2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

The initial kinetic energy of the car is: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 3.0 \times 4.0^2 = 24\text{ J}\). The work done by the braking force to stop the car must equal this kinetic energy: \(W = F \times d = 24\text{ J}\). Since the force \(F = 8.0\text{ N}\), we can solve for distance \(d\): \(d = \frac{24}{8.0} = 3.0\text{ m}\). Alternatively, using equations of motion, the acceleration is: \(a = -\frac{F}{m} = -\frac{8.0}{3.0}\text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), we get: \(0 = 4.0^2 + 2 \times \left(-\frac{8.0}{3.0}\right) \times s\), which simplifies to \(16 = \frac{16}{3} s\), giving \(s = 3.0\text{ m}\).

1 mark for the correct answer B. Correct calculation of initial kinetic energy and equating it to work done (or using Newton's second law and equations of motion) to obtain the braking distance.

At \(30^\circ\text{C}\), the kinetic energy is lower, meaning fewer successful collisions occur, so it takes longer (\(120\text{ s}\)) compared to \(40^\circ\text{C}\) (\(40\text{ s}\)). At \(60^\circ\text{C}\), the high temperature causes the enzyme to denature. This permanent change in the shape of the active site means the starch (substrate) can no longer fit, preventing any breakdown from taking place.

1 mark for the correct answer C. Identifies that high temperatures denature enzymes by altering the shape of their active sites, preventing substrate binding.

The resistance of a wire is given by \(R = \rho \frac{l}{A}\), where \(A = \pi r^2\). For the second wire, the length is \(3l\) and the radius is \(2r\), meaning its cross-sectional area is \(A_2 = \pi (2r)^2 = 4 \pi r^2 = 4A\). Therefore, the new resistance is: \(R_2 = \rho \frac{3l}{4A} = \frac{3}{4} \left(\rho \frac{l}{A}\right) = 0.75 R\).

1 mark for the correct answer B. Correctly scales length by 3 and area by 4 (since area is proportional to the square of the radius), resulting in a factor of 3/4.

According to the kinetic theory, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass. Since neon has the smallest relative mass (\(M_r = 20\)), its particles have the highest average speed at this temperature and will diffuse out most rapidly. All gases at the same temperature have the same average kinetic energy, so D is incorrect.

1 mark for the correct answer B. Understands that lower relative molecular/atomic mass results in a faster rate of diffusion at a given temperature.

Substance X has a high melting point and does not conduct in either state, which is characteristic of a giant covalent structure (e.g., silicon(IV) oxide). Substance Y has a high melting point and conducts electricity only when molten (due to mobile ions), which is typical of a giant ionic structure. Substance Z has a low melting point and does not conduct, typical of a simple covalent substance.

1 mark for the correct answer A. Correctly links physical properties (melting point, electrical conductivity in solid/liquid states) to the appropriate type of structure (giant covalent, giant ionic, simple covalent).

Protein digestion begins in the stomach, where the acidic environment is optimal for pepsin, a protease enzyme. The chemical digestion of proteins is completed in the small intestine, and the final products absorbed into the blood are amino acids.

1 mark for the correct answer A. Accurately identifies the stomach as the start of protein digestion, pepsin as the starting enzyme, and amino acids as the final absorbed products.

The refractive index \(n\) is related to the angle of incidence \(i\) and angle of refraction \(r\) by Snell's law: \(n = \frac{\sin i}{\sin r}\). Rearranging to find \(r\): \(r = \sin^{-1}\left(\frac{\sin i}{n}\right) = \sin^{-1}\left(\frac{\sin 45^\circ}{1.5}\right) \approx \sin^{-1}(0.4714) \approx 28^\circ\).

1 mark for the correct answer A. Correctly uses Snell's law equation and trigonometric functions to calculate the angle of refraction.

First, subtract the background count rate to find the activity due to the sample alone. Initial sample count rate = \(408 - 24 = 384\text{ counts/minute}\). Final sample count rate = \(72 - 24 = 48\text{ counts/minute}\). Determine how many half-lives have passed: \(384 \rightarrow 192 \rightarrow 96 \rightarrow 48\), which is exactly \(3\) half-lives. Since \(3\) half-lives elapsed over \(3.0\text{ hours}\), one half-life is equal to \(\frac{3.0\text{ hours}}{3} = 1.0\text{ hour}\).

1 mark for the correct answer B. Correctly accounts for background radiation by subtraction and determines the number of half-lives elapsed to find the half-life.

At extreme pH values away from the optimum (pH 8.0), enzymes undergo denaturation. At pH 3.0, the hydrogen bonds and other bonds maintaining the three-dimensional structure of the enzyme are disrupted. This permanently alters the shape of the active site, meaning the substrate can no longer fit, and the reaction ceases. Substrates are not denatured, and kinetic energy changes are related to temperature, not pH.

1 mark for identifying that the active site shape is permanently altered due to denaturation, preventing the substrate from fitting.

According to the principles of gas diffusion, lighter molecules (lower relative molecular mass, Mr) diffuse faster than heavier molecules at the same temperature. Ammonia (Mr = 17) has a lower relative molecular mass than hydrogen chloride (Mr = 36.5), so ammonia molecules diffuse faster and travel a greater distance. Hydrogen chloride molecules are heavier and diffuse slower, travelling a shorter distance. Therefore, the white ring forms closer to the hydrogen chloride end of the tube.

1 mark for selecting the correct explanation that HCl is heavier and diffuses slower, resulting in the ring forming closer to the HCl end.

First, calculate the acceleration of the object using Newton's second law: a = F / m = 12.0 N / 4.0 kg = 3.0 m/s^2. Next, find the final velocity after 3.0 s using the equation: v = u + at = 5.0 m/s + (3.0 m/s^2 * 3.0 s) = 14.0 m/s. Finally, calculate the final kinetic energy: Ek = 0.5 * m * v^2 = 0.5 * 4.0 kg * (14.0 m/s)^2 = 2.0 * 196 = 392 J.

1 mark for the correct calculation of final velocity (14.0 m/s) and kinetic energy (392 J).

The number of protons (17) defines the element as chlorine. Since the number of electrons (18) is one greater than the number of protons (17), the particle has a net charge of 1-, making it an anion. The mass number is the sum of protons and neutrons (17 + 18 = 35), and chlorine is in Group VII, not Group VIII.

