CBSE Class 10 Science: 2024 board paper, set 31/4/1

1. Zinc is more reactive than hydrogen, so it displaces hydrogen from ethanoic acid, forming zinc ethanoate and ${\mathrm{H}}_2$. This is a displacement reaction.

   1 mark

Answer: (B) Displacement reaction

1. A salt that turns blue litmus red is acidic, so it comes from a strong acid and a weak base.

2. HCl is a strong acid and $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$ is a weak base, so they give $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$, which is acidic.

   1 mark

Answer: (D) HCl and NH4OH

1. Acids give ${\mathrm{H}}^{+}$ ions in water and so conduct electricity. Glucose and alcohol do not ionise, so they do not conduct. The bulb glows with HCl and ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$.

   1 mark

Answer: (D) Hydrochloric acid and sulphuric acid

1. Metals in the middle of the reactivity series, such as copper and silver, occur both free and as sulphide or oxide ores. Gold and platinum occur only in the free state.

   1 mark

Answer: (C) Copper and silver

1. The benzene ring has alternate bonds: 3 C=C double bonds and 3 C–C single bonds.

2. There are also 6 C–H single bonds, so single bonds $=3+6=9$ and double bonds $=3$.

   1 mark

Answer: (B) 9 and 3

1. At the end of digestion, proteins become amino acids, carbohydrates become glucose, and fats become fatty acids and glycerol.

   1 mark

Answer: (B)

1. Abscisic acid is the growth-inhibiting hormone; it causes wilting of leaves.

   1 mark

Answer: (C) Abscisic acid

1. Sugarcane, roses and grapes are grown by layering or grafting. Mustard and papaya are grown from seeds, so the other options are ruled out.

   1 mark

Answer: (A) Sugarcane, roses, grapes

1. Cotyledons store food that the embryo uses during germination.

   1 mark

Answer: (A) Cotyledon

1. Each human gamete has 23 chromosomes, so the zygote has $23+23=46$ chromosomes.

   1 mark

Answer: (C) 46

1. Since $n=c/v$, the product $nv=c$ is the same for both media: $n_g{v}_g=n_w{v}_w$.

2. $\frac{3}{2}\times 2\times 10^8=\frac{4}{3}{v}_w$, so $v_w=\frac{9}{4}\times 10^8$ m/s.

   1 mark

Answer: (A) 9/4 × 10^8 m/s

1. Very fine particles scatter shorter wavelengths the most, so blue light is mainly scattered.

   1 mark

Answer: (C) Blue

1. I: $2\parallel 2=1\Omega$. II: $3\parallel 3\parallel 3=1\Omega$.

   1 mark

2. III: $6\parallel 6=3\Omega$. IV: $0.5\parallel 0.5=0.25\Omega$. So only I and II give $1\Omega$.

Answer: (C) I and II

1. $H=I^{2}Rt=5^2\times 20\times 30=15000$ J

   1 mark

Answer: (A) 15000 J

1. XY and AB carry current in the same direction (upward), so they attract each other. The opposite current in CD is farther away, so its repulsion is weaker.

   1 mark

2. The net force is attractive, so XY moves towards side AB.

Answer: (B) move towards side AB

1. Plastic refills and glass bottles are not broken down by microbes. Garden waste, fruit and vegetable peels and cotton are biodegradable.

   1 mark

Answer: (B) (ii) and (iii)

1. A is true: sulphide ores are roasted before reduction.

2. R is false: roasting converts sulphide ores into metal oxides, which are then reduced. It does not directly give the metal in general.

   1 mark

Answer: (C)

1. Ventricles pump blood to the lungs and the whole body, so they need thick muscular walls. R correctly explains A.

   1 mark

Answer: (A)

1. At a point of intersection, the field would have two directions, which is impossible. So field lines never intersect, and R correctly explains A.

   1 mark

Answer: (A)

1. A is true: aerobic organisms need ${\mathrm{O}}_2$ for respiration. R is also true ($O+O_2\to O_3$), but ozone formation does not explain why oxygen is essential.

   1 mark

Answer: (B)

1. (i) Concentrated sulphuric acid reacts with sodium chloride to give hydrogen chloride (HCl) gas: $\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{NaHS}{\mathrm{O}}_4+\mathrm{HCl}$.

   ½ mark

2. (ii)(I) With dry blue litmus paper there is no change in colour, because dry HCl gas does not release ${\mathrm{H}}^{+}$ ions.

