2025 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

(a) Mitochondria (or mitochondrion) are the cellular organelles responsible for aerobic respiration.
(b) A root hair cell possesses a long, thin extension / projection which significantly increases the surface area to volume ratio, allowing rapid uptake of water by osmosis.

(a) mitochondrion / mitochondria; [1]
(b) elongated extension / projection / hair-like structure; [1]
large(r) surface area (for uptake / osmosis); [0.5]

(a) Osmosis is the diffusion of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane.
(b) The concentrated sucrose solution has a lower water potential than the plant cell sap. Therefore, water leaves the cells by osmosis down the water potential gradient, causing a loss of mass.

(a) osmosis; [1]
(b) water leaves / moves out of the cells / tissue (by osmosis); [1]
(because) water potential inside the cell is higher than outside / down a water potential gradient; [0.5]

(a) Arteries transport oxygenated or deoxygenated blood under high pressure away from the ventricles of the heart.
(b) Capillary walls consist of a single layer of flattened endothelial cells (one cell thick) and are narrow, creating a short diffusion pathway for nutrients and gases to move between blood plasma and tissue fluid.

(a) artery / arteries; [1]
(b) wall is one cell thick / very thin; [1]
short diffusion distance / pathway (for dissolved gases / nutrients); [0.5]

(a) Phloem is the vascular tissue responsible for translocation of sucrose and amino acids.
(b) As temperature rises, water particles gain kinetic energy, which accelerates the rate of evaporation from mesophyll cell walls into air spaces and speeds up diffusion out through the stomata, thus increasing the transpiration rate.

(a) phloem; [1]
(b) (rate of transpiration) increases; [1]
(water molecules have) more kinetic energy / faster evaporation / faster diffusion; [0.5]

(a) Discontinuous variation produces discrete phenotypic categories (e.g. ABO blood groups) with no intermediates, unaffected by environmental conditions.
(b) In the presence of an antibiotic, a selective pressure is applied. Mutant bacteria possessing the resistance allele survive (differential survival), while susceptible bacteria are killed. The resistant survivors reproduce, passing on the allele for resistance to subsequent generations.

(a) distinct / discrete categories OR no intermediates / overlapping ranges OR controlled by single gene / few genes OR not influenced by environment; [1]
(b) resistant bacteria survive (and non-resistant die); [1]
survivors reproduce and pass on the (resistance) allele / gene to offspring; [0.5]

(a) The cell wall is a rigid outer structure composed of cellulose that resists internal turgor pressure, preventing lysis in hypotonic solutions.
(b) Water moves from an area of higher water potential (soil solution) to an area of lower water potential (inside the root hair cell cytoplasm and vacuole) by osmosis across the partially permeable cell surface membrane.

(a) cell wall [1];
(b) by osmosis [0.5]; down a water potential gradient / from higher to lower water potential [0.5]; across / through a partially permeable membrane [0.5].

(a) The pulmonary vein delivers oxygenated blood returning from the pulmonary circulation into the left atrium.
(b) The left ventricle has a thicker muscular wall (myocardium) to create greater contractile force, generating higher hydrostatic pressure necessary to propel blood through systemic circulation to all body organs, whereas the right ventricle only pumps blood against lower resistance to the adjacent lungs.

(a) pulmonary vein [1];
(b) (left ventricle) generates higher pressure / greater force [0.5]; to pump blood around the whole body / further distance / systemic circulation [0.5]; (whereas) right ventricle only pumps blood to the lungs [0.5].

(a) A mutation is an unpredictable alteration in the genetic sequence of DNA, forming novel alleles.
(b) Genetic variation within a population means some individuals possess traits that offer a selective advantage. These individuals have higher survival rates (survival of the fittest) and reproduce more successfully, transmitting their advantageous alleles to their offspring, thereby increasing the frequency of the adaptive trait over generations.

(a) mutation [1];
(b) individuals with advantageous alleles / traits have a survival advantage / are better adapted [0.5]; (these individuals) reproduce [0.5]; passing on their (advantageous) alleles to their offspring [0.5].

