UPSC NDA-2 Archives

1. Consider the following statements regarding Sanchi and its archaeologi

Consider the following statements regarding Sanchi and its archaeological findings :

1. When Sanchi was discovered in 1818, three of its four gateways were still standing, the fourth was lying on the spot where it had fallen
2. Animals depicted at Sanchi include elephants, horses, monkeys and cattle

Which of the statements given above is/are correct ?

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Only 1

Only 2

Both 1 and 2

Neither 1 nor 2

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This question was previously asked in

UPSC NDA-2 – 2023

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Statement 1: Sanchi was discovered in 1818 by General Taylor. Historical accounts confirm that at the time of discovery, three of the four gateways (Toranas) of the Great Stupa (Stupa 1) were standing, while the fourth (the northern gateway) had fallen. So, statement 1 is correct.
Statement 2: The carvings on the gateways and railings at Sanchi, particularly Stupa 1, are renowned for depicting various scenes from Jataka tales and events from the life of Buddha. These carvings include numerous representations of animals such as elephants, horses, monkeys, cattle, deer, lions, etc., often used symbolically or as part of narrative scenes. So, statement 2 is correct. Since both statements are correct, the answer is C.

Both the condition of Sanchi gateways upon discovery in 1818 and the variety of animals depicted in its sculptures are historically and archaeologically accurate facts.

Sanchi is a UNESCO World Heritage Site in Madhya Pradesh, India, famous for its Great Stupa built by Emperor Ashoka. The gateways are particularly noted for their intricate and detailed carvings.

2. The book ‘Nation First : The Golden Saga of 82 Years’ is the history o

The book ‘Nation First : The Golden Saga of 82 Years’ is the history of which one of the following organizations ?

Indo-Tibetan Border Police

Border Security Force

Central Reserve Police Force

Central Industrial Security Force

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UPSC NDA-2 – 2023

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The book ‘Nation First: The Golden Saga of 82 Years’ chronicles the history of the Central Reserve Police Force (CRPF). The book was released in 2021, commemorating 82 years since the raising of the CRPF in 1939 as the Crown Representative’s Police.

‘Nation First: The Golden Saga of 82 Years’ is a book about the history of the Central Reserve Police Force (CRPF).

The Central Reserve Police Force is one of the Central Armed Police Forces of India, under the authority of the Ministry of Home Affairs. It was established as the Crown Representative’s Police on 27 July 1939 and became the Central Reserve Police Force upon enactment of the CRPF Act on 28 December 1949.

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3. Which one of the following combinations of events represents the corre

Which one of the following combinations of events represents the correct sequence during reproduction in flowering plants ?

Ovule → fruit; egg → embryo

Embryo → egg; ovary → fruit

Ovule → zygote; zygote → seed

Ovary → fruit; zygote → egg; embryo → ovule; seed → fruit

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This question was previously asked in

UPSC NDA-2 – 2023

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This question is potentially flawed as multiple options contain incorrect pairings or sequences. However, evaluating the given options based on common transformations and developmental steps in flowering plant reproduction:
– Ovary develops into fruit (Ovary → fruit): This is a correct transformation that occurs after fertilization.
– Ovule develops into seed (Ovule → seed): This is a correct transformation.
– Egg (after fertilization, becomes zygote) develops into embryo (Egg → embryo, simplified lineage): This represents a correct developmental lineage, though it skips the zygote stage.
– Zygote develops into embryo (Zygote → embryo): This is a correct developmental step.

Let’s examine the options:
A) Ovule → fruit (Incorrect); egg → embryo (Correct lineage, simplified). Contains one incorrect and one correct (simplified) pairing.
B) Embryo → egg (Incorrect sequence); ovary → fruit (Correct transformation). Contains one incorrect and one correct pairing.
C) Ovule → zygote (Incorrect); zygote → seed (Incorrect). Contains two incorrect pairings.
D) Ovary → fruit (Correct transformation); zygote → egg (Incorrect sequence); embryo → ovule (Incorrect transformation); seed → fruit (Incorrect transformation). Contains one correct and multiple incorrect pairings.

Comparing options A and B, both contain one correct/partially correct statement paired with an incorrect one. Option B contains the direct and standard transformation “ovary → fruit”. While “egg → embryo” in A is a correct developmental lineage, “ovary → fruit” in B is a more direct transformation outcome commonly paired in such questions. Given the potential flaw in the question structure where combinations might mix correct and incorrect statements, option B is selected as the most likely intended answer because it contains the key correct transformation “ovary → fruit”.

Key transformations in flowering plant reproduction include Ovary developing into Fruit and Ovule developing into Seed. The egg cell participates in the formation of the embryo (Egg → Zygote → Embryo). Option B includes the correct transformation “ovary → fruit”.

