CBSE Class 10 Science 2025 Question Paper (Set 31/1/1) with Solutions

1. $2\mathrm{H_2O} \rightarrow 2\mathrm{H_2} + \mathrm{O_2}$. Twice as many molecules of hydrogen as oxygen are formed (the volume ratio is $2 : 1$).
2. Mass: $2 \times 2 = 4$ g of $\mathrm{H_2}$ for $32$ g of $\mathrm{O_2}$, so $M_H : M_O = 4 : 32 = 1 : 8$.

Answer: (D) 1 : 8

Common mistake: Giving $2 : 1$. That is the ratio of volumes, not masses; oxygen atoms are $16$ times heavier than hydrogen atoms.

1. Aluminium is $\mathrm{Al^{3+}}$ and oxygen is $\mathrm{O^{2-}}$: balance the charges, $2 \times (+3) = 3 \times (-2)$, so the oxide is $\mathrm{Al_2O_3}$.
2. Magnesium is $\mathrm{Mg^{2+}}$, which balances one $\mathrm{O^{2-}}$: $\mathrm{MgO}$.

Answer: (B) $\mathrm{Al_2O_3}$ and $\mathrm{MgO}$

1. Read the pH values: $A = 5$, $B = 7$, $C = 13$, $D = 3$.
2. $\mathrm{pH} < 7$ is acidic, the lower the stronger: $A$ ($5$) is a weak acid and $D$ ($3$) is a strong acid. $\mathrm{pH} = 7$ is neutral ($B$). $\mathrm{pH} > 7$ is basic: $C$ ($13$) is a strong base.

Answer: (D) Weak acid, neutral, strong base, strong acid

Common mistake: Reading the solutions in the order of the scale (1 to 14) instead of in the order of the letters A, B, C, D.

1. (i) $\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$ (neutralisation).
2. (ii) $\mathrm{MgO + 2HCl \rightarrow MgCl_2 + H_2O}$ (a metal oxide is basic, so it neutralises the acid).
3. (iii) $\mathrm{CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O}$ (a non-metal oxide is acidic, so it reacts with the base).
4. In all three, a salt and water are formed. No hydrogen gas appears, and the salt in (iii) is a basic salt, not neutral or acidic.

Answer: (A) Salt and water is formed.

1. When A gives electrons to B, an ionic compound forms.
2. Ionic compounds have strong electrostatic forces between the oppositely charged ions. That is why they have high melting points, dissolve in water and conduct electricity when molten or dissolved.
3. So the one property it does not show is “weak” forces of attraction.

Answer: (C) It has weak electrostatic forces of attraction between its oppositely charged ions.

1. Metals high in the reactivity series (sodium, calcium, magnesium and so on) hold on to their oxygen or chlorine too strongly for carbon to reduce them. They are extracted by electrolysis of their molten chlorides.
2. Gold and silver are very unreactive and occur native or are easily obtained, and iron is extracted by reduction of its oxide with carbon. So the pair is calcium and magnesium.

Answer: (B) Calcium and magnesium

1. Magnesium has $2$ outer electrons (2, 8, 2) and oxygen has $6$ (2, 6) and needs $2$ more to complete its octet.
2. So one Mg atom transfers both its outer electrons to one O atom, giving $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ in the ratio $1 : 1$: $\mathrm{MgO}$.
3. Option A shows exactly this. The other options show a wrong number of electrons moved or a wrong ratio of ions.

Answer: (A)

1. Saliva contains the enzyme salivary amylase, which starts breaking down starch in the mouth into simple sugars (maltose).
2. With less saliva, this conversion is affected. Protein and fat digestion happen later, in the stomach and small intestine.

Answer: (C) starch into simple sugars

1. Auxin collects on the shaded side of the shoot and makes the cells there grow longer. The shoot then bends towards the light.
2. Adrenaline is an animal hormone, and cytokinins and gibberellins do not cause this bending.

Answer: (D) Auxins

1. A bisexual flower has both male (stamen) and female (pistil) parts, so (i) is true and (ii) is false.
2. Because it has both parts, it can be self-pollinated or cross-pollinated, so (iii) is true. It can form fruits on its own after pollination and fertilisation, so (iv) is false.