1 mark for identifying X as a 1- anion of chlorine.

When light intensity on an LDR increases, its resistance decreases. Since the LDR and the 6.0 ohm resistor are in series, the total resistance of the circuit decreases, which increases the current flowing through the circuit. According to V = IR, an increased current through the fixed 6.0 ohm resistor results in a higher voltage across it, so the voltmeter reading increases.

1 mark for identifying that LDR resistance decreases and voltmeter reading increases.

The relationship between refractive index (n) and critical angle (c) is given by sin(c) = 1 / n. Substituting n = 1.50 gives sin(c) = 1 / 1.50 = 0.667. Solving for c, we get c = arcsin(0.667) = 41.8 degrees.

1 mark for correctly applying the critical angle formula to find 41.8 degrees.

A decrease in mass is caused by water leaving the potato cells via osmosis. Water moves down its potential gradient, from a higher water potential (inside the potato cells) to a lower water potential (outside in the sucrose solution). The 1.0 mol/dm^3 sucrose solution has the lowest water potential, creating the steepest concentration gradient, which results in the greatest net loss of water and the largest percentage decrease in mass.

1 mark for selecting 1.0 mol/dm^3 as the solution causing the greatest mass loss due to osmosis.

During beta-minus decay, a neutron inside the nucleus decays into a proton and an electron (beta particle). The proton remains in the nucleus, increasing the atomic (proton) number by 1 (from 83 to 84), while the mass number remains unchanged (212). The emitted electron is a beta particle.

1 mark for correctly identifying that beta emission results in an increase of 1 in the proton number with no change in mass number.

Pepsin is adapted to the highly acidic conditions of the stomach, giving it an optimum pH of around 2. Trypsin works in the slightly alkaline environment of the small intestine, giving it an optimum pH of around 8.

1 mark for identifying the correct optimum pH values for both enzymes.

The decrease in gravitational potential energy is equal to the kinetic energy gained. The change in height is \(15\text{ m} - 5.0\text{ m} = 10\text{ m}\). The kinetic energy is calculated as: \(E_k = mg\Delta h = 2.0\text{ kg} \times 10\text{ m/s}^2 \times 10\text{ m} = 200\text{ J}\).

1 mark for the correct calculation of kinetic energy.

The rate of diffusion of a gas is inversely related to its relative molecular mass (\(M_r\)). Lighter molecules diffuse faster. The molecular masses are: \(\text{CO}_2 = 44\), \(\text{CH}_4 = 16\), \(\text{N}_2 = 28\), and \(\text{SO}_2 = 64\). Methane has the smallest molecular mass and therefore diffuses the fastest.

1 mark for identifying methane as the gas with the lowest molecular mass that diffuses fastest.

For an ideal transformer, the ratio of currents is inversely proportional to the ratio of turns: \(I_p / I_s = N_s / N_p\). Substituting the values: \(I_p / 0.40\text{ A} = 900 / 150 = 6\). Thus, \(I_p = 6 \times 0.40\text{ A} = 2.4\text{ A}\).

1 mark for the correct calculation of primary current using the transformer power/current relationship.

The proton number (atomic number) is the lower subscript, which is 19. A neutral atom has the same number of electrons as protons, which is 19. The number of neutrons is the mass number minus the proton number: \(41 - 19 = 22\).

1 mark for correctly identifying the number of protons, neutrons, and electrons.

Water moves down a water potential gradient from a region of higher water potential (-300 kPa) to a region of lower water potential (-800 kPa). This causes water to leave the cell by osmosis. The loss of water causes the vacuole and cytoplasm to shrink, pulling the cell membrane away from the cell wall, which makes the cell plasmolysed.

1 mark for identifying the correct direction of water movement and the resulting state of the plant cell.

To decrease from 800 to 50 counts per minute, the sample must halve four times: \(800 \rightarrow 400 \rightarrow 200 \rightarrow 100 \rightarrow 50\). Since four half-lives occur in 24 hours, the half-life is \(24 / 4 = 6\text{ hours}\).

1 mark for determining the number of half-lives and calculating the duration of one half-life.

Alkenes like propene are unsaturated and undergo addition reactions with halogens. Bromine atoms add across the double bond, breaking it and forming dibromopropane with the formula \(\text{C}_3\text{H}_6\text{Br}_2\).

1 mark for identifying addition reaction and the correct molecular formula of the product.

As temperature increases, the kinetic energy of the molecules increases, which increases the frequency of effective collisions between the enzyme and substrate. However, at temperatures above the optimum, the high thermal energy disrupts the bonds maintaining the three-dimensional structure of the enzyme's active site. This permanent change in shape is called denaturation, preventing the substrate from fitting into the active site. Denaturation is an irreversible process.

1 mark for identifying the correct statement regarding denaturation and the permanent shape change of the active site.

The rate of diffusion of a gas depends on its relative molecular mass (Mr). The lower the relative molecular mass, the faster the gas molecules move and diffuse. Let us calculate the Mr values: CO2 = 12 + (2 * 16) = 44; NO = 14 + 16 = 30; SO2 = 32 + (2 * 16) = 64; CH4 = 12 + (4 * 1) = 16. Since CH4 (methane) has the lowest Mr, it diffuses the fastest.

1 mark for identifying that the gas with the lowest relative molecular mass diffuses fastest, and correctly identifying methane (CH4).

First, calculate the work done (which is equal to the gain in gravitational potential energy): Work done = weight * height = m * g * h = 400 kg * 10 N/kg * 15 m = 60000 J. Next, calculate the power: Power = Work done / time = 60000 J / 20 s = 3000 W.

1 mark for using the correct formulas (Work = m * g * h and Power = Work / t) to obtain 3000 W.

The atomic number is determined by the number of protons, which is 19. Looking at the periodic table, element 19 is potassium (K). Since the number of protons (19) is greater than the number of electrons (18) by one, the particle has a net charge of 1+. Therefore, it is a positively charged potassium ion.

1 mark for correctly determining that 19 protons corresponds to potassium and 18 electrons yields a 1+ charge.

