Solved 2025 Sample Question Paper CBSE Class 10 Science

Identify ‘p’, ‘q’ and ‘r’ in the following balanced reaction

$\text{pPb}(\text {NO}_3)_2 (\text s) \xrightarrow{\text {Heat}} \text q\text{PbO(s)} + \text r \text {NO}_2(\text g) + \text O_2 (\text g)$

options

1. 2, 2, 4
2. 2, 4, 2
3. 2, 4, 4
4. 4, 2, 2

Answer

2, 2, 4

Reason — On heating, lead nitrate ($\text{Pb}(\text {NO}_3)_2$) undergoes thermal decomposition, releasing brown fumes of nitrogen dioxide ($\text {NO}_2$) and colourless oxygen gas ($\text O_2$). The solid residue left is yellow lead oxide ($\text{PbO}$).

The balanced chemical equation is :

$2\text{Pb}(\text {NO}_3)_2 (\text s) \xrightarrow{\text{Heat}} 2\text{PbO(s)} + 4 \text {NO}_2(\text g) + \text O_2 (\text g)$

Here,

• p = 2
• q = 2
• r = 4

Thus, the coefficients are 2, 2 and 4.

Match column I with column II and select the correct option using the given codes.

options

1. a – (ii), b – (i), c – (iii), d – (iv)
2. a – (iii), b – (i), c – (iv), d – (ii)
3. a – (iv), b – (ii), c – (iii), d – (i)
4. a – (iii), b – (ii), c – (i), d – (iv)

Answer

a – (iii), b – (i), c – (iv), d – (ii)

Reason —

a. Amphoteric oxides → (iii) Al.

• Aluminum forms amphoteric oxide Al₂O₃, which reacts with both acids and bases.

b. Melts on the palm → (i) Ga.

• Gallium has a very low melting point (~29.7 °C), so it melts with body heat.

c. Reacts with nitric acid → (iv) Mn.

• Manganese is a reactive metal and readily reacts with HNO₃.

d. Cannot displace hydrogen from acids → (ii) Au.

• Gold is a noble metal (below H in the activity series), so it does not liberate H₂ from acids.

The solution in the given figure is likely to be

options

1. HNO₃
2. C₂H₅OH
3. H₂SO₄
4. CO₂ in water

Answer

C₂H₅OH

Reason — The circuit is a simple conductivity tester in which the bulb glows only if the solution contains enough ions to conduct current.

• HNO₃ and H₂SO₄ are strong acids; they dissociate completely in water to give many ions, so the bulb would glow brightly.
• CO₂ in water produces weak carbonic acid (H₂CO₃), a weak electrolyte that may give at most a very faint glow.
• Ethanol (C₂H₅OH), however, is a covalent, non-electrolyte in water— it does not ionize, so the solution does not conduct and the bulb does not glow.

Hence, the solution is ethanol (C₂H₅OH).

Which among the following is considered as the strongest electrolyte?

options

1. Dilute acid
2. Dilute sugar solution
3. Glucose solution
4. Ethanol in water

Answer

Dilute acid

Reason — A strong electrolyte is a substance that completely dissociates into ions when dissolved in water, allowing electric current to pass easily through the solution.

• Dilute acids, such as hydrochloric acid (HCl) or sulphuric acid (H₂SO₄), dissociate completely into hydrogen (H⁺) and corresponding anions, making them strong electrolytes.
• In contrast, sugar solution and glucose solution are non-electrolytes — they do not produce ions in water and hence do not conduct electricity.
• Ethanol in water is also a covalent compound and does not ionize, so it is a poor conductor.

Thus, among the given options, the dilute acid is considered the strongest electrolyte because it produces the maximum number of ions in solution, resulting in the highest electrical conductivity.

An aqueous solution ‘A’ turns the phenolphthalein solution pink. On addition of an aqueous solution ‘B’ to ‘A’, the pink colour disappears. Which of the following statement is true for the solutions ‘A’ and ‘B’.

options

1. A is strongly basic and B is a weak base.
2. A is strongly acidic and B is a weak acid.
3. A has a pH greater than 7 and B has a pH less than 7.
4. A has a pH less than 7 and B has a pH greater than 7.

Answer

A has a pH greater than 7 and B has a pH less than 7.

Reason — Phenolphthalein is colourless in acidic/neutral solutions and turns pink in basic solutions (typically pH ≳ 8.2). Since solution A turns phenolphthalein pink, A must be basic (pH > 7).

When solution B is added, the pink colour disappears, meaning B neutralizes the base and shifts the pH to acidic/neutral; therefore, B must be acidic (pH < 7).

So, A has a pH greater than 7 and B has a pH less than 7.

When 50 g of lead powder is added to 300 ml of blue copper sulphate solution, after a few hours, the solution becomes colourless. This is an example of

options

1. Combination reaction
2. Decomposition reaction
3. Displacement reaction
4. Double displacement reaction

Answer

Displacement reaction

Reason — When lead (Pb) powder is added to copper sulphate (CuSO₄) solution, a chemical reaction takes place in which lead displaces copper from its compound because lead is more reactive than copper.

The blue colour of the CuSO₄ solution gradually fades and finally becomes colourless, as lead sulphate (PbSO₄), which is white, is formed, and reddish-brown copper (Cu) metal gets deposited.

The reaction is:

Pb(s) + CuSO₄(aq) ⟶ PbSO₄(s) + Cu(s)

This is an example of a single displacement reaction, where a more reactive metal (Pb) displaces a less reactive metal (Cu) from its salt solution.

The electronic configuration of three elements X, Y and Z are X - 2, 8, 7; Y - 2, 8, 2; and Z - 2, 8

options

1. Y and Z are metals
2. Y and X are non-metals
3. X is a non-metal and Y is a metal
4. Y is a non-metal and Z is a metal

Answer

X is a non-metal and Y is a metal

Reason —

• X : 2, 8, 7 → 7 valence electrons (like chlorine) so it tends to gain 1 electron to complete the octet, a typical non-metal behaviour.
• Y : 2, 8, 2 → 2 valence electrons (like magnesium) so it readily loses 2 electrons to form Y²⁺, which is characteristic of metals.
• Z : 2, 8 → complete octet (neon) so it is an inert noble gas.

Thus, X is a non-metal and Y is a metal.

Which of the following is an endothermic reaction?

options

1. Burning of candle.
2. Cooking of food.
3. Decomposition of Vegetable matter.
4. Reaction of Sodium with air

Answer

Cooking of food.

Reason — An endothermic reaction is a chemical process that absorbs heat energy from the surroundings for the reaction to take place.

During the cooking of food, ingredients such as rice, vegetables or dough take in heat energy to undergo various chemical and physical changes, including the breaking of molecular bonds and the formation of new substances.

Since this process requires a continuous supply of heat, the reaction will stop if the external heat source is removed.

In contrast, processes like the burning of a candle and the reaction of sodium with air are exothermic reactions, as they release heat energy into the surroundings.

Similarly, the decomposition of vegetable matter (rotting) is also a slow exothermic biological process.

Therefore, cooking of food is a clear example of an endothermic reaction, as it absorbs heat to proceed.

During cellular oxidation of Glucose, ATP is produced along with formation of other products in this reaction. Which of the following events is associated with production of maximum ATP molecules per molecule of Glucose during this process? Synthesis of

options

1. Ethanol in yeast
2. Lactic acid in muscle cells
3. Carbon dioxide in yeast cells
4. Carbon dioxide in human cells

Answer

Carbon dioxide in human cells

Reason — The breakdown of glucose inside cells to release energy is called cellular respiration and it can occur either aerobically (with oxygen) or anaerobically (without oxygen).

• The aerobic respiration of glucose occurs in human cells (and most animal cells) and involves the complete oxidation of glucose into carbon dioxide (CO₂) and water (H₂O), releasing a large amount of energy. This process yields about 38 molecules of ATP per molecule of glucose, which is the maximum possible ATP output.
• In contrast, anaerobic respiration produces far less energy as in yeast, glucose is converted into ethanol and CO₂, yielding only 2 ATP molecules.
• In muscle cells under low oxygen, glucose is converted to lactic acid, also producing just 2 ATP molecules.

Thus, the formation of carbon dioxide in human cells occurs during aerobic respiration, which is associated with the maximum production of ATP per glucose molecule.

During which of the following stages of the circulation of blood in a normal human being, the oxygenated blood is pumped to all parts of the body?

options

1. Contraction of the left atrium
2. Contraction of left ventricle
3. Relaxation of the right atrium
4. Relaxation of the right ventricle

Answer

Contraction of left ventricle

Reason —

• Oxygenated blood enters the left atrium from the lungs, then moves to the left ventricle.
• When the left ventricle contracts, it pumps this blood into the aorta, sending it to all parts of the body.
• The right side (right atrium/ventricle) deals with deoxygenated blood and sends it to the lungs, not the body.

Therefore, it is specifically during the contraction of the left ventricle that oxygenated blood is distributed to all parts of the body.