   ½ mark

3. (ii)(II) Wet blue litmus paper turns red, because HCl dissolves in the water on the paper and forms ${\mathrm{H}}^{+}$ ions.

   ½ mark

4. Conclusion: HCl gas is acidic in nature. It shows its acidic behaviour only in the presence of water.

   ½ mark

Answer: (i) Hydrogen chloride (HCl) gas; (ii) (I) dry blue litmus shows no change, (II) wet blue litmus turns red; the gas is acidic.

1. Example: zinc granules react with dilute sulphuric acid to form the salt zinc sulphate and hydrogen gas:
   $\mathrm{Zn}(\mathrm{s})+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{ZnS}{\mathrm{O}}_4(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

2. Test: bring a burning matchstick or candle near the mouth of the tube. The gas burns with a pop sound, which confirms that it is hydrogen.

   1 mark

Answer: $\mathrm{Zn}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{ZnS}{\mathrm{O}}_4+{\mathrm{H}}_2$; hydrogen burns with a pop sound when a burning matchstick is brought near it.

1. A saturated hydrocarbon burns with a clean blue flame, because combustion is complete.

   ½ mark

2. An unsaturated hydrocarbon burns with a yellow, sooty flame that gives a lot of black smoke, because combustion is incomplete.

   ½ mark

3. Products of combustion of a saturated hydrocarbon: carbon dioxide and water, along with heat and light. For example, $\mathrm{C}{\mathrm{H}}_4+2{\mathrm{O}}_2\to \mathrm{C}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$ + heat + light.

   1 mark

Answer: Saturated: clean blue flame; unsaturated: yellow sooty flame. Products: CO₂ and H₂O (with heat and light).

1. The roots absorb water from the soil. This water enters the xylem vessels, which form a continuous channel from the roots through the stem to the leaves.

   1 mark

2. Water evaporates from the leaves through the stomata (transpiration). This creates a suction force called the transpiration pull, which pulls the water column up through the xylem from the roots to the leaves.

   1 mark

Answer: Water absorbed by the roots moves up through the xylem, pulled by the suction (transpiration pull) created by loss of water from the stomata of the leaves.

1. Amoeba shows heterotrophic (holozoic) nutrition.

   1 mark

2. Amoeba pushes out temporary finger-like projections called pseudopodia. These surround the food particle and fuse over it to form a food vacuole. Inside the food vacuole, complex food is broken down into simpler substances, which diffuse into the cytoplasm. The undigested material is thrown out through the cell surface.

   1 mark

Answer: Holozoic (heterotrophic) nutrition. Food is engulfed by pseudopodia into a food vacuole, where it is digested into simpler substances.

1. Variations arise during reproduction, so the individuals of a population are not all identical. Some of them may carry a variation that helps them cope with a changed environment.

   1 mark

2. Example: in a population of bacteria living in temperate water, suppose a few bacteria have a variation that lets them tolerate heat. If the water temperature rises sharply (for example in a heat wave due to global warming), most bacteria die but the heat-resistant variants survive and reproduce. In this way the species survives, so variation promotes its survival.

   1 mark

Answer: Example: heat-tolerant variant bacteria survive a rise in water temperature while the others die, so the species continues.

1. Distant vision is corrected by a concave (diverging) lens, which has negative power. So the lens of power **$-4.0$ D** is for distant vision.

   1 mark

2. Focal length: $f=\frac{1}{P}=\frac{1}{-4.0\text{D}}=-0.25\text{m}=-25\text{cm}$

   1 mark

Answer: The −4.0 D lens corrects distant vision; its focal length is −0.25 m (−25 cm).

1. Wire A (diameter 0.2 mm) offers more resistance.

   1 mark

2. Justification: $R=\rho \frac{l}{A}$. For the same material and length, $R\propto \frac{1}{A}$, and $A=\frac{\pi d^2}{4}$, so $R\propto \frac{1}{d^2}$. Hence $\frac{R_A}{R_B}={\left(\frac{0.3}{0.2}\right)}^2=2.25$. The thinner wire A has the smaller area of cross-section, so it has the greater resistance.

   1 mark

Answer: Wire A, because resistance is inversely proportional to area of cross-section (R_A = 2.25 R_B).