First, calculate the equivalent resistance of the parallel combination: 1/R_p = 1/6.0 + 1/12.0 = 2/12.0 + 1/12.0 = 3/12.0. So, R_p = 12.0 / 3 = 4.0 ohms. Next, add the series resistor: R_total = R_p + R_series = 4.0 ohms + 4.0 ohms = 8.0 ohms.

1 mark for calculating the parallel resistance as 4.0 ohms, and then adding 4.0 ohms to get 8.0 ohms.

Human sperm (male gametes) are very small in size, highly mobile due to their flagellum, and are produced continuously in millions. Human eggs (female gametes) are much larger because they contain energy stores for the initial stages of development, are immobile, and are released only once a month during ovulation.

1 mark for correctly identifying the complementary size and mobility traits of human male and female gametes.

All electromagnetic waves travel at the exact same speed in a vacuum (approximately 3 * 10^8 m/s). Radio waves actually have the longest wavelength and lowest frequency. Sound waves are longitudinal mechanical waves, not transverse electromagnetic waves.

1 mark for identifying that all electromagnetic waves travel at the same speed in a vacuum.

Reacting vigorously with cold water indicates that Metal Z is highly reactive (like sodium or calcium). Reacting with steam but not cold water means Metal Y has medium-high reactivity (like iron or zinc). Being unable to react with steam and having an oxide reduced by carbon indicates Metal X has lower reactivity (like lead or copper). Therefore, the order of decreasing reactivity is Z -> Y -> X.

1 mark for analyzing the physical/chemical tests to order the metals correctly from most to least reactive.

First, find the final velocity \(v\) using the conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting the values: \((2.0 \times 6.0) + (3.0 \times 0) = (2.0 + 3.0) v\), which simplifies to \(12 = 5.0 v\), hence \(v = 2.4\text{ m/s}\). Next, calculate the initial kinetic energy: \(E_{k,\text{initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times 6.0^2 = 36.0\text{ J}\). Now calculate the final kinetic energy: \(E_{k,\text{final}} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 5.0 \times 2.4^2 = 14.4\text{ J}\). Finally, calculate the kinetic energy lost: \(\Delta E_k = 36.0 - 14.4 = 21.6\text{ J}\).

1 mark for calculating the final velocity of \(2.4\text{ m/s}\), finding the difference between initial kinetic energy (\(36.0\text{ J}\)) and final kinetic energy (\(14.4\text{ J}\)), yielding the correct loss of \(21.6\text{ J}\).

At temperatures above the optimum, the excessive thermal energy breaks the weak bonds holding the enzyme's specific three-dimensional structure together. This denatures the enzyme, permanently changing the shape of its active site so that the substrate molecule can no longer fit and bind to form an enzyme-substrate complex.

1 mark for correctly identifying that high temperatures denature enzyme molecules by permanently altering the shape of their active sites, preventing substrate binding.

During the electrolysis of concentrated aqueous \(\text{NaCl}\): At the anode (+), halide chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions because they are in high concentration, producing chlorine gas. At the cathode (-), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions because hydrogen is less reactive, producing hydrogen gas. As \(\text{H}^+\) and \(\text{Cl}^-\) are removed, \(\text{Na}^+\) and \(\text{OH}^-\) ions remain in the solution, forming sodium hydroxide, which is basic and causes the pH of the remaining electrolyte to increase.

1 mark for correctly identifying chlorine gas at the anode, hydrogen gas at the cathode, and an increase in the pH of the remaining electrolyte.

First, calculate the equivalent parallel resistance \(R_p\) of the circuit: \(1/R_p = 1/R_1 + 1/R_2 = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0\), which gives \(R_p = 3.0\ \Omega\). Next, calculate the total power using \(P = V^2 / R_p\): \(P = 6.0^2 / 3.0 = 36.0 / 3.0 = 12.0\text{ W}\).

1 mark for calculating the correct total equivalent resistance of \(3.0\ \Omega\) and using the electrical power formula to find \(12.0\text{ W}\).

Blood group alleles \(I^A\) and \(I^B\) are codominant, while \(I^O\) is recessive. Since the first child has blood group O (genotype \(I^O I^O\)), both parents must carry one copy of the recessive allele \(I^O\). Thus, the father's genotype is \(I^A I^O\) and the mother's genotype is \(I^B I^O\). A genetic cross (\(I^A I^O \times I^B I^O\)) results in a 1 in 4 chance (\(25\%\)) of an offspring having the genotype \(I^O I^O\) (blood group O).

1 mark for correctly identifying the heterozygous genotypes of both parents and finding the correct probability of \(25\%\).

First, find the moles of \(\text{CO}_2\) gas produced: \(\text{moles} = 2.4\text{ dm}^3 / 24\text{ dm}^3/\text{mol} = 0.1\text{ mol}\). According to the balanced equation, \(1\text{ mol}\) of \(\text{CaCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\). Therefore, \(0.1\text{ mol}\) of \(\text{CaCO}_3\) is required. Calculate the mass of \(\text{CaCO}_3\): \(\text{mass} = 0.1\text{ mol} \times 100\text{ g/mol} = 10.0\text{ g}\).

1 mark for calculating the amount of gas (0.1 mol), recognizing the 1:1 mole ratio, and converting this to 10.0 g of calcium carbonate.

Use Snell's Law: \(n = \frac{\sin i}{\sin r}\), where \(n = 1.5\) and \(i = 45^\circ\). Rearranging the formula gives \(\sin r = \frac{\sin 45^\circ}{1.5} = \frac{0.7071}{1.5} \approx 0.4714\). Calculating the inverse sine: \(r = \arcsin(0.4714) \approx 28.1^\circ\). Rounded to the nearest degree, the angle of refraction is \(28^\circ\).

1 mark for correctly applying Snell's Law, solving for the sine of the angle of refraction, and calculating the angle as \(28^\circ\).

Water moves by osmosis down a water potential gradient from a region of higher water potential (inside the plant cell sap) to a region of lower water potential (the concentrated external sucrose solution) across a partially permeable membrane. The loss of water causes the vacuole and cytoplasm to shrink, pulling the cell membrane away from the cell wall, leaving the cell in a plasmolysed state.

1 mark for identifying that water net movement is out of the cells and that this results in plasmolysed cells.