Answer: (C) (i) and (iii)

1. The parents give only one kind of gamete each: $Ry$ from $RRyy$ and $rY$ from $rrYY$.
2. Every $F_1$ plant is $RrYy$. Round ($R$) is dominant over wrinkled and yellow ($Y$) is dominant over green, so all the seeds are round and yellow.

Answer: (C) 100% round and yellow

Common mistake: Writing the 9 : 3 : 3 : 1 ratio. That is for the F₂ generation, not F₁.

1. The first step, glucose to pyruvate, happens in the cytoplasm of every cell.
2. Lactic acid forms when pyruvate is broken down without oxygen in our muscle cells (yeast makes ethanol instead). So ‘b’ is oxygen-deficient muscle cells.

Answer: (D) Cytoplasm and Oxygen deficient muscle cells

1. Mirror X concentrates sunlight, so it is a concave mirror. Sunlight comes from far away (infinity), and a concave mirror focuses it to a real, highly diminished point image at its focus. So (i) is true and (iii) is false.
2. Mirror Y is a rear-view mirror, which is a convex mirror. A convex mirror always forms a virtual, erect and diminished image. So (ii) is true and (iv) is false.

Answer: (A) (i) and (ii)

1. Weak ciliary muscles and a stiffer lens with age cause presbyopia: the person cannot focus on both near and far objects.
2. It is corrected with a bifocal lens (concave on the top for distance, convex at the bottom for reading).

Answer: (B) presbyopia and bifocal lens

Common mistake: Choosing hypermetropia. Presbyopia is the age-related defect; hypermetropia is caused by a short eyeball or a weak lens at any age.

1. A food chain starts with a producer and each organism is eaten by the next one. (i) grass $\rightarrow$ rabbit $\rightarrow$ wolf $\rightarrow$ lion and (iii) grass $\rightarrow$ grasshopper $\rightarrow$ frog $\rightarrow$ snake $\rightarrow$ hawk can both be put in a proper order, so they are chains.
2. In (ii), man does not eat plankton in a chain with a grasshopper and fish, and in (iv) a snake does not eat grass, so neither can be arranged into a chain. These two do not constitute a food chain.

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

1. Green plants in a terrestrial ecosystem capture only about $1\%$ of the solar energy falling on their leaves and convert it into food energy.
2. So about $100\% - 1\% = 99\%$ is not converted.

Answer: (D) 99%

Common mistake: Answering 1%. That is the energy that is converted; the question asks for the part that is not.

1. Assertion: most decomposition reactions need heat, light or electricity to break bonds, so they absorb energy. True.
2. Reason: composting releases heat, so it is exothermic. True.
3. But the reason is an exception to the rule in the assertion, so it does not explain it.

Answer: (B) Both A and R are true, but R is not the correct explanation of A

1. Assertion: reproduction copies the basic body design, so a child has all the basic human features. True.
2. Reason: a child does not look exactly like its parents; DNA copying is not perfect and sexual reproduction mixes the genes of two parents, so there are variations. False.

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

1. The direction of the field at a point is the direction a compass needle points there. A needle can point only one way, so a point cannot have two field directions.
2. So two field lines can never cross. The reason is true and it is exactly why the assertion is true.

Answer: (A) Both A and R are true and R is the correct explanation of A

1. The ozone layer thinned sharply in the 1980s, mainly because of chlorofluorocarbons (CFCs). Assertion: true.
2. Ozone forms when an oxygen atom joins an oxygen molecule ($\mathrm{O + O_2 \rightarrow O_3}$). Reason: true. But this is how ozone is made, not why it dropped, so it does not explain the assertion.

Answer: (B) Both A and R are true, but R is not the correct explanation of A

1. The reaction is $\mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2}$.
2. Observation 1: bubbles of a gas (hydrogen) are given off from the surface of the zinc granules, so a new substance forms.
3. Observation 2: the flask becomes warm, because the reaction gives out heat (a rise in temperature). The zinc granules also slowly get smaller and disappear.

