CBSE Class 10 Science: 2025 board paper, set 31/3/1

1. Balance Al and O: $2\mathrm{Al}+3{\mathrm{H}}_2\mathrm{O}\to \mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2$. Check: 2 Al, 3 O and 6 H on each side.

   1 mark

Answer: p, q, r, s = 2, 3, 1, 3; option (C)

1. Salts with the same acid radical form a family. ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4$, $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$ and $\mathrm{CaS}{\mathrm{O}}_4$ all contain the sulphate ion, so they form the sulphate family.

   1 mark

Answer: Option (B)

1. Metals of moderate reactivity, such as Zn, Fe and Pb, are obtained from their oxides by heating them with carbon (coke). This is the most widely used and cheapest method, e.g. $\mathrm{ZnO}+\mathrm{C}\to \mathrm{Zn}+\mathrm{CO}$.

   1 mark

Answer: Option (A)

1. Structure (ii) is n-butane and structure (iii) is isobutane (2-methylpropane). Both have the formula ${\mathrm{C}}_4{\mathrm{H}}_{10}$ but different structures, so they are structural isomers.

   1 mark

Answer: Option (C), (ii) and (iii)

1. Hydrogenation is the addition of ${\mathrm{H}}_2$ across C=C bonds, which gives a saturated product. It needs a nickel or palladium catalyst, not an acid catalyst, so statement (D) is incorrect.

   1 mark

Answer: Option (D)

1. The matches are:

   • Fertilisation takes place in the oviduct: a-(iii).
   • Implantation takes place in the uterus: b-(ii).
   • Sperms enter through the vagina: c-(i).
   • Wastes of the embryo are removed through the placenta: d-(iv).
   1 mark

Answer: Option (D)

1. The cerebellum, part of the hindbrain, controls posture and the balance of the body.

   1 mark

Answer: Option (C), Cerebellum

1. Cytokinins promote cell division, so they are found in higher concentration where cells divide rapidly, such as in fruits and seeds. Option (B).

   1 mark

Answer: (B) Cytokinins

1. Hibiscus and mustard flowers have both stamens and pistil, so they are bisexual. Papaya and watermelon have unisexual flowers.

   1 mark

Answer: Option (B)

1. Gastric glands secrete HCl, pepsin (which digests proteins) and mucus (which protects the stomach lining). Trypsin comes from the pancreas and salivary amylase from the salivary glands. Option (C).

   1 mark

Answer: (C) Mucus and Pepsin

1. Since $n=c/v$, we have $\frac{v_w}{v_g}=\frac{n_g}{n_w}=\frac{3/2}{4/3}=\frac{9}{8}$.

   ½ mark

2. $v_w=\frac{9}{8}\times 2\times 10^8=\frac{9}{4}\times 10^8=2.25\times 10^8$ m/s. Option (A) shows $\frac{9}{4}$ but leaves out the factor $10^8$.

   ½ mark

Answer: $\frac{9}{4}\times 10^8$ m/s (none of the printed options is exactly correct; (A) is intended)

1. Very fine particles scatter light of shorter wavelength more strongly. Blue has the shortest wavelength among the options, so blue light is scattered the most. Option (C).

   1 mark

Answer: (C) blue

1. Stretching does not change the volume, so when the length becomes $3l$ the area becomes $A/3$.

   ½ mark

2. $R^{\prime }=\rho \frac{3l}{A/3}=9\rho \frac{l}{A}=9R$

   ½ mark

Answer: 9R, option (C)

1. Inside a bar magnet the field lines run from the south pole to the north pole. Statement (D) says the opposite, so it is not true.

   1 mark

Answer: Option (D)

1. The field inside a long solenoid depends on the number of turns, the current and the core material. It does not depend on the radius of the coil. A change in the direction of current reverses the field but does not change its strength.

   1 mark

Answer: Option (D)

1. A self-sustaining aquarium needs aquatic plants and animals and decomposers.

   ½ mark

2. Terrestrial organisms are not needed. Clown fishes and sea urchins are marine organisms and are not required for small herbivorous fishes. Hence (ii) and (iv) are not required.