The structure of this question with multiple pairings in options, some of which are incorrect, suggests it might be testing the identification of *any* correct transformation or sequence within the combination, despite the combination as a whole not necessarily representing a strict sequence of pairs. In typical questions, “Ovary → fruit” and “Ovule → seed” are standard correct transformations.

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4. Which one of the following is the correct sequence during sexual repro

Which one of the following is the correct sequence during sexual reproduction in Angiosperms ?

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Egg → zygote → embryo → seed

Embryo → egg → zygote → seed

Egg → embryo → zygote → seed

Egg → seed → zygote → embryo

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This question was previously asked in

UPSC NDA-2 – 2023

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During sexual reproduction in Angiosperms (flowering plants), the sequence of development starting from the female gamete is as follows: The haploid egg cell fuses with a haploid sperm cell during fertilization to form a diploid zygote. The zygote undergoes mitotic divisions and differentiation to develop into a multicellular embryo. The embryo is contained within the seed, which develops from the ovule after fertilization. Therefore, the correct sequence is Egg → zygote → embryo → seed.

The developmental sequence after fertilization in flowering plants is Egg + Sperm → Zygote → Embryo → Seed.

The seed is a mature ovule, and the fruit is a mature ovary, both developing after fertilization and containing the embryo within the seed.

5. In flowering plants, DNA content of the parent plant gets halved

In flowering plants, DNA content of the parent plant gets halved during

Seed germination

Fruit formation

Flower bud formation

Pollen formation

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UPSC NDA-2 – 2023

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In flowering plants, DNA content gets halved during meiosis. Meiosis is a type of cell division that produces gametes (sex cells) with half the number of chromosomes (and thus half the DNA content) of the parent cell. In flowering plants, pollen grains contain the male gametes (sperm), and their formation (microsporogenesis) involves meiosis. Similarly, the formation of the female gametophyte and egg cell (megasporogenesis) also involves meiosis, halving the DNA content. Seed germination, fruit formation, and flower bud formation involve mitotic division and growth, not the halving of DNA content for gamete production.

Meiosis reduces the chromosome number (and DNA content) by half to produce gametes. Pollen formation involves meiosis.

Fertilization, the fusion of male and female gametes, restores the diploid number of chromosomes (and doubles the DNA content compared to the gametes) in the zygote, which then develops into the embryo.

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6. Which one of the following is NOT a requirement for photosynthesis ?

Which one of the following is NOT a requirement for photosynthesis ?

Chlorophyll

CO₂

Water

O₂

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UPSC NDA-2 – 2023

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D) O₂ is NOT a requirement for photosynthesis.

– Photosynthesis is the process by which plants, algae, and some bacteria convert light energy into chemical energy in the form of glucose.
– The main reactants required for photosynthesis are carbon dioxide (CO₂), water (H₂O), and light energy. Chlorophyll is the pigment that captures light energy.
– The general equation for photosynthesis is: 6CO₂ + 6H₂O + Light Energy → C₆H₁₂O₆ (glucose) + 6O₂.

– Chlorophyll is essential as it absorbs light energy necessary to drive the process.
– Carbon dioxide is the source of carbon for synthesizing glucose.
– Water provides hydrogen atoms and electrons, and is also a source of oxygen which is released as a byproduct.
– Oxygen (O₂) is a product of photosynthesis, not a requirement for the process itself. Plants release oxygen into the atmosphere during photosynthesis.

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7. Match List I with List II and select the correct answer using the code

Match List I with List II and select the correct answer using the code given below the Lists :

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List I (Process)	List II (Separation method)
A. Separation of acetone and water from their mixture	1. Chromatography
B. Separation of water and kerosene oil from their mixture	2. Centrifugation
C. Separation of cream from milk	3. Distillation
D. Separation of pigments from plant extract	4. Separating Funnel

A-3, B-2, C-4, D-1

A-4, B-2, C-1, D-3

A-3, B-4, C-2, D-1

A-4, B-1, C-2, D-3

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C) The correct matching is A-3, B-4, C-2, D-1.

– This question requires matching separation processes with appropriate methods based on the properties of the mixtures and their components.
– A. Separation of acetone and water from their mixture: Acetone and water are miscible liquids with different boiling points. Distillation is used to separate miscible liquids based on differences in volatility. Acetone boils at a lower temperature (around 56°C) than water (100°C), allowing it to be vaporized and condensed separately. This matches with 3. Distillation.
– B. Separation of water and kerosene oil from their mixture: Water and kerosene oil are immiscible liquids with different densities. They form two distinct layers when left undisturbed. A separating funnel is used to separate immiscible liquids, allowing the denser liquid (water) to be drained out from the bottom. This matches with 4. Separating Funnel.
– C. Separation of cream from milk: Milk is a suspension or emulsion of fat globules (cream) in a watery phase. Centrifugation is used to separate components of a mixture based on density differences using centrifugal force. Spinning milk rapidly forces the denser skim milk to the periphery while the lighter cream concentrates towards the center. This matches with 2. Centrifugation.
– D. Separation of pigments from plant extract: Plant extracts contain a mixture of different colored pigments. Chromatography is a technique used to separate components of a mixture based on their differential partitioning between a stationary phase and a mobile phase. Different pigments travel at different rates, allowing them to be separated. This matches with 1. Chromatography.

