Karnataka SSLC question papers, with every answer worked out

Correct option: (d)

D) 182. Explanation: Formula: HCF(a, b) × LCM(a, b) = a × b. Substitution: 13 × LCM = 26 × 91. Calculation: LCM = (26 × 91) / 13 = 182.

Correct option: (a)

A) 2² × 5 × 7. Formula: Prime factor tree. Substitution: 140 ÷ 2 = 70, 70 ÷ 2 = 35, 35 ÷ 5 = 7, 7 ÷ 7 = 1. Calculation: 2 × 2 × 5 × 7 = 2² × 5 × 7. Answer: 2² × 5 × 7.

Correct option: (c)

C) It factors into 17 × 92, revealing factors other than 1 and itself. Formula: A composite number has more than two distinct positive divisors. Substitution: N = 7 × 13 × 17 + 17. Calculation: N = 17(7 × 13 + 1) = 17 × 92 = 17 × 2² × 23. It has factors 2, 4, 17, 23, 46, 68, 92, etc. Answer: C) It factors into 17 × 92, revealing factors other than 1 and itself.

Correct option: (a)

A) Assume √5 = a/b where a, b are coprime integers and b ≠ 0. Explanation: Formula: Rational definition requires √5 = a/b with coprime integers. Substitution: Square both sides: 5b^2 = a^2. Calculation: 5|a^2 ⇒ 5|a, then 5|b. Conclusion: Contradicts coprime assumption, proving irrationality.

Correct option: (b)

B) unique. Formula: Fundamental Theorem of Arithmetic. Substitution: N/A. Calculation: N/A. Answer: The theorem states that factorisation is unique, apart from the order of the prime factors.

Correct option: (b)

B) 22338. Formula: HCF(a, b) × LCM(a, b) = a × b. Substitution: 9 × LCM = 306 × 657. Calculation: LCM = (306 × 657) / 9 = 201042 / 9 = 22338. Answer: 22338.

Correct option: (c)

C) It contains only the prime factor 2, lacking the necessary factor 5. Explanation: Formula: To end in 0, a number's prime factorisation must contain both 2 and 5. Substitution: 4^n = (2^2)^n = 2^(2n). Calculation: Prime factors present = {2} only. Conclusion: By uniqueness of factorisation, 5 cannot appear, so it never ends in 0.

Correct option: (a)

A) 36 minutes, because LCM(18, 12) gives the smallest common multiple of their lap times. Formula: LCM(a, b) = Product of greatest powers of all primes. Substitution: 18 = 2¹ × 3², 12 = 2² × 3¹. Calculation: LCM = 2² × 3² = 4 × 9 = 36. Answer: 36 minutes.

Isolating √3 yields a rational difference, contradicting the proven irrationality of √3.

Unique prime factorization ensures any prime in the square's factors must originate from the base integer.

If a prime p divides a², then p must also divide the integer a.

It is the product of the smallest power of each common prime factor involved.

The product of the two integers equals the product of their HCF and LCM.

Extending to three numbers introduces overlapping prime powers, breaking the direct multiplicative relationship.

Prime factorisation is strictly unique, while composite factorisation permits multiple non-unique combinations.

Factoring gives 13 × 78 (or 13² × 6), confirming it has factors beyond 1 and itself.

Prime factorisation gives 18 = 2 × 3^2 and 24 = 2^3 × 3. HCF = 2^1 × 3^1 = 6 and LCM = 2^3 × 3^2 = 72. Verification: HCF × LCM = 6 × 72 = 432, and 18 × 24 = 432. The relation holds true.

Assume √5 is rational, so √5 = a/b where a and b are coprime integers. Squaring gives 5b² = a², meaning 5 divides a², and by Theorem 1.2 (if prime p divides a², then p divides a), 5 must divide a. Substituting a = 5c reveals 5 also divides b, contradicting coprimality and proving √5 is irrational.

The sum or difference of a rational number and an irrational number is always irrational. Additionally, the product or quotient of a non-zero rational number and an irrational number is always irrational.

The runners meet simultaneously when each has completed a whole number of laps, which occurs at the smallest time divisible by both lap durations. Prime factors are 18 = 2 × 3² and 12 = 2² × 3, yielding an LCM of 2² × 3² = 36. Therefore, they will meet again at the starting point after exactly 36 minutes.