(a) (i) Use the formula: \(\text{Magnification} = \frac{\text{size of image}}{\text{actual size of specimen}}\)
\(\text{Magnification} = \frac{48\text{ mm}}{0.06\text{ mm}} = 800\) or \(\times 800\).

(ii) The organelle responsible for absorbing light for photosynthesis is the chloroplast, which contains the green pigment chlorophyll.

(b) Plant cells have a cellulose cell wall and a permanent central vacuole, which are absent in animal cells.

(a)(i) correct formula or substitution: \(\frac{48}{0.06}\) ;
800 / \(\times 800\) ; [2]

(a)(ii) chloroplast ;
chlorophyll ; [2]

(b) (cellulose) cell wall / permanent vacuole / large central vacuole ; [1]
[Reject: 'cell membrane', 'chloroplast' (as excluded by prompt)]

(a) As sucrose concentration increases, the percentage change in mass decreases (or changes from positive to negative).

(b) The concentration is \(0.4\text{ mol/dm}^3\) because at this point the percentage change in mass is \(0.0\%\), showing that the water potential inside the potato cells is equal to the external solution (no net osmosis).

(c) The \(0.8\text{ mol/dm}^3\) sucrose solution has a lower water potential than the potato cells (it is hypertonic). Water moves out of the cells by osmosis from an area of higher water potential to an area of lower water potential across a partially permeable membrane, resulting in a loss of mass.

(a) as sucrose concentration increases, (percentage) change in mass decreases / ORA ; [1]

(b) 0.4 (\(\text{mol/dm}^3\)) ;
(because) there is no change in mass / net movement of water is zero ; [2]

(c) water potential of solution is lower than inside the cell / potato cells have higher water potential ;
water moves out (of cells) by osmosis / down a water potential gradient ; [2]

(a) Increase in heart rate = \(165 - 68 = 97\text{ beats per min}\).
\(\text{Percentage increase} = \frac{97}{68} \times 100 = 142.647...\%\).
To two significant figures, this is \(140\%\).

(b) During exercise, muscle contraction requires more energy released via aerobic respiration. To meet this demand, the heart beats faster to transport more oxygen and glucose to active muscle tissues and to remove carbon dioxide/lactic acid more rapidly.

(c) The left ventricle has thick muscular walls to pump oxygenated blood at high pressure into the aorta.

(a) correct calculation of difference: \(165 - 68 = 97\) OR \(\frac{165-68}{68} \times 100\) ;
140 (\(\%\)) [allow 143(%) if 3 s.f. not penalised, but 140 for 2 s.f.] ; [2]

(b) (muscles require) more energy / increased aerobic respiration ;
to deliver more oxygen / more glucose (to contracting muscles) / remove carbon dioxide faster ; [2]

(c) left ventricle ; [1]

(a) Shell banding falls into distinct categories without intermediates, which is an example of discontinuous variation.

(b) The percentage of banded snails is much higher in the woodland (82%) than in the grassy meadow (24%), a difference of 58%.

(c) In the woodland habitat, banded snails are better camouflaged against the dark leaf litter and shadows. Consequently, they suffer less predation by birds/predators. Snails with banded shells are more likely to survive and reproduce (survival of the fittest), passing on the alleles for banded shells to their offspring over generations.

(a) discontinuous (variation) ; [1]

(b) higher in woodland (82%) than meadow (24%) / 58(%) higher in woodland / correct comparative data quote ; [1]

(c) banded snails are better camouflaged / blend in with dark background / leaf litter ;
banded snails are less likely to be eaten by predators / have higher survival rate ;
surviving snails reproduce and pass on their alleles / genes (for banding) to offspring ; [3]