1. Common name: Plaster of Paris. Chemical name: calcium sulphate hemihydrate.

   1 mark

2. Preparation: gypsum is heated at 373 K, and it loses part of its water of crystallisation:
   $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{373\mathrm{K}}{\to }\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

   1 mark

3. Reaction with water: Plaster of Paris mixed with water sets into a hard mass of gypsum:
   $\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\to \mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: Plaster of Paris; calcium sulphate hemihydrate. It is made by heating gypsum at 373 K. $\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\to \mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$

1. It is an oxidation reaction because oxygen is added to ethanol (and hydrogen is removed) to form ethanoic acid.

   ½ mark

2. Oxidising agent: alkaline potassium permanganate (or acidified potassium dichromate).

   ½ mark

3. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{alkaline }\mathrm{KMn}{\mathrm{O}}_4+\text{heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   1 mark

4. Difference: this is a controlled oxidation in which ethanol changes into another organic compound, ethanoic acid. When ethanol burns in oxygen, it undergoes complete combustion to give carbon dioxide and water with a lot of heat and light: ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+3{\mathrm{O}}_2\to 2C{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$ + heat + light.

   1 mark

Answer: Oxygen is added; oxidising agent is alkaline KMnO₄ or acidified K₂Cr₂O₇; $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$; burning is complete combustion giving CO₂, H₂O, heat and light.

1. In Hydra, a small outgrowth called a bud develops at one specific site on the body, due to repeated cell division.

   ½ mark

2. The bud grows into a tiny individual with tentacles and a mouth. When fully mature, it detaches from the parent and becomes a new, independent Hydra.

   1 mark

3. Labelled diagram (in words): a row of stages showing the parent Hydra with tentacles; a small bud on the body wall; the bud with developing tentacles; and the bud separating as a young Hydra. Labels: parent Hydra, tentacles, bud, new Hydra.

   1 mark

4. The cells used are regenerative cells.

   ½ mark

Answer: A bud forms by repeated cell division at one site, grows into a tiny Hydra and detaches. Regenerative cells are used.

1. (i) Seminal vesicles and prostate gland: their secretions provide nourishment to the sperms, and make the transport of sperms easier by forming a fluid medium.

   1 mark

2. (ii) Oviduct: it carries the egg from the ovary to the uterus, and it is the site of fertilisation.

   1 mark

3. (iii) Testis: it produces sperms, and it secretes the hormone testosterone.

   1 mark

Answer: (i) Nourish sperms and ease their transport; (ii) carries the egg to the uterus and is the site of fertilisation; (iii) produces sperms and secretes testosterone.

1. Difference 1: when both factors are present together (as in Yy), the dominant trait is expressed, whereas the recessive trait is not expressed.

   1 mark

2. Difference 2: a single copy of the dominant factor is enough for the trait to show, whereas the recessive trait shows only when both copies are recessive (yy).

   1 mark

3. Cross: YY × yy gives F₁ all Yy (yellow). Selfing F₁ (Yy × Yy) gives F₂ in the ratio YY : Yy : yy = 1 : 2 : 1. Yellow (YY + Yy) = 3/4, so 75% of the plants had yellow seeds.

   1 mark

Answer: 75% of the F₂ plants had yellow seeds.

1. Power of accommodation is the ability of the eye lens to adjust its focal length so that both near and distant objects can be seen clearly.

   1 mark

2. The image distance remains unchanged. The image is always formed on the retina, at a fixed distance from the lens.

   1 mark

3. The part responsible is the ciliary muscles.

   ½ mark

4. When the object is moved farther away, the ciliary muscles relax. The eye lens becomes thinner and its focal length increases, so the image still forms on the retina.

   ½ mark

Answer: Power of accommodation is the ability of the eye lens to adjust its focal length. The image distance remains the same. The ciliary muscles relax, making the lens thin and increasing its focal length.

1. (a) A magnetic field exists around the bar magnet. It exerts forces on the iron filings, which align along the field.

   ½ mark

2. (b) The crowding of filings at the ends shows that the magnetic field is strongest near the poles.

   1 mark

3. (c) The lines represent magnetic field lines.

   ½ mark

4. (d) Inside the solenoid, the filings arrange in straight, parallel, equally spaced lines.

   ½ mark

5. Conclusion: the magnetic field inside a current-carrying solenoid is uniform (the same at all points).

   ½ mark

Answer: (a) Magnetic field around the magnet; (b) field is strongest at the poles; (c) magnetic field lines; (d) parallel equidistant straight lines, so the field inside the solenoid is uniform.

1. (i) Food chain (b) belongs to a terrestrial (grassland/cropland) ecosystem.

   ½ mark

2. (i) Food chain (c) belongs to an aquatic (pond) ecosystem.