(a) The enzyme has a specifically shaped region called the active site. The substrate has a complementary shape that fits precisely into this active site, much like a key fits into a lock. They bind together to form an enzyme-substrate complex, where the reaction takes place.
(b) Amylase has an optimum pH (typically around neutral pH 7). Extremely low pH (pH 3) or extremely high pH (pH 11) denatures the enzyme. Denaturation permanently alters the shape of the enzyme's active site, meaning the substrate is no longer complementary and cannot bind. Therefore, no enzyme-substrate complexes can form, and the reaction rate falls to near zero.
(c) Decreasing the temperature will decrease the rate of reaction. This is because molecules have less kinetic energy at lower temperatures, meaning they move more slowly. This reduces the frequency of collisions between enzyme and substrate molecules, resulting in fewer successful collisions per unit time.

(a) 1 mark: Mention of active site.
1 mark: Substrate shape is complementary to active site / fits like lock-and-key.
1 mark: Formation of enzyme-substrate complex.

(b) 1 mark: Extreme pH values cause the enzyme to denature.
1 mark: The shape of the active site is permanently altered.
1 mark: The substrate can no longer fit / bind into the active site.
1 mark: No enzyme-substrate complexes are formed.

(c) 1 mark: Reaction rate decreases.
1 mark: Molecules have less kinetic energy / move slower.
1 mark: Less frequent / fewer collisions per second between enzyme and substrate.

(a)(i) Acceleration is calculated using: \(a = \frac{v - u}{t} = \frac{4.0 - 0}{2.5} = 1.6\text{ m/s}^2\).
(a)(ii) Kinetic energy is: \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times 4.0^2 = 0.25 \times 16 = 4.0\text{ J}\).
(b)(i) Using Newton's second law: \(F_{\text{resultant}} = m \times a = 0.5 \times 1.6 = 0.8\text{ N}\).
(b)(ii) Since \(F_{\text{resultant}} = F_{\text{forward}} - F_{\text{friction}}\), we have \(0.8 = F_{\text{forward}} - 0.3\). Therefore, \(F_{\text{forward}} = 0.8 + 0.3 = 1.1\text{ N}\).
(c) The kinetic energy of the moving car is transferred into thermal (internal) energy of the wheels, track, and surroundings. This is due to work done by the frictional forces opposing the motion.

(a)(i) 1 mark: Correct formula or substitution: \(a = \frac{4.0}{2.5}\).
1 mark: \(1.6\text{ m/s}^2\) (must include correct unit).

(a)(ii) 1 mark: Correct formula or substitution: \(E_k = \frac{1}{2} \times 0.5 \times 4.0^2\).
1 mark: \(4.0\text{ J}\) (must include correct unit).

(b)(i) 1 mark: Correct formula or substitution: \(F = 0.5 \times 1.6\).
1 mark: \(0.8\text{ N}\).

(b)(ii) 1 mark: \(1.1\text{ N}\) (accept calculation: \(0.8 + 0.3\)).

(c) 1 mark: Identifies initial energy as kinetic energy of the car.
1 mark: Identifies final energy as thermal/internal energy (of wheels/track/surroundings).
1 mark: Mentions work done against friction as the cause.

(a) To measure the rate of reaction using gas collection, connect the flask containing the reaction mixture to a gas syringe (or an inverted measuring cylinder filled with water) via a delivery tube. Measure the volume of carbon dioxide gas collected in the syringe at regular time intervals (using a stopwatch). The rate can be expressed as the volume of gas produced per unit time.
(b)(i) Increasing the concentration of hydrochloric acid increases the number of acid particles per unit volume. This makes the particles closer together, which increases the frequency of collisions (number of collisions per unit time) between the acid particles and the calcium carbonate, leading to a higher rate of successful collisions and thus a faster reaction rate.
(b)(ii) Replacing large chips with powder of the same mass greatly increases the surface area of the calcium carbonate. This means more calcium carbonate particles are exposed to the acid on the surface, which increases the frequency of collisions between the reactants per unit time, resulting in an increased reaction rate.
(c) The temperature of the reaction mixture can be increased (or a catalyst could be used) to increase the rate of reaction.

(a) 1 mark: Use of a gas syringe (or inverted measuring cylinder over water).
1 mark: Measure the volume of carbon dioxide gas produced.
1 mark: Record the measurements at regular time intervals using a timer/stopwatch.

(b)(i) 1 mark: More reactant particles per unit volume.
1 mark: Increased collision frequency (collisions per unit time).
1 mark: Leads to more successful collisions per unit time (increased rate).

(b)(ii) 1 mark: Powder has a larger surface area (to volume ratio).
1 mark: More particles are exposed to the acid.
1 mark: Increased frequency of collisions.

(c) 1 mark: Temperature (accept catalyst).

(a) For a parallel circuit, the equivalent resistance \(R_p\) is calculated by:
\(\frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{6.0} + \frac{1}{4.0} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}\text{ }\Omega^{-1}\).
Therefore, \(R_p = \frac{12}{5} = 2.4\text{ }\Omega\).

(b)(i) The total current in the circuit is:
\(I = \frac{V}{R_p} = \frac{12}{2.4} = 5.0\text{ A}\).

(b)(ii) In a parallel circuit, the potential difference across each branch is equal to the supply voltage. Therefore, the voltage across Resistor B is \(12\text{ V}\):
\(I_B = \frac{V}{R_B} = \frac{12}{4.0} = 3.0\text{ A}\).

(c) First, convert time to seconds: \(t = 5.0 \times 60 = 300\text{ s}\).
Electrical energy is: \(E = V \times I \times t = 12 \times 5.0 \times 300 = 18000\text{ J}\) (or \(18\text{ kJ}\)).

(d) Adding another resistor in parallel decreases the overall equivalent resistance of the circuit. Therefore, the total current drawn from the supply will increase.

(a) 1 mark: Correct formula or substitution: \(\frac{1}{R_p} = \frac{1}{6} + \frac{1}{4}\).
1 mark: \(2.4\text{ }\Omega\).

(b)(i) 1 mark: Correct use of Ohm's law: \(I = \frac{12}{\text{their (a)}}\).
1 mark: \(5.0\text{ A}\) (accept correct unit).