– Distillation is used for miscible liquids with different boiling points.
– Separating funnel is used for immiscible liquids.
– Centrifugation is used to separate components based on density differences, often for solid-liquid suspensions or liquid-liquid emulsions.
– Chromatography is a versatile technique for separating components based on differential affinities to stationary and mobile phases, widely used for separating colored substances (pigments).

8. Which one of the following metal is NOT an essential component in stai

Which one of the following metal is NOT an essential component in stainless steel ?

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Iron

Nickel

Chromium

Tin

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D) Tin is NOT an essential component in stainless steel.

– Stainless steel is primarily an alloy based on Iron (Fe).
– The key element that provides its characteristic corrosion resistance (“stainless” property) is Chromium (Cr), which must be present in a minimum amount (typically 10.5% by mass). Chromium forms a passive, protective oxide layer on the surface.
– Nickel (Ni) is another common and essential alloying element in many widely used types of stainless steel, particularly austenitic stainless steels (like the 300 series, e.g., 304 and 316). Nickel enhances corrosion resistance, strength, and ductility.

– Iron, Chromium, and often Nickel are the primary essential metallic components of stainless steel. Other elements like Carbon (controlled for strength and ductility), Manganese, Silicon, Molybdenum (for increased corrosion resistance), etc., are also present, but Iron, Chromium, and Nickel are central to its structure and properties.
– Tin (Sn) is not typically used as an alloying element in stainless steel. It is used in other alloys (like bronze, pewter, solder) or as a coating (tin plating), but it is not considered an essential component of stainless steel formulations.

9. What is the action of litmus on ethanol ?

What is the action of litmus on ethanol ?

Litmus is neutral towards ethanol

Ethanol turns blue litmus to red; confirming acidic nature of ethanol

Ethanol turns red litmus to blue; confirming basic nature of ethanol

Ethanol decolorizes litmus through bleaching action

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A) Litmus is neutral towards ethanol.

– Litmus is an acid-base indicator. It turns red in acidic solutions (pH < 7) and blue in basic solutions (pH > 7).
– Ethanol (C₂H₅OH) is an alcohol. The -OH group in alcohols is covalently bonded, unlike the ionic hydroxide in strong bases (e.g., NaOH).
– Alcohols are generally considered neutral compounds in aqueous solutions. While they can act as very weak acids (donating a proton from the -OH group, especially when reacting with strong bases like alkali metals) or very weak bases (accepting a proton from a strong acid at the oxygen atom), their acidity or basicity is negligible in the context of affecting litmus paper pH indicators.

– Ethanol does not dissociate significantly in water to produce H⁺ ions (acidic) or OH⁻ ions (basic).
– Therefore, when litmus paper is placed in ethanol or an aqueous solution of ethanol, its color remains unchanged (red litmus stays red, blue litmus stays blue), indicating a neutral pH.
– Ethanol does not possess bleaching properties like chlorine-based compounds.

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10. Which one of the following acid is used by goldsmith for cleaning of g

Which one of the following acid is used by goldsmith for cleaning of gold and silver articles ?

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Sulphuric acid

Nitric acid

Hydrochloric acid

Phosphoric acid

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UPSC NDA-2 – 2023

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B) Nitric acid is commonly used by goldsmiths for cleaning gold and silver articles.

– Goldsmiths often need to clean gold and silver items to remove tarnish, dirt, and oxidation.
– Tarnish on silver is typically silver sulfide (Ag₂S), and tarnish on gold can involve sulfides or oxides depending on the alloy.
– Nitric acid (HNO₃) is effective in dissolving many metal oxides and sulfides, including silver sulfide tarnish, while not significantly affecting the noble metals like gold and pure silver (though alloys might react).

– Hydrochloric acid (HCl) can also be used for cleaning certain metal surfaces but is less effective against silver sulfide than nitric acid.
– Sulphuric acid (H₂SO₄) is a strong acid but not typically preferred for cleaning jewellery tarnish compared to nitric or hydrochloric acid.
– Phosphoric acid (H₃PO₄) is a weaker acid used in some cleaning agents (like rust removers) but not commonly by goldsmiths for gold and silver tarnish.
– A mixture of concentrated nitric acid and concentrated hydrochloric acid (in a 1:3 ratio), known as Aqua Regia, is capable of dissolving gold and platinum. This is used for refining or testing gold, NOT for cleaning finished jewellery.