Answer: Evolution of gas (hydrogen) and a rise in temperature

1. Budding in Hydra goes in stages:
2. 1. A regenerative (dividing) cell group on the side of the parent divides and forms a small outgrowth, the bud.
3. 2. The bud grows, develops a mouth and tentacles, and stays attached to the parent for a while.
4. 3. The fully formed bud detaches from the parent and lives as a new, independent Hydra.

Answer: Bud forms, bud grows with tentacles, bud detaches (diagram above)

1. (a) A leak would lower the blood pressure and so reduce the efficiency of the pumping system (the heart), and the body would get less oxygen. The component that helps is the platelets, which gather at the injured spot and help the blood to clot, sealing the leak.
2. (b)(i) Plants have low energy needs: they do not move and have a large proportion of dead cells in many tissues, so a slow transport system is enough.
3. (ii) Phloem carries (translocates) the soluble products of photosynthesis, such as sucrose, from the leaves to the other parts of the plant, including storage organs of roots, fruits, seeds and growing organs. It also carries amino acids.

Answer: (a) To keep the blood pressure up; platelets clot the blood   (b) low energy needs; phloem translocates food from leaves

1. With the new Cartesian sign convention, the object is on the left: $u = -60$ cm. A concave lens has a negative focal length: $f = -30$ cm.
2. Lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$, so $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = -\dfrac{1}{30} - \dfrac{1}{60} = -\dfrac{3}{60} = -\dfrac{1}{20}$.
3. $v = -20$ cm. The image is $20$ cm from the lens on the same side as the object (it is virtual, erect and diminished: $m = \dfrac{v}{u} = \dfrac{1}{3}$).

Answer: $v = -20$ cm: the image is $20$ cm in front of the lens, on the object's side

Common mistake: Dropping the sign of $f$ or $u$. Both are negative here, and a sign slip gives a wrong (positive, real) image position.

1. (a) Each part has one third of the length, so its resistance is $\dfrac{R}{3}$.
2. In parallel: $\dfrac{1}{R_p} = \dfrac{3}{R} + \dfrac{3}{R} + \dfrac{3}{R} = \dfrac{9}{R}$, so $R_p = \dfrac{R}{9}$.
3. (b) Electric power is the rate at which electrical energy is consumed (or dissipated) in a circuit: $P = \dfrac{W}{t} = VI$.
4. The power is $1$ watt when $1$ joule of energy is used in $1$ second, that is when a current of $1$ A flows at a potential difference of $1$ V ($1\ \mathrm{W} = 1\ \mathrm{V} \times 1\ \mathrm{A}$).

Answer: (a) $\dfrac{R}{9}$   (b) rate of consuming electrical energy; $1$ W $= 1$ J per second

1. Chemicals and pesticides used by farmers get washed from the fields into the soil and into water bodies, and they affect the living (biotic) and non-living (abiotic) parts of the ecosystem.
2. Most of these chemicals are not biodegradable, so they are taken in by plants and small organisms and pass up the food chain. They get concentrated at each trophic level (biological magnification), so the top consumers, including humans, carry the most.
3. Thus the health of organisms at all trophic levels is affected, and soil fertility and water quality fall.

Answer: Non-biodegradable chemicals wash into soil and water and build up along the food chain

1. (a) A more reactive metal such as aluminium can displace a less reactive metal from its oxide, and the reaction gives out so much heat that the metal is obtained in the molten state. Examples:
2. $\mathrm{Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + heat}$ (the thermite reaction, used to weld railway tracks)
3. $\mathrm{3MnO_2 + 4Al \rightarrow 3Mn + 2Al_2O_3 + heat}$
4. (b) Metals high in the reactivity series (K, Na, Ca, Mg, Al) have more affinity for oxygen than carbon has. So carbon cannot take the oxygen away from their oxides. They are obtained by electrolysis of their molten compounds.

Answer: Thermite reactions of aluminium; high metals bind oxygen more strongly than carbon does

1. (a) Take three test tubes A, B and C and put clean iron nails in each.
2. A: pour ordinary water and cork it (air and water both present). B: pour boiled distilled water, add about 1 mL of oil on top and cork it (oil stops air from dissolving, so water but no air). C: put anhydrous calcium chloride and cork it (it absorbs moisture, so air but no water).
3. After a few days the nails rust in test tube A only. So iron rusts only when it is exposed to both air (oxygen) and water.
4. (b)(i) Sodium and potassium (lithium also reacts, less violently). Observations: a violent reaction takes place; a large amount of heat is given out; the metal moves on the water and the gas evolved may catch fire.
5. (ii) The gas is hydrogen ($\mathrm{2Na + 2H_2O \rightarrow 2NaOH + H_2}$). Bring a burning matchstick or splinter near it: it burns with a pop sound.

