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Detailed calculations:

(a)(i) Using \(a = \frac{v - u}{t}\):
\(a = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\).

(ii) Using \(F = m a\):
\(F = 1200\text{ kg} \times 2.5\text{ m/s}^2 = 3000\text{ N}\).

(b)(i) Using \(d = v \times t\):
\(d = 15\text{ m/s} \times 10\text{ s} = 150\text{ m}\).

(ii) Using \(E_k = \frac{1}{2} m v^2\):
\(E_k = 0.5 \times 1200\text{ kg} \times (15\text{ m/s})^2 = 600 \times 225 = 135\,000\text{ J}\) (or \(135\text{ kJ}\)).

(c) As the car brakes, work is done by friction between the brake pads and the discs. This transfers the kinetic energy of the car into thermal energy, which increases the temperature of the brakes and the surroundings.

(a)(i)
- 1 mark for correct formula or substitution: \((15 - 0) / 6\)
- 1 mark for correct final answer with units: \(2.5\text{ m/s}^2\)

(a)(ii)
- 1 mark for correct formula or substitution: \(1200 \times 2.5\)
- 1 mark for correct final answer with units: \(3000\text{ N}\) (accept consequential error from a(i))

(b)(i)
- 1 mark for correct formula or substitution: \(15 \times 10\)
- 1 mark for correct final answer with units: \(150\text{ m}\)

(b)(ii)
- 1 mark for correct formula or substitution: \(0.5 \times 1200 \times 15^2\)
- 1 mark for correct final answer with units: \(135\,000\text{ J}\) or \(135\text{ kJ}\)

(c)
- 1 mark for mentioning work is done by friction / brakes.
- 1 mark for stating kinetic energy is transferred to thermal / heat energy.

(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by the diffusion of water vapour through the stomata.

(b)(i) Not all water taken up by the plant is lost through transpiration. Some water is used inside the plant cells for photosynthesis, and some is used to maintain cell turgidity.

(b)(ii) Environmental factors that increase transpiration rate: increased temperature, increased wind speed, increased light intensity, or decreased humidity.
Explanation: For example, higher temperature increases the kinetic energy of water molecules, which increases the rate of evaporation from the mesophyll surfaces and speeds up the diffusion of water vapour out of the stomata.

(c) Structural differences:
1. Xylem vessels consist of dead cells, whereas phloem sieve tubes consist of living cells.
2. Xylem vessels are hollow tubes with no cytoplasm or cross-walls, whereas phloem sieve tubes contain cytoplasm and are separated by perforated sieve plates.
3. Xylem walls are thickened with lignin to provide support, whereas phloem cell walls do not contain lignin.

(a)
- 1 mark for: loss of water vapour from leaves / parts of plant.
- 1 mark for: by evaporation (from mesophyll cell surfaces) followed by diffusion (through stomata).

(b)(i)
- 1 mark for: some water is used in photosynthesis.
- 1 mark for: some water is used to keep cells turgid / support the plant.

(b)(ii)
- 1 mark for stating two correct factors (e.g., high temperature and high wind speed).
- 1 mark for explaining the effect of one chosen factor (e.g., temperature increases kinetic energy).
- 1 mark for linking this explanation to faster diffusion / evaporation.

(c)
- 1 mark per correct structural difference (max 3):
- Dead cells (xylem) vs living cells (phloem).
- Lignified walls (xylem) vs non-lignified walls (phloem).
- No cross-walls / hollow (xylem) vs sieve plates present (phloem).
- Companion cells present in phloem vs absent in xylem.

(a) In the mouth, teeth chew and grind large pieces of food into smaller pieces (mechanical digestion). This increases the surface area-to-volume ratio of the food, allowing chemical digestion enzymes (such as amylase) to break down food molecules much more rapidly.

(b)(i) Substrate: protein. Product: amino acids (or peptides).

(b)(ii) Salivary amylase works best in neutral conditions (around pH 7). The highly acidic conditions in the stomach (pH 2) denature the enzyme. This happens because the low pH alters the shape of the enzyme's active site, meaning starch molecules can no longer bind to it to form enzyme-substrate complexes, stopping amylase activity.