(a) Balance carbon atoms first: there are 2 carbons in ethanol, so 2 molecules of \(\text{CO}_2\) are formed. Balance hydrogen atoms next: there are 6 hydrogens in ethanol (5+1), so 3 molecules of \(\text{H}_2\text{O}\) are formed. Finally, count the oxygen atoms on the right-hand side: \((2 \times 2) + (3 \times 1) = 7\) oxygen atoms. Ethanol provides 1 oxygen atom, leaving 6 oxygen atoms needed from \(\text{O}_2\), which requires \(3\text{O}_2\).
Balanced equation: $$\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$$

(b) The displayed formula must show two carbon atoms bonded by a single covalent bond (\(\text{C}-\text{C}\)), with three hydrogen atoms bonded to the first carbon (\(\text{C}-\text{H}\)), two hydrogen atoms bonded to the second carbon (\(\text{C}-\text{H}\)), and an oxygen atom bonded to the second carbon (\(\text{C}-\text{O}\)) which is in turn bonded to a hydrogen atom (\(\text{O}-\text{H}\)).

(a) 3 (before \(\text{O}_2\)) [0.5]; 2 (before \(\text{CO}_2\)) and 3 (before \(\text{H}_2\text{O}\)) [1.0];
(b) Correct displayed formula showing all 8 single covalent bonds (\(\text{C}-\text{C}\), five \(\text{C}-\text{H}\), \(\text{C}-\text{O}\), and \(\text{O}-\text{H}\)) explicitly drawn with no grouped atoms (e.g. not \(-\text{OH}\) without the \(\text{O}-\text{H}\) bond) [1.0].

(a)(i) The reaction is an addition reaction (or hydration).

(ii) Step 1: Calculate the theoretical maximum moles of ethene:
\(\text{moles of }\text{C}_2\text{H}_4 = \frac{56.0\times 10^3\text{ g}}{28\text{ g/mol}} = 2000\text{ mol}\)

Step 2: Molar ratio \(\text{C}_2\text{H}_4 : \text{C}_2\text{H}_5\text{OH} = 1 : 1\), so theoretical moles of ethanol \(= 2000\text{ mol}\).

Step 3: Calculate theoretical mass of ethanol:
\(\text{mass} = 2000\text{ mol} \times 46\text{ g/mol} = 92000\text{ g} = 92.0\text{ kg}\)

Step 4: Calculate percentage yield:
\(\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100\% = \frac{73.6\text{ kg}}{92.0\text{ kg}} \times 100\% = 80.0\%\)

(b) The energy of the products is lower than the reactants because more energy is released when making bonds than is absorbed when breaking bonds.

(a)(i) addition / hydration [1];
(ii) moles of ethene \(= 2000\text{ mol}\) OR theoretical mass of ethanol \(= 92.0\text{ kg}\) [1];
\(\frac{73.6}{92.0} \times 100\) [1];
\(80.0\%\) (accept \(80\%\)) [0.5];
(b) lower AND energy released forming bonds > energy taken in breaking bonds [1].

(a) Relative atomic mass calculation:
\(A_r = \frac{(28 \times 92.2) + (29 \times 4.7) + (30 \times 3.1)}{100}\)
\(A_r = \frac{2581.6 + 136.3 + 93.0}{100} = \frac{2810.9}{100} = 28.109\)
Rounded to two decimal places: \(28.11\).

(b) Step 1: Calculate \(M_r\) of \(\text{SiO}_2\):
\(M_r(\text{SiO}_2) = 28.0 + (2 \times 16.0) = 60.0\text{ g/mol}\)

Step 2: Calculate moles of \(\text{SiO}_2\):
\(\text{moles of }\text{SiO}_2 = \frac{15.0\text{ g}}{60.0\text{ g/mol}} = 0.250\text{ mol}\)

Step 3: Molar ratio \(\text{SiO}_2 : \text{C} = 1 : 2\):
\(\text{moles of }\text{C} = 0.250 \times 2 = 0.500\text{ mol}\)

Step 4: Calculate mass of carbon:
\(\text{mass of C} = 0.500\text{ mol} \times 12.0\text{ g/mol} = 6.0\text{ g}\)

(a) correct substitution into weighted average formula: \(\frac{(28 \times 92.2) + (29 \times 4.7) + (30 \times 3.1)}{100}\) [1];
\(28.11\) [1];
(b) \(M_r\) of \(\text{SiO}_2 = 60.0\) AND moles of \(\text{SiO}_2 = 0.250\text{ mol}\) [1];
moles of \(\text{C} = 0.500\text{ mol}\) [1];
\(6.0\text{ g}\) [0.5].