Find LCM of 30 and 45: 30 = 2 × 3 × 5 and 45 = 3^2 × 5. LCM = 2 × 3^2 × 5 = 90 seconds. Adding 90 seconds (1 min 30 sec) to 9:00:00 AM gives 9:01:30 AM. They will next change together at 9:01:30 AM.

For 12^n to end with 0, its prime factorisation must contain both 2 and 5. Since 12^n = (2^2 × 3)^n = 2^(2n) × 3^n, it only contains the primes 2 and 3. By the uniqueness of the Fundamental Theorem of Arithmetic, 5 cannot appear as a factor, so it never ends in 0.

Assume √3 is rational, so √3 = a/b where a and b are coprime. Squaring gives 3b² = a², meaning 3 divides a². By the prime divisibility theorem, 3 divides a, leading to a common factor of 3, which contradicts that a and b are coprime. Hence, √3 is irrational.

Proving √2 irrational relies on prime divisibility, whereas 3√2 leverages algebraic isolation of a known irrational. Assuming 3√2 = a/b rearranges to √2 = a/(3b), where a/(3b) is a rational ratio of integers. This directly contradicts the established irrationality of √2, confirming 3√2 cannot be rational.

For a number to end with zero, it must be divisible by both 2 and 5, meaning 5 must be present in its prime factorisation. The prime factorisation of the base 14 is 2 × 7, so 14ⁿ = 2ⁿ × 7ⁿ for any natural number n. By the Fundamental Theorem of Arithmetic, this factorisation is unique and contains no prime factor other than 2 and 7. Since the prime 5 never appears in the factorisation of 14ⁿ, it cannot be divisible by 5. Consequently, 14ⁿ cannot end with the digit zero for any natural number n.

Suppose 6 - 3√2 is rational, meaning it can be expressed as p/q where p and q are coprime integers and q ≠ 0. Rearranging the equation gives 3√2 = 6 - p/q, which simplifies to √2 = (6q - p) / 3q. Since p, q, and 6 are integers, the right-hand side (6q - p) / 3q is a rational number. This implies √2 is rational, which directly contradicts the established fact that √2 is irrational. Therefore, the initial assumption is false, and 6 - 3√2 is irrational.

Assume √7 is rational, so it can be written as a/b where a and b are coprime positive integers. Squaring both sides gives 7 = a²/b², which rearranges to a² = 7b², proving that 7 divides a². By the theorem that if a prime divides a square it divides the base, 7 must also divide a, so we substitute a = 7c. This yields 49c² = 7b², simplifying to b² = 7c², which proves 7 also divides b. Since both a and b are divisible by 7, this contradicts the coprime assumption, proving √7 is irrational.

First, we state the initial assumption that √5 is rational and can be written as a fraction a/b where a and b are coprime integers. We substitute this into the squared equation 5b² = a² to deduce that 5 divides a, allowing us to write a = 5c for some integer c. Then, we calculate the resulting substitution 5b² = 25c² which simplifies to b² = 5c², proving that 5 also divides b. The final answer reveals a direct contradiction to the coprime condition. Therefore, the assumption is false and √5 must be irrational.

Step 1: Identify that the meeting time is the LCM of their individual round times: 18 = 2^1 × 3^2 and 12 = 2^2 × 3^1. Step 2: Apply LCM formula: Take greatest powers of all primes involved = 2^2 × 3^2. Step 3: Calculate: LCM = 4 × 9 = 36 minutes. Step 4: Interpret: They will meet exactly when both complete an integer number of rounds. Conclusion: They will meet again at the starting point after 36 minutes.

The divisibility rule for 10 states that a number must contain both prime factors 2 and 5 to end in 0. Substituting 6ⁿ into the prime factorisation formula gives (2 × 3)ⁿ = 2ⁿ × 3ⁿ. The calculation clearly shows that only the primes 2 and 3 are present, while the prime factor 5 is entirely missing. By the Fundamental Theorem of Arithmetic, this unique factorisation cannot spontaneously generate a new prime like 5 for any value of n. Hence, 6ⁿ will never end with the digit 0.

Step 1: State the fundamental relationship for two positive integers: HCF(a, b) × LCM(a, b) = a × b. Step 2: Substitute known values into the formula: 9 × LCM(306, 657) = 306 × 657. Step 3: Rearrange and calculate: LCM = (306 × 657) / 9 = 34 × 657. Step 4: Final multiplication: 34 × 657 = 22338. Conclusion: The LCM of 306 and 657 is 22338.