(c) Adaptations of villi:
1. They provide a very large surface area for absorption.
2. They have walls that are only one cell thick, which shortens the diffusion distance.
3. They contain a rich network of blood capillaries to maintain a steep concentration gradient for rapid absorption.

(a)
- 1 mark for: teeth chew / grind / tear food (or tongue rolls food into a bolus).
- 1 mark for: breaks food into smaller pieces / increases surface area.
- 1 mark for: allows enzymes more access / increases rate of chemical digestion.

(b)(i)
- 1 mark for: substrate is protein.
- 1 mark for: product is amino acids / peptides.

(b)(ii)
- 1 mark for: amylase is denatured.
- 1 mark for: due to active site changing shape / pH being far from optimum.
- 1 mark for: substrate (starch) can no longer fit/bind into active site.

(c)
- 1 mark per correct adaptation (max 2):
- Large surface area (due to finger-like shapes / microvilli).
- Thin wall / one-cell-thick epithelium (short pathway).
- Good blood supply / network of capillaries (maintains concentration gradient).
- Presence of lacteals (for absorbing fats).

(a)(i) Isotopes are atoms of the same element with the same number of protons (atomic number) but different numbers of neutrons (mass number).

(a)(ii) In a neutral atom of \(^{37}\text{Cl}\):
- Protons = 17 (given by the atomic number)
- Electrons = 17 (equal to the number of protons in a neutral atom)
- Neutrons = 37 - 17 = 20

(b)(i) Sodium is in Group I and has 1 outer electron. It transfers this 1 electron to a chlorine atom, which is in Group VII and has 7 outer electrons. This forms a positive sodium ion (\(\text{Na}^+\)) and a negative chloride ion (\(\text{Cl}^-\)). The strong electrostatic attraction between these oppositely charged ions forms the ionic bond.

(b)(ii) Hydrogen and chlorine share a pair of electrons to form a single covalent bond. By sharing, the hydrogen atom achieves a stable outer shell of 2 electrons (like helium), and the chlorine atom achieves a stable octet of 8 outer electrons.

(a)(i)
- 1 mark for: same number of protons / same atomic number.
- 1 mark for: different number of neutrons / different mass number.

(a)(ii)
- 1 mark for: protons = 17 and electrons = 17.
- 1 mark for: neutrons = 20 (calculated as 37 - 17).

(b)(i)
- 1 mark for: sodium loses 1 electron (to form \(\text{Na}^+\)).
- 1 mark for: chlorine gains 1 electron (to form \(\text{Cl}^-\)).
- 1 mark for: electrostatic attraction between oppositely charged ions.

(b)(ii)
- 1 mark for: sharing of one pair of electrons / single covalent bond.
- 1 mark for: hydrogen has 2 electrons in its outer shell.
- 1 mark for: chlorine has 8 electrons in its outer shell.

(a) To construct this circuit diagram:
1. Draw the symbol for a d.c. power supply / battery.
2. In series with the battery in the main branch, draw a closed or open switch and an ammeter (symbol A inside a circle).
3. Split the main branch into two parallel branches. In one branch, draw a \(6.0\ \Omega\) resistor, and in the other branch, draw a \(3.0\ \Omega\) resistor. Then rejoin the branches back to the negative terminal of the power supply.

(b)(i) Combined parallel resistance \(R_p\):
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0\ \Omega} + \frac{1}{3.0\ \Omega}\)
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0} = \frac{1}{2.0}\)
\(R_p = 2.0\ \Omega\).

(b)(ii) Using Ohm's Law \(I = \frac{V}{R_p}\):
\(I = \frac{12\text{ V}}{2.0\ \Omega} = 6.0\text{ A}\).

(b)(iii) The voltage across each resistor in parallel is equal to the supply voltage, which is \(12\text{ V}\).
Power \(P = \frac{V^2}{R}\):
\(P = \frac{12^2}{3.0} = \frac{144}{3.0} = 48\text{ W}\).