(a)(i) Step 1: Calculate moles of \(\text{Zn}\):
\(\text{moles of Zn} = \frac{0.130\text{ g}}{65.0\text{ g/mol}} = 0.00200\text{ mol}\)

Step 2: Molar ratio \(\text{Zn} : \text{H}_2 = 1 : 1\), so moles of \(\text{H}_2 = 0.00200\text{ mol}\).

Step 3: Calculate volume of \(\text{H}_2\):
\(\text{Volume in dm}^3 = 0.00200\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 0.0480\text{ dm}^3\)
\(\text{Volume in cm}^3 = 0.0480 \times 1000 = 48.0\text{ cm}^3\)

(ii) \(\text{Average rate} = \frac{\text{volume of gas}}{\text{time taken}} = \frac{48.0\text{ cm}^3}{40\text{ s}} = 1.2\text{ cm}^3\text{/s}\)

(b) The rate increases because zinc powder has a larger surface area (per unit mass), which increases the collision frequency between reactant particles.

(a)(i) moles of \(\text{Zn} = 0.00200\text{ mol}\) [1];
conversion to \(\text{cm}^3\) via multiplying by \(24.0\) and \(1000\) [1];
\(48.0\text{ cm}^3\) (or \(48\text{ cm}^3\)) [0.5];
(ii) \(1.2\text{ cm}^3\text{/s}\) (allow ecf from (a)(i)) [1];
(b) increases AND greater surface area / higher collision frequency [1].

(a) Step 1: Calculate the relative formula mass \(M_r\) of \(\text{Fe}_2\text{O}_3\):
\(M_r(\text{Fe}_2\text{O}_3) = (2 \times 56.0) + (3 \times 16.0) = 112.0 + 48.0 = 160.0\text{ g/mol}\)

Step 2: Calculate moles (or tonne-moles) of \(\text{Fe}_2\text{O}_3\):
\(\text{amount of }\text{Fe}_2\text{O}_3 = \frac{320\text{ tonnes}}{160.0} = 2.00\text{ million moles (or }2.00\text{ tonne-mol)}\)

Step 3: Molar ratio \(\text{Fe}_2\text{O}_3 : \text{Fe} = 1 : 2\):
\(\text{amount of Fe} = 2.00 \times 2 = 4.00\text{ tonne-mol}\)

Step 4: Calculate mass of \(\text{Fe}\):
\(\text{mass of Fe} = 4.00\text{ tonne-mol} \times 56.0 = 224\text{ tonnes}\)

(b)(i) Carbon monoxide / CO is oxidized because it gains oxygen (or oxidation state of C increases from +2 to +4).
(ii) Activation energy is the minimum energy required by reacting particles to start a reaction.

(a) \(M_r(\text{Fe}_2\text{O}_3) = 160.0\) [1];
correct mole ratio application (multiplying by 2) [1];
\(224\text{ (tonnes)}\) [0.5];
(b)(i) carbon monoxide / \(\text{CO}\) [1];
(ii) minimum energy needed by colliding particles to react / initiate a reaction [1].

(a) Aqueous bromine is reddish-brown/orange and turns colourless (is decolourised) in the presence of an alkene.

(b) An addition reaction occurs when the unsaturated \(\text{C}=\text{C}\) double bond opens up into a single bond, allowing bromine atoms to bond across the carbon atoms so that only one product (1,2-dibromoethane) is formed from the two reactants.