Which of the following adaptations in herbivores helps in digestions of cellulose?

options

1. Longer large intestine
2. Smaller large intestine
3. Smaller small intestine
4. Longer small intestine

Answer

Longer small intestine

Reason — Herbivores have a longer small intestine because their diet consists mainly of plant material rich in cellulose, which is difficult to digest.

The increased length allows more time and surface area for digestion and absorption which also provides space for symbiotic bacteria that help break down cellulose into simpler, absorbable substances, enabling herbivores to extract maximum nutrients from their plant-based food.

There was a cerebellar dysfunction in a patient. Which of the following activities will get disturbed in this patient as a result of this?

options

1. Salivation
2. Hunger control
3. Posture and balance
4. Regulation of blood pressure

Answer

Posture and balance

Reason — The cerebellum is the part of the brain responsible for maintaining balance, posture and coordination of voluntary movements so it ensures that body movements are smooth, precise and well-controlled.

If there is cerebellar dysfunction, the person may experience loss of balance, unsteady gait, poor coordination and difficulty in maintaining posture.

The other functions—salivation, hunger control and blood pressure regulation—are managed by different parts of the brain such as the medulla and hypothalamus, not the cerebellum.

Hence, posture and balance are the activities most affected by cerebellar dysfunction.

In snails individuals can begin life as male and depending on environmental conditions they can become female as they grow. This is because

options

1. Male snails have dominant genetic makeup.
2. Female snails have dominant genetic makeup.
3. Expression of sex chromosomes can change in a snail’s life time.
4. Sex is not genetically determined in snails.

Answer

Sex is not genetically determined in snails.

Reason — In snails, sex determination depends on environmental conditions rather than fixed sex chromosomes. Many snail species are protandric hermaphrodites, meaning individuals start life as males and later develop female reproductive organs as they grow or when environmental factors—such as population density, temperature, or availability of mates—change.

This shows that the expression of sex is flexible and not genetically predetermined as in humans or other animals with distinct sex chromosomes.

Hence, the correct explanation is that sex is not genetically determined in snails.

In the following cases, a ray is incident on a concave mirror. In which case is the angle of incidence equal to zero?

options

1. A ray parallel to the principal axis.
2. A ray passing through the centre of curvature and incident obliquely.
3. A ray passing through the principal focus and incident obliquely.
4. A ray incident obliquely to the principal axis, at the pole of the mirror.

Answer

A ray passing through the centre of curvature and incident obliquely.

Reason — For a concave (spherical) mirror, the normal at any point on the surface is the radius drawn to that point. A ray that passes through the centre of curvature (C) travels along the radius, so it meets the mirror normally at the point of incidence; hence the angle of incidence is zero.

Rays parallel to the principal axis or striking at the pole obliquely or through the focus, all meet the mirror at a non-zero angle to the local normal.

Choose the correct option for the colour of rays for A and B.

options

1. A
2. B
3. C
4. D

Answer

C

Reason — In a glass prism, white light disperses into its colours since violet has the highest refractive index in glass so it deviates the most, while red has the lowest refractive index so it deviates the least toward the base of the prism.

In the diagram, ray B emerges closer to the base (more deviated), so it is violet and ray A is less deviated and therefore it is red.

Identify the incorrect statement

‘The energy available to the producers is maximum’ because:

options

1. It is the first trophic level which absorbs 1% of light energy directly from the source.
2. It utilizes the most of the chemical energy for its own respiration, growth, reproduction, movement etc.
3. It utilizes 10% of light energy and transfers the rest to the next trophic level.
4. It transfers only 10% of light energy to the next trophic level

Answer

It utilizes 10% of light energy and transfers the rest to the next trophic level.

Reason — Producers (green plants) form the first trophic level in a food chain so they absorb about 1% of the sunlight that reaches them and convert it into chemical energy through photosynthesis which is stored in plant biomass and is the maximum energy available to any trophic level in the ecosystem.

However, producers use most of this energy for their own metabolic activities such as respiration, growth and reproduction and only about 10% of the energy is transferred to the next trophic level (herbivores).

Therefore, the statement claiming that producers “utilize 10% and transfer the rest” is incorrect — it’s actually the opposite.

Which of the following is not a role of decomposers in the ecosystem?

options

1. Natural replenishment of soil.
2. Enrichment of oxygen in atmosphere.
3. Waste decomposition.
4. Break-down of dead remains.

Answer

Enrichment of oxygen in atmosphere.

Reason — Decomposers such as bacteria and fungi play a vital role in maintaining the balance of the ecosystem as they break down dead plants and animals into simpler substances, a process called decomposition.

This helps in the recycling of nutrients, leading to the natural replenishment of soil with minerals and organic matter and decomposers also help in waste decomposition, keeping the environment clean.

However, they do not enrich the atmosphere with oxygen. In fact, during decomposition, decomposers use up oxygen for respiration and release carbon dioxide instead.

Hence, enrichment of oxygen in the atmosphere is not a role of decomposers.

Assertion (A) : On adding dil. HCl to a test tube containing a substance ‘X’, a colourless gas is produced which gives a pop sound when a burning match stick is brought near it.

Reason (R) : In this reaction metal ‘X’ is displaced by Hydrogen.

options

1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, and R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

A is true but R is false.

Reason —

• Assertion (A) is true because when dilute hydrochloric acid (HCl) is added to a substance ‘X’, a colourless gas is evolved that produces a pop sound when a burning matchstick is brought near it. The gas is hydrogen (H₂), which is released when a metal reacts with an acid and the pop sound confirms the presence of hydrogen gas.

• Reason (R) is false because in this reaction the metal 'X' displaces hydrogen from the acid as metals are more reactive than hydrogen and the displaced hydrogen is released as a gas.

Hence, (A) is true, but (R) is false.

Assertion (A) : Generally, the number of chromosomes in a cell and in a germ cell is not the same in species.

Reason (R) : When two germ cells combine, they restore the normal number of chromosomes in a species.

options

1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, and R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true, and R is the correct explanation of A.

Reason —

• Assertion (A) is true because in most species, the number of chromosomes in somatic cells (body cells) and germ cells (sex cells) is not the same. Somatic cells are diploid (2n), meaning they have pairs of chromosomes, while germ cells are haploid (n), containing only one chromosome from each pair. This difference results from meiosis, where the chromosome number is halved to ensure stability of chromosome count across generations.

• Reason (R) is also true because when two germ cells (a sperm and an egg) combine during fertilization, the normal diploid number of chromosomes is restored in the zygote which maintains the species’ chromosome number from one generation to the next.

Hence, both A and R are true, and R is the correct explanation of A.

Assertion (A) : A convex mirror always forms an image behind it and the image formed is virtual.

Reason (R) : According to the sign convention, the focal length of a convex mirror is positive.

options

1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, and R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true, and R is not the correct explanation of A.

Reason —

• Assertion (A) is true because the reflected rays from a convex mirror diverge and when extended backward, they appear to meet at a point behind the mirror. Therefore, the image formed is always virtual, erect, and diminished, regardless of the object’s position.

• Reason (R) is true because by convention, distances measured in the direction of the incident light are taken as positive and the focus of a convex mirror lies on the same side as the incident light, hence its focal length is positive.

Hence, both A and R are true, and R is not the correct explanation of A.

Assertion (A) : If the lions are removed from a food chain it will not affect the food chain, however if the plants are removed from a food chain it will disturb the ecosystem.

Reason (R) : Plants are producers who can make food using sunlight, while lions are consumers.

options

1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, and R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

A is false but R is true.

Reason —

Assertion (A) is false because if lions (top carnivores) are removed from a food chain, the immediate effect may not collapse the entire chain, but over time it will disturb the ecological balance—for example, herbivore populations may increase uncontrollably.

However, if plants are removed, the entire food chain breaks down, as they are the primary source of energy for all organisms.

Therefore, while the statement recognizes plants' importance, it is incorrect to say removing lions has no effect on the food chain.

Reason (R) is true because plants are producers, capable of photosynthesis, converting sunlight into food energy and serving as the foundation for all trophic levels while lions, on the other hand, are consumers, dependent on other organisms for energy.

Hence, (A) is false, but (R) is true.

Identify the type of each of the following reactions stating the reason for your answers.

(a) Fe₂O₃ + 2Al → Al₂O₃ + 2Fe + heat

(b) Pb(NO₃)₂ + 2KI → PbI₂(↓)+ 2KNO₃

Answer

(a) This is an exothermic displacement reaction, which is also a redox reaction in which aluminium (a stronger reducing agent) reduces iron(III) oxide to iron, while itself getting oxidized to aluminium oxide and the large heat released is characteristic of the thermite reaction.

(b) This is a double displacement reaction that is specifically a precipitation reaction in which ions exchange partners to form insoluble lead(II) iodide, PbI₂, which appears as a yellow precipitate, while KNO₃ remains in solution.

Differentiate between alveoli and nephron on the basis of the following points:

Answer

List the steps for the synthesis of glucose by the plants. What special feature is found in desert plants related to this process?