(a)
- 1 mark for: correct symbol for d.c. supply / battery connected to the main circuit.
- 1 mark for: switch and ammeter in series in the main branch (before the split).
- 1 mark for: two resistors shown in parallel branches (clearly showing the split and rejoin).

(b)(i)
- 1 mark for: correct formula \(1/R = 1/R_1 + 1/R_2\) or substitution.
- 1 mark for: correct final answer with units: \(2.0\ \Omega\).

(b)(ii)
- 1 mark for: correct use of \(I = V / R\).
- 1 mark for: correct final answer with units: \(6.0\text{ A}\) (accept consequential error from b(i)).

(b)(iii)
- 1 mark for: identifying voltage across parallel branch is \(12\text{ V}\).
- 1 mark for: correct power formula (e.g., \(P = V^2 / R\) or finding current in branch as \(4\text{ A}\) and using \(P = I^2 R\)).
- 1 mark for: correct final answer with units: \(48\text{ W}\).

(a)(i) Cracking of decane:
\(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{16} + \text{C}_2\text{H}_6\)
(The other product is ethane, which balances the carbon and hydrogen atoms).

(a)(ii) Catalytic cracking requires a high temperature (typically \(450\text{--}800\ ^\circ\text{C}\)) and a catalyst (such as silica, alumina, or zeolites).

(b)(i) The chemical test is adding bromine water (or aqueous bromine).
- Alkane (saturated): no reaction, the solution remains orange / brown / yellow.
- Alkene (unsaturated): addition reaction occurs, the solution turns colourless (decolourises).

(b)(ii) Propene has a chain of three carbon atoms. Its bonds include:
- One carbon-carbon double covalent bond (\(\text{C}=\text{C}\)).
- One carbon-carbon single covalent bond (\(\text{C}-\text{C}\)).
- Six carbon-hydrogen single covalent bonds (\(\text{C}-\text{H}\)).

(a)(i)
- 1 mark for identifying the second product as \(\text{C}_2\text{H}_6\).
- 1 mark for balancing the overall equation correctly.

(a)(ii)
- 1 mark for: high temperature (accept range \(450\text{--}800\ ^\circ\text{C}\)).
- 1 mark for: catalyst / alumina / silica / zeolite.

(b)(i)
- 1 mark for: bromine water / aqueous bromine.
- 1 mark for: alkane stays orange / brown / no change.
- 1 mark for: alkene decolourises / turns colourless.

(b)(ii)
- 1 mark for: identifying the double covalent bond between carbon atoms (\(1 \times \text{C}=\text{C}\)).
- 1 mark for: identifying the single covalent bond between carbon atoms (\(1 \times \text{C}-\text{C}\)).
- 1 mark for: identifying six single carbon-hydrogen covalent bonds (\(6 \times \text{C}-\text{H}\)).

(a)(i) Region A between Infrared and Radio waves is Microwaves.

(a)(ii) Use of X-rays: medical imaging of bones, scanning luggage at airport security.
Hazard of ultraviolet: sunburn, skin cancer, premature aging of skin, eye damage (cataracts).

(a)(iii) The speed of all electromagnetic waves in a vacuum is \(3.0 \times 10^8\text{ m/s}\).

(b)(i) Refraction is the change in direction (bending) of a wave when it passes from one medium to another. This change in direction is caused by a change in the speed of the wave.

(b)(ii) Using Snell's Law:
\(n = \frac{\sin i}{\sin r}\)
\(1.5 = \frac{\sin 30^\circ}{\sin r}\)
\(\sin r = \frac{\sin 30^\circ}{1.5} = \frac{0.5}{1.5} = 0.3333\)
\(r = \sin^{-1}(0.3333) \approx 19.5^\circ\) (accept \(19^\circ\) or \(20^\circ\) to significant figures).

(a)(i)
- 1 mark for: microwaves.
-
(a)(ii)
- 1 mark for correct use of X-rays (e.g. medical imaging / looking for fractures / security scan).
- 1 mark for correct hazard of UV (e.g. skin cancer / sunburn / cataracts).

(a)(iii)
- 1 mark for: \(3.0 \times 10^8\text{ m/s}\) (must include unit for mark).