(a) (orange / red-brown / brown to) colourless / decolourised; [1] (Reject: clear)

(b) \(\text{C}=\text{C}\) / carbon-carbon double bond breaks / opens up; [1]
(Two molecules react to) form only one product / a single product / bromine atoms add across the double bond; [1]

(a) Both isotopes have atomic number 17, so each contains 17 protons and 17 electrons. Chlorine-35 has \(35 - 17 = 18\) neutrons, whereas chlorine-37 has \(37 - 17 = 20\) neutrons.

(b) Chemical properties depend on electron arrangements, specifically valence electrons. Since both have 7 valence electrons (configuration 2,8,7), their chemical reactions are identical.

(a) Similarity: same number of protons / same number of electrons / 17 protons / 17 electrons; [1]
Difference: different number of neutrons / Cl-35 has 18 neutrons and Cl-37 has 20 neutrons / Cl-37 has 2 more neutrons; [1]

(b) Same number of outer shell electrons / same electronic configuration / both have 7 outer electrons; [1]

When temperature increases:
1. Particles gain thermal energy which converts to kinetic energy, causing them to move faster.
2. The frequency of collisions between reactant particles increases.
3. A greater fraction of colliding particles have energy equal to or greater than the activation energy (\(E_a\)), resulting in a higher proportion of successful/effective collisions per unit time.

(Particles) have more kinetic energy / move faster; [1]
More frequent collisions / higher collision rate / collisions occur more often; [1] (Reject: 'more collisions' without reference to time/rate)
Greater proportion / fraction of particles have energy equal to or greater than activation energy / more collisions are successful / effective; [1]

(a) \(\text{Fe}_2\text{O}_3\) is reduced because reduction is the loss of oxygen, and \(\text{Fe}_2\text{O}_3\) loses oxygen to form \(\text{Fe}\).

(b)
1. Calculate relative formula mass of \(\text{Fe}_2\text{O}_3\):
\(M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 112 + 48 = 160\)
2. Relative mass of iron produced per mole of \(\text{Fe}_2\text{O}_3\):
\(2 \times A_r(\text{Fe}) = 2 \times 56 = 112\)
3. Mass of \(\text{Fe}\) produced:
\(\text{Mass} = \frac{112}{160} \times 320\text{ tonnes} = 224\text{ tonnes}\)

(a) \(\text{Fe}_2\text{O}_3\) / iron(III) oxide AND it loses oxygen; [1]

(b) \(M_r\) of \(\text{Fe}_2\text{O}_3 = 160\) OR moles of \(\text{Fe}_2\text{O}_3 = \frac{320}{160} = 2\text{ (mol/Mtonnes)}\); [1]
\(224\text{ (tonnes)}\); [1]

Step 1: Calculate the work done (change in gravitational potential energy) in lifting the load:
\[ W = \Delta E_p = mgh = 45\text{ kg} \times 9.8\text{ N/kg} \times 16\text{ m} = 7056\text{ J} \]

Step 2: Calculate the power output using \( P = \frac{W}{t} \):
\[ P = \frac{7056\text{ J}}{12\text{ s}} = 588\text{ W} \]

Step 3: State the correct unit of power:
Unit = \( \text{W} \) (watts) or \( \text{J/s} \).

• (Useful energy / work =) \( mgh = 45 \times 9.8 \times 16 \) OR \( 7056\text{ [J]} \) [1];
• (Power =) \( \frac{7056}{12} = 588 \) [1];
• \( \text{W} \) / \( \text{J/s} \) / \( \text{watts} \) [1];
(Note: If \( g = 10\text{ N/kg} \) is used, accept \( 600\text{ W} \) for max 2 marks if formula and unit are correct).

Step 1: Calculate the total thermal energy supplied by the heater:
\[ t = 3.0\text{ minutes} = 3.0 \times 60 = 180\text{ s} \]
\[ E = P \times t = 250\text{ W} \times 180\text{ s} = 45\,000\text{ J} \]

Step 2: Calculate specific heat capacity \( c \) using \( E = mc\Delta T \):
\[ c = \frac{E}{m \Delta T} = \frac{45\,000}{0.80 \times 52} = \frac{45\,000}{41.6} = 1081.7\text{ J/(kg }^\circ\text{C)} \approx 1080\text{ J/(kg }^\circ\text{C)} \text{ (or } 1100\text{ to 2 s.f.)} \]

Step 3: State the unit:
Unit = \( \text{J/(kg }^\circ\text{C)} \) or \( \text{J/(kg K)} \) or \( \text{J kg}^{-1\text{ }^\circ\text{C}^{-1} \).

