the solenoid
2. The changing magnetic field passes through the aluminium ring, which is a conductor. By Faraday's law, this changing flux induces an e.m.f. and hence a current in the ring
3. By Lenz's law, the induced current must flow in a direction that opposes the change causing it. The change is an increasing upward magnetic field (assuming north pole at top), so the ring's induced current creates its own magnetic field that opposes the increase
4. The ring's magnetic field is directed opposite to the solenoid's field, making the ring behave like a magnet with the same pole facing the solenoid's top. Like poles repel, so the ring experiences an upward repulsive force that launches it off the solenoid

(b) Plastic is an electrical insulator. No current can be induced in an insulator because it has no free electrons to carry a current. Without an induced current, there is no magnetic field produced by the ring, so there is no magnetic force and the ring does not move.

(c) If the power supply voltage is increased, the current in the solenoid rises to a larger value and does so more rapidly. This produces a greater rate of change of magnetic field, which induces a larger current in the ring and a stronger repulsive force. The ring therefore jumps higher.

(d) When the switch is opened suddenly after being held closed for a long time:

• The current in the solenoid drops rapidly to zero, causing the magnetic field to decrease rapidly
• This decreasing field again induces a current in the ring (by Faraday's law), but now Lenz's law dictates that the ring's field opposes the decrease, so the ring is attracted downward towards the solenoid rather than repelled upward

Part (a): Mass and weight of the concrete column

Volume of the column:

\(V = 0.30 \times 0.30 \times 2.5 = 0.225\text{ m}^3\)

Mass:

\(m = \rho V = 2400 \times 0.225 = 540\text{ kg}\)

Weight:

\(W = mg = 540 \times 9.8 = 5292\text{ N}\)

Part (b): Pressure on the ground

The base area of the square column:

\(A = 0.30 \times 0.30 = 0.090\text{ m}^2\)

\(p = \frac{F}{A} = \frac{5292}{0.090} = 58\,800\text{ Pa} \approx 5.9 \times 10^4\text{ Pa}\)

Part (c): Total pressure at the bottom of the pool

Pressure due to the water column:

\(p_{\text{water}} = \rho g h = 1000 \times 9.8 \times 3.0 = 29\,400\text{ Pa}\)

Total pressure includes atmospheric pressure pressing on the water surface:

\(p_{\text{total}} = 1.0 \times 10^5 + 29\,400 = 129\,400\text{ Pa} \approx 1.29 \times 10^5\text{ Pa}\)

Part (d): Force from the large piston

First find the pressure in the oil. Convert the small piston area: 2.0 cm\(^2\) = 2.0 \(\times\) 10\(^{-4}\) m\(^2\).

\(p = \frac{F}{A} = \frac{50}{2.0 \times 10^{-4}} = 2.5 \times 10^5\text{ Pa}\)

This pressure is transmitted through the oil to the large piston (area = 80 cm\(^2\) = 80 \(\times\) 10\(^{-4}\) m\(^2\)):

\(F = p \times A = 2.5 \times 10^5 \times 80 \times 10^{-4} = 2000\text{ N}\)

The force multiplication ratio is \(\frac{80}{2.0} = 40\), turning 50 N into 2000 N.

Part (e): Why the column exerts higher pressure despite the pool covering more area

Pressure is force per unit area, not total force. The concrete column concentrates a large weight (5292 N) on a small base (0.09 m\(^2\)), producing about 59 000 Pa. The swimming pool water at only 3.0 m deep produces just 29 400 Pa from the water alone, because the liquid pressure formula \(p = \rho g h\) depends only on depth, density and \(g\), not on the width or total area of the pool. A wider pool has more total weight but also more area, so the pressure per unit area stays the same.

Marking Scheme

• (a) [2 marks] Matt black is a good absorber of radiation / absorbs the most solar radiation [1]; this allows the base plate to absorb the maximum possible amount of energy from sunlight [1]
• (b) [2 marks] Glass is transparent to the short-wavelength visible light from the Sun, which passes through to the base [1]; the warm base emits longer-wavelength infrared radiation, which is absorbed by the glass and cannot escape, trapping thermal energy inside [1]
• (c) [1 mark] The reflective panel directs additional sunlight through the glass cover onto the base, increasing the total solar energy entering the cooker [1]
• (d) [1 mark] To reduce thermal energy loss by conduction through the walls to the cooler surroundings [1]
• (e) [1 mark] On an overcast day, clouds block and scatter much of the direct sunlight, so much less solar energy reaches the cooker and the temperature inside does not rise high enough to cook food [1]
• (f) [1 mark] It uses no fuel, so there is no cost for energy and no smoke or harmful emissions [1]

Explanation

(a) The matt black surface has the highest absorptivity of any surface type. When solar radiation (a mixture of visible light and infrared) strikes the base plate, the matt black finish absorbs nearly all of it, converting the radiant energy into thermal energy that heats the plate. A shiny or light-coloured surface would reflect a large fraction of the incoming radiation, reducing the energy available for cooking. The matt finish is important because a glossy black surface, while still dark, reflects more light at certain angles than a matt surface does.