(b)(ii) 1 mark: Recognising voltage across B is \(12\text{ V}\) or showing substitution \(\frac{12}{4}\).
1 mark: \(3.0\text{ A}\) (accept correct unit).

(c) 1 mark: Correct unit conversion of time: \(5.0 \times 60 = 300\text{ s}\).
1 mark: Correct formula or substitution: \(E = 12 \times 5.0 \times 300\) (allow ecf from b(i)).
1 mark: \(18000\text{ J}\) or \(18\text{ kJ}\) (must include correct unit).

(d) 1 mark: (Total current) increases.

(a)(i) Vitamin C is needed to maintain healthy skin, blood vessels, and bones, and to prevent scurvy.
(a)(ii) Iron is a crucial component of haemoglobin, which transports oxygen in red blood cells. Deficiency leads to anaemia.
(a)(iii) Dietary fibre adds bulk to the undigested food in the intestines, facilitating peristalsis and preventing constipation.
(b) In the mouth, teeth chew and break food down mechanically (physical digestion) into smaller pieces. This increases the total surface area of the food, making it easier and faster for chemical digestive enzymes (like amylase) to access and chemically break down the food molecules.
(c)(i) The stomach walls secrete gastric juice containing hydrochloric acid and pepsin (a protease). The acid creates a highly acidic environment (around pH 1.5–2.0), which kills pathogens and provides the optimum pH for pepsin to digest proteins into smaller peptides.
(c)(ii) In the small intestine, protease enzymes (such as trypsin) continue the digestion of proteins/peptides. They break down these peptides into simple, soluble amino acids, which are small enough to be absorbed through the intestinal wall into the bloodstream.

(a)(i) 1 mark: Prevents scurvy / maintains healthy skin, tissues, or gums.
(a)(ii) 1 mark: Component of haemoglobin / prevents anaemia.
(a)(iii) 1 mark: Prevents constipation / aids peristalsis by providing bulk.

(b) 1 mark: Teeth mechanically break down / chew food into smaller pieces.
1 mark: This action increases the surface area of the food.
1 mark: Larger surface area allows chemical enzymes to digest food more rapidly.

(c)(i) 1 mark: Secretes hydrochloric acid, which provides the optimum acidic pH for enzymes.
1 mark: Pepsin (protease) starts breaking down proteins into peptides.

(c)(ii) 1 mark: Proteases (e.g. trypsin) break down peptides into amino acids.
1 mark: These amino acids are small/soluble enough to be absorbed.

(a) The iron compound in hematite is iron(III) oxide (\(\text{Fe}_2\text{O}_3\)). The other raw materials added are coke (carbon) and limestone (calcium carbonate).
(b)(i) Coke burns in oxygen to form carbon dioxide, which then reacts with more hot coke to form carbon monoxide: \(\text{C(s)} + \text{CO}_2\text{(g)} \rightarrow 2\text{CO(g)}\).
(b)(ii) Carbon monoxide reduces iron(III) oxide to iron, forming carbon dioxide: \(\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(l)} + 3\text{CO}_2\text{(g)}\).
(c)(i) Limestone (calcium carbonate, \(\text{CaCO}_3\)) decomposes when heated in the furnace to form calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)).
(c)(ii) Calcium oxide is a basic oxide, and the sand (silicon dioxide, \(\text{SiO}_2\)) is an acidic impurity. They react together in a neutralization reaction to form molten slag (calcium silicate, \(\text{CaSiO}_3\)): \(\text{CaO(s)} + \text{SiO}_2\text{(s)} \rightarrow \text{CaSiO}_3\text{(l)}\).

(a) 1 mark: Iron(III) oxide (accept \(\text{Fe}_2\text{O}_3\)).
1 mark: Coke (accept carbon).
1 mark: Limestone (accept calcium carbonate).

(b)(i) 2 marks: Correct balanced equation: \(\text{C} + \text{CO}_2 \rightarrow 2\text{CO}\). Give 1 mark if formulas are correct but unbalanced.

(b)(ii) 2 marks: Correct balanced equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). Give 1 mark if formulas are correct but unbalanced.

(c)(i) 1 mark: Explains that high heat in furnace causes calcium carbonate to split into calcium oxide and carbon dioxide (or states \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)).

(c)(ii) 1 mark: Explanation that basic calcium oxide reacts with acidic silicon dioxide to form slag (calcium silicate).
1 mark: Balanced equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).

(a) A wavefront is a line or surface connecting all adjacent points on a wave that are vibrating in phase (e.g., all peak crests of water waves).
(b) In a transverse wave, the direction of vibration of the particles (or fields) is perpendicular (at right angles) to the direction of energy transfer (propagation of the wave). An example is light waves (or any electromagnetic wave). In a longitudinal wave, the direction of vibration is parallel to the direction of energy transfer. An example is sound waves.
(c)(i) Using the wave equation: \(v = f \lambda = 2.5\text{ Hz} \times 0.12\text{ m} = 0.30\text{ m/s}\).
(c)(ii) The period \(T\) is the inverse of frequency: \(T = \frac{1}{f} = \frac{1}{2.5} = 0.40\text{ s}\).
(d) Since \(v = f \lambda\) and \(f\) is constant, if speed \(v\) decreases, then the wavelength \(\lambda\) must also decrease.

(a) 1 mark: A line joining points of a wave that are in the same phase / crests of waves.

(b) 1 mark: Transverse wave: oscillation/vibration is perpendicular to wave propagation direction.
1 mark: Transverse example: light / electromagnetic wave / water waves.
1 mark: Longitudinal wave: oscillation/vibration is parallel to wave propagation direction.
1 mark: Longitudinal example: sound wave.

(c)(i) 1 mark: Correct substitution: \(v = 2.5 \times 0.12\).
1 mark: \(0.30\text{ m/s}\) (must include unit).

(c)(ii) 1 mark: Correct formula or substitution: \(T = \frac{1}{2.5}\).
1 mark: \(0.40\text{ s}\) (must include unit).

(d) 1 mark: Wavelength decreases.