The proof begins with the assumption that √5 is rational, expressed as a/b where a and b are coprime integers. Squaring both sides gives 5b² = a², which implies that 5 divides a², and by Theorem 1.2, 5 must also divide a. Substituting a = 5c back into the equation yields 5b² = 25c², which simplifies to b² = 5c², proving 5 divides b. The answer reveals that both a and b share a common factor of 5, directly contradicting the initial coprime condition. This logical contradiction confirms that √5 cannot be rational.

The Fundamental Theorem of Arithmetic states that every composite number can be uniquely expressed as a product of prime factors, apart from their order. To express 140 as a product of primes, we begin by dividing it by the smallest prime number, 2. Dividing 140 by 2 gives 70. Dividing 70 by 2 again yields 35. Next, we divide 35 by the prime number 5 to get 7. Finally, 7 is itself a prime number. Therefore, combining these factors, we get 140 = 2 × 2 × 5 × 7, which is written as 2² × 5¹ × 7¹. This representation is unique as guaranteed by the theorem.

Suppose, for the sake of contradiction, that 5 – √3 is a rational number. By definition, it can be written in the form a/b, where a and b are integers with no common factors and b ≠ 0. Setting up the equation 5 – √3 = a/b, we rearrange it to isolate the square root term as √3 = 5 – (a/b). Since 5, a, and b are all integers, the expression 5 – (a/b) represents a rational number. However, this result contradicts the established mathematical fact that √3 is an irrational number. The contradiction arises solely from our initial false assumption. Therefore, we conclude that 5 – √3 must be irrational.

The proof begins with the assumption that √5 is rational, expressed as a/b where a and b are coprime. Squaring the equation gives the relation 5b² = a², which implies 5 divides a². Applying the prime divisibility theorem, since 5 divides a², it must also divide a, so we substitute a = 5c. Calculating the substitution yields 5b² = 25c², which simplifies to b² = 5c², proving 5 also divides b. This derivation shows 5 is a common factor of both a and b, contradicting the coprime assumption. Evaluating the contradiction confirms the initial rational assumption is false. Thus, √5 is strictly an irrational number.

The fundamental relationship formula is HCF(a, b) × LCM(a, b) = a × b. Substituting the given numbers, we first find prime factors: 120 = 2³ × 3¹ × 5¹ and 90 = 2¹ × 3² × 5¹. Calculating the HCF by taking the lowest common powers gives 2¹ × 3¹ × 5¹ = 30. Calculating the LCM by taking the highest powers gives 2³ × 3² × 5¹ = 360. Verifying the left side yields 30 × 360 = 10800. Verifying the right side yields 120 × 90 = 10800. Both products match exactly, confirming the HCF and LCM relationship holds true for 120 and 90.

The standard formula for constructing a quadratic polynomial from its roots is p(x) = k[x² - (α + β)x + αβ]. We substitute the given sum 3/5 and product -2/5 into this expression. This yields p(x) = k[x² - (3/5)x - 2/5]. To obtain integer coefficients, we assign the scaling constant k = 5. Distributing the constant across the terms results in p(x) = 5x² - 3x - 2. We verify the sum of zeroes using -b/a, which gives -(-3)/5 = 3/5. We verify the product using c/a, which gives -2/5. The derived polynomial perfectly satisfies the initial conditions in standard algebraic units.

Since x = -3 is a root, substituting it into the polynomial equation must yield a value of zero. We calculate 5(-3)² + 13(-3) + k = 0, which simplifies to 45 - 39 + k = 0. Solving this linear equation directly provides the value k = -6. With k determined, the polynomial becomes 5x² + 13x - 6. To find the second zero, we apply the sum of zeroes formula α + β = -b/a. Substituting the known values gives -3 + β = -13/5. Isolating β yields the second zero as -13/5 + 3 = 2/5. The missing coefficient and the remaining root are thus accurately calculated in standard mathematical units.

Correct option: (b)

B) The water must be in the form of steam, indicating the reaction requires high temperature to proceed. Step 1: Interpret (g) as gaseous/steam state. Step 2: Recall that iron does not react with cold/boiling liquid water to form Fe₃O₄; it only reacts with steam. Step 3: The notation specifies necessary reaction conditions (high T) for this displacement/redox process to occur.