   ½ mark

Answer: Option (D), (ii) and (iv)

1. The Assertion is true: metals generally do not give $H_2$ with nitric acid.

2. The Reason is false: $HN{O}_3$ is a strong oxidising agent, not a reducing agent. It oxidises the $H_2$ produced to water and is itself reduced to nitrogen oxides. So the answer is option (C).

   1 mark

Answer: (C) A is true, R is false

1. Both statements are true. Voluntary actions like writing or talking involve the nervous system signalling the muscles, and the peripheral nervous system does consist of cranial and spinal nerves.

   ½ mark

2. However, R only describes how the nervous system is organised. It does not explain the communication described in A.

   ½ mark

Answer: Option (B)

1. Both statements are true: field lines never cross, and the magnetic field has both magnitude and direction.

   ½ mark

2. Field lines do not cross because the field can have only one direction at any point. R states only that the field is a vector, so the scheme treats it as not the correct explanation.

   ½ mark

Answer: Option (B)

1. Jute is biodegradable, so it does not persist as waste the way plastic bags do. Jute bags can also be reused many times, which cuts down on waste.

2. Both statements are true, and the Reason explains why jute bags reduce pollution. The answer is option (A).

   1 mark

Answer: (A) Both true; R explains A

1. Decomposition reactions need energy from heat (thermal decomposition), light (photolytic decomposition) or electricity (electrolytic decomposition).

   1 mark

2. Example (light): silver bromide decomposes in sunlight into silver and bromine, which is why it is used in black-and-white photography: $2AgBr(s)\stackrel{\text{Sunlight}}{\to }2Ag(s)+B{r}_2(g)$.

   1 mark

Answer: Heat, light, electricity; e.g. $2AgBr(s)\stackrel{\text{Sunlight}}{\to }2Ag(s)+B{r}_2(g)$

1. Observation: water droplets collect on the cooler upper part of the tube and the green crystals turn white (anhydrous $FeS{O}_4$).

   ½ mark

2. On strong heating, the white solid turns reddish-brown ($F{e}_2{O}_3$) and a pungent smell of burning sulphur is given off ($S{O}_2$, $S{O}_3$).

   ½ mark

3. Loss of water of crystallisation: $FeS{O}_4\cdot 7H_{2}O(s)\stackrel{\text{Heat}}{\to }FeS{O}_4(s)+7H_{2}O(g)$

   ½ mark

4. Decomposition: $2FeS{O}_4(s)\stackrel{\text{Heat}}{\to }F{e}_2{O}_3(s)+S{O}_2(g)+S{O}_3(g)$

   ½ mark

Answer: Water droplets form, green crystals turn white then reddish-brown with a burning-sulphur smell; $FeS{O}_4\cdot 7H_{2}O\to FeS{O}_4+7H_{2}O$, $2FeS{O}_4\to F{e}_2{O}_3+S{O}_2+S{O}_3$

1. (a)(i) Acids give hydrogen ions, $H^{+}$, which exist as hydronium ions $H_3{O}^{+}$ in water.

   ½ mark

2. (a)(ii) Bases give hydroxide ions, $O{H}^{-}$, in water.

   ½ mark

3. (b) Dry HCl gas does not dissociate into $H^{+}$ ions without water, so no acidic behaviour is shown and dry litmus does not change colour.

   1 mark

Answer: (a)(i) $H^{+}$ / $H_3{O}^{+}$; (ii) $O{H}^{-}$; (b) no $H^{+}$ ions are formed without water

1. Veins collect deoxygenated blood from the different organs of the body and carry it back to the heart.

   1 mark

2. Blood in veins flows under low pressure, so thick elastic walls are not needed; valves keep the blood flowing in one direction.

   1 mark

Answer: Veins return deoxygenated blood from organs to the heart; their walls are thin because blood in them is at low pressure.