Answer: (a) Iron rusts in the presence of both air and water   (b) Na, K; the gas is hydrogen and burns with a pop

1. (a) Plant cells use electrical and chemical means to pass the information that a touch has occurred from cell to cell.
2. (b) The cells change their shape by changing the amount of water in them: some cells lose water and shrink, so the leaf folds and droops.
3. (c) Touch-me-not (a nastic movement) is growth independent and does not take place in the direction of the stimulus. The tendril of a pea plant (a tropic movement, thigmotropism) is growth dependent: the cells on the side away from the support grow faster, and it moves in the direction of the stimulus.

Answer: Electro-chemical signals; cells change shape as water moves out; nastic (growth independent) vs tropic (growth dependent)

1. (a) Chromosomes are thread-like structures in the nucleus. They carry the genes that control the traits of an organism, and they contain the information for inheritance of features from the parents to the next generation, in the form of DNA (deoxyribonucleic acid) molecules.
2. (b) Each body cell has two copies of every chromosome, one from the male parent and one from the female parent. During the formation of germ cells (gametes) the number is halved. When the male and female gametes fuse in fertilisation, the zygote gets back the normal number of chromosomes and the same amount of DNA. So the DNA content stays the same from one generation to the next.

Answer: Chromosomes carry genes (DNA); gametes have half the number and fertilisation restores it

1. (i) Object at infinity: rays arrive parallel to the principal axis. After reflection they diverge and look as if they come from the focus F behind the mirror. The image is formed at F, virtual and erect, and highly diminished (point-sized).
2. (ii) Object between infinity and the pole: the image is formed behind the mirror, between P and F. It is virtual, erect and diminished.
3. Mark arrows on the rays and show the reflected rays going back as dashed lines to the image.

Answer: (i) at F, virtual, point-sized   (ii) between P and F, virtual, erect, diminished

1. (a) The 10 Ω and 15 Ω pair in parallel: $\dfrac{1}{R_1} = \dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3 + 2}{30} = \dfrac{1}{6}$, so $R_1 = 6\ \Omega$.
2. The 60 Ω and 40 Ω pair in parallel: $\dfrac{1}{R_2} = \dfrac{1}{60} + \dfrac{1}{40} = \dfrac{2 + 3}{120} = \dfrac{1}{24}$, so $R_2 = 24\ \Omega$.
3. The two parts are in series: $R = R_1 + R_2 = 6 + 24 = 30\ \Omega$.
4. (b) $I = \dfrac{V}{R} = \dfrac{15}{30} = 0.5$ A.
5. (c) The same $0.5$ A passes through the first parallel pair, so the potential difference across it is $V_1 = I R_1 = 0.5 \times 6 = 3$ V.

Answer: (a) $30\ \Omega$   (b) $0.5$ A   (c) $3$ V

Common mistake: Adding 10 + 15 + 60 + 40 as if all four were in series, or forgetting that the parallel pair is reduced first.

1. (a) $R = \rho\dfrac{l}{A}$, so $\rho = \dfrac{RA}{l}$.
2. Unit of $\rho$ $= \dfrac{\Omega \times \mathrm{m^2}}{\mathrm{m}} = \Omega\,\mathrm{m}$ (ohm metre).
3. (b) Alloys have a higher resistivity than pure metals, so more heat is produced for the same current. They also do not oxidise (burn) readily at high temperatures, so they last long.