Correct option: (c)

C) Nitrogen dioxide. Step 1: Identify the reactant undergoing thermal decomposition: 2Pb(NO₃)₂(s). Step 2: Write the balanced equation: 2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g). Step 3: Match the physical observation (brown fumes) to the gaseous product NO₂, as lead oxide is solid and oxygen is colourless.

Correct option: (b)

The temperature increases. Explanation: Step 1: Activity 1.3 asks 'Touch the conical flask or test tube. Is there any change in its temperature?' Step 2: The reaction between zinc and acid releases heat. Step 3: This temperature change is one of the four observations that indicate a chemical reaction has occurred.

Correct option: (c)

C) Copper metal gets deposited on the iron nail due to displacement. Explanation: Iron is more reactive than copper and displaces it from the solution (Fe + CuSO₄ → FeSO₄ + Cu). The displaced copper metal deposits as a brownish layer on the iron nail, while the blue colour of the solution fades.

Correct option: (c)

To remind us not to change anything inside the boxes while balancing. Explanation: Step 1: Step I of balancing states 'draw boxes around each formula. Do not change anything inside the boxes while balancing the equation.' Step 2: This prevents altering subscripts in chemical formulas. Step 3: Only coefficients can be changed, not the molecular formulas themselves.

Correct option: (a)

A) Net energy is absorbed from the surroundings to break reactant bonds, making it endothermic. Explanation: A noticeable drop in temperature indicates that the reaction system absorbs thermal energy from its surroundings. This occurs when the energy required to break the bonds in Ba(OH)₂ and NH₄Cl exceeds the energy released when new bonds form in the products, classifying it as an endothermic reaction.

Correct option: (a)

A) Oxidised. Explanation: In classical redox terminology, oxidation is defined as the gain of oxygen or loss of hydrogen. Therefore, any substance that loses hydrogen atoms during a reaction is undergoing oxidation.

Correct option: (d)

D) Iron is more reactive than copper and displaces it from the salt solution. Explanation: In the reactivity series, iron is placed above copper. Therefore, iron atoms lose electrons more readily and displace copper ions from CuSO₄ solution, forming green FeSO₄ (fading blue colour) and depositing brown copper metal on the nail. The reaction is Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s).

Cu + ZnSO₄ → no reaction occurs because copper is less reactive than zinc.

Reaction conditions like heat (Δ) or catalysts required for the reaction.

Paint forms a protective layer that physically blocks oxygen and water from contacting iron surface.

It ensures mass conservation by equating atom counts, like balancing 2H₂ + O₂ → 2H₂O.

Photolytic energy splits 2AgCl into metallic Ag(s) and Cl₂(g), creating permanent dark images.

When ions exchange to form insoluble products, they precipitate as visible solids.

It is a thermal decomposition reaction producing ferric oxide and sulphur oxides.

It releases heat energy along with the formation of carbon dioxide and water.

The blue colour fades because the more reactive iron displaces copper ions from the solution, forming pale green iron sulphate. The displaced copper metal deposits as a brownish layer on the surface of the iron nail. This process is a displacement reaction governed by the relative reactivity of metals.

A combination reaction is a chemical change in which two or more reactants combine to form a single new product. The reaction between calcium oxide and water is exothermic because it releases a large amount of heat energy while forming calcium hydroxide. This noticeable temperature increase is a defining characteristic of exothermic processes.

The Law of Conservation of Mass states that mass can neither be created nor destroyed during a chemical reaction. This law requires balancing to ensure the number of atoms of each element remains identical on both the reactant and product sides. For iron and steam, the balanced equation is 3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g), which satisfies this law.

Two observable changes are a change in colour and the evolution of a gas. These changes occur because the original reactant molecules break their chemical bonds and rearrange to form entirely new product molecules with different physical and chemical properties.

Manganese dioxide (MnO₂) is reduced because it loses oxygen atoms to form manganese chloride (MnCl₂). Hydrogen chloride (HCl) is oxidised because it loses hydrogen atoms (which combine with oxygen to form water), and chlorine atoms combine to form chlorine gas (Cl₂).

The balanced chemical equation is 2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g). The reaction is confirmed by the immediate emission of reddish-brown fumes of nitrogen dioxide gas. Additionally, the white lead nitrate powder converts into a pale yellow solid residue of lead oxide.

Exothermic reactions release thermal energy into the surroundings, resulting in a temperature rise, such as the cellular respiration of glucose. Endothermic reactions absorb energy from the surroundings, causing a temperature drop, like the thermal decomposition of calcium carbonate in a kiln.