1. (a) The height of a pea plant depends on the amount of a plant hormone it makes. This hormone is produced by an enzyme (a protein). If the gene gives an efficient enzyme, more hormone is made and the plant is tall; if the gene is altered and the enzyme is less efficient, less hormone is made and the plant is short. So proteins (enzymes) coded by genes control characteristics.

   1 mark

2. (b) The section of DNA that provides information for a protein and controls a characteristic is called a gene.

   1 mark

Answer: (a) Gene → enzyme (protein) → amount of growth hormone → tallness; (b) Gene

1. The student can see distant objects clearly but not nearby ones, so the defect is hypermetropia (far-sightedness).

   ½ mark

2. Reason 1: the focal length of the eye lens is too long.

   ½ mark

3. Reason 2: the eyeball has become too small (shortened).

   ½ mark

4. Correction: a convex (converging) lens of suitable power.

   ½ mark

Answer: Hypermetropia; focal length of eye lens too long, eyeball too small; convex (converging) lens

1. Diagram: draw triangular prism ABC. Incident ray PE strikes face AB obliquely at E; draw the normal at E. The ray bends towards the normal inside the glass along EF to face AC; draw the normal at F. The ray bends away from the normal and emerges along FS, bending towards the base BC.

   1 mark

2. Mark arrows on the incident ray, refracted ray and emergent ray to show the direction of light.

   ½ mark

3. Produce the incident ray forward and the emergent ray backward (dotted) until they meet; the angle between them is the angle of deviation $\angle D$; mark it.

   ½ mark

Answer: Ray diagram through prism with incident ray, refracted ray, emergent ray bent towards base, and angle of deviation D between extended incident and emergent rays.

1. Given $P=1000W$, $V=230V$. Using $P=\frac{V^2}{R}$, so $R=\frac{V^2}{P}$.

   ½ mark

2. $R=\frac{(230{)}^2}{1000}=\frac{52900}{1000}$

   1 mark

3. $R=52.9\Omega$

   ½ mark

Answer: $R=52.9\Omega$

1. (a) The reactivity series is a list of metals arranged in order of decreasing reactivity.

   ½ mark

2. It is developed by performing displacement reactions – a more reactive metal displaces a less reactive metal from its salt solution.

   ½ mark

3. Order: Calcium > Aluminium > Lead > Copper.

   1 mark

4. (b) Thermit reaction: $F{e}_2{O}_3(s)+2Al(s)\to A{l}_2{O}_3(s)+2Fe(l)+\text{Heat}$

   1 mark

Answer: (a) Metals in decreasing order of reactivity, found by displacement reactions; Ca > Al > Pb > Cu; (b) $F{e}_2{O}_3+2Al\to A{l}_2{O}_3+2Fe+\text{heat}$

1. (i) Sodium hydroxide: electricity is passed through brine (aqueous NaCl) – the chlor-alkali process. Chlorine forms at the anode, hydrogen at the cathode and NaOH solution near the cathode.

   ½ mark

2. $2NaCl(aq)+2H_{2}O(l)\to 2NaOH(aq)+C{l}_2(g)+H_2(g)$

   1 mark

3. (ii) Sodium hydrogen carbonate: a cold solution of brine is treated with ammonia and carbon dioxide; sodium hydrogen carbonate separates out and ammonium chloride forms.

   ½ mark

4. $NaCl+H_{2}O+C{O}_2+N{H}_3\to N{H}_{4}Cl+NaHC{O}_3$

   1 mark

Answer: (i) Electrolysis of brine: $2NaCl+2H_{2}O\to 2NaOH+C{l}_2+H_2$; (ii) $NaCl+H_{2}O+C{O}_2+N{H}_3\to N{H}_{4}Cl+NaHC{O}_3$

1. Set-up: fix two iron nails on a rubber cork in a 100 mL beaker; connect them through a bulb, a switch and a 6 V battery. Pour dilute HCl into the beaker and switch on – the bulb glows. Repeat with dilute $H_{2}S{O}_4$ – the bulb glows.

   1 mark

2. Now repeat separately with glucose solution and alcohol solution – the bulb does not glow, so these solutions do not conduct electricity.