(b) This is the greenhouse effect applied to cooking. The Sun emits radiation with a peak in the visible spectrum (short wavelengths, around 500 nm). Glass is transparent to these wavelengths, so the sunlight passes through the glass cover and reaches the black base plate. The base absorbs this energy and heats up to 130 °C. At this temperature, the base re-emits radiation, but because it is much cooler than the Sun, the emitted radiation has much longer wavelengths (peaking in the infrared, around 8000-10000 nm). Glass is largely opaque to these longer wavelengths, so it absorbs the outgoing infrared radiation instead of transmitting it. The energy is effectively trapped inside the box, raising the temperature well above what would be achieved without the glass.

(c) The reflective aluminium panel acts as a mirror, redirecting sunlight that would otherwise miss the cooker. By angling the panel, the user can bounce additional solar radiation through the glass cover and onto the base plate. This effectively increases the amount of solar energy entering the cooker beyond what the glass opening alone would collect. The more reflective the panel, the more additional energy is directed into the cooker.

(d) The insulated walls reduce the rate at which thermal energy escapes from the hot interior (130 °C) to the cooler surroundings (32 °C). The temperature difference of 98 °C would drive rapid conduction through uninsulated walls. The insulation (which contains trapped air pockets) has a very low thermal conductivity, creating a high thermal resistance. This keeps the interior hot enough to cook food by ensuring that most of the solar energy absorbed by the base stays inside the box.

(e) Solar cookers rely on direct sunlight to function. On an overcast day, the cloud cover scatters and absorbs a large proportion of the incoming solar radiation before it reaches the ground. The diffuse light that does reach the cooker is much less intense than direct sunlight and arrives from many directions, so it cannot be effectively focused by the reflective panel. The reduced energy input means the temperature inside the cooker does not rise high enough to cook food.

(f) A solar cooker uses free solar energy and produces no smoke, soot, or harmful emissions during operation. A wood-burning stove requires fuel (wood or charcoal), which costs money, contributes to deforestation, and produces smoke and particulate matter that cause respiratory health problems, especially when used indoors. The solar cooker also produces no carbon dioxide from combustion, making it more environmentally friendly.

Marking Scheme

• (a) e.m.f. = W / Q (or: e.m.f. = work done per unit charge) [1]
• (b) e.m.f. = 540 / 100 [1]; e.m.f. = 5.4 V [1]
• (c) \(I = V/R = 5.4 / 5.4\) [1]; \(I = 1.0\) A [1]
• (d) volt (V) [1]

Explanation

(a) Electromotive force (e.m.f.) is defined as the work done by a source of electrical energy per unit charge that passes through it:
\(\text{e.m.f.} = \frac{W}{Q}\)
where \(W\) is the work done (energy transferred) in joules and \(Q\) is the charge in coulombs. Although it has "force" in its name, e.m.f. is actually measured in volts and represents an energy-per-charge quantity, not a force.

(b) Substituting the given values:
\(\text{e.m.f.} = \frac{540 \text{ J}}{100 \text{ C}} = 5.4\) V
This means the cell does 5.4 J of work on every coulomb of charge that passes through it.

(c) Since the cell has negligible internal resistance, the terminal voltage equals the e.m.f. (5.4 V). All this voltage appears across the external resistor. Applying Ohm's law:
\(I = \frac{V}{R} = \frac{5.4}{5.4} = 1.0\) A
The numerical coincidence that \(V = R\) gives \(I = 1.0\) A is simply because the resistance was chosen to equal the e.m.f. value.

(d) The unit of e.m.f. is the volt (V), the same as for potential difference. One volt equals one joule per coulomb (1 V = 1 J/C).

(a) Recoil speed of the cart in trial 1 [2]

Before release, both the cart and block are stationary, so total momentum = 0. By conservation of momentum, total momentum after release must also be zero:

\(0 = m_{\text{block}} v_{\text{block}} + m_{\text{cart}} v_{\text{cart}}\)

\(0 = (0.20 \times 2.0) + (0.80 \times v_{\text{cart}})\)

\(0 = 0.40 + 0.80 v_{\text{cart}}\)

\(v_{\text{cart}} = \dfrac{-0.40}{0.80} = -0.50\) m/s

The negative sign means the cart moves to the left (opposite to the block). The cart recoils at 0.50 m/s. The cart is four times heavier than the block, so it moves at one-quarter of the block's speed.