(b)(i)
- 1 mark for: change in direction / bending of light.
- 1 mark for: caused by change in speed / when entering a different medium.

(b)(ii)
- 1 mark for correct formula: \(n = \sin i / \sin r\).
- 1 mark for correct substitution: \(1.5 = \sin 30^\circ / \sin r\).
- 1 mark for finding \(\sin r = 0.3333\).
- 1 mark for correct angle calculation: \(19.5^\circ\) (accept range \(19\text{ to } 20^\circ\)).

(a)(i) Phenotype.

(a)(ii) Heterozygous.

(b)(i) The heterozygous red plant must have one dominant and one recessive allele, so its genotype is \(Rr\).
The white plant is homozygous recessive because white is the recessive trait, so its genotype is \(rr\).

(b)(ii) The genetic cross can be shown using a Punnett square:
- Gametes from heterozygous parent: \(R\) and \(r\)
- Gametes from white parent: \(r\) and \(r\)

| | R | r |
|---|---|---|
| r | Rr | rr |
| r | Rr | rr |

Offspring genotypes are:
- \(Rr\) (heterozygous)
- \(rr\) (homozygous recessive)

Offspring phenotypes are:
- \(Rr\): Red flowers
- \(rr\): White flowers

(b)(iii) Out of the 4 possibilities, 2 are \(rr\) (white). Therefore, the probability is \(\frac{2}{4} = 0.5\) or \(50\%\) (or a 1 in 2 ratio).

(a)(i)
- 1 mark for: phenotype.

(a)(ii)
- 1 mark for: heterozygous.

(b)(i)
- 1 mark for: \(Rr\) (heterozygous red).
- 1 mark for: \(rr\) (white).

(b)(ii)
- 1 mark for correct gametes of parents identified (\(R, r\) and \(r\)).
- 1 mark for correct setup of Punnett square.
- 1 mark for correct offspring genotypes in the grid (\(Rr\) and \(rr\)).
- 1 mark for matching genotypes to correct phenotypes (\(Rr\) = red, \(rr\) = white).

(b)(iii)
- 1 mark for identifying that white-flowered offspring are \(rr\).
- 1 mark for correct probability: \(0.5\) / \(50\%\) / \(1/2\) / \(1:1\).

(a) Evaporation of water occurs from the cell walls of mesophyll cells into the air spaces inside the leaf. This water vapour then diffuses out of the leaf through the stomata. This loss of water creates a tension or transpiration pull that draws water up the xylem vessels. Cohesion represents the attraction between water molecules themselves, maintaining a continuous column of water, while adhesion is the attraction between water molecules and the walls of the xylem vessels. (b)(i) Cut the shoot underwater to prevent air bubbles entering the xylem; seal all joints with petroleum jelly to ensure the system is airtight. (b)(ii) Some of the absorbed water is used by the plant for photosynthesis, and some is used to maintain cell turgidity (turgor pressure) within the plant cells rather than being lost to the atmosphere. (c) Transpiration rate decreases because high external humidity decreases the concentration gradient of water vapour between the inside of the leaf and the outside air, reducing the rate of diffusion.

(a) Evaporation from mesophyll cell walls into air spaces [1]; diffusion of water vapour through stomata [1]; transpiration pull/tension created in xylem [1]; cohesion attraction between water molecules AND adhesion attraction between water molecules and xylem walls [1]. (b)(i) Cut underwater [1]; seal joints with petroleum jelly / airtight [1]. (b)(ii) Water used in photosynthesis [1]; water used to maintain turgidity/cell volume [1]. (c) Transpiration decreases [1]; because the concentration gradient of water vapour between leaf interior and exterior is reduced [1].