This reaction is classified as endothermic because it requires the continuous absorption of light energy from the sun to break the chemical bonds within the silver chloride crystals. Without this external energy input, the compound would not split into silver metal and chlorine gas.

In a displacement reaction, a more reactive element actively replaces a less reactive element from its aqueous compound through a single exchange mechanism. Conversely, a double displacement reaction involves the mutual swapping of positive and negative ions between two distinct compounds to form new substances. For example, zinc displaces copper from solution via Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s). In contrast, mixing barium chloride and sodium sulphate triggers complete ion exchange: BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq). The fundamental difference is single-element substitution versus complete ionic partner exchange.

Copper is positioned higher than silver in the reactivity series, enabling it to displace silver ions from the aqueous salt solution. The initially colourless silver nitrate solution will gradually turn blue due to the formation of dissolved copper(II) nitrate. Simultaneously, shiny greyish-white crystalline deposits of metallic silver will visibly coat the surface of the copper strip. The balanced chemical equation representing this single displacement process is Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s). This visual transformation confirms that a more reactive metal can successfully replace a less reactive metal from its ionic compound.

Both processes are classified as exothermic because they continuously release thermal and chemical energy into the surrounding environment during molecular breakdown. In cellular respiration, glucose undergoes controlled oxidation with oxygen to generate ATP, carbon dioxide, water, and heat. The balanced chemical equation is C6H12O6(aq) + 6O2(aq) -> 6CO2(g) + 6H2O(l) + Energy. Similarly, microbial action during composting breaks down complex organic polymers, releasing measurable heat as a byproduct. This consistent outward energy transfer confirms their classification as exothermic chemical changes.

Combination reactions involve two or more reactants merging to produce a single new product, whereas decomposition reactions involve a single reactant splitting into two or more simpler substances. These reaction types are fundamentally opposite in their chemical progression and structural outcomes. A classic example of combination is the burning of magnesium ribbon, represented by the balanced equation 2Mg(s) + O2(g) -> 2MgO(s). Conversely, the thermal decomposition of limestone follows the equation CaCO3(s) -> CaO(s) + CO2(g). The primary distinction lies in reactant consolidation versus reactant fragmentation under energy input.

Manganese dioxide undergoes reduction while hydrochloric acid undergoes oxidation during this chemical transformation. Oxidation is identified by the loss of hydrogen, which occurs when HCl molecules release hydrogen atoms to form chlorine gas. Reduction is identified by the loss of oxygen, which occurs when MnO2 molecules shed oxygen atoms to become manganese chloride. Consequently, MnO2 functions as the oxidising agent by supplying oxygen, and HCl acts as the reducing agent by accepting it. This simultaneous transfer of oxygen and hydrogen defines the overall process as a redox reaction.

In this reaction, manganese dioxide undergoes reduction as manganese changes from +4 oxidation state in MnO₂ to +2 in MnCl₂, gaining electrons. Hydrogen chloride undergoes oxidation as chloride ions (Cl⁻, oxidation state -1) lose electrons to form chlorine gas (Cl₂, oxidation state 0). Using traditional definitions: reduction involves loss of oxygen (MnO₂ loses O to form MnCl₂ and H₂O), while oxidation involves gain of oxygen or loss of hydrogen. Here, HCl does not lose hydrogen in the traditional sense; rather, the chloride ions are oxidized. This simultaneous oxidation and reduction makes it a redox reaction.

Rusting is an electrochemical oxidation process where iron reacts with atmospheric oxygen and moisture to produce hydrated iron(III) oxide. The chemical transformation can be summarised by the equation 4Fe + 3O2 + xH2O -> 2Fe2O3.xH2O. Applying a thick layer of paint prevents corrosion by establishing an impermeable barrier that physically blocks oxygen and water molecules from contacting the metal surface. Galvanisation prevents rusting by coating the iron with zinc, which sacrificially oxidises first due to its higher reactivity. Both strategies effectively halt the electron transfer necessary for iron degradation.

Mixing these solutions initiates a double displacement reaction where lead cations and potassium anions completely swap their ionic partners in the aqueous medium. The lead(II) ions immediately bond with iodide ions to create lead(II) iodide, which exceeds its solubility limit and separates from the solution. The balanced chemical equation for this process is Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq). Potassium nitrate remains fully dissolved because of its high aqueous solubility and low lattice energy. The immediate appearance of a bright yellow solid confirms successful precipitation through ionic recombination.