   ½ mark

3. (ii) Producers (green plants) are the first trophic level because only they can capture solar energy and convert it into chemical energy (food) by photosynthesis. All other organisms depend on them for food.

   ½ mark

4. (ii) Producers capture about 1% of the solar energy falling on the leaves.

   ½ mark

5. (iii) The arrows point in one direction because the flow of energy is unidirectional.

   ½ mark

6. (iii) Justification: energy that passes from one trophic level to the next (for example, from plants to herbivores) does not come back to the previous level.

   ½ mark

Answer: (i) (b) terrestrial/grassland, (c) aquatic/pond; (ii) producers convert solar energy into chemical energy, capturing about 1%; (iii) energy flow is unidirectional and cannot revert.

1. A chemical reaction is a process in which bonds between atoms in the reactants are broken and new bonds are formed, producing new substances (products) with different properties.

   1 mark

2. (i) Change of colour, activity: take potassium iodide solution in a test tube and add lead nitrate solution to it. The colourless solutions give a yellow precipitate of lead iodide.

   1 mark

3. $\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2(\mathrm{aq})+2KI(\mathrm{aq})\to \mathrm{Pb}{\mathrm{I}}_2(\mathrm{s})+2KN{\mathrm{O}}_3(\mathrm{aq})$. The yellow precipitate shows that a chemical change has occurred.

   1 mark

4. (ii) Change in temperature, activity: take a small amount of calcium oxide (quicklime) in a beaker and slowly add water. The reaction is vigorous and the beaker becomes hot.

   1 mark

5. $\mathrm{CaO}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Ca}(\mathrm{OH}{)}_2(\mathrm{aq})$ + heat. The release of heat (rise in temperature) shows a chemical change; slaked lime is formed.

   1 mark

Answer: Chemical reaction: reactants change into new substances by breaking and forming bonds. (i) Pb(NO₃)₂ + KI gives a yellow PbI₂ precipitate; (ii) CaO + H₂O gives Ca(OH)₂ with heat released.

1. (i) A decomposition reaction is one in which a single reactant breaks down into two or more simpler products.

   1 mark

2. (I) In electrolysis, water (a single reactant) splits into hydrogen and oxygen: $2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$. So it is a decomposition reaction.

   ½ mark

3. Energy involved: electrical energy.

   ½ mark

4. (II) Silver bromide breaks down into silver (which is black, hence the blackening) and bromine: $2AgBr(\mathrm{s})\stackrel{\mathrm{sunlight}}{\to }2Ag(\mathrm{s})+\mathrm{B}{\mathrm{r}}_2(\mathrm{g})$. So it is a decomposition reaction.

   ½ mark

5. Energy involved: light energy.

   ½ mark

6. (ii)(I) Formation of calcium oxide: $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\stackrel{\mathrm{heat}}{\to }\mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$

   ½ mark

7. This is a decomposition reaction, and it is endothermic (heat is absorbed).

   ½ mark

8. (II) Formation of calcium hydroxide: $\mathrm{CaO}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Ca}(\mathrm{OH}{)}_2(\mathrm{aq})$ + heat

   ½ mark

9. This is a combination reaction, and it is exothermic (heat is released). Decomposition is the opposite of combination, and endothermic is the opposite of exothermic, so the two reactions are opposite to each other.

   ½ mark

Answer: (i) Decomposition: a single reactant breaks into simpler products. (I) H₂O → H₂ + O₂ using electrical energy; (II) AgBr → Ag + Br₂ using light energy. (ii) CaCO₃ → CaO + CO₂ is decomposition/endothermic; CaO + H₂O → Ca(OH)₂ + heat is combination/exothermic, so they are opposite.

1. (i) A reflex arc is the pathway along which nerve impulses travel during a reflex action, from the receptor through the spinal cord to the effector.

   1 mark

2. Reflex arcs have evolved because the thinking process of the brain is not fast enough. A reflex arc gives a quick, automatic response that protects the body from injury.

   ½ mark

3. Sequence on touching a hot object: receptors in the skin detect the heat → sensory neuron carries the impulse to the spinal cord → relay neuron in the spinal cord passes it on → motor neuron carries the impulse to the effector (arm muscle) → the muscle contracts and the hand is withdrawn.

   1½ marks

4. (ii) The peripheral nervous system carries communication between the central nervous system and other parts of the body.

   1 mark

5. Its two components are the cranial nerves (arising from the brain) and the spinal nerves (arising from the spinal cord).