Answer

Steps for the synthesis of glucose (Photosynthesis) :

• Absorption of sunlight : The chlorophyll present in the leaves absorbs light energy from Sun.
• Conversion of light energy into chemical energy : The absorbed light energy splits water molecules into hydrogen and oxygen.
• Reduction of carbon dioxide : Carbon dioxide is reduced using hydrogen to form glucose.

Special feature in desert plants :

Desert plants have their stomata open at night to take in carbon dioxide and store it as an organic acid. During the day, when the stomata are closed to prevent water loss, this stored carbon dioxide is used to carry out photosynthesis which helps desert plants conserve water and survive in hot, dry conditions.

Explain the role of the following enzymes in the process of digestion of food in humans:

(a) Salivary amylase

(b) Pepsin

(c) Trypsin

(d) Lipase

Answer

(a) Salivary amylase: This enzyme is present in the saliva secreted by the salivary glands and acts on starch (a complex carbohydrate) and breaks it down into simpler sugars like maltose. Thus, it helps in the digestion of carbohydrates in the mouth.

(b) Pepsin: Pepsin is a protein-digesting enzyme present in the stomach and acts in an acidic medium and breaks down proteins into smaller peptides.

(c) Trypsin: Trypsin acts in the small intestine in an alkaline medium which helps in the digestion of proteins by converting them into peptides and amino acids.

(d) Lipase: Lipase acts in the small intestine and helps in the digestion of fats by breaking them down into fatty acids and glycerol.

The below figure shows the formation of an image by a lens shown by a thick line.

Analyse the figure and answer the following questions.

(a) What is the type of lens used?

(b) What is the nature of the image?

(c) If the image is formed at a distance of 30 cm from the lens and the image is twice the size of the object, then where is the object placed?

Answer

(a) Convex lens

Explanation — The lens shown in the figure is a convex lens (converging lens) because the rays passing through the lens seems to be converged at a point on the same side of the lens, forming a virtual and magnified image.

(b) The image formed is virtual, erect and magnified.

(c) Given,

• Image distance (v) = -30 cm ∵ (image is on the same side of the object)
• Size of the image ($\text h_i$) = 2 x Size of the object ($\text h_o$)

Let, object distance be 'u'.

As, magnification of a lens is given by,

$\text {Magnification (m)} = \dfrac {\text h_i}{\text h_o} = \dfrac {\text v}{\text u} \\[1em] \Rightarrow \dfrac {2\times \text h_o}{\text h_o} = \dfrac {\text v}{\text u} \\[1em] \Rightarrow \dfrac {2}{1} = \dfrac {\text v}{\text u} \\[1em] \Rightarrow \dfrac {\text u}{\text v} = \dfrac {1}{2} \\[1em] \Rightarrow \text u = \dfrac {1}{2} \times \text v \\[1em] \Rightarrow \text u = \dfrac {1}{2} \times -30 \\[1em] \Rightarrow \text u = -15 \text { cm}$

This means the object is placed 15 cm in front of the lens.

(a) What type of lens always forms a virtual erect and diminished image?

(b) List two uses of such a lens.

Answer

(a) Concave lens

Explanation — The lens that always forms a virtual, erect and diminished image is a concave lens (diverging lens). A concave lens causes parallel rays of light to diverge outward, making them appear to come from a point behind the lens.

(b) Two uses of concave lens:

1. Correction of myopia (short-sightedness).
2. Used in peepholes or door eye to provide a wider field of view.

The electric circuit (below figure) in a clothes dryer contains two heaters X and Y in parallel. The below figure shows the circuit connected to a 230 V power supply. When both switches are closed, the current in X is 3.5 A.

Analyse the circuit given above and answer the following questions.

(a) Calculate the power developed in heater X.

(b) If the resistance of X is double that of Y calculate the current in heater Y.

Answer

Given,

Voltage supply ($\text V$) = 230 V Current flowing through X ($\text I_\text X$) = 3.5 A

(a) Power developed ($\text P$) in heater X is given by,

$\text P = \text V \text I_\text X \\[1em] = 3.5 \times 230 \\[1em] \Rightarrow \text P = 805 \text{ W}$

Hence, the power developed in heater X is 805 W.

(b) Given,

Resistance of the heater X ($\text R_\text X$) = 2 x Resistance of the heater Y ($\text R_\text Y$)

Let, current flowing through the heater Y be '$\text I_\text Y$'.

Since, heaters X and Y are connected in parallel combination then potential drop across them should be equal to 230 V i.e.,

$\text V_\text X = \text V_\text Y$ = 230 V ............... (i)

From Ohm's law,

$\text V = \text {IR}$

On putting in equation (i),

$\Rightarrow \text I_\text X\text R_\text X = \text I_\text Y\text R_\text Y \\[1em] \Rightarrow \text I_\text X \times 2\text R_\text Y = \text I_\text Y\text R_\text Y \\[1em] \Rightarrow \text I_\text Y = \dfrac{2\text I_\text X \text R_\text Y}{\text R_\text Y} \\[1em] \Rightarrow \text I_\text Y = \dfrac{2\text I_\text X}{1} \\[1em] \Rightarrow \text I_\text Y = 2 \times 3.5 \\[1em] \Rightarrow \text I_\text Y = 7.0\ \text A$

Hence, the current in heater Y is 7.0 A.

The below figure shows two resistors X and Y connected in series to a battery. The power dissipated for this combination is $\text P_1$. When these resistors are connected in parallel to the same battery then the power dissipated is given by $\text P_2$. Find out the ratio $\dfrac{\text P_1}{\text P_2}$.

Answer

From the figure,

Resistance of resistor X ($\text R_\text X$) = $\text R$
Resistance of resistor Y ($\text R_\text Y$) = $2\text R$

Let, voltage supply be '$\text V$'.

Case 1 : Series connection of X and Y

When X and Y are connected in series then net resistance of the circuit is given by,

$\text R_\text S = \text R_\text X + \text R_\text Y \\[1em] = \text R + 2\text R \\[1em] \Rightarrow \text R_\text S = 3\text R$

Now,

$\text P_1 = \dfrac{\text V^2}{\text R_\text S} \\[1em] = \dfrac{\text V^2}{3\text R}$

Case 2 : Parallel connection of X and Y

When X and Y are connected in parallel then net resistance of the circuit is given by,

$\dfrac {1}{\text R_\text P} = \dfrac {1}{\text R_\text X} + \dfrac {1}{\text R_\text Y} \\[1em] = \dfrac {1}{\text R} + \dfrac {1}{2\text R} \\[1em] = \dfrac {2 + 1}{2\text R} \\[1em] \Rightarrow \dfrac {1}{\text R_\text P} = \dfrac {3}{2\text R} \\[1em] \Rightarrow \text R_\text P = \dfrac {2\text R}{3}$

Now,

$\text P_2 = \dfrac{\text V^2}{\text R_\text P} \\[1em] = \dfrac{\text V^2}{\dfrac {2\text R}{3}}\\[1em] = \dfrac{3\text V^2}{2\text R}$

So,

$\dfrac{\text P_1}{\text P_2} = \dfrac{\dfrac{\text V^2}{3\text R}}{\dfrac{3\text V^2}{2\text R}} \\[1em] = \dfrac{2 \times 1}{3 \times 3} \\[1em] \Rightarrow \dfrac{\text P_1}{\text P_2} = \dfrac{2}{9}$

Hence, the ratio $\dfrac{\textbf P_\bold 1}{\textbf P_\bold 2} \textbf { is } \dfrac{\bold 2}{\bold 9}$.

We have four resistors A, B, C and D of resistance 3 Ω, 6 Ω, 9 Ω and 12 Ω respectively. Find out the lowest resistance which can be obtained by combining these four resistors.

Answer

Given,

• Resistance of resistor A ($\text R_\text A$) = 3 Ω
• Resistance of resistor B ($\text R_\text B$) = 6 Ω
• Resistance of resistor C ($\text R_\text C$) = 9 Ω
• Resistance of resistor D ($\text R_\text D$) = 12 Ω

To minimize equivalent resistance, connect all the resistors in parallel (parallel combinations always give a resistance less than the smallest individual resistor).

When resistors A, B, C and D are connected in parallel then net resistance of the circuit is given by,

$\dfrac {1}{\text R_\text P} = \dfrac {1}{\text R_\text A} + \dfrac {1}{\text R_\text B} + \dfrac {1}{\text R_\text C} + \dfrac {1}{\text R_\text D} \\[1em] = \dfrac {1}{3} + \dfrac {1}{6} + \dfrac {1}{9} + \dfrac {1}{12} \\[1em] = \dfrac {12 + 6 + 4 + 3}{36} \\[1em] \Rightarrow \dfrac {1}{\text R_\text P} = \dfrac {25}{36} \\[1em] \Rightarrow \text R_\text P = \dfrac {36}{25} \\[1em] \Rightarrow \text R_\text P = 1.44\ \text Ω$

Hence, the lowest resistance of the circuit is 1.44 Ω.

You are given 2 fuse wires A and B with current ratings 2 A and 5 A, respectively. Justify with reason which of the two would you use with a 1000 W, 220 V room heater?