Oxidation is the gain of oxygen or loss of hydrogen. Reduction is the loss of oxygen or gain of hydrogen. In the reaction CuO + H₂ → Cu + H₂O: CuO loses oxygen and is reduced to Cu, so CuO is the substance reduced. H₂ gains oxygen and is oxidised to H₂O, so H₂ is the substance oxidised. This simultaneous oxidation and reduction makes it a redox reaction. This principle explains rancidity: when fats and oils are exposed to air, they undergo oxidation (gain of oxygen), producing unpleasant smell and taste. This is prevented by storing in airtight containers.

Combination reaction: Two or more reactants combine to form a single product. Example: Burning of magnesium - 2Mg + O₂ → 2MgO (exothermic). Decomposition reaction: A single reactant breaks down into two or more simpler substances. Example: Thermal decomposition of ferrous sulphate - 2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃(g) (endothermic). Combination reactions release heat (exothermic) while decomposition reactions require heat energy (endothermic).

Rusting of iron is an oxidation process where iron reacts with oxygen and moisture to form hydrated iron(III) oxide [4Fe + 3O₂ + 2nH₂O → 2Fe₂O₃·nH₂O]. Rancidity is oxidation of fats/oils by atmospheric oxygen. In both cases, the substances undergo oxidation (gain oxygen). Prevention methods: For iron - painting, oiling, galvanization (coating with zinc) to prevent contact with air/moisture. For food - storing in air-tight containers, nitrogen flushing, adding antioxidants like BHT. These methods prevent oxidation by eliminating contact with oxygen.

When acids react with metal carbonates: Metal carbonate + Acid → Salt + Water + Carbon dioxide. Example: Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g). When acids react with metal hydrogencarbonates: Metal hydrogencarbonate + Acid → Salt + Water + Carbon dioxide. Example: NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g). The CO₂ gas is tested by passing through lime water [Ca(OH)₂ solution], which turns milky due to formation of insoluble white precipitate of calcium carbonate: Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l).

Tooth decay starts when the pH in the mouth falls below 5.5. Below this pH, tooth enamel begins to corrode. Bacteria in the mouth produce acids from food remains, lowering the pH. Using basic toothpaste helps neutralize the excess acid and prevent decay. In agriculture, when soil becomes too acidic, plants do not grow well. Farmers add slaked lime [Ca(OH)₂] to acidic fields to neutralize the acidity. This is a neutralisation reaction where acid in the soil reacts with the base (slaked lime) to form salt and water, bringing the soil pH to a suitable level for crops.

The chlor-alkali process involves passing electricity through a concentrated aqueous solution of sodium chloride (brine). At the cathode, hydrogen gas is liberated. At the anode, chlorine gas is evolved. Sodium hydroxide solution forms near the cathode. The overall reaction is: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g). The chlorine gas produced is used for making bleaching powder. When baking soda (sodium hydrogencarbonate) is heated during cooking, it decomposes: 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(l) + CO₂(g). The carbon dioxide gas released makes the food soft, spongy and easy to digest.

Correct option: (a)

A) Pondicherry. Originally Valikondapuram, it was developed by the French into a major trade center and became their capital in India.

Correct option: (a)

Dupleix, who arrived in 1746 in Pondicherry as the Governor General of the French, had high ambitions of establishing French political power in India. He sought to interfere in Indian political affairs and use the policy of 'subsidiary alliance' to expand French influence.

Correct option: (c)

C) Vasco da Gama. He left Lisbon and reached Kappad near Calicut in 1498, successfully discovering the new sea route to India.

Correct option: (d)

D) Black Hole Tragedy. Siraj-ud-Daulah imprisoned 146 Englishmen in a small room, causing 123 deaths, which enraged Robert Clive and led to the war.

Correct option: (a)

The Battle of Plassey marked the beginning of British political control in India. It established the British as a major power in Bengal and paved the way for their territorial expansion.

Correct option: (a)

A) It prioritized naval dominance and maritime trade control over establishing large inland territorial empires. Explanation: Francisco de Almeida's Blue Water Policy specifically aimed to establish Portuguese supremacy over the sea rather than pursuing land-based territorial conquests, which contrasted with Albuquerque's later approach of capturing strategic land holdings like Goa.