   1 mark

Answer: (i) A reflex arc is the pathway of impulses in a reflex action; it evolved for a quick response because the brain is slow. Sequence: receptor → sensory neuron → spinal cord (relay neuron) → motor neuron → muscle (effector). (ii) Peripheral nervous system: cranial nerves and spinal nerves.

1. (i) The stimulus is touch.

   ½ mark

2. Cause: the plant cells change shape by changing the amount of water in them. Cells near the base of the leaves lose water, which causes the leaves to fold and droop. Electrical-chemical signals carry the information from cell to cell.

   1 mark

3. No, there is no growth involved in this movement.

   ½ mark

4. (ii) Geotropism is the growth movement of a plant part in response to gravity.

   1 mark

5. Positive geotropism: growth of a plant part towards gravity. Example: roots grow downwards.

   1 mark

6. Negative geotropism: growth of a plant part away from gravity. Example: shoots grow upwards.

   1 mark

Answer: (i) Touch; change in water content of cells changes their shape; no growth. (ii) Geotropism is growth in response to gravity: positive (towards gravity), e.g. roots; negative (away from gravity), e.g. shoots.

1. (i) In observation S. No. 3, $u=-50$ cm and $v=+50$ cm. The object distance and the image distance are equal in size, which happens only when the object is at $2F_1$; the image then forms at $2F_2$.

   1 mark

2. (i) So $2f=50$ cm, giving focal length $f=25$ cm.

   1 mark

3. (ii) Observation S. No. 6 ($u=-15$ cm, $v=+37.5$ cm) is not correct.

   ½ mark

4. (ii) Reason: $u=15$ cm is less than $f=25$ cm, so the object lies between $F_1$ and the optical centre $O$. The image is then virtual, erect and on the same side as the object, so $v$ must be negative. Check: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{25}-\frac{1}{15}=-\frac{2}{75}$, so $v=-37.5$ cm.

   ½ mark

5. (ii) Ray diagram, described in words:

   • Draw a convex lens with its principal axis, and mark $F_1$ and $F_2$ at 25 cm on either side of $O$.
   • Place the object AB upright at 15 cm from $O$, between $F_1$ and $O$.
   • Ray 1 leaves B parallel to the principal axis and, after refraction, passes through $F_2$.
   • Ray 2 leaves B, passes through $O$ and goes on undeviated.
   • The two refracted rays diverge, so produce them backwards as dotted lines. They meet at B′ on the object side, at about 37.5 cm from the lens.
   • A′B′ is virtual, erect and enlarged.
   • Show arrowheads on all rays; ½ mark is deducted if ray directions are missing.
   1 mark

6. (iii) For S. No. 5: $m=\frac{v}{u}=\frac{+150\text{ cm}}{-30\text{ cm}}=-5$. The image is real, inverted and 5 times the size of the object.

   1 mark

Answer: (i) f = 25 cm (u = v = 50 cm ⇒ object at 2F); (ii) S. No. 6 is wrong – object inside F gives a virtual image, v ≈ −37.5 cm; (iii) m = −5

1. (i) Principal axis: the imaginary straight line passing through the two centres of curvature of the lens. It passes through the optical centre.

   1 mark

2. (i) Ray diagram, described in words:

   • Draw a concave lens with its principal axis, and mark the optical centre $O$ and the principal focus $F_1$ on the same side as the incoming light.
   • A ray parallel to the principal axis strikes the lens and, after refraction, bends away from the axis (diverges).
   • Produce the refracted ray backwards as a dotted line; it appears to come from $F_1$.
   • Show arrows on the rays.
   1 mark

3. (ii) For a concave lens, $f=-20$ cm. The image is virtual and on the object side, so $v=-15$ cm. Object height $h=+5$ cm.

   Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$

   ½ mark

4. $\frac{1}{u}=\frac{1}{-15}-\frac{1}{-20}=\frac{-4+3}{60}$

   ½ mark

5. $\frac{1}{u}=-\frac{1}{60}$, so $u=-60$ cm. The object must be placed 60 cm in front of the lens.

   1 mark

6. Magnification: $m=\frac{h^{\prime }}{h}=\frac{v}{u}\Rightarrow h^{\prime }=\frac{v}{u}\times h=\frac{-15}{-60}\times 5=+1.25$ cm. The image is 1.25 cm tall, erect and diminished.

   1 mark

Answer: (ii) Object at 60 cm from the lens (u = −60 cm); image height = +1.25 cm (erect, diminished)

1. (a) In the test tube containing magnesium. Magnesium is the most reactive of the three metals, so it gives hydrogen bubbles fastest and releases the most heat.