Answer

Given,

• Current rating of the fuse wire A ($\text I_\text A$) = 2 A
• Current rating of the fuse wire B ($\text I_\text B$) = 5 A
• Power rating of the heater (P) = 1000 W
• Voltage rating of the heater (V) = 220 V

Then,

Current drawn by the heater (I) is given by,

$\text I = \dfrac{\text P}{\text V} = \dfrac{1000}{220} \approx$ 4.54 A

Here,

$\text I \gt \text I_\text A$ and $\text I \lt \text I_\text B$

Thus,

A fuse should have a rating just above the normal working current. As 2 A fuse would melt during normal use, cutting off the supply but 5 A fuse will allow the heater to operate normally and will blow only if the current exceeds about 5 A, protecting the circuit from overload.

Hence, fuse wire B with current rating 5 A should be used with the heater.

The cartoon below addresses a growing concern:

What impact will the process shown in the image have on Humans if they occupy the last trophic level? Explain

Answer

Pesticides are non-biodegradable, so they persist and keep accumulating at each trophic level. Their concentration rises up the food chain (bio-magnification), making the last trophic level, humans, the most contaminated. This leads to the highest pesticide load in humans and related health risks.

Create a food chain with more than 2 trophic levels that exists in the cabbage farm. If Humans occupy the last trophic level, then how would spraying pesticide affect the humans? Explain.

Answer

Food chain in a cabbage farm:

Cabbage → Rabbit → Snake → Hawk

Pesticides are non-biodegradable and persist in the ecosystem. When humans eat plants or animals that fed on those plants, the pesticide enters the food chain and accumulates at each trophic level. Hence, the last trophic level (humans) gets the highest concentration of pesticide — a process called biological magnification which leads to adverse effects on human health.

(a) Anirudh took two metal oxides; aluminium oxide and magnesium oxide as shown in the pictures given below. But he forgot to label them. How will you guide/ help Anirudh to identify the oxides and label them?

(b) In an activity Aishu was given two substances; Copper Sulphide (Cu₂S) and Copper Oxide (Cu₂O) to obtain copper from these compounds. She was able to extract Copper successfully. Illustrate with the help of chemical equations how Aishu might have completed the activity.

Answer

(a) Anirudh can prepare aqueous solutions of both oxides. (MgO forms a mildly basic solution; Al₂O₃ is insoluble.) Dip red litmus into each sample.
The solution that turns red litmus to blue is Magnesium oxide (MgO) and the solution with which red litmus does not show any colour change is Aluminium oxide (Al₂O₃).

(b) When copper sulphide is heated in air, it undergoes roasting, reacting with oxygen to form copper oxide (Cu₂O) and sulphur dioxide gas then, the freshly formed copper oxide reacts further with more copper sulphide, producing copper metal and sulphur dioxide gas again.

Hence, copper is obtained in the metallic form — by roasting and reduction as shown below :

2Cu₂S(s) + 3O₂(g) $\xrightarrow{\text {Heat}}$ 2Cu₂O(s) + 2SO₂(g)

2Cu₂O(s) + Cu₂S(s) $\xrightarrow{\text {Heat}}$ 6Cu(s) + SO₂(g)

Give reasons for the following :

(a) Certain metals are used for making cooking utensils.

(b) Hydrogen gas does not evolve when certain metals except Mg & Mn react with nitric acid.

Answer

(a) Certain metals such as copper and aluminium are used for making cooking utensils because they are good conductors of heat, allowing uniform heating of food. These metals also have high melting points, so they do not melt or deform while cooking.

(b) Hydrogen gas does not evolve when most metals react with nitric acid because nitric acid is a strong oxidizing agent. Instead of releasing hydrogen, it oxidizes hydrogen to water and gets reduced to nitrogen oxides such as NO, NO₂, or N₂O. However, magnesium (Mg) and manganese (Mn) are exceptions because they are highly reactive and can displace hydrogen from very dilute nitric acid before it gets oxidized. Therefore, hydrogen gas is evolved only in their reactions with nitric acid.

(i) In the given series of reactions, name the compounds X and Z.

(ii) Which type of reaction is X to Z?

(iii) You are given 3 unknown solutions A, B, and C with pH values of 6, 8 and 9.5 respectively. In which solution will the maximum number of hydronium ions be present? Arrange the given samples in the increasing order of H⁺ ion concentration.

Answer

(i) X → Sodium hydrogen carbonate (NaHCO₃), Z = Sodium carbonate (Na₂CO₃)

Explanation — When NaCl is treated with CO₂ and NH₃:

NaCl + H₂O + CO₂(↑) + NH₃(↑) → NaHCO₃ + NH₄Cl

Here, X = Sodium hydrogen carbonate (NaHCO₃) and Y = Ammonium chloride (NH₄Cl).

On heating X :

$2 \text {NaHCO}_3 \xrightarrow{\text {Heat}} \text {Na}_2 \text {CO}_3 + \text H_2 \text O + \text {CO}_2(\uparrow)$

So, Z = Sodium carbonate (Na₂CO₃)

Hydration of Z gives :

Na₂CO₃ + 10H₂O → Na₂CO₃.10H₂O

So, here Q = Washing soda (Na₂CO₃.10H₂O)

(ii) Thermal decomposition because on heating, sodium hydrogen carbonate decomposes to sodium carbonate, water and carbon dioxide.

(iii) C (pH 9.5) < B (pH 8) < A (pH 6)

Explanation — The pH of a solution is inversely related to its hydrogen ion concentration such that a lower pH value means a higher concentration of hydrogen ions (or hydronium ions) and vice versa.

Among the given solutions — A (pH 6), B (pH 8) and C (pH 9.5) — solution A has the lowest pH therefore, it contains the maximum number of hydronium ions, making it the most acidic of the three.

Comment on the following statements :

(a) Bee sting is treated with baking soda paste whereas wasp sting is treated with dilute vinegar.

(b) Farmers treat soil with quicklime when tilling.

(c) Ancient sculptures and marble structures are conserved by treating them with certain chemicals.

Answer

(a) Bee stings inject formic acid into the skin so to neutralize this acid, a mild base such as baking soda (sodium bicarbonate) is applied. Wasp stings, on the other hand, are alkaline in nature therefore, they are treated with a mild acid like dilute vinegar (acetic acid) to neutralize the alkali.

(b) Some soils become too acidic due to continuous use of fertilizers or natural processes. Quicklime (calcium oxide) is a basic substance and when added to acidic soil, it neutralizes the excess acid, restoring the soil’s pH to a level suitable for crop growth. Thus, it helps to improve the fertility and productivity of the soil.

(c) Marble and limestone are mainly composed of calcium carbonate (CaCO₃). Over time, they react with acidic gases like sulphur dioxide (SO₂) and carbon dioxide (CO₂) in the atmosphere, forming calcium sulphate or calcium bicarbonate, which damage the surface.
To protect them, they are treated with chemicals such as ammonium bicarbonate or calcium hydroxide, which neutralize acids and restore the protective layer. This treatment prevents further corrosion and helps preserve historical monuments.

Water is used by the leaves of the plants for photosynthesis but rather than watering the leaves, we water the plant through the soil. How does this water reach the leaves of the plant?

Answer

• In plants, water is absorbed from the soil through the roots. The xylem tissue present in roots, stems and leaves forms a continuous system of water-conducting channels that allows the upward movement of water.
• During the daytime, when stomata on the leaves are open, transpiration (loss of water in the form of vapour) takes place which creates a transpiration pull, which becomes the major driving force for the upward movement of water through the xylem.
• As water evaporates from the stomata, it creates a suction pressure that pulls water upward from the roots to the stem and finally to the leaves, where it is used in photosynthesis.

(a) In a family of four individuals, the father possessed long ears and the mother possessed short ears. If the parents had pure dominant and recessive traits respectively and F1 individual is married to an unrelated individual of the same genotype, then calculate the ratio of genetic makeup of F2 generation. Show a suitable cross.

(b) If father had short ears and the mother had long ears, and both the parents are homozygous for the allelic pair of genes, explain what effect it will have on the ratio of genetic makeup in F2 generation, if F1 individual is married to an unrelated individual of the same genotype.

Answer

(a) Let L = long ears (dominant), l = short ears (recessive).

Father (pure dominant) = LL; Mother (pure recessive) = ll.

P : LL × ll

F1 : all Ll (long)

F1 individual marries an unrelated individual of same genotype → Ll × Ll

F2 :

F2 genotypic ratio: 1 LL : 2 Ll : 1 ll (Phenotype: 3 long : 1 short)

(b) Father short = ll, Mother long = LL (both homozygous).

P : ll × LL → F1 : all Ll (long), same as in (a).

Again Ll × Ll → F2 genotypic ratio: 1 LL : 2 Ll : 1 ll (phenotype 3 : 1).

Effect : No change in the F2 generation and the ratio will still be same i.e., 1 LL : 2 Ll : 1 ll because the cross is still between a pure dominant and recessive allele as shown in the cross above.

(a) What is the fundamental difference between hypermetropia and myopia in terms of the optical experience of a person?