(a) The proton number (atomic number) is the lower number, which is 53. The number of neutrons is the nucleon number (mass number) minus the proton number: \(131 - 53 = 78\). So the nucleus has 53 protons and 78 neutrons.
(b) In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting an electron (beta particle). This increases the proton number by 1 (to 54) while the nucleon number remains 131. The equation is:
\(^{131}_{\hphantom{0}53}\text{I} \rightarrow ^{131}_{\hphantom{0}54}\text{Xe} + ^{\hphantom{-}0}_{-1}\beta\) (or \(^{\hphantom{-}0}_{-1}\text{e}\))
Both mass numbers (131 = 131 + 0) and atomic numbers (53 = 54 - 1) balance.
(c)(i) Half-life is the time taken for half the radioactive nuclei in a sample to decay (or the time taken for the activity of a sample to decrease to half its original value).
(c)(ii) The number of half-lives that pass in \(24.0\text{ days}\) is \(\frac{24.0}{8.0} = 3\) half-lives.
After 1 half-life: \(800\text{ Bq} \div 2 = 400\text{ Bq}\)
After 2 half-lives: \(400\text{ Bq} \div 2 = 200\text{ Bq}\)
After 3 half-lives: \(200\text{ Bq} \div 2 = 100\text{ Bq}\).
So, the activity after \(24.0\text{ days}\) is \(100\text{ Bq}\).
(d) Safety precautions include: use long-handled tongs to keep distance from the source; store sources in thick lead-lined containers when not in use; do not point sources towards anyone; wear protective lab clothing / gloves / safety glasses.

(a) 1 mark: 53 protons.
1 mark: 78 neutrons.

(b) 1 mark: Correct Xenon symbol showing nucleon number 131 and proton number 54 (\(^{131}_{54}\text{Xe}\)).
1 mark: Correct beta particle symbol (\(^{0}_{-1}\beta\) or \(^{0}_{-1}\text{e}\)).
1 mark: Completely balanced equation with reactant on left and products on right.

(c)(i) 1 mark: Time taken for half the nuclei of a sample to decay / time for activity to halve.

(c)(ii) 1 mark: Working showing 3 half-lives (e.g. \(24 \div 8 = 3\) or progressive halving: \(800 \rightarrow 400 \rightarrow 200\)).
1 mark: \(100\text{ Bq}\) (must include unit).

(d) 2 marks: Any two valid precautions (1 mark each), e.g., use long-handled tongs, store in lead-lined containers, wear safety equipment, limit exposure time, keep a safe distance.

(a) (i) Pepsin is an acidic protease. As pH increases from 1, the rate of reaction increases to its optimum (around pH 1.5 to 2.0) and then decreases rapidly, reaching zero by pH 7.
(ii) Extreme pH levels alter the ionic charges on the amino acids making up the enzyme. This changes the tertiary structure and deforms the shape of the active site. The substrate can no longer fit into the active site, meaning the enzyme has been denatured.
(b) (i) Proteins are polymers made of amino acids, which are released upon digestion.
(ii) All proteins contain Carbon (C), Hydrogen (H), Oxygen (O), and Nitrogen (N).
(c) Heating to 70 °C increases the kinetic energy of the atoms in the enzyme. The atoms vibrate violently, breaking the weak hydrogen and ionic bonds maintaining the enzyme's specific 3D structure. The active site loses its shape permanently (denaturation), preventing substrate binding.

(a) (i) Max 2 marks:
- Peak/optimum activity at low pH / pH 1.5 to 2.0 [1]
- Rate decreases as pH increases above the optimum / activity is zero at pH 7 [1]
(ii) Max 3 marks:
- active site changes shape [1]
- substrate is no longer complementary / cannot bind [1]
- enzyme is denatured [1]
(b) (i) Max 1 mark:
- amino acids [1]
(b) (ii) Max 2 marks:
- Carbon, hydrogen, oxygen, nitrogen [2] (allow 1 mark if 2 or 3 are correct)
(c) Max 2 marks:
- atoms/molecules gain kinetic energy and vibrate more [1]
- hydrogen/ionic/chemical bonds are broken, destroying the shape of the active site / denaturing the enzyme [1]

(a) (i) The soft iron core is easily magnetised and demagnetised. It concentrates the magnetic field lines, ensuring maximum magnetic flux linkage between the primary and secondary coils.
(ii) Using the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{12}{V_s} = \frac{200}{50}\)
\(V_s = 12 \times \frac{50}{200} = 3.0\text{ V}\)
(iii) It is a step-down transformer because the output voltage is less than the input voltage (or \(N_s < N_p\)).
(b) An alternating current (a.c.) in the primary coil creates a continuously changing magnetic field in the iron core. This changing magnetic field cuts through the secondary coil, inducing an alternating electromotive force (e.m.f.) across it. A direct current (d.c.) would produce a steady, unchanging magnetic field, which does not induce any e.m.f. because there is no relative motion or change in magnetic flux.
(c) Using Ohm's law:
\(I_s = \frac{V_s}{R}\)
\(I_s = \frac{3.0\text{ V}}{3.0\ \Omega} = 1.0\text{ A}\)

(a) (i) Max 2 marks:
- easily magnetised and demagnetised [1]
- concentrates/links the magnetic field from primary to secondary coil [1]
(a) (ii) Max 2 marks:
- formula or substitution: \(V_s = V_p \times \frac{N_s}{N_p}\) or \(\frac{12}{V_s} = \frac{200}{50}\) [1]
- correct calculation: \(3.0\text{ V}\) [1]
(a) (iii) Max 1 mark:
- step-down [1]
(b) Max 3 marks:
- alternating current creates a changing/varying magnetic field [1]
- this changing magnetic field cuts the secondary coil [1]
- inducing an electromotive force / voltage (due to electromagnetic induction) [1]
- direct current produces a constant magnetic field, so no voltage is induced [1]
(c) Max 2 marks:
- substitution: \(I = \frac{3.0}{3.0}\) [1]
- correct value with unit: \(1.0\text{ A}\) (or \(1.0\text{ amperes}\)) [1] (accept 1 A, reject 1 without unit)

(a) (i) The main iron compound in hematite is iron(III) oxide (or ferric oxide, \(\text{Fe}_2\text{O}_3\)).
(ii) The incomplete combustion of carbon (coke) with oxygen is represented by:
\(2\text{C} + \text{O}_2 \rightarrow 2\text{CO}\)
(iii) Carbon monoxide acts as a reducing agent because it removes oxygen from the iron(III) oxide, reducing the iron ions to metallic iron, while itself being oxidised to carbon dioxide. The chemical equation is:
\(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)
(b) (i) Limestone undergoes thermal decomposition in the hot furnace:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
The basic calcium oxide (\(\text{CaO}\)) then reacts with the acidic silicon dioxide (\(\text{SiO}_2\)) impurity in a neutralisation reaction to form calcium silicate (slag):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)
(ii) Slag is used as a material for road building or in the manufacture of cement.