(b) Showing consistency with 0.50 J of spring energy [2]

The total kinetic energy after release should equal the elastic potential energy stored in the spring:

\(KE_{\text{total}} = \frac{1}{2} \times 0.20 \times 2.0^2 + \frac{1}{2} \times 0.80 \times 0.50^2\)

\(= 0.40 + 0.10 = 0.50\) J

This equals the spring's stored energy of 0.50 J, confirming consistency. All the elastic potential energy has been converted to kinetic energy (no losses on the frictionless surface).

(c) Trial 2: block mass doubled to 0.40 kg [4]

The spring compression is the same, so total KE = 0.50 J. Two equations are needed (momentum conservation and energy conservation).

Momentum conservation (total momentum = 0):

\(0 = 0.40 v_b + 0.80 v_c\)

\(v_c = -0.50 v_b\)

Energy conservation:

\(\frac{1}{2} \times 0.40 \times v_b^2 + \frac{1}{2} \times 0.80 \times v_c^2 = 0.50\)

Substituting \(v_c = -0.50 v_b\):

\(0.20 v_b^2 + 0.40 \times (0.50 v_b)^2 = 0.50\)

\(0.20 v_b^2 + 0.40 \times 0.25 v_b^2 = 0.50\)

\(0.20 v_b^2 + 0.10 v_b^2 = 0.50\)

\(0.30 v_b^2 = 0.50\)

\(v_b^2 = 1.667\)

\(v_b = 1.29\) m/s (accept 1.3 m/s) - block speed to the right

\(v_c = -0.50 \times 1.29 = -0.645\) m/s, so cart recoils at 0.65 m/s to the left

Doubling the block mass reduces the block's launch speed (from 2.0 to 1.3 m/s) but increases the cart's recoil speed (from 0.50 to 0.65 m/s). The heavier block takes more of the spring's energy as its own KE, leaving less for the cart, but the momentum constraint forces the cart to move faster to compensate for the heavier block's momentum.

Marking Scheme

• (a) [1 mark]: The image is inverted.
• (b) [1 mark]: Light travels in straight lines through the pinhole, so rays from the top of the candle travel downward through the pinhole to the bottom of the screen, and rays from the bottom travel upward to the top of the screen.
• (c) [2 marks]: \(\dfrac{\text{image height}}{20} = \dfrac{15}{60}\) [1]. Image height = \(20 \times \dfrac{15}{60} = 5.0\) cm [1].
• (d) [2 marks]: The image becomes brighter because more light enters through the larger hole [1]. The image also becomes blurred (less sharp) because light from each point on the object can pass through a wider opening, spreading over a larger area on the screen [1].
• (e) [1 mark]: Any one valid similarity, e.g.: both form inverted images; both have a small aperture (pupil/pinhole); both detect light on a surface at the back (retina/screen).

Explanation

A pinhole camera works because light travels in straight lines. Each point on the object sends out light in all directions, but the tiny pinhole selects only a narrow cone of rays. A ray from the top of the candle passes through the pinhole and continues in a straight line to the bottom of the screen; a ray from the bottom of the candle ends up at the top. This crossing of rays produces an inverted image.

The image size is found using similar triangles. The triangle formed by the object and the pinhole is similar to the triangle formed by the image and the pinhole:

\(\dfrac{\text{image height}}{\text{object height}} = \dfrac{\text{image distance (box length)}}{\text{object distance}}\)

\(\dfrac{h_i}{20} = \dfrac{15}{60} = 0.25\)

\(h_i = 20 \times 0.25 = 5.0\) cm

Making the pinhole larger allows more light through (brighter image) but destroys sharpness. With a larger hole, multiple rays from the same point on the object can enter at slightly different angles, each hitting a different spot on the screen. The point becomes a disc of confusion, and the image blurs. The pinhole camera trades brightness for sharpness.

The human eye works on the same principle: light enters through a small aperture (the pupil) and forms an inverted image on the retina at the back. The eye also has a lens to focus the light more precisely, which a basic pinhole camera lacks.

Marking Scheme and Explanation:

(a) Efficiency of each power station [3]

Efficiency is calculated using: \( \text{efficiency} = \frac{\text{useful output}}{\text{total input}} \)

• Coal-fired: \( \frac{320}{800} = 0.40 \text{ (40\%)} \) [1]
• Gas turbine: \( \frac{300}{600} = 0.50 \text{ (50\%)} \) [1]
• Nuclear: \( \frac{525}{1500} = 0.35 \text{ (35\%)} \) [1]

Each station wastes a significant fraction of the input energy, mostly as thermal energy in cooling water and exhaust gases.