   1 mark

3. Reason: acids dissociate in water to give $H^{+}$ ions which carry current; glucose and alcohol contain hydrogen but do not release $H^{+}$ ions in water, so they are not acids.

   1 mark

Answer: Bulb glows with acid solutions but not with glucose/alcohol solutions, since only acids release $H^{+}$ ions in water.

1. (a) Aerobic respiration: pyruvate is broken down in mitochondria into carbon dioxide and water (with much energy).

   ½ mark

2. (a) Anaerobic respiration in human muscles: pyruvate is converted into lactic acid.

   ½ mark

3. (b) Respiration in plants releases carbon dioxide.

   ½ mark

4. (b) Photosynthesis releases oxygen.

   ½ mark

5. (c) Terrestrial animals breathe through lungs (or skin/tracheae).

   ½ mark

6. (c) Fishes use gills to take oxygen dissolved in water.

   ½ mark

Answer: (a) CO₂ + water vs lactic acid; (b) CO₂ vs O₂; (c) lungs vs gills

1. (a) Cross RRYY × rryy gives all RrYy, so all $F_1$ seeds are round and yellow (round and yellow are dominant).

   1 mark

2. (b) $F_2$ combinations: round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1.

   1 mark

3. (c) New combinations (round green, wrinkled yellow) appear because the two traits are inherited independently of each other (independent assortment).

   1 mark

Answer: (a) Round, yellow; (b) RY : Ry(green) : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1; (c) independent inheritance of traits

1. A rainbow is a natural spectrum of seven colours appearing in the sky after a rain shower, seen in the direction opposite to the sun. It is caused by dispersion of sunlight by tiny water droplets in the atmosphere.

   1 mark

2. Diagram: draw a spherical raindrop with sunlight (white light) entering it. At entry the light is refracted and dispersed into colours; the rays then undergo internal reflection at the back surface of the drop, and refract again on leaving the drop. Show red and violet separating, with the observer's eye on the side of the incoming sunlight (sun behind observer).

   1½ marks

3. Label: sunlight, raindrop, refraction and dispersion, internal reflection, refraction, red, violet; mark arrows on all rays.

   ½ mark

Answer: A rainbow is a natural spectrum after rain due to dispersion, refraction and internal reflection of sunlight by water droplets; labelled diagram of a raindrop.

1. (a) Point 1 (near P): field lines are small concentric circles around the wire at P, direction given by the right-hand thumb rule for the current there.

   ½ mark

2. Point 3 (near Q): concentric circles around the wire at Q in the opposite sense, since current at Q flows in the opposite direction (down vs up through the board).

   ½ mark

3. Point 2 (centre): the field line is nearly a straight line, perpendicular to the plane of the loop, as fields from all parts of the loop add up in the same direction.

   ½ mark

4. (b) Factor 1: strength of current in the loop (field ∝ current).

   ½ mark

5. Factor 2: number of turns of the loop (also radius – field decreases as radius increases).

   ½ mark

6. (c) Right-hand thumb rule.

   ½ mark

Answer: (a) Circles around P and Q in opposite senses, straight line at centre; (b) current and number of turns (also radius); (c) right-hand thumb rule

1. (a) Reason 1: energy captured by plants does not go back to the sun, and energy passed to herbivores does not come back to the autotrophs.

   1 mark

2. Reason 2: as energy moves through successive trophic levels, much is lost as heat at each level, so it is no longer available to the previous level.

   1 mark

3. (b) Producer 10000 J → primary consumer 1000 J → secondary consumer 100 J.

   ½ mark

4. Reason: by the 10% law, only about 10% of the energy at one trophic level is transferred to the next.

   ½ mark

Answer: (a) Energy does not return to earlier levels and is lost at each level; (b) 100 J (10% law)

1. (i) I. Silver (Ag) does not react with oxygen even at high temperature.