(a)(i) Using the equation \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\), we can rearrange to find \(N_s = N_p \times \frac{V_s}{V_p} = 800 \times \frac{12}{240} = 40\text{ turns}\). (a)(ii) The alternating current in the primary coil produces an alternating/changing magnetic field in the soft iron core. The soft iron core links this changing magnetic field to the secondary coil. As the changing magnetic field cuts through the secondary coil, it induces an alternating electromotive force (e.m.f.) across its terminals. (b)(i) For a given transmission power, increasing the voltage reduces the current because \(P = V \times I\). The rate of thermal energy loss (power dissipation) in the transmission cables is given by \(P_{\text{loss}} = I^2 \times R\). A lower current drastically reduces this power loss, making the transmission much more efficient. (b)(ii) High voltages require large, expensive pylons and thick, high-quality insulation, and present a higher danger of electrocution/safety hazards.

(a)(i) \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) or correct substitution [1]; \(40\text{ turns}\) [1]. (a)(ii) Alternating current in primary produces changing magnetic field [1]; soft iron core guides/links magnetic field [1]; changing magnetic field cuts secondary coil [1]; induces alternating e.m.f. in secondary [1]. (b)(i) Higher voltage means lower current [1]; power loss in cables is proportional to current squared (\(P = I^2 R\)) [1]; less thermal energy/heat lost to surroundings [1]. (b)(ii) High cost of insulation/pylons OR increased electrocution risk [1].

(a)(i) The essential conditions are: 1. Yeast (to act as the catalyst/source of enzymes); 2. A temperature between \(25\ ^\circ\text{C}\) and \(35\ ^\circ\text{C}\) (which is the optimum for yeast enzymes); 3. Anaerobic conditions (absence of oxygen, to prevent oxidation of ethanol to ethanoic acid). (a)(ii) The balanced equation is \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\). (a)(iii) The required conditions for catalytic hydration of ethene are: a temperature of \(300\ ^\circ\text{C}\), a pressure of \(60\text{ atm}\), and a phosphoric acid (\(H_3PO_4\)) catalyst. (b) In a molecule of ethanoic acid, there is a methyl group (\(-CH_3\)) where a carbon atom is single-bonded to three hydrogen atoms. This carbon is single-bonded to a second carbon atom. The second carbon atom is double-bonded to one oxygen atom (\(=O\)) and single-bonded to another oxygen atom, which in turn is single-bonded to a hydrogen atom (\(-OH\)).

(a)(i) Yeast/enzymes [1]; temperature \(25\text{ to }35\ ^\circ\text{C}\) [1]; anaerobic/absence of oxygen [1]. (a)(ii) Correct formulas [1]; balanced equation [1]. (a)(iii) Temperature of \(300\ ^\circ\text{C}\) [1]; pressure of \(60\text{ atm}\) [1]; phosphoric acid catalyst [1]. (b) Two carbon atoms with correct connectivity (\(-CH_3\) and \(-COOH\) groups) [1]; all bonds drawn/described correctly, including C=O and O-H bonds [1].

(a)(i) Acceleration is given by \(a = \frac{v - u}{t} = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\). (a)(ii) Using Newton's second law, \(F = m \times a = 1200\text{ kg} \times 2.5\text{ m/s}^2 = 3000\text{ N}\). (a)(iii) Kinetic energy is \(KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (15\text{ m/s})^2 = 0.5 \times 1200 \times 225 = 135000\text{ J}\) (or \(135\text{ kJ}\)). (b)(i) The total resistive force is \(400\text{ N}\). Because the car is travelling at a constant speed, its acceleration is zero, which means the resultant force is zero. Thus, the forward driving force is exactly balanced by the backward resistive force. (b)(ii) Power is \(P = F \times v = 400\text{ N} \times 15\text{ m/s} = 6000\text{ W}\) (or \(6.0\text{ kW}\)).

(a)(i) \(2.5\) [1]; \(\text{m/s}^2\) [1]. (a)(ii) Formula \(F = ma\) or correct substitution [1]; \(3000\text{ N}\) [1]. (a)(iii) Formula \(KE = \frac{1}{2}mv^2\) or correct substitution [1]; \(135000\text{ J}\) (or \(135\text{ kJ}\)) [1]. (b)(i) \(400\text{ N}\) [1]; explanation that at constant speed, acceleration is zero / forces are balanced [1]. (b)(ii) Formula \(P = Fv\) or correct substitution [1]; \(6000\text{ W}\) (or \(6.0\text{ kW}\)) [1].