(b) Most efficient power station [1]

The gas turbine station is the most efficient at 50%. Modern gas turbine (combined-cycle) plants can exceed 50% because they recover energy from hot exhaust gases using a secondary steam cycle.

(c) Power wasted by the coal-fired station [1]

\( P_{\text{wasted}} = P_{\text{input}} - P_{\text{output}} = 800 - 320 = 480 \text{ MW} \)

This is 60% of the input power.

(d) Two ways energy is wasted in a coal-fired power station [2]

1. Thermal energy (heat) is lost in the exhaust gases that go up the chimney. These gases carry away energy that was not transferred to the steam. [1]
2. Thermal energy is lost in the cooling water discharged from the condenser. The steam must be cooled back to water after passing through the turbine, and this rejected heat is carried away by cooling water to a river, lake, or cooling tower. [1]

Part (a) [1 mark]

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume [1].

Mathematically: \(pV = \text{constant}\), or equivalently \(p \propto \frac{1}{V}\). Doubling the volume halves the pressure, because the same number of molecules now has twice the space and strikes the walls half as often per unit area.

Part (b) [2 marks]

• W is the Bourdon gauge (pressure gauge) [1]
• X is the plunger (piston) of the gas syringe [1]

The Bourdon gauge contains a curved metal tube that straightens under increased pressure, moving a pointer across a scale. The plunger slides within the syringe barrel to change the volume of the trapped gas.

Part (c) [2 marks]

Close the three-way tap to the syringe so that only the flask is connected to the Bourdon gauge [1]. Heat the water bath slowly, stir to ensure uniform temperature, and record the pressure reading at regular temperature intervals after allowing each reading to stabilise [1].

Stirring ensures all the water (and therefore the flask) reaches a uniform temperature. Waiting for the reading to stabilise ensures the gas inside the flask has reached thermal equilibrium with the water bath.

Part (d) [4 marks]

(i) Boyle's law calculation:

\(p_1 V_1 = p_2 V_2\) [1]

\(V_2 = \frac{p_1 V_1}{p_2} = \frac{100 \times 80}{200} = 40\) cm³ [1]

To double the pressure, the volume must be halved (from 80 cm³ to 40 cm³).

(ii) Pressure law calculation:

Convert to kelvin: \(T_1 = 20 + 273 = 293\) K, \(T_2 = 80 + 273 = 353\) K [1]

\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)

\(p_2 = \frac{p_1 \times T_2}{T_1} = \frac{100 \times 353}{293} = 120.5\) kPa (accept 120 kPa) [1]

Heating the gas by 60 °C increases the absolute temperature by about 20%, so the pressure also rises by about 20%. The pressure law requires absolute temperatures because pressure is proportional to the average kinetic energy of the particles, which is measured from absolute zero.

Part (e) [2 marks]

Compressing the gas may cause a slight temperature rise even if done slowly, affecting the Boyle's law results [1]. The gas may not behave perfectly as an ideal gas, there may be small gas leaks at the joints, or the gauge may have a zero error [1].

Part (f) [2 marks]

Any two valid improvements, for example: use a data logger for more precise and frequent readings [1]; repeat each measurement three times and calculate an average to reduce random errors [1].

Other acceptable improvements include insulating the syringe to maintain constant temperature, using a more precise digital pressure sensor, or allowing longer equilibration time between readings.

(a) Definition of half-life [2]

The half-life of a radioactive isotope is the time taken for the activity of the sample to decrease to half its original value. [1] Equivalently, it is the time taken for half the undecayed nuclei in a sample to decay. [1]

Both definitions are equivalent: as half the nuclei decay, the number of decays per second (activity) also halves.

(b) Why a gamma-emitting isotope is chosen [2]

Gamma rays are highly penetrating and can pass through body tissue to be detected by the gamma camera positioned outside the patient's neck. [1] Alpha particles would be absorbed within millimetres of tissue and would never reach the external detector. Additionally, alpha particles are highly ionising and would cause excessive damage to the surrounding cells. [1]

(c) Activity after 15 days [3]

The effective half-life inside the body is 5.0 days (shorter than the physical half-life of 8.0 days because the body also excretes the iodine).

Number of half-lives in 15 days:

\(n = \frac{15}{5.0} = 3\) [1]

After 3 half-lives, the activity is:

\(A = 4.0 \times \left(\frac{1}{2}\right)^3 = \frac{4.0}{8}\) [1]

\(A = \mathbf{0.50 \text{ MBq}}\) [1]

Step by step: 4.0 \(\rightarrow\) 2.0 \(\rightarrow\) 1.0 \(\rightarrow\) 0.50 MBq.