(b) The diagram below shows a special case of an eye defect.

(i) What is the defect that is shown in the figure?

(ii) State one cause for such a defect?

(iii) Explain with reason if a concave lens can be used to correct the defect.

Answer

(a) Hypermetropia is a defect that causes difficulty in focusing on near objects, with clearer vision observed for distant objects while, in Myopia distant objects appear blurry while near objects are seen clearly.

(b)

(i) The diagram shows that light rays from a nearby object are converging behind the retina hence, the defect is Hypermetropia (long-sightedness).

(ii) One cause of hypermetropia is shortening of the eye ball.

(iii) No, a concave lens would diverge the rays entering the eye and make the image form further away from the retina. A convex lens should be used instead, as it converges the rays and helps to form the image exactly on the retina.

(a) What is the fundamental difference between hypermetropia and myopia in terms of the optical experience of a person?

(b) What are the causes of myopia in the human eye?

Answer

(a) Hypermetropia is a defect that causes difficulty in focusing on near objects, with clearer vision observed for distant objects while, in Myopia distant objects appear blurry while near objects are seen clearly.

(b) Causes of myopia in the human eye :

• Elongation of the eyeball : The eyeball becomes too long from front to back, causing light rays from distant objects to focus in front of the retina.
• Excessive curvature of the eye lens : The lens becomes too powerful and converges light rays more than required, forming the image before the retina.

(a) State the relationship between the resistance R of a wire to its length l and cross sectional area A. Use the mathematical symbols to arrive at the final formula.

(b) Using the formula define the resistivity of a material.

Answer

(a) The resistance ($\text R$) of a wire depends on three factors :

1. Length ($\text l$) of the wire,
2. Cross-sectional area ($\text A$) of the wire, and
3. Nature (material) of the wire, expressed by a constant called resistivity ($\text ρ$).

The relationship is as follows:

• Resistance is directly proportional to the length of the wire.
  $\text R \propto \text l$

• Resistance is inversely proportional to the cross-sectional area of the wire.
  $\text R \propto \dfrac{1}{\text A}$

Combining these two proportionalities :

$\text R \propto \dfrac {\text l}{\text A}$

Introducing a constant of proportionality (ρ), we get the formula:

$\text R = \text ρ\dfrac {\text l}{\text A}$

(b) From the above equation,

$\text ρ = \text R\dfrac {\text A}{\text l}$

On putting $\text l = 1\ \text m$ and $\text A = 1\ \text m^2$ in the above relation,

$\text ρ = \text R\dfrac {1}{1} \\[1em] \Rightarrow \text ρ = \text R$

So, resistivity (ρ) of a material is defined as the resistance offered by a wire of that material having unit length and unit cross-sectional area.

Mona was doing an experiment with a magnetic compass and a straight current-carrying wire. She observed that as she moved the compass away from the current-carrying wire, the deflection of the compass needle reduced.

(a) Explain why the deflection of the compass needle reduced as Mona moved away the compass needle from the current carrying wire.

(b) Mention one thing that could have changed in the circuit of the wire that could increase the deflection of the needle.

(c) Explain with reason what will be the direction of the magnetic field associated with the wire for the case described by the above figure.

Answer

(a) When an electric current flows through a straight wire, it produces a magnetic field around it and the strength of this magnetic field is inversely proportional to the distance from the wire. Therefore, when Mona moved the compass needle away from the current-carrying wire, the magnetic field at that point became weaker. This reduced the force acting on the compass needle and hence, the deflection of the compass needle also decreased.

(b) As the strength of a magnetic field of a wire is directly proportional to the current flowing through it so a higher current produces a stronger magnetic field around the wire, which causes a greater deflection of the compass needle.

(c) As per the battery connection shown, current is flowing from the top of the wire downwards. Hence, by Right-Hand Thumb rule we get magnetic field to be in the clockwise direction.

(a) Explain why the deflection of the compass needle reduced as Mona moved away the compass needle from the current carrying wire.

(b) Mention one thing that could have changed in the circuit of the wire that could increase the deflection of the needle.

(c) Explain with reason how the direction of the magnetic field associated with the wire changes if the polarity of battery reversed.

Answer

(a) When an electric current flows through a straight wire, it produces a magnetic field around it and the strength of this magnetic field depends inversely on the distance from the wire. Therefore, when Mona moved the compass needle away from the current-carrying wire, the magnetic field at that point became weaker which reduces the force acting on the compass needle and hence, the deflection of the compass needle also decreased.

(b) As the strength of a magnetic field of a wire depends directly on its current so a higher current produces a stronger magnetic field around the wire, which causes a greater deflection of the compass needle.

(c) If the polarity of the battery is reversed, the direction of the current in the wire also reverses which in turn reverses the direction of the magnetic field around the wire.

(a) “Keerthi thinks that Substitution reaction occurs in saturated Hydrocarbons, on the contrary Krishi thinks, it occurs in unsaturated Hydrocarbons.” Justify with valid reasoning whose thinking is correct.

(b) “Methane and Propane and their Isomers are used as fuels” Comment. Draw the electron dot structure of the immediate lower homologue of Propane. Give any two characteristics of homologues of a given homologous series.

(c) A mixture of oxygen and ethyne is burnt for welding. Can you predict why a mixture of ethyne and air is not used?

Answer

(a) Keerthi's thinking is correct as substitution reactions are characteristic of saturated hydrocarbons (alkanes) where all bonds are single (C–C, C–H). With no reactive double/triple bond, an incoming atom/group can replace a hydrogen atom.
By contrast, unsaturated hydrocarbons (alkenes/alkynes) contain reactive double and triple bonds. These double and triple bonds readily open and add atoms across it, so they usually undergo addition, not substitution.

(b) Methane and propane (and their isomers) are used as fuels because they are saturated hydrocarbons that burn easily in air, releasing a large amount of heat energy and producing mainly carbon dioxide and water without leaving any residue. This makes them clean fuels suitable for domestic and industrial purposes.

The immediate lower homologue of propane (C₃H₈) is ethane (C₂H₆) and ethane has the following electron dot structure :

Characteristics of homologous series :

• All members of a homologous series have the same general formula.
• Each member of the series differs from the preceding one by the addition of CH₂ group and by molecular mass of 14 a.m.u.

(c) A mixture of oxygen and ethyne is used for welding because this mixture produces a very high flame temperature of about 3000 °C, sufficient to melt metals. However, a mixture of ethyne and air is not used, because air contains nitrogen and less oxygen, which leads to incomplete combustion of ethyne and this incomplete combustion produces sooty flame with a lower temperature, making it unsuitable for welding purposes.

(a) ‘A’ & ‘B’ are sodium salts of long-chain carboxylic acid and long chain Sulphonic acid respectively. Which one of A or B will you prefer as a cleansing agent while using underground water (hand pump water)? Give the reason for your answer.

(b) Elaborate on the process of cleansing action. Illustrate micelle with the help of labelled diagram.

(c) Write the chemical equation of the preparation of soap from an ester CH₃COOCH₃. What is the name of this process?

Answer

(a) ‘A’ is the sodium salt of a long-chain carboxylic acid, which means it is a soap, while ‘B’ is the sodium salt of a long-chain sulphonic acid, which is a synthetic detergent.
When underground water (hard water) is used, it contains calcium and magnesium ions and these ions react with soaps to form insoluble scum, which reduces the cleansing efficiency of soap.
However, detergents do not form scum with the ions present in hard water because the sulphonate group of the detergent does not react with calcium or magnesium ions.
Hence, ‘B’ (detergent) is preferred as a cleansing agent in underground (hard) water because it remains effective and forms more lather even in hard water.

(b) Cleansing action of soap :

• Soaps are molecules in which the two ends have differing properties, one is hydrophilic, that is, it dissolves in water, while the other end is hydrophobic, that is, it dissolves in hydrocarbons.
• The molecules of soap are sodium or potassium salts of long-chain carboxylic acids. The ionic end of soap dissolves in water while the carbon chain dissolves in oil. The soap molecules, thus form structures called micelles where one end of the molecules is towards the oil droplet while the ionic end faces outside. This forms an emulsion in water. The soap micelle thus helps in dissolving the dirt in water and we can wash our clothes clean.

(c) When an ester reacts with a strong base (NaOH), it forms a salt of carboxylic acid (soap) and an alcohol.

Chemical equation :

CH₃COOCH₃ + NaOH → CH₃COONa + CH₃OH

This process is called saponification.

The image below shows a banana plant which is growing with the help of suckers. These suckers are small plant stem outgrowths which can be separated from the main plant and planted separately and they will grow into a new plant subsequently.

(a) Give the name and type of reproduction that is shown in the image above.

(b) List two advantages the farmer will have on using the type of reproduction mentioned above.

(c) The above plant produces male flowers. Explain how this plant will be involved in the process of pollination.

(d) Why is the offspring of this banana plant not absolutely identical to its parent plant?

Answer

(a) The name of reproduction shown in the image is vegetative propagation by suckers and it is a type of asexual reproduction.