   ½ mark

2. II. Aluminium (Al) forms a protective oxide layer at ordinary temperature.

   ½ mark

3. III. Potassium (K) catches fire when kept in the open, so it is stored in kerosene.

   ½ mark

4. IV. Copper (Cu) does not burn but gets coated with black copper(II) oxide when heated.

   ½ mark

5. (ii) Amphoteric oxides are metal oxides that react with both acids and bases to form salt and water.

   1 mark

6. With acid: $A{l}_2{O}_3+6HCl\to 2AlC{l}_3+3H_{2}O$

   ½ mark

7. With base: $A{l}_2{O}_3+2NaOH\to 2NaAl{O}_2+H_{2}O$ (sodium aluminate). Since it reacts with both, $A{l}_2{O}_3$ is amphoteric.

   ½ mark

8. (iii) Alkalis are bases that are soluble in water.

   ½ mark

9. Example: sodium hydroxide (NaOH) or potassium hydroxide (KOH).

   ½ mark

Answer: (i) I Ag, II Al, III K, IV Cu; (ii) oxides reacting with both acids and bases; $A{l}_2{O}_3+6HCl\to 2AlC{l}_3+3H_{2}O$, $A{l}_2{O}_3+2NaOH\to 2NaAl{O}_2+H_{2}O$; (iii) water-soluble bases, e.g. NaOH

1. (i)(I) Cinnabar (HgS) is roasted in air to convert it into mercuric oxide: $2HgS(s)+3O_2(g)\stackrel{\text{Heat}}{\to }2HgO(s)+2S{O}_2(g)$

   1 mark

2. On further heating, mercuric oxide is reduced to mercury: $2HgO(s)\stackrel{\text{Heat}}{\to }2Hg(l)+O_2(g)$

   1 mark

3. (II) Copper glance ($C{u}_{2}S$) is partially roasted in air: $2C{u}_{2}S+3O_2(g)\stackrel{\text{Heat}}{\to }2C{u}_{2}O+2S{O}_2(g)$

   1 mark

4. The copper(I) oxide formed reacts with the remaining sulphide (self-reduction): $2C{u}_{2}O+C{u}_{2}S\stackrel{\text{Heat}}{\to }6Cu(s)+S{O}_2(g)$

   1 mark

5. (ii)(I) Silver articles get a black coating of silver sulphide, $A{g}_{2}S$.

   ½ mark

6. (II) Copper articles get a green coating of basic copper carbonate, $CuC{O}_3\cdot Cu(OH{)}_2$.

   ½ mark

Answer: (i) $2HgS+3O_2\to 2HgO+2S{O}_2$, $2HgO\to 2Hg+O_2$; $2C{u}_{2}S+3O_2\to 2C{u}_{2}O+2S{O}_2$, $2C{u}_{2}O+C{u}_{2}S\to 6Cu+S{O}_2$; (ii) Ag₂S; basic copper carbonate

1. (i) X – Stigma; Y – Anther.

   1 mark

2. (ii) The anther produces yellowish pollen grains.

   1 mark

3. (iii) Transfer of pollen from anther to stigma is called pollination.

   1 mark

4. (iv) After pollination, a pollen tube grows out of the pollen grain and travels through the style to reach the ovule in the ovary. The male germ cell fuses with the female germ cell (egg) – fertilisation – forming a zygote.

   1 mark

5. The zygote divides many times to form an embryo inside the ovule; the ovule develops a tough coat and becomes the seed, while the ovary grows into the fruit.

   1 mark

Answer: (i) X stigma, Y anther; (ii) pollen grains; (iii) pollination; (iv) pollen tube → fertilisation → zygote → embryo → ovule becomes seed

1. (i) Binary fission (longitudinal, in a definite plane).

   ½ mark

2. (ii) The flagellated organism is Leishmania.

   ½ mark

3. (iii) Advantage 1: a large number of offspring are produced in a short time.

   1 mark

4. Advantage 2: only one parent is needed and no gametes are formed (offspring are genetically similar).

   1 mark

5. (iv) Mode: Budding in Hydra.

   1 mark

6. A bud forms as an outgrowth due to repeated cell division at a specific site; it develops into a tiny individual and, on maturing, detaches from the parent to become a new independent Hydra.