Answer: (a) $R = \rho l/A$; unit $\Omega\,\mathrm{m}$   (b) higher resistivity and they do not burn readily

1. (a)(i) The two isomers of butene are but-1-ene, $\mathrm{CH_2{=}CH{-}CH_2{-}CH_3}$, and but-2-ene, $\mathrm{CH_3{-}CH{=}CH{-}CH_3}$ (drawn above). (2-methylpropene is a third isomer.)
2. (ii) (I) $\mathrm{CH_3CH_2CH_2Cl}$ is chloropropane (1-chloropropane). (II) The four-carbon chain with a $\mathrm{C{=}O}$ on the third carbon (a ketone) is butanone (butan-2-one), $\mathrm{CH_3COCH_2CH_3}$.
3. (iii) (I) $\mathrm{CH_3CH_2OH + 2[O] \xrightarrow{\text{alk. KMnO}_4 \text{ or acidified K}_2\text{Cr}_2\text{O}_7,\ \text{heat}} CH_3COOH + H_2O}$.
4. (II) $\mathrm{CH_3CH{=}CH_2 + H_2 \xrightarrow{\text{Ni / Pd catalyst}} CH_3CH_2CH_3}$.
5. (III) $\mathrm{CH_3COOH + C_2H_5OH \xrightarrow{\text{conc. H}_2\text{SO}_4} CH_3COOC_2H_5 + H_2O}$ (ethyl ethanoate, an ester).
6. (b)(i) $X$ is ethanol ($\mathrm{C_2H_5OH}$), $Y$ is sodium ethoxide ($\mathrm{C_2H_5ONa}$) and $Z$ is hydrogen gas ($\mathrm{H_2}$), which turns vegetable oil into ghee.
7. $\mathrm{2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2}$.
8. (ii) (I) $\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + heat + light}$. (II) $\mathrm{CH_3CH_2OH \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 443\ \text{K}} CH_2{=}CH_2 + H_2O}$ (dehydration gives ethene). (III) $\mathrm{CH_3CH_2OH \xrightarrow{\text{acidified K}_2\text{Cr}_2\text{O}_7} CH_3COOH}$.

Answer: (a) but-1-ene and but-2-ene; chloropropane; butanone; equations above   (b) X ethanol, Y sodium ethoxide, Z hydrogen

Common mistake: Writing the conditions as “catalyst” without naming it. Say nickel or palladium for hydrogenation, and concentrated sulphuric acid for esterification.

1. (a)(i) (I) Ovary: produces the female gametes (eggs) and the female hormones (oestrogen and progesterone). (II) Fallopian tube (oviduct): the site of fertilisation, where the egg and sperm meet. (III) Uterus: the site of implantation of the embryo and its development into a baby.
2. (ii) Methods used by males: a mechanical barrier, the condom, which stops the sperms from reaching the egg; and a surgical method, vasectomy, in which the vas deferens is blocked or cut so sperms are not released.
3. (b)(i) Self-pollination: pollen grains move from the anther of a flower to the stigma of the same flower. Cross-pollination: pollen grains move from the anther of one flower to the stigma of another flower of the same species.
4. (ii) A is the stigma: it receives the pollen and provides a suitable surface for it to germinate. B is the pollen tube: it carries the male germ cells (gametes) down to the female gamete in the ovary. C is the egg cell (female germ cell): it fuses with the male gamete to form the zygote.

Answer: (a) ovary: eggs and hormones; fallopian tube: fertilisation; uterus: implantation; condoms, vasectomy   (b) A stigma, B pollen tube, C egg cell

1. (a)(i) A negative power means a concave (diverging) lens. $f = \dfrac{1}{P} = \dfrac{1}{-2.5\ \mathrm{D}} = -0.4$ m $= -40$ cm. A concave lens is prescribed for myopia (short-sightedness).
2. (ii) For a lens $m = \dfrac{v}{u}$. With $m = -2$ and $u = -20$ cm: $v = m u = (-2)(-20) = +40$ cm. (This lens is a convex lens, since a concave lens never forms a magnified, inverted image.)
3. (I) The image is real and inverted. (II) It is magnified: twice the size of the object. (III) It is on the other side of the lens, $40$ cm from it, beyond $2F$. (IV) The height of the image is negative.
4. (iii) Power $P = \dfrac{1}{f}$, so the lens with the smaller focal length has the larger power. Lens A ($10$ cm) shows the higher degree of convergence/divergence, because a shorter focal length means light is bent more.
5. (b)(i) A ray bends towards the normal on entering the glass, and away from the normal on leaving, so the emergent ray is parallel to the incident ray but shifted sideways (diagram above).
6. (ii) Snell’s law: for light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant: $\dfrac{\sin i}{\sin r} = \text{constant}$.
7. (iii) Convex lens: the virtual image forms only when the object is between $O$ and $F$, and the image is magnified. Concave lens: the virtual image forms for an object anywhere in front of the lens, and the image is diminished.