(b) The advantages of using this method of reproduction are :

• Quick and uniform propagation : A large number of banana plants can be produced in a shorter time without the need of pollinators and all offspring show similar characteristics as the parent plant.
• Assured fruit quality : Since the new plants are genetically similar to the parent, they retain all desirable traits such as good taste, size, and disease resistance of the original plant.

(c) The banana plant shown produces male flowers, which contain pollen grains. During pollination, these pollen grains from anther are transferred (by wind, water, or insects) to the stigma of female flowers of another banana plant. This helps in the process of sexual reproduction, allowing the fusion of male and female gametes.

(d) The offspring of this banana plant is not absolutely identical to its parent plant because after pollination, some changes might occur in the process of copying of DNA from the male and female gametes to the offspring.

A banana plant is growing with the help of suckers. These suckers are small plant stem outgrowths which can be separated from the main plant and planted separately and they will grow into a new plant subsequently.

(a) Give the name and type of reproduction.

(b) List two advantages the farmer will have on using the type of reproduction mentioned above.

(c) The above plant produces only male flower. Explain how this plant will be involved in the process of pollination.

(d) Why is the offspring of this banana plant not absolutely identical to its parent plant?

Answer

(a) The name of reproduction shown in the image is vegetative propagation by suckers and it is a type of asexual reproduction.

(b) The advantages of using this method of reproduction are :

• Quick and uniform propagation : A large number of banana plants can be produced in a shorter time and all offspring show similar characteristics as the parent plant.
• Assured fruit quality : Since the new plants are genetically similar to the parent, they retain all desirable traits such as good taste, size, and disease resistance of the original plant.

(c) The banana plant shown produces male flowers, which contain pollen grains. During pollination, these pollen grains from anther are transferred (by wind, water, or insects) to the stigma of female flowers of another banana plant. This helps in the process of sexual reproduction, allowing the fusion of male and female gametes.

(d) The offspring of this banana plant is not absolutely identical to its parent plant because during pollination, minor DNA/genetic variations can occur due to mutations or environmental influences during cell division and these small differences make the new plant slightly different from the parent.

The image below shows a developing fetus in the mother's womb. The developing fetus is connected to the placenta by means of umbilical cord. The Umbilical vein and artery run inside the umbilical cord.

(a) Name two substance that moves through the blood vessels.

(b) If the placenta has less villi how will it affect the baby’s growth?

(c) Name the region where the embryo develops inside the female body. Explain how this region is adapted for nourishing the baby.

(d) Some of the fetal cells fall off into the amniotic fluid and can be collected by careful procedure. The cells were screened and found to contain XY chromosome.

(i) What is the sex of the foetus?
(ii) How is this prenatal sex determination misused?

Answer

(a) The two main substances that move through the blood vessels in the umbilical cord are oxygen and nutrients.

(b) If the placenta has fewer villi, the surface area for exchange of materials between the mother and the fetus is reduced which means that less oxygen and nutrients can reach the developing baby and less waste can be removed. As a result, the growth and development of the baby will be slowed down or stunted and it may lead to poor health or low birth weight.

(c) The embryo develops inside the uterus of the female body. The uterus is well adapted for nourishing the baby as its inner lining (endometrium) is thick, soft, and richly supplied with blood vessels, which provide nutrients and oxygen to the developing embryo.

(d)

(i) The presence of XY chromosomes indicates that the foetus is male.

(ii) Prenatal sex determination is misused in some cases for female foeticide where female foetuses are aborted due to a preference for male children.

A developing fetus is connected to the placenta by means of umbilical cord. The Umbilical vein and artery run inside the umbilical cord.

(a) Name two substance that moves through the blood vessels.

(b) If the placenta has less villi how will it affect the baby’s growth?

(c) Name the region where the embryo develops inside the female body. Explain how this region is adapted for nourishing the baby.

(d) Some of the fetal cells fall off into the amniotic fluid and can be collected by careful procedure. The cells were screened and found to contain XY chromosome.

(i) What is the sex of the foetus?
(ii) How is this prenatal sex determination misused?

Answer

(a) The two main substances that move through the blood vessels in the umbilical cord are oxygen and nutrients.

(b) If the placenta has fewer villi, the surface area for exchange of materials between the mother and the fetus is reduced which means that less oxygen and nutrients can reach the developing baby and less waste can be removed. As a result, the growth and development of the baby will be slowed down or stunted and it may lead to poor health or low birth weight.

(c) The embryo develops inside the uterus of the female body. The uterus is well adapted for nourishing the baby as its inner lining (endometrium) is thick, soft, and richly supplied with blood vessels, which provide nutrients and oxygen to the developing embryo.

(d)

(i) The presence of XY chromosomes indicates that the foetus is male.

(ii) Prenatal sex determination is misused in some cases for female foeticide where female foetuses are aborted due to a preference for male children.

The below circuit is a part of an electrical device. Use the information given in the question to calculate the following.

(a) Potential Difference across R₂.

(b) Value of the resistance R₂.

(c) Value of resistance R₁.

Answer

From the diagram,

• Voltage of battery (V) = 12 V
• Net current through the circuit (I) = 2.0 A
• Current through 4 Ω resistor (i) = 1.5 A

(a) As, 4 Ω and R₂ are in parallel then the potential difference across them should be equal i.e.,

Potential difference across R₂ = Potential difference across 4 Ω

= i x 4

= 1.5 x 4

= 6.0 V

Hence, the potential Difference across R₂ is 6.0 V.

(b) Now,

Current through R₂ = I - i = 2.0 - 1.5 = 0.5 A

Then,

Potential difference across R₂ = (I - i) x R₂

⇒ (I - i) x R₂ = 6

⇒ 0.5 x R₂ = 6

⇒ R₂ = $\dfrac{6}{0.5} = \dfrac{60}{5}$

⇒ R₂ = 12 Ω

Hence, value of the resistance R₂ is 12 Ω.

(c) As,

Potential drop across all series resistor = V

⇒ Potential drop across R₁ + Potential drop across R₂ + Potential drop across 2.0 Ω = 12

⇒ I x R₁ + 6 + I x 2 = 12

⇒ 2 x R₁ + 2 x 2 = 12 - 6

⇒ 2 x R₁ + 4 = 6

⇒ 2 x R₁ = 6 - 4

⇒ 2 x R₁ = 2

⇒ R₁ = $\dfrac {2}{2}$

⇒ R₁ = 1 Ω

Hence, value of resistance R₁ is 1 Ω.

(a) State the law and write the formula connecting the electric current flowing through a conductor and voltage applied across it.

(b) In a household five fans each of 100 W are used for 4 hours and an electric press of 500 W for 2 hours every day. Calculate the cost of using the fans and electric press for 60 days if the cost of 1 unit of electrical energy is Rs. 6.5

Answer

(a) The relationship between the electric current flowing through a conductor and the voltage applied across it is given by Ohm’s Law.

Ohm’s Law : At constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across its ends, provided the physical conditions of the conductor remain constant.

Mathematically,

V ∝ I

or

V = IR

where:

• V = Potential difference (in volts, V)
• I = Current (in amperes, A)
• R = Resistance of the conductor (in ohms, Ω)

(b) Given,

• Number of fans (n) = 5
• Power rating of each fan (P₁) = 100 W
• Daily usage time for fan (t₁) = 4 hours
• Power rating of electric press (P₂) = 500 W
• Daily usage time for the press (t₂) = 2 hours
• Total number of days = 60
• Cost of 1 unit of electricity = Rs. 6.5

Total power consumed by the fans = n x P₁ = 5 x 100 = 500 W

Total electrical energy consumed by the fans each day = Total power consumed by the fans x t₁ = 500 x 4 = 2000 Wh

Similarly,

Total electrical energy consumed by the electric press each day = P₂ x t₂ = 500 x 2 = 1000 Wh

Now,

Total energy consumed by the fans and the press per day = 2000 + 1000 = 3000 Wh = 3 KWh

Then,

Total energy consumed by the fans and the press for 60 days = 3 x 60 = 180 KWh

As,

1 KWh = 1 unit

Then,

Total cost after 60 days = Total energy consumed by the fans and the press for 60 days x Cost of 1 unit of electricity

= 180 x 6.5

= ₹ 1170

Hence, the cost of using the fans and electric press for 60 days is ₹ 1170.

As shown in the figure below A and B are two lamps. Lamp A is rated at 12 V, 24W. Lamp B is rated at 6.0 V. When lamp B operates at its rated voltage, the current in it is 3.0 A. The values of R₁ and R₂ are chosen so that both lamps operate at their rated voltages.

Based on the information given, answer the following.

(a) Calculate the current in Lamp A.

(b) State and give reason for the reading of the Voltmeter.

(c) Calculate the resistance of R₂.

(d) Find the value of the resistance R₁.