Correct option: (b)

B) Constantinople. The text states that in 1453, the Ottoman Turks captured Constantinople, bringing all connecting trade routes under their control and disrupting Asian-European trade.

Correct option: (b)

B) Alfonso de Albuquerque. He defeated the Sultan of Bijapur in 1510, captured Goa, and made it the administrative centre of Portuguese power.

The Ottoman Turks captured Constantinople in 1453, disrupting Asian trade routes.

He unified local principalities, controlled key trading centers, and decisively defeated the Dutch forces in the 1741 battle.

Although Pondicherry was returned, France lost all territorial bases and military influence, permanently ceding Indian supremacy to Britain.

Raja Marthanda Varma successfully defeated the Dutch forces and secured Travancore independence.

Mir Jafar betrayed Siraj-ud-Daula by remaining neutral during the Battle of Plassey.

European rivalries between trading companies often reflected broader political conflicts in Europe, leading to proxy wars in India through alliances with local rulers.

The Treaty of Pondicherry officially concluded the Second Carnatic War.

Raja Marthanda Varma successfully defeated the Dutch forces in 1741.

The Dutch formed the Dutch East India Company in 1602 and concentrated on establishing fortified warehouses in coastal areas like Surat and Kochin to monopolize the spice trade. In contrast, the French established a state-owned company in 1664 and prioritized developing strategic political centers like Pondicherry under Governor Dupleix to challenge British territorial ambitions.

By the end of the 17th century, the English established three Presidencies: Bombay (acquired 1668, given by Charles II on annual rent of £10), Madras (taken 1639, St. George Fort built), and Calcutta (purchased 1690, Fort William built). These became centers of British administration. By the later 18th century, Calcutta became the capital where they implemented their own civil and criminal procedure codes, laying foundation for colonial administrative control.

Viceroy Francisco de Almeida implemented the Blue Water Policy to secure naval supremacy in maritime routes rather than pursuing territorial conquests on land. This strategy focused on controlling strategic sea lanes and ports, enabling the Portuguese to monopolize Asian trade and effectively restrict rival European merchants.

The English East India Company received royal charter from Queen Elizabeth in 1600. They established their first factory at Surat in 1613 with permission from Mughal Emperor Jahangir. Sir Thomas Roe obtained permission to establish more factories at Agra, Ahmedabad and Broach in 1617. They acquired Madras in 1639 and built St. George Fort, received Bombay in 1668, and purchased Calcutta villages in 1690 building Fort William.

Vasco da Gama, a Portuguese sailor, discovered the new sea route to India. He left Lisbon and successfully reached Kappad near Calicut on the West coast of India in the year 1498.

Alfonso de Albuquerque is considered the real founder of Portuguese power in India. In 1510, he defeated the Sultan of Bijapur and captured Goa, which became the administrative centre of Portuguese rule in India.

The Portuguese under Almeida followed the Blue Water Policy focusing on naval supremacy and controlling sea routes rather than territorial conquest on land. In contrast, the French under Governor Dupleix who arrived in 1746 pursued political ambitions and territorial influence, establishing Pondicherry as a political center to challenge British power, representing a land-based political approach rather than purely maritime commercial strategy.

The Portuguese sailor Vasco da Gama discovered the new direct sea route to India. He successfully sailed from Lisbon and reached Kappad near Calicut on the west coast of India in the year 1498.

Francisco de Almeida implemented the Blue Water Policy to establish Portuguese supremacy over the sea rather than over land. This strategy successfully maintained their monopoly over Indian Ocean trade for a century. However, their power declined due to: (1) arrival of English and French in India; and (2) rise of English in India combined with Dutch acquisition of islands in Southeast Asia. These factors gradually eroded Portuguese commercial advantages and political influence.

The English established three Presidency towns: (1) Madras - taken from King of Chandragiri in 1639, with Fort St. George built; (2) Bombay - given by Charles II to East India Company in 1668 on annual rent of ten pounds; (3) Calcutta - purchased three villages (Sutanauti, Calcutta, Govindapura) in 1690, with Fort William built. By end of 17th century, these became centers of English Presidencies, with Calcutta becoming capital by later 18th century.

Dupleix arrived in Pondicherry in 1746 as the Governor General of the French with high ambitions of establishing French supremacy in India.