• (Energy =) \( P \times t = 250 \times 180 \) OR \( 45\,000\text{ [J]} \) [1];
• \( c = \frac{45\,000}{0.80 \times 52} = 1080 \) / \( 1100 \) / \( 1082 \) [1];
• \( \text{J/(kg }^\circ\text{C)} \) / \( \text{J/(kg K)} \) / \( \text{J}\text{ kg}^{-1\text{ }^\circ\text{C}^{-1} \) [1];

Step 1: Use the principle of conservation of energy for an ideal transformer (\( P_{\text{in}} = P_{\text{out}} \)):
\[ P_{\text{in}} = 24\text{ W} \]

Step 2: Relate power, voltage, and current for the primary coil using \( P = V_p \times I_p \):
\[ I_p = \frac{P}{V_p} = \frac{24\text{ W}}{230\text{ V}} = 0.1043...\text{ A} \approx 0.104\text{ A} \text{ (or } 0.10\text{ A)} \]

Step 3: State the unit of electric current:
Unit = \( \text{A} \) (amperes) or \( \text{mA} \) (if given as \( 104\text{ mA} \)).

• (Formula/Method =) \( I_p = \frac{P}{V_p} \) OR \( I_s = \frac{24}{12} = 2.0\text{ [A]} \) AND \( I_p = \frac{12 \times 2.0}{230} \) [1];
• (Evaluation =) \( 0.104 \) OR \( 0.10 \) [1];
• \( \text{A} \) / \( \text{amperes} \) (or \( 104 \) with \( \text{mA} \)) [1];

Step 1: Calculate the number of half-lives that have elapsed:
\[ \text{Number of half-lives} = \frac{24\text{ hours}}{6.0\text{ hours}} = 4 \]

Step 2: Calculate the remaining activity after 4 half-lives:
\[ \text{Activity} = 960 \times \left(\frac{1}{2}\right)^4 = \frac{960}{16} = 60\text{ kBq} \]
(or \( 60\,000\text{ Bq} \))

Step 3: State the appropriate unit:
Unit = \( \text{kBq} \) (kilobecquerels) or \( \text{Bq} \) (becquerels) matching the value.

• (Determining half-lives =) \( \frac{24}{6.0} = 4 \) half-lives OR halving activity 4 times [1];
• (Calculation =) \( 60 \) (if unit given as \( \text{kBq} \)) OR \( 60\,000 \) (if unit given as \( \text{Bq} \)) [1];
• \( \text{kBq} \) / \( \text{Bq} \) / \( \text{becquerels} \) matching calculated value [1];

Step 1: Calculate the total distance travelled by the ultrasound pulse (to the seabed and back):
\[ \text{Total distance} = v \times t = 1500\text{ m/s} \times 0.36\text{ s} = 540\text{ m} \]

Step 2: Calculate the depth by dividing the total distance by 2:
\[ \text{Depth} = \frac{540\text{ m}}{2} = 270\text{ m} \]
(Alternatively: one-way time \( = \frac{0.36}{2} = 0.18\text{ s} \); \( \text{Depth} = 1500 \times 0.18 = 270\text{ m} \))

Step 3: State the unit:
Unit = \( \text{m} \) (metres).