Answer

Given,

• Voltage rating of the lamp A ($\text V_ \text A$) = 12 V
• Power rating of the lamp A ($\text P_ \text A$) = 24 W
• Voltage rating of the lamp B ($\text V_ \text B$) = 6 V
• Current passing through the lamp B ($\text I_ \text B$) = 3 A
• Battery voltage ($\text V$) = 15 V

(a) Current through the lamp A is given by,

$\text I_ \text A = \dfrac{\text P_ \text A}{\text V_ \text A} \\[1em] = \dfrac{24}{12} \\[1em] \Rightarrow \text I_ \text A = 2\ \text A$

Hence, the current in Lamp A is 2 A.

(b) Both branches (lamp A alone and lamp B in series with R₂) are in parallel, so they must have the same potential difference. For lamp A to run at its rated condition, the branch voltage must be 12 V. Hence the voltmeter connected across the branches reads 12 V.

(c) As, potential drop across lamp B branch is 12 V and is given by,

$\text V_ \text B + \text I_ \text B\text R_ \text 2 = 12 \\[1em] \text 6 + 3 \times \text R_ \text 2 = 12 \\[1em] \Rightarrow 3 \times \text R_2 = 12 - 6 \\[1em] \Rightarrow 3 \times \text R_2 = 6 \\[1em] \Rightarrow \text R_2 = \dfrac{6}{3} \\[1em] \Rightarrow \text R_2 = 2\ \text Ω$

Hence, the resistance of R₂ is 2 Ω.

(d) Current through R₁ is given by

$\text I = \text I_ \text A + \text I_ \text B \\[1em] = 2 + 3 \\[1em] = 5\ \text A \\[1em]$

And,

Potential drop across R₁ ($\text V'$) = $\text V - \text V_ \text A$ = 15 - 12 = 3 V

Also,

$\text V' = \text I \times \text R_1 \\[1em] \Rightarrow \text R_1 = \dfrac{\text V'}{\text I} \\[1em] \Rightarrow \text R_1 = \dfrac{3}{5} \\[1em] \Rightarrow \text R_1 = 0.6\ \text Ω$

Hence, the resistance of R₁ is 0.6 Ω.

(a) State Joules law of heating and write its mathematical expression.

(b) Two resistors of resistances 2 Ω and 4 Ω are connected in

1. series
2. parallel

with a battery of given potential difference. Compute the ratio of total quantity of heat produced in the combination in the two cases if the total voltage and time are kept the same for both.

Answer

(a) Joule's Law of Heating states that the amount of heat produced in a conductor is directly proportional to the square of the electric current passing through it, the resistance of the conductor and the time for which the current flows.

Mathematically, it can be expressed as

$\text H = \text I^2 \text {Rt}$

• $\text H$ is the heat produced (in joules),
• $\text I$ is the electric current (in amperes),
• $\text R$ is the resistance of the conductor (in ohms),
• $\text t$ is the time for which the current flows (in seconds).

(b) Given,

• $\text R_1$ = 2 Ω
• $\text R_2$ = 4 Ω
• $\text t_1 = \text t_2 = \text t$(let)
• $\text V_1 = \text V_2 = \text V$(let)

1. Series connection

Net resistance of the circuit is given by,

$\text R_\text S = \text R_1 + \text R_2 \\[1em] = 2 + 4 \\[1em] = 6\ \text Ω$

Let current through the combination is $\text I_\text S$.

Then,

$\text I_\text S = \dfrac{\text V_1}{\text R_\text S} \\[1em] = \dfrac{\text V}{6}$

So, heat produced by this combination is given by,

$\text H_\text S = \text I_\text S^2 \text R_\text S \text t_1 \\[1em] = \left(\dfrac{\text V}{6}\right)^2 \times 6 \times \text t \\[1em] = \dfrac{\text V^2}{36} \times 6 \times \text t \\[1em] = \dfrac{\text V^2}{6} \times \text t \\[1em] \Rightarrow \text H_\text S = \dfrac{\text V^2 \text t}{6}$

2. Parallel connection

Net resistance of the circuit is given by,

$\dfrac {1}{\text R_\text P} = \dfrac {1}{\text R_1} + \dfrac {1}{\text R_2} \\[1em] = \dfrac{1}{2} + \dfrac{1}{4} \\[1em] = \dfrac{2 + 1}{4} \\[1em] = \dfrac{3}{4} \\[1em] \Rightarrow \text R_\text P = \dfrac{4}{3}\ \text Ω$

Let current through the combination is $\text I_\text P$.

Then,

$\text I_\text P = \dfrac{\text V_2}{\text R_\text P} \\[1em] = \dfrac{\text V}{\dfrac{4}{3}} \\[1em] = \dfrac{3 \times \text V}{4} \\[1em] = \dfrac{3\text V}{4}$

So, heat produced by this combination is given by,

$\text H_\text P = \text I_\text P^2 \text R_\text P \text t_2 \\[1em] = \left(\dfrac{3\text V}{4}\right)^2 \times \dfrac{4}{3} \times \text t \\[1em] = \dfrac{9\text V^2}{16} \times \dfrac{4}{3} \times \text t \\[1em] = \dfrac{3\text V^2}{4} \times \text t \\[1em] \Rightarrow \text H_\text P = \dfrac{3\text V^2 \text t}{4}$

Now,

$\dfrac{\text H_\text S}{\text H_\text P} = \dfrac {\dfrac{\text V^2 \text t}{6}}{\dfrac{3\text V^2 \text t}{4}} \\[1em] = \dfrac{4\text V^2 \text t}{18\text V^2 \text t} \\[1em] = \dfrac {4}{18} \\[1em] = \dfrac{2}{9}$

Hence, the ratio of heat energy produced by series connection to the parallel connection is 2 : 9.

$\text{1.} \space \boxed{\text{A}} + \text{H}_2\text{O} \xrightarrow{\text{ELECTROLYSIS}} \underset{\text{CAUSTIC SODA}}{\boxed{\text{x}}} + \underset{\text{ANODE}}{\boxed{\text{Y} \uparrow}} + \underset{\text{CATHODE}}{\boxed{\text{Z} \uparrow}} \\[1em] \text{2.} \space \underset{\text{MOLTEN}}{\boxed{\text{A}}} \xrightarrow{\text{ELECTROLYSIS}} \underset{\text{CATHODE}}{\boxed{\text{M}}} + \underset{\text{ANODE}}{\boxed{\text{Y} \uparrow}} \\[1em] \text{3.} \space \boxed{\text{M}} + \text{CH}_3\text{COOH} \longrightarrow \boxed{\text{P}} + \boxed{\text{Z} \uparrow} \\[1em] \text{4.} \space \text{CH}_3\text{COOH} + \boxed{\text{Q}} \xrightarrow[\text{H}_2\text{SO}_4]{\text{Conc.}} \boxed{\text{R}} \\[1em] \text{5.} \space \boxed{\text{R}} + \boxed{\text{X}} \longrightarrow \boxed{\text{P}} + \boxed{\text{Q}}$

(a) Derive the names of A, Y, Z, M, P & R

Attempt either subpart B or C.

(b) Improvise an activity to test Z.

OR

(c) Name the process in which compounds X, Y & Z are formed from A. Justify your response.

Answer

(a)

$1. \space \boxed{\text{A}} + \text{H}_2\text{O} \xrightarrow{\text{ELECTROLYSIS}} \underset{\text{CAUSTIC SODA}}{\boxed{\text{x}}} + \underset{\text{ANODE}}{\boxed{\text{Y} \uparrow}} + \underset{\text{CATHODE}}{\boxed{\text{Z} \uparrow}}$

The only electrolysis that gives caustic soda (NaOH), hydrogen gas (H₂) at the cathode and chlorine gas (Cl₂) at the anode is the electrolysis of aqueous Sodium Chloride (NaCl).

$\text{NaCl} + \text{H}_2\text{O} \xrightarrow{\text{ELECTROLYSIS}} \underset{\text{CAUSTIC SODA}}{\text{NaOH}} + \underset{\text{ANODE}}{\text{Cl}_2 \uparrow} + \underset{\text{CATHODE}}{\text{H}_2 \uparrow}$

Hence, A → Sodium Chloride (NaCl)

$2. \space \underset{\text{MOLTEN}}{\boxed{\text{A}}} \xrightarrow{\text{ELECTROLYSIS}} \underset{\text{CATHODE}}{\boxed{\text{M}}} + \underset{\text{ANODE}}{\boxed{\text{Y} \uparrow}} \\[1em] \text{NaCl} \xrightarrow{\text{ELECTROLYSIS}} \underset{\text{CATHODE}}{\text{Na}} + \underset{\text{ANODE}}{\text{Cl}_2}$

Hence, Y → Chlorine gas (Cl₂)

$3. \space \boxed{\text{M}} + \text{CH}_3\text{COOH} \longrightarrow \boxed{\text{P}} + \boxed{\text{Z} \uparrow}$

Sodium metal (M) reacts with acetic acid (CH₃COOH) to form sodium acetate and hydrogen gas:

2Na + 2CH₃COOH → 2CH₃COONa + H₂(↑)

Hence, M → Sodium metal (Na) and Z → Hydrogen gas (H₂)

$4. \space \text{CH}_3\text{COOH} + \boxed{\text{Q}} \xrightarrow[\text{H}_2\text{SO}_4]{\text{Conc.}} \boxed{\text{R}}$

Acetic acid (CH₃COOH) reacts with ethanol (C₂H₅OH) in the presence of conc. H₂SO₄ to form ethyl ethanoate (CH₃COOC₂H₅) and water.