   1 mark

2. (b) None. Magnesium, aluminium and iron all react with dilute HCl, because all three are more reactive than hydrogen and lie above it in the reactivity series. Iron reacts the slowest. (Credit is also given for naming a less reactive metal such as copper, which lies below hydrogen and cannot displace it from the acid.)

   1 mark

3. (c)(i) Dilute nitric acid is a strong oxidising agent. It oxidises the hydrogen produced to water, so no hydrogen gas is released.

   1 mark

4. (c)(i) The ultimate products are water and oxides of nitrogen (N${}_2$O, NO or NO${}_2$), along with the metal nitrate. (Exception: Mg and Mn give H${}_2$ with very dilute HNO${}_3$.)

   1 mark

5. (c)(ii) Reactivity of metals is decided by displacement reactions.

   1 mark

6. (c)(ii) Dip metal X in a solution of a salt of Y.

   • If X displaces Y (Y gets deposited or the solution changes colour), then X is more reactive than Y.
   • If no change occurs, Y is more reactive. You can confirm this by dipping Y in a salt solution of X, where Y will displace X.

   Example: $\text{Fe(s)}+{\text{CuSO}}_4\text{(aq)}\to {\text{FeSO}}_4\text{(aq)}+\text{Cu(s)}$, so Fe is more reactive than Cu.

   1 mark

Answer: (a) Magnesium; (b) none – all three are above hydrogen in the reactivity series; (c)(i) HNO₃ is a strong oxidiser and oxidises H₂ to water; products: water and oxides of nitrogen; (c)(ii) displacement reaction – X is more reactive if it displaces Y from Y's salt solution

1. (a)(i) The renal artery brings oxygenated blood to the kidney.

   ½ mark

2. (a)(ii) The cluster of thin-walled capillaries inside Bowman's capsule is the glomerulus.

   ½ mark

3. (b) Urine is stored in the urinary bladder.

   ½ mark

4. (b) The bladder is muscular and under nervous control, so we can control the urge to pass urine.

   ½ mark

5. (c)(i) Step 1, glomerular filtration: in the glomerulus, blood is filtered under pressure into Bowman's capsule.

   ½ mark

6. (c)(i) Function: it removes nitrogenous wastes such as urea and uric acid from the blood, along with water, glucose, amino acids and salts.

   ½ mark

7. (c)(i) Step 2, selective (tubular) reabsorption: the filtrate flows through the tubule.

   ½ mark

8. (c)(i) Function: useful substances such as glucose, amino acids, salts and a major amount of water are taken back into the surrounding blood capillaries. What remains becomes urine.

   ½ mark

9. (c)(ii) Selective reabsorption takes place in the tubular part of the nephron.

   1 mark

10. (c)(ii) The amount of water reabsorbed depends on:

    • how much excess water there is in the body;
    • how much dissolved waste has to be excreted.
    1 mark

Answer: (a) (i) Renal artery (ii) Glomerulus; (b) Urinary bladder, nervous control; (c)(i) Filtration – removes urea and other wastes from blood; reabsorption – returns glucose, amino acids, salts and water to blood; (c)(ii) Tubular part of the nephron; depends on excess water in the body and the amount of dissolved waste to be excreted

1. (a) Between A and B, the 4 Ω, 6 Ω and 16 Ω resistors are in series: $R_s=4+6+16=26\Omega$

   1 mark

2. (b) Between B and C, the two 8 Ω resistors are in parallel: $\frac{1}{R_p}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}$, so $R_p=4\Omega$

   1 mark

3. (c)(i) The series part and the parallel part are in series with each other: total resistance $R=26+4=30\Omega$

   1 mark

4. (c)(i) By Ohm's law, $I=\frac{V}{R}=\frac{6\text{ V}}{30\Omega }$

   ½ mark

5. (c)(i) $I=0.2$ A

   ½ mark

6. (c)(ii) The 16 Ω resistor has the greater potential difference. Both are in series, so the same current flows through them, and by Ohm's law ($V=IR$) the larger resistance has the larger potential difference.

   1 mark

7. (c)(ii) Check:

   • $V_{16}=0.2\times 16=3.2$ V
   • $V_{BC}=0.2\times 4=0.8$ V

   Since 3.2 V > 0.8 V, the 16 Ω resistor has more potential difference.

   1 mark

Answer: (a) 26 Ω; (b) 4 Ω; (c)(i) 0.2 A; (c)(ii) 16 Ω resistor (3.2 V vs 0.8 V)