   1 mark

Answer: (i) Binary fission; (ii) Leishmania; (iii) many offspring quickly, single parent/no gametes; (iv) budding

1. (i) Why the emergent ray is parallel to the incident ray: At the first face, light bends towards the normal as it goes from air into glass. At the second face, it bends away from the normal as it goes from glass into air. The two faces are parallel and the same pair of media is involved, so the bending at the second face is equal and opposite to the bending at the first face. The angle of emergence therefore equals the angle of incidence ($\angle e=\angle i$). The emergent ray is parallel to the incident ray, only shifted sideways (lateral displacement).

   1 mark

2. Ray diagram (glass slab): Draw a rectangle ABCD (the slab).

   • Incident ray PE meets face AB at E; draw normal $N_{1}E{N}_1^{\prime }$ and mark $\angle i$.
   • Refracted ray EF inside the glass bends towards the normal; mark $\angle r$.
   • At face CD, draw normal $N_{2}F{N}_2^{\prime }$. The emergent ray FS bends away from the normal; mark $\angle e$.
   • Show FS parallel to PE: extend PE as a dotted line and mark the lateral displacement between it and FS.
   • Put arrows on all rays.
   1 mark

3. Normal incidence: If a ray falls normally on a face (angle of incidence $=0^{\circ }$), it enters the glass without bending, because $\angle r=0^{\circ }$. Diagram: a ray drawn perpendicular to one face passes straight through the glass with no deviation at that face, with an arrow showing its direction.

   1 mark

4. (ii) For a concave lens, $f=-20$ cm. The object is on the left, so $u=-30$ cm.

   Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, so $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-20}+\frac{1}{-30}$

   1 mark

5. $\frac{1}{v}=\frac{-3-2}{60}=-\frac{5}{60}=-\frac{1}{12}$, so $v=-12$ cm

   ½ mark

6. The image is formed 12 cm from the optical centre, on the same side of the lens as the object. It is virtual and erect.

   ½ mark

Answer: (i) Bending at the two parallel faces is equal and opposite, so the emergent ray is parallel to the incident ray; a ray incident normally passes undeviated. (ii) v = −12 cm, i.e. 12 cm from the lens on the same side as the object (virtual, erect).

1. (i) Type of image:

   • When the candle is beyond F, the image is real and inverted.
   • When the candle is between P and F, the image is virtual and erect.
   1 mark

2. Change observed: As the candle is moved gradually away from the pole, the real image moves closer towards F and becomes smaller (diminished).

   1 mark

3. Ray diagram (object at C): Draw the concave mirror with its principal axis, and mark P, F and C.

   • Place the candle AB at C.
   • Draw a ray from A parallel to the axis; after reflection it passes through F.
   • Draw a ray from A through F; after reflection it travels parallel to the axis.
   • The two reflected rays meet below the axis at A′. The image A′B′ is at C: real, inverted and of the same size as the object.
   • Put arrows on all rays.
   1 mark

4. (ii) For a convex mirror, $f=+3.00$ m and $u=-6.00$ m.

   Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, so $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{3.00}-\frac{1}{(-6.00)}$

   1 mark

5. $\frac{1}{v}=\frac{2+1}{6.00}=\frac{1}{2.00}$, so $v=+2.0$ m

   ½ mark

6. The image is 2.0 m behind the mirror. It is virtual, erect and diminished.

   ½ mark

Answer: (i) Real and inverted (object beyond F) or virtual and erect (object between P and F); the image becomes smaller as the candle moves away; with the object at C, the image is at C, real, inverted and same size. (ii) v = +2.0 m, i.e. 2.0 m behind the mirror.

1. (a) Test tube III (tubewell water + salt A) forms the least foam. Tubewell water is hard, and salt A is soap, which does not lather well in hard water.

   1 mark

2. (b) Salt A is a soap: a sodium or potassium salt of a long-chain carboxylic acid, which does not lather well with hard water. Salt B is a detergent: a sodium salt of a sulphonic acid (or an ammonium salt with chloride or bromide ions), which lathers even with hard water.