Answer: (a) concave lens, $f = -40$ cm, myopia; real, inverted, magnified, $v = +40$ cm, negative; lens A   (b) see diagram; Snell’s law; table

Common mistake: Taking $v = -40$ cm in (ii). For a lens the magnification is $\dfrac{v}{u}$ with signs, so a negative $m$ with a negative $u$ gives a positive $v$.

1. (a) $\mathrm{2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)}$. Chlorine forms at the anode and hydrogen at the cathode; sodium hydroxide stays in the solution.
2. (b) Uses of sodium hydroxide: making soaps and detergents, paper making (also degreasing metals, artificial fibres). (Or hydrogen: fuel, and making margarine and ammonia; or chlorine: water treatment, PVC, bleach.) Any two uses of one product.
3. (c)(i) A is sodium hydrogencarbonate (baking soda), $\mathrm{NaHCO_3}$; B is sodium carbonate (washing soda is its hydrated form), $\mathrm{Na_2CO_3}$. $\mathrm{2NaHCO_3 \xrightarrow{heat} Na_2CO_3 + H_2O + CO_2}$.
4. (c)(ii) Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt. Examples: $\mathrm{CuSO_4{\cdot}5H_2O}$ (blue vitriol) and $\mathrm{CaSO_4{\cdot}2H_2O}$ (gypsum); also $\mathrm{Na_2CO_3{\cdot}10H_2O}$.

Answer: (a) $2\mathrm{NaCl} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{Cl_2} + \mathrm{H_2}$   (c) A $= \mathrm{NaHCO_3}$, B $= \mathrm{Na_2CO_3}$ (OR) fixed water molecules per formula unit

1. (a) Photosynthesis. It is the process by which green plants capture sunlight and convert it into chemical energy (food), with the help of chlorophyll, by turning carbon dioxide and water into carbohydrates.
2. (b) $\mathrm{6CO_2 + 6H_2O \xrightarrow{\text{sunlight, chlorophyll}} C_6H_{12}O_6 + 6O_2}$.
3. (c)(i) Desert plants: (1) they open their stomata at night and take in carbon dioxide, which they store as an intermediate acid; (2) in the day, chlorophyll absorbs light energy; (3) light energy is converted to chemical energy; (4) the stored intermediate is acted on by this energy and carbon dioxide is reduced to carbohydrate.
4. (c)(ii) (I) In cloudy weather the rate of photosynthesis decreases, because less sunlight (light energy) reaches the leaves. (II) When the stomata are blocked by dust, the rate decreases, because carbon dioxide cannot enter and gas exchange is reduced.

Answer: (a) photosynthesis   (b) $6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}$   (c) see steps

1. (a) The live wire has red insulation and the neutral wire has black insulation (the earth wire is green).
2. (b) $P = 1\ \mathrm{kW} = 1000$ W and $V = 220$ V. From $P = VI$: $I = \dfrac{1000}{220} = 4.54$ A. The nearest standard rating above this is a $5$ A circuit.
3. (c)(i) The earth wire gives a low-resistance path to the ground for any leakage current. If the metal body of the iron becomes live, the current flows to the earth and the body stays at the potential of the earth, so the user does not get an electric shock.
4. (c)(ii) Two precautions: use a fuse wire in the circuit (it melts and breaks the circuit when the current is too high, preventing damage from overloading), and connect the metallic body of appliances to the earth wire (it prevents electric shock from leakage of current).

Answer: (a) red (live) and black (neutral)   (b) $4.54$ A, so a $5$ A rating   (c) low-resistance path to earth, so no shock (OR) fuse and earth wire

Common mistake: Writing 4 A. The current is 4.54 A, which is more than 4, so the rating has to be the next standard value, 5 A.