(a) (i) Max 1 mark:
- iron(III) oxide / ferric oxide / \(\text{Fe}_2\text{O}_3\) [1]
(a) (ii) Max 2 marks:
- correct reactant and product formulas: \(\text{C} + \text{O}_2 \rightarrow \text{CO}\) [1]
- correctly balanced equation: \(2\text{C} + \text{O}_2 \rightarrow 2\text{CO}\) [1]
(a) (iii) Max 3 marks:
- explanation: carbon monoxide removes oxygen from iron(III) oxide (or iron ions gain electrons / carbon oxidation state increases) [1]
- equation formula: \(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\) [1]
- balanced equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) [1]
(b) (i) Max 3 marks:
- thermal decomposition equation: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1]
- neutralisation equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\) [1]
- explanation that \(\text{CaO}\) is a basic oxide reacting with acidic \(\text{SiO}_2\) [1]
(b) (ii) Max 1 mark:
- road building / cement manufacture / foundations [1]

(a) (i) An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons.
(ii) During alpha decay, the nucleon (mass) number decreases by 4 and the proton (atomic) number decreases by 2. Thus:
\(A = 220 - 4 = 216\)
\(Z = 86 - 2 = 84\)
(b) (i) Half-life is defined as the time taken for half the radioactive nuclei in a sample to decay, or the time taken for the activity (count rate) of a source to decrease to half of its initial value.
(ii) Number of half-lives elapsed:
\(n = \frac{224\text{ s}}{56\text{ s}} = 4\)
After 1 half-life: \(1600 / 2 = 800\text{ cpm}\)
After 2 half-lives: \(800 / 2 = 400\text{ cpm}\)
After 3 half-lives: \(400 / 2 = 200\text{ cpm}\)
After 4 half-lives: \(200 / 2 = 100\text{ cpm}\)
(c) Alpha particles have a larger mass and charge (+2) compared to beta particles (charge -1, negligible mass), making them highly ionizing but very easily absorbed. Therefore, alpha radiation has higher ionizing ability but lower penetrating power than beta radiation.

(a) (i) Max 2 marks:
- 2 protons [1]
- 2 neutrons [1]
(a) (ii) Max 2 marks:
- \(A = 216\) [1]
- \(Z = 84\) [1]
(b) (i) Max 2 marks:
- time taken for half the (unstable) nuclei to decay / time taken for activity/count rate to halve [2]
- (allow 1 mark for incomplete definition, e.g., 'time for radiation to halve')
(b) (ii) Max 2 marks:
- determination of 4 half-lives (e.g., \(224 / 56 = 4\)) [1]
- correct calculation of final count rate: 100 counts per minute / cpm [1]
(c) Max 2 marks:
- alpha is more ionizing than beta [1]
- alpha is less penetrating than beta / stopped by paper, whereas beta passes through paper (but stopped by aluminium) [1]

Detailed experimental procedure to investigate the effect of pH on amylase activity:
1. Identify the indicators: Iodine turns blue-black with starch and remains yellow-brown when starch is completely digested.
2. Setup buffers of varying pH (e.g., pH 4 to 8) to vary the independent variable.
3. Keep biological/physical parameters constant: temperature (using a thermostatically controlled water-bath or a water-bath kept at a constant temperature, e.g., 35 \(^{\circ}\)C), volumes and concentrations of starch and amylase solutions.
4. Method of sampling: Use a spotting tile with iodine solution. Take samples at fixed intervals (e.g., 30 seconds) until the color change no longer occurs (reaches end-point).
5. Compare the time taken for starch to disappear across the different pH values; the shortest time corresponds to the fastest rate of reaction and hence the optimum pH.

Maximum 8.57 marks:
- (a)(i) Blue-black / black [1 mark]
- (a)(ii) Brown / orange-brown / yellow-brown [1 mark]
- (b) Method points (up to 4 marks):
- Use of buffer solutions to obtain different pH values [1 mark]
- Mixing starch and amylase, and starting timing immediately [1 mark]
- Sampling mixture at regular time intervals and adding to iodine solution on a spotting tile [1 mark]
- Repeating the procedure for at least 3 different pH values [1 mark]
- (b) Control variables & reliability (up to 2.57 marks):
- Mentioning temperature must be controlled (e.g., using a water bath) [1 mark]
- Mentioning volume or concentration of amylase/starch must be kept constant [1 mark]
- Explaining that the reaction is complete when iodine no longer changes color (remains orange/brown) [0.57 mark]

Detailed chemical tests:
1. Ammonium test: Reaction of ammonium salts with warm alkali (NaOH) releases ammonia gas, which is alkaline and turns damp red litmus paper blue.
2. Copper(II) test: Aqueous ammonia reacts with copper(II) ions to initially precipitate copper(II) hydroxide, which is a light blue solid. In excess ammonia, a soluble tetraamine copper complex forms, yielding a deep, dark blue solution.
3. Anion analysis: Acidified silver nitrate is used to precipitate silver chloride (white precipitate). Acidified barium chloride or nitrate is used to precipitate barium sulfate (white precipitate).