CH₃COOH + C₂H₅OH → CH₃COOC₂H₅

Hence, R → Ethyl ethanoate (CH₃COOC₂H₅)

$5. \space \boxed{\text{R}} + \boxed{\text{X}} \longrightarrow \boxed{\text{P}} + \boxed{\text{Q}}$

Ethyl ethanoate reacts with NaOH (alkali) to form sodium ethanoate (CH₃COONa) and ethanol (C₂H₅OH).

CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH

Hence, P → Sodium ethanoate (CH₃COONa)

(b) Activity to test Z (Hydrogen gas):

Aim : To test for hydrogen gas.

Procedure :

1. Set the apparatus as shown in below figure.

2. Take about 5 mL of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.

3. Pass the gas being evolved through the soap solution.

4. Take a burning candle near a gas filled bubble.

Observation: The gas burns with a pop sound.

Inference: The gas is hydrogen.

(c) The process is called Chlor-Alkali Process.
The name of the process comes from its products — Chlor from Chlorine and Alkali from sodium hydroxide which is an alkali.

(a) Distinguish between ethanol and ethanoic acid experimentally.

Attempt either subpart B or C.

(b) Give the IUPAC name of the first member of Alkene which is formed by addition of conc. sulphuric acid to it. Illustrate the change with the help of a chemical equation.

OR

(c) “All combustion reactions are oxidation but all oxidation reactions are not combustion.” Justify.

Answer

(a) We can distinguish ethanol (C₂H₅OH) and ethanoic acid (CH₃COOH) by the following test:

Reaction with Sodium Carbonate (Na₂CO₃) : Add a small amount of Na₂CO₃ or NaHCO₃ to both liquids.

• Ethanol shows no effervescence.
• Ethanoic acid produces brisk effervescence with carbon dioxide (CO₂) gas as shown below :

2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂(↑)

(b) The first member of the alkene series is Ethene (C₂H₄).

When ethanol is treated with concentrated sulphuric acid (H₂SO₄) and heated at about 170°C, it undergoes dehydration to form ethene.

$\text C_2 \text H_5 \text {OH} \xrightarrow[\text{H}_2\text{SO}_4]{\text{Conc.}} \text C_2 \text H_4 + \text H_2\text O$

(c) In combustion reactions oxygen is added hence all combustion reactions are oxidation, whereas in oxidation reactions, energy may or may not be released (along with the products), hence all oxidation reactions are not combustion reactions.

Mohan and Rohit observed that shoots of a plant growing in shade bend towards the sunlight. Whereas, leaves of ‘Touch me not’ plant fold and droop soon after touching. They were curious to know how these movements occur in plants.

In order to help them understand the movements in the plants, answer the following questions:

Attempt either subpart A or B.

(a) What causes the bending of shoots in the plants as shown in figure A?

OR

(b) What causes the folding of the leaves in ‘Touch me not’ plant as shown in figure B?

(c) Compare the movement of growth of the pollen tube towards ovule with the movements shown in part A of the above figure.

(d) Compare the movement shown in figure B with the movement of body parts in the animals.

Answer

(a) The bending of shoots in the plants :

• Bending of shoots of plants is a response to the stimulus and a directional, growth-related movement.
• When growing plants detect sunlight, a hormone called auxin, synthesized at shoot tip helps the cells to grow longer.
• When light is coming from one side of the plant, auxin diffuses to the shady side of the shoot.
• This concentration of auxin stimulates the cells of the shoot to grow longer on the side of the shoot which is away from the light. Thus, plant appears to bend towards light.

(b) The folding of the leaves in ‘Touch me not’ plant :

• Leaves of ‘Touch me not’ plant respond to the stimulus by showing growth independent movement.
• These plants use electrical–chemical means to convey the information from cell to cell.
• Movement happens at a point different from the point of touch.
• Plant cells change shape by changing the amount of water in them, resulting in swelling or shrinking, and therefore in changing shape.

(c) Growth of pollen tubes towards the ovule is an example of chemotropism whereas bending of shoots towards sunlight is an example of phototropism.

(d)

• Although both plants and animals show electrical–chemical means to convey the information from cell to cell but unlike nerve cells in animals there is no specialized tissue in plants for conduction of information.
• In animal cells, change in shape occurs because of the specialized proteins found in muscle cells; plant cells change shape by changing the amount of water in them.

During a field trip, Mohan and Rohit observed that shoots of sunflower plants bend towards the sunlight. Whereas, leaves of ‘Touch me not’ plant begin to fold and droop soon after touching even during the day. They were curious to know how these movements occur in plants.

Attempt either subpart A or B.

(a) What causes the bending of shoots in the sunflower plants towards sunlight?

OR

(b) What causes the folding of the leaves in ‘Touch me not’ plants when touched by hand?

(c) Compare the movement of growth of the pollen tube towards ovule with the bending of shoots of sunflower plant towards sunlight.

(d) Compare the movement in folding of leaves of ‘Touch me not’ plants with the movement of body parts in the animals.

Answer

(a) The bending of shoots in the plants :

• Bending of shoots of plants is a response to the stimulus and a directional, growth-related movement.
• When growing plants detect sunlight, a hormone called auxin, synthesized at shoot tip helps the cells to grow longer.
• When light is coming from one side of the plant, auxin diffuses to the shady side of the shoot.
• This concentration of auxin stimulates the cells of the shoot to grow longer on the side of the shoot which is away from the light. Thus, plant appears to bend towards light.

(b) The folding of the leaves in ‘Touch me not’ plant :

• Leaves of ‘Touch me not’ plant respond to the stimulus by showing growth independent movement.
• These plants use electrical–chemical means to convey the information from cell to cell.
• Movement happens at a point different from the point of touch.
• Plant cells change shape by changing the amount of water in them, resulting in swelling or shrinking, and therefore in changing shape.

(c) Growth of pollen tubes towards the ovule is an example of chemotropism whereas bending of shoots towards sunlight is an example of phototropism.

(d)

• Although both plants and animals show electrical–chemical means to convey the information from cell to cell but unlike nerve cells in animals there is no specialized tissue in plants for conduction of information.
• In animal cells, change in shape occurs because of the specialized proteins found in muscle cells; plant cells change shape by changing the amount of water in them.

The below image is that of a reflecting telescope. Reflecting telescopes revolutionised our ways of looking into the sky. They employ mirrors to gather and focus light, rather than relying solely on lenses as in their refracting counterparts. These telescopes utilise precisely shaped and polished mirrors to capture incoming light and reflect it to a focal point, where it forms an image for observation.

(a) What kind of image of the star is seen by the observer at the eyepiece?

(b) What kind of mirror is used in this reflecting telescope?

Attempt either subpart C or D.

(c) Explain with reason what kind of optical device (type of lens or mirror) that is used at the eyepiece.

OR

(d) What is the role of the plane mirror in the telescope?

Answer

(a) The image of the star seen by the observer at the eyepiece is a real image, as the final image is formed by the converging lens at the eyepiece.

(b) The curved mirror used in this reflecting telescope is a concave mirror, which gathers and focuses the incoming parallel rays of light from the distant star.

(c) A converging lens (convex lens) is used at the eyepiece to collect the rays reflected from the plane mirror and to allow the observer to view a real and erect image of the star.

(d) The plane mirror in the telescope laterally inverts the image formed by the curved mirror and its position helps to redirect the rays towards the eyepiece, enabling comfortable viewing of the image.

Azim Taraporewala was a traveller and science enthusiast. During one of his travels he found himself on the edge of an island without any mode of communication. As he had read in many stories, he thought he would light a fire on the beach and travelling boats or ships could see that fire and come to give him a ride. He had run out of lighters and match-sticks but had a reading glass. Being a science enthusiast he knew some tricks and used that lens and a scrap of paper to light a fire, with the help of scorching rays from the sun.

(a) Which lens can be used by Azim to create the fire?

(b) What property of the lens helps Azim to create the fire?

Attempt either subpart C or D.

(c) List two more uses of this kind of lens.

OR

(d) Explain with reason the condition under which the lens can form both real as well as virtual images.

Answer

(a) The lens used by Azim to create the fire is a convex lens.

(b) The convex lens has the property of converging parallel rays of sunlight to a single point known as the principal focus.

(c) Two more uses of a convex lens are:

1. It is used in a magnifying glass to view small objects clearly by producing a magnified virtual image.
2. It is used in optical instruments such as microscopes, cameras, projectors and telescopes to focus light and form images.

(d) A convex lens can form both real and virtual images depending on the position of the object:

• When the object is placed beyond the focus (F), the lens forms a real and inverted image on the opposite side of the lens.
• When the object is placed between the optical center (O) and the focus (F), the lens forms a virtual, erect and magnified image on the same side as the object.