• \( \text{distance} = 1500 \times 0.36 \) OR \( 540\text{ [m]} \) OR \( \text{time} = \frac{0.36}{2} = 0.18\text{ [s]} \) [1];
• \( \frac{540}{2} = 270 \) OR \( 1500 \times 0.18 = 270 \) [1];
• \( \text{m} \) / \( \text{metres} \) [1];

Step 1: Calculate the work done (gain in gravitational potential energy, \(\Delta E_p\)):
\[ \Delta E_p = mgh = 85\text{ kg} \times 9.8\text{ N/kg} \times 14\text{ m} = 11\,662\text{ J} \]

Step 2: Calculate the power output:
\[ P = \frac{\text{work done}}{\text{time taken}} = \frac{11\,662\text{ J}}{28\text{ s}} = 416.5\text{ W} \approx 417\text{ W} \]

Step 3: State the unit:
Watts (\(\text{W}\)) or joules per second (\(\text{J/s}\)).

1. `(W = ) mgh` OR `85 × 9.8 × 14` [1]
2. `(P = W / t = ) 11662 / 28 = 416.5` / `417` / `420` [1]
3. `W` / `watt(s)` / `J/s` [1]

(a) \( E_p = mgh = 0.50 \times 9.8 \times 0.80 = 3.92\text{ J} \).

(b) By conservation of mechanical energy, as the trolley descends, gravitational potential energy is converted entirely into kinetic energy (since friction is negligible). Therefore, the kinetic energy at the bottom is equal to the initial gravitational potential energy, \( 3.92\text{ J} \).

(a) \( (E_p =) mgh \) OR \( 0.50 \times 9.8 \times 0.80 \); [1]
3.92 (J) / 3.9 (J) (accept 4.0 J if \( g = 10\text{ N/kg} \) used); [1]

(b) gravitational potential (energy) transferred to kinetic (energy); [1]
3.92 (J) / ecf from (a); [1]

(a) The graph starts at \( 90^\circ\text{C} \) and slopes downwards until it reaches \( 55^\circ\text{C} \). At \( 55^\circ\text{C} \), there is a horizontal plateau where the temperature stays constant during freezing. After complete solidification, the curve slopes downwards again asymptotically towards room temperature (\( 20^\circ\text{C} \)).

(b) Temperature is proportional to the average kinetic energy of the particles. During freezing, thermal energy continues to be lost to the surroundings, but latent heat is released as attractive forces/bonds form between the particles. This energy release balances the heat loss, keeping the average kinetic energy (and thus temperature) constant.

(a) temperature decreases to \( 55^\circ\text{C} \) AND levels off / horizontal plateau at \( 55^\circ\text{C} \); [1]
temperature decreases from \( 55^\circ\text{C} \) towards \( 20^\circ\text{C} \) / room temperature; [1]

(b) bonds form / attractive forces between particles strengthen; [1]
energy is released (latent heat) which balances heat loss / average kinetic energy of particles does not change; [1]

(a) Beta decay involves the emission of an electron \( ^{0}_{-1}\beta \) (or \( ^{0}_{-1}\text{e} \)). Conservation of nucleon number gives: \( 234 = A + 0 \implies A = 234 \). Conservation of proton number gives: \( 90 = Z - 1 \implies Z = 91 \). Thus: \( ^{234}_{90}\text{Th} \rightarrow ^{234}_{91}\text{Pa} + ^{0}_{-1}\beta \).

(b) Number of half-lives \( n = \frac{72\text{ days}}{24\text{ days}} = 3 \).
After 1 half-life (24 days): \( 120 / 2 = 60\text{ mg} \).
After 2 half-lives (48 days): \( 60 / 2 = 30\text{ mg} \).
After 3 half-lives (72 days): \( 30 / 2 = 15\text{ mg} \).

(a) \( ^{234}_{91}\text{Pa} \) (both nucleon and proton number correct); [1]
\( ^{0}_{-1}\beta \) OR \( ^{0}_{-1}\text{e} \); [1]

(b) 3 half-lives / shows division by 2 three times / \( 120 \times (\frac{1}{2})^3 \); [1]
15 (mg); [1]

(a) When the coil is parallel to the magnetic field, the sides of the coil cut magnetic field lines at the maximum rate, producing maximum induced e.m.f. As the coil rotates to the perpendicular position (\( 90^\circ \)), the rate of cutting drops to zero (e.m.f. = 0). As it continues rotating (\( 180^\circ \)), the sides cut field lines in opposite directions, producing a maximum negative e.m.f., returning to zero at \( 270^\circ \) and returning to the initial positive maximum at \( 360^\circ \).