   1 mark

3. (c)(i) Esters are sweet-smelling compounds used in perfumes and flavouring agents. They are formed when a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (esterification).

   ½ mark

4. When an ester is treated with an alkali such as NaOH, it is converted back into an alcohol and the sodium salt of the carboxylic acid. This reaction is called saponification.

   ½ mark

5. $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}$

   1 mark

6. (c)(ii) (Alternative to (c)(i), also worth 2 marks.)

   • Cause of hardness: water contains dissolved calcium and magnesium salts (bicarbonates, chlorides, sulphates).
   • Insoluble substance: scum.
   • Cause of its formation: soap molecules react with the $\mathrm{C}{\mathrm{a}}^{2+}$ and $\mathrm{M}{\mathrm{g}}^{2+}$ ions in the water and form an insoluble precipitate (scum), so little foam is formed.

Answer: (a) Test tube III; (b) A is a soap (Na/K salt of long-chain carboxylic acid, poor lather in hard water), B is a detergent (sulphonate salt, lathers in hard water); (c)(i) esters are sweet-smelling compounds; with NaOH they give alcohol + sodium salt of the acid (saponification): CH₃COOC₂H₅ + NaOH → C₂H₅OH + CH₃COONa; (c)(ii) Ca/Mg salts cause hardness; the insoluble substance is scum, formed by soap reacting with Ca²⁺/Mg²⁺ ions.

1. (a)(i) The hormone secreted is adrenaline.

   1 mark

2. (a)(ii) (Alternative to (a)(i).) The source gland is the adrenal gland, which lies on top of each kidney.

3. (b) Response 1: the heart beats faster, so more oxygen is supplied to the muscles.

   1 mark

4. (b) Response 2: the breathing rate increases because the diaphragm and rib muscles contract more. Blood supply to the digestive system and skin is reduced and diverted to the skeletal muscles.

   1 mark

5. (c) A hormone (chemical signal) travels through the bloodstream and can reach a wide range of target cells all over the body; its action is slower and longer-lasting.

   An electrical impulse travels only along nerve cells, reaches only the cells connected to those nerves, and acts quickly and briefly.

   1 mark

Answer: (a)(i) Adrenaline / (a)(ii) Adrenal gland; (b) faster heartbeat (more O₂ to muscles) and faster breathing (blood diverted from digestive system and skin to skeletal muscles); (c) hormones travel via blood to many target cells (slower, longer-lasting), whereas nerve impulses travel only along nerve cells (fast, localised).

1. (a) The graphs have I on the y-axis and V on the x-axis, so slope $=\frac{I}{V}=\frac{1}{R}$. The series combination has the largest resistance, so it has the smallest slope: Graph A.

   1 mark

2. (b) The parallel combination has the least resistance (less than both $R_1$ and $R_2$), so it has the steepest slope: Graph D.

   1 mark

3. (c)(i) Arrangement: connect two $10\Omega$ resistors in parallel, and connect this pair in series with the third $10\Omega$ resistor.

   Parallel part: $R_p=\frac{R_1{R}_2}{R_1+R_2}=\frac{10\times 10}{10+10}=5\Omega$

   1 mark

4. Total resistance: $R=R_p+R_3=5\Omega +10\Omega =15\Omega$

   1 mark

5. (c)(ii) (Alternative to (c)(i), also worth 2 marks.)

   • Total resistance in series: $R_s=0.1+0.2+0.3+0.4+0.5=1.5\Omega$
   • Current: $I=\frac{V}{R_s}=\frac{6\text{ V}}{1.5\Omega }=4.0$ A
   • Justification: in a series circuit the same current flows through every resistor, so 4.0 A flows through the $0.3\Omega$ resistor.

Answer: (a) Graph A; (b) Graph D; (c)(i) two 10 Ω resistors in parallel (5 Ω) in series with the third 10 Ω gives 15 Ω; (c)(ii) 4.0 A, because the same current flows through all resistors in series.