Maximum 8.57 marks:
- (a) Test for ammonium:
- Add sodium hydroxide and heat / warm [1 mark]
- Test gas with damp red litmus paper [1 mark]
- Litmus paper turns blue [1 mark]
- (b) Test for copper(II) with ammonia:
- Add aqueous ammonia dropwise: light blue precipitate [1 mark]
- Add excess ammonia: precipitate dissolves to form a deep blue / dark blue solution [1 mark]
- (c) Anion tests:
- For chloride: Add dilute nitric acid and silver nitrate; positive result is a white precipitate [1.57 marks]
- For sulfate: Add dilute hydrochloric (or nitric) acid and barium chloride (or nitrate); positive result is a white precipitate [2 marks]

Measurement accuracy:
1. Aligning the ruler vertically and ensuring it is close to the spring prevents misalignment errors. Parallax error is minimized by looking perpendicular (90 degrees) to the scale.
Calculations:
2. Extension \(x = l_1 - l_0 = 8.6 - 3.2 = 5.4\text{ cm}\).
3. Spring constant \(k = F/x = 4.5 / 5.4 = 0.83\text{ N/cm}\) (or \(4.5 / 0.054 = 83.3\text{ N/m}\)).
Proportionality limit:
4. Up to the limit of proportionality, the graph is a straight line through the origin (Hooke's Law is obeyed). Beyond this point, the relationship is non-linear, and the line curves.

Maximum 8.57 marks:
- (a) Measurement techniques:
- Ruler placed parallel and close to the spring [1 mark]
- Use of a set square / horizontal pointer to align ruler and spring [1 mark]
- Eye level level with the pointer/reading to prevent parallax error [1 mark]
- (b)(i) Correct calculation of extension: \(5.4\text{ cm}\) [1 mark]
- (b)(ii) Calculation of spring constant: \(0.83\text{ N/cm}\) or \(83.3\text{ N/m}\) (accept 0.83 to 0.833) [1 mark]
- Correct unit consistent with calculation (\(\text{N/cm}\) or \(\text{N/m}\)) [1 mark]
- (c) Limit of proportionality / elastic limit [1 mark]
- Graph: line is no longer straight / line starts to curve [1.57 marks]

Circuit and procedures for resistance-length experiment:
1. Standard circuit symbols: Ammeter must be in series with the test wire; Voltmeter must be in parallel across the length of the test wire.
2. Collecting readings systematically by moving a crocodile clip along the wire. Disconnecting the circuit between readings prevents heat buildup.
3. Keeping temperature constant ensures that any change in resistance is due to the change in length alone, confirming a fair test.

Maximum 8.57 marks:
- (a) Circuit diagram:
- Correct symbols for cell/battery, ammeter, voltmeter, and wire [1 mark]
- Ammeter in series with test wire [1 mark]
- Voltmeter connected in parallel across the test wire [1 mark]
- (b) Procedure:
- Connect crocodile clip at different lengths along the ruler [1 mark]
- Record current \(I\) and potential difference \(V\) for each length [1 mark]
- (c)(i) Explanation: Temperature change affects the resistance of the wire / temperature is a control variable [1.57 marks]
- (c)(ii) Prevention: Switch off circuit between readings / use low current [2 marks]

The rate of reaction can be measured by monitoring the volume of carbon dioxide gas evolved over time:
1. Set up a gas syringe connected to a conical flask containing a known mass of marble chips.
2. Add a specific concentration of HCl. Record volume vs. time.
3. Vary the concentration of HCl systematically. To make \(0.5\text{ mol/dm}^3\) from \(1.0\text{ mol/dm}^3\), mix equal volumes of acid and distilled water.
4. Ensure temperature and the surface area of marble chips are constant, as both strongly influence reaction kinetics.

Maximum 8.57 marks:
- Apparatus: Flask, gas syringe (or inverted cylinder/water trough), stopwatch, balance [1 mark]
- Method (Acid dilution): Describes how to vary concentration by mixing different ratios of acid and water [1 mark]
- Method (Execution): Adds acid to calcium carbonate, seals flask, starts stopwatch [1 mark]
- Measurements: Measures volume of gas at regular time intervals (or measures time to collect a fixed volume) [1.57 marks]
- Repetition: Repeats with at least 3 different concentrations of acid [1 mark]
- Comparison/Analysis: Graph of volume vs. time plotted; gradient compared OR reciprocal of time to reach a fixed volume used [1 mark]
- Control variables (any two): Temperature of reactants, mass of marble chips, surface area of chips, total volume of acid [2 marks]

Osmosis investigation analysis:
1. Surface liquid: Removing excess liquid ensures only the mass of the tissue itself is weighed, avoiding overestimation of final mass.
2. Normalization: Percentage change (\(\frac{\text{final} - \text{initial}}{\text{initial}} \times 100\)) standardizes the data across cylinders of slightly different initial masses.
3. Equilibrium point: At the concentration where the line crosses the x-axis (0% change), the water potential of the solution equals the water potential inside the potato cells, so no net osmosis occurs. This point identifies the isotonic concentration.

Maximum 8.57 marks:
- (a) To remove surface liquid / water [1 mark]
- ...which would add extra mass / lead to inaccurate final mass readings [1 mark]
- (b) Initial masses of cylinders are not identical [1.57 marks]
- ...allows direct comparison between different samples [1 mark]
- (c) Graph description:
- Curve/line going from positive percentage change down to negative percentage change [1 mark]
- Intersects the x-axis (0% change) [1 mark]
- Explanation: At the x-intercept, there is no net movement of water [1 mark]
- ...meaning the concentration of the sucrose solution is equal to the concentration inside the potato cells [1 mark]

Refraction through a glass block:
1. Pin alignment method: Looking through the block allows the observer to align the physical pins with the refracted image of the previous pins, defining the entry and exit points.
2. Drawing normal: Both \(i\) and \(r\) are defined with respect to the normal (perpendicular to the glass boundary).
3. Precision: Pin separation is critical. Small errors in alignment translate to large angular errors if pins are close.

Maximum 8.57 marks:
- (a) Pin method details:
- Draw outline of the glass block [1 mark]
- Place two pins on the entry side to define the incident path [1 mark]
- View from opposite side and place two more pins in line with the images of the first two [1 mark]
- (b) Angles measurement:
- Draw normal line perpendicular to the block's surface at the entry point [1 mark]
- Use a protractor to measure angle of incidence \(i\) between normal and incident ray [1 mark]
- Draw the refracted ray inside the block and measure angle of refraction \(r\) between normal and refracted ray [1.57 marks]
- (c) Error & Improvement:
- Identifies error: thickness of pin lines / pins placed too close [1 mark]
- Suggests improvement: place pins far apart (at least 5 cm) / use thin sharp pencil/pins [1 mark]