(b) According to Faraday's law of electromagnetic induction, induced e.m.f. is directly proportional to the rate of cutting of magnetic flux. To increase peak e.m.f.: rotate the coil faster, use stronger magnets, increase the number of turns of wire on the coil, or use a coil with larger cross-sectional area.

(a) e.m.f. starts at maximum / peak value (at \( 0^\circ \)); [1]
produces an alternating / sinusoidal waveform / reaches zero when coil is perpendicular to field (\( 90^\circ \) and \( 270^\circ \)) AND reverses polarity / direction every half turn; [1]

(b) any two from:
- increase the speed / frequency of rotation of the coil;
- use stronger magnets / increase magnetic field strength;
- increase the number of turns / loops on the coil;
- increase the area of the coil / use an iron core; [2]

(a) The distance between two consecutive crests is the wavelength \( \lambda = 0.080\text{ m} \).
Using the wave equation \( v = f \lambda \):
\( v = 25\text{ Hz} \times 0.080\text{ m} = 2.0\text{ m/s} \).

(b) Transverse waves have oscillations/vibrations perpendicular (\( 90^\circ \)) to the direction of wave propagation / energy transfer (e.g. water waves, electromagnetic waves). Longitudinal waves have oscillations/vibrations parallel to the direction of wave propagation / energy transfer (e.g. sound waves).

(a) \( (v =) f \lambda \) OR \( 25 \times 0.080 \); [1]
2.0 (accept 2) AND \( \text{m/s} \) / \( \text{m}\text{ s}^{-1} \); [1]

(b) transverse: vibrations / oscillations are perpendicular / at right angles to direction of travel / energy transfer; [1]
longitudinal: vibrations / oscillations are parallel to direction of travel / energy transfer; [1]

(a) The phenomenon where an e.m.f. is induced across a conductor when it cuts through magnetic field lines (or when magnetic flux linkage changes) is called electromagnetic induction.

(b) When the coil is parallel to the magnetic field lines at \(t = 0\), the rate of cutting magnetic field lines is at a maximum, so the induced e.m.f. begins at a peak value (\(+V_0\) or \(-V_0\)).
- At a quarter turn (\(t = T/4\)), the coil is perpendicular to the field lines, so the induced e.m.f. is \(0\).
- At half a turn (\(t = T/2\)), the coil is parallel again but inverted relative to the magnetic field, producing a peak in the opposite direction (\(-V_0\)).
- At three-quarter turn (\(t = 3T/4\)), the e.m.f. is \(0\).
- At one full rotation (\(t = T\)), the e.m.f. returns to the initial peak value (\(+V_0\)).
This forms a standard cosine waveform for one full period \(T\).

(c) According to Faraday's law of electromagnetic induction, the induced e.m.f. is proportional to the rate of change of magnetic flux linkage. To increase the peak e.m.f.:
1. Rotate the coil faster (increase rotational speed / angular frequency).
2. Increase the strength of the magnetic field (use stronger magnets).
3. Increase the number of turns on the coil.
4. Increase the cross-sectional area of the coil.

(a)
* electromagnetic induction [1]
(accept: electromagnetic / induction)

(b)
* Smooth alternating / wave-like curve shown with alternating positive and negative values AND starts at non-zero peak (\(+V_0\) or \(-V_0\)), crosses zero at \(\frac{1}{4}T\) and \(\frac{3}{4}T\), and completes exactly one full cycle at \(T\) [1]
(reject: straight-line triangle/square waves)

(c) Any two from [2]:
* increase the speed of rotation / rotate the coil faster / increase frequency of rotation;
* use a stronger magnet / increase magnetic field strength;
* increase the number of turns / loops in the coil;
* increase the area of the coil / use a larger coil;
* add a soft iron core inside the coil;