The English established administrative control through: (1) building fortified settlements - Fort St. George at Madras (1639), Fort William at Calcutta (after 1690); (2) implementing their own civil and criminal procedure codes in controlled areas; (3) making Calcutta their capital city by later 18th century; and (4) securing territories through purchase (Calcutta villages), rent (Bombay), or treaty (Madras). This gradual establishment of fortified presidencies with independent legal systems laid groundwork for their expanding political authority.

The English initially constructed coastal warehouses purely to store and protect commercial merchandise from European rivals. Over time, they fortified these locations by building strong defensive structures like Fort William and St. George Fort. These secured enclaves gradually evolved into major Presidencies that housed military garrisons and administrative offices. The British further consolidated authority by implementing their own civil and criminal procedure codes in these territories. This systematic transition from simple trading outposts to fortified administrative hubs laid the groundwork for territorial conquest.

Vasco da Gama, a Portuguese sailor, successfully discovered the new sea route to India in 1498 when he landed at Kappad near Calicut. This voyage allowed the Portuguese to become the first Europeans to re-establish direct maritime trade relations with India. The newly discovered route served as the primary commercial corridor between India and Europe for several centuries. It remained the standard shipping path until the construction of the Suez Canal in 1869 drastically shortened travel distances. Ultimately, this discovery ended the long-standing monopoly of Arab and Italian merchants over the Asian spice trade.

By the end of the 17th century, the English established Bombay, Madras and Calcutta as centers of their Presidencies. By the later part of 18th century, they made Calcutta their capital city. They implemented their own civil and criminal procedure codes in the areas that were under their control.

The Ottoman Turks captured Constantinople in 1453, bringing vital overland trade routes under their strict control. They imposed heavy taxes on merchandise, which rendered traditional Asian-European commerce highly unprofitable. Consequently, Spanish and Portuguese rulers began sponsoring adventurous sailors to discover alternative maritime paths. This geopolitical disruption, combined with navigational inventions like the compass and astrolabe, directly motivated maritime exploration. Ultimately, this shift successfully bypassed Arab and Italian monopolies and opened direct European trade with India.

Dupleix arrived in Pondicherry in 1746 as the Governor General of the French with high ambitions of establishing French political power in India. His arrival marked a shift from purely commercial interests to political ambitions, which threatened the English East India Company's position. The French had already established their capital at Pondicherry and developed it as a major trade center since 1674. Dupleix's political ambitions set the stage for future conflicts between the English and French in India, as both European powers sought to expand their influence beyond trade into territorial and political dominance.

Dupleix arrived in Pondicherry in 1746 as the Governor General of the French with high ambitions of establishing French political power in India. His arrival marked a significant escalation in French ambitions from mere trade to political dominance. He sought to challenge British influence and expand French territorial control, transforming the commercial rivalry into political and military confrontation. This shift led to the Carnatic Wars between the British and French for supremacy in South India.

Portuguese decline: Their absolute monopoly over trade with India lasted only a century. Their power declined with the arrival of English and French in India. Dutch decline: The rise of the English in India and acquisition of islands of South-East Asia by the Dutch were the main factors. While Portuguese lost to European competitors, Dutch redirected focus to Southeast Asia while English consolidated in India. Both were ultimately overshadowed by British commercial and political expansion.

Lord Dalhousie implemented the Doctrine of Lapse in 1848 as a policy to integrate Indian princely states with the British Empire. According to this policy, if any Indian ruler died childless, his adopted male child had no legal right over the throne. Such a state was merged with the British Empire. This policy denied traditional Indian succession customs and allowed the British to acquire territories without military confrontation. For example, Satara was annexed under this rule. The implementation of this policy created resentment among ruling families and the common people who remained loyal to their traditional kings.

Lord Dalhousie introduced the Doctrine of Lapse in 1848 as a policy to annex Indian territories. According to this policy, if any Indian ruler died childless, his adopted male child had no legal right over the throne, and such states were merged with the British Empire. This denied succession rights to adopted children and allowed British to take over kingdoms without natural heirs. The policy led to annexation of several states and caused resentment among royal families who lost their ancestral power and kingdoms. It was a final attempt to integrate Indian princely states with the British Empire.

Lord Dalhousie introduced the Doctrine of Lapse in 1848 to integrate Indian princely states with the British Empire. According to this policy, if any Indian ruler died childless, his adopted male child had no legal right over the throne. Such states were merged with the British Empire. This policy denied traditional succession rights to royal families and allowed British annexation without military confrontation. It served as a cunning political strategy to expand British territory systematically.