UPSC NDA-1

31. 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
(Molecule)		 List II
(Product of digestion)
A. Proteins 1. Nitrogenous bases and pentose sugars
B. Carbohydrates 2. Fatty acids and glycerol
C. Nucleic acids 3. Monosaccharides
D. Lipids 4. Amino acids
2 3 1 4
2 1 3 4
4 1 3 2
4 3 1 2
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The correct matching is A-4, B-3, C-1, D-2.
Digestion breaks down complex molecules into simpler units that can be absorbed by the body.
– Proteins are polymers of amino acids, and digestion breaks them down into individual amino acids. (A -> 4)
– Carbohydrates (like starch, disaccharides) are broken down into monosaccharides (simple sugars like glucose, fructose, galactose). (B -> 3)
– Nucleic acids (DNA, RNA) are broken down into nucleotides, and further breakdown yields nitrogenous bases, pentose sugars, and phosphate groups. Option 1 lists nitrogenous bases and pentose sugars, which are components resulting from nucleic acid digestion. (C -> 1)
– Lipids (fats and oils) are broken down into fatty acids and glycerol. (D -> 2)
Digestion is a catabolic process carried out by enzymes. For example, proteases digest proteins, amylases digest carbohydrates, nucleases digest nucleic acids, and lipases digest lipids. These broken-down monomers (amino acids, monosaccharides, fatty acids, glycerol, nitrogenous bases, pentose sugars, phosphate) are then absorbed across the intestinal lining into the bloodstream or lymphatic system for use by the body’s cells.

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32. Which one of the following carbon compounds will not give a sooty fl

Which one of the following carbon compounds will not give a sooty flame ?

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Benzene
Hexane
Naphthalene
Anthracene
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Hexane will not give a sooty flame.
Sooty flames are a result of incomplete combustion, where there is insufficient oxygen to fully oxidize the fuel. This often occurs with hydrocarbons that have a relatively high carbon content compared to hydrogen. Aromatic hydrocarbons (like Benzene, Naphthalene, Anthracene) and unsaturated aliphatic hydrocarbons have a higher carbon-to-hydrogen ratio and tend to burn with sooty flames due to incomplete combustion and the formation of carbon particles (soot). Saturated aliphatic hydrocarbons (alkanes like Hexane, C6H14) have a lower carbon-to-hydrogen ratio and typically undergo more complete combustion, burning with a clean, blue flame, assuming sufficient oxygen is available.
Benzene (C6H6), Naphthalene (C10H8), and Anthracene (C14H10) are all aromatic hydrocarbons with high carbon content. Hexane (C6H14) is a saturated alkane. The general formula for alkanes is CnH2n+2, while aromatics have proportionally less hydrogen for the same number of carbon atoms. This difference in composition affects how they burn.

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33. A given conductor carrying a current of 1 A produces an amount of heat

A given conductor carrying a current of 1 A produces an amount of heat equal to 2000 J. If the current through the conductor is doubled, the amount of heat produced will be

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2000 J
4000 J
8000 J
1000 J
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The amount of heat produced will be 8000 J.
According to Joule’s Law of Heating, the heat produced (H) in a resistor is directly proportional to the square of the current (I), the resistance (R), and the time (t) for which the current flows: H = I²Rt. Assuming the resistance of the conductor and the time duration are constant, the heat produced is proportional to the square of the current (H ∝ I²).
Given that H1 = 2000 J for I1 = 1 A.
When the current is doubled, I2 = 2 * I1 = 2 A.
The new heat produced H2 will be proportional to (I2)²:
H2 / H1 = (I2)² / (I1)²
H2 / 2000 = (2I1)² / (I1)²
H2 / 2000 = 4 * (I1)² / (I1)²
H2 / 2000 = 4
H2 = 4 * 2000 J = 8000 J.
Joule’s law explains why wires and electrical components heat up when current passes through them. This principle is used in various applications like electric heaters, toasters, and fuses. The power dissipated as heat is given by P = I²R, and the total heat produced over time t is H = Pt = I²Rt.

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34. A man weighing 70 kg is coming down in a lift. If the cable of the lif

A man weighing 70 kg is coming down in a lift. If the cable of the lift breaks suddenly, the weight of the man would become

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70 kg
35 kg
140 kg
Zero
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The weight of the man would become zero.
Weight is the force of gravity acting on an object’s mass (W = mg). Apparent weight is the normal force exerted by the supporting surface on the object. When the lift cable breaks, the lift and everything inside it are in free fall, accelerating downwards at the acceleration due to gravity (g). In a state of free fall, there is no supporting force counteracting gravity. Therefore, the apparent weight of the man (the force exerted by the lift floor on him) becomes zero. This is the sensation of weightlessness. His actual weight (mass * g) remains the same, but his apparent weight is zero.
If the lift were accelerating upwards, the apparent weight would be greater than his actual weight. If the lift were accelerating downwards at a rate less than g, his apparent weight would be less than his actual weight but greater than zero. If the lift were stationary or moving at constant velocity, his apparent weight would equal his actual weight.

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35. A racing car accelerates on a straight road from rest to a speed of 50

A racing car accelerates on a straight road from rest to a speed of 50 m/s in 25 s. Assuming uniform acceleration of the car throughout, the distance covered in this time will be

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625 m
1250 m
2500 m
50 m
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The distance covered by the racing car is 625 m.
We are given the initial velocity (u = 0 m/s, since it starts from rest), final velocity (v = 50 m/s), and time (t = 25 s). Assuming uniform acceleration (a), we can use the kinematic equations. First, find the acceleration using v = u + at: 50 = 0 + a * 25, which gives a = 50/25 = 2 m/s². Then, find the distance (s) using s = ut + (1/2)at²: s = (0 * 25) + (1/2) * 2 * (25)² = 0 + 1 * 625 = 625 m. Alternatively, the average velocity is (u+v)/2 = (0+50)/2 = 25 m/s. Distance = Average velocity * time = 25 m/s * 25 s = 625 m.
This problem involves basic kinematics under constant acceleration. The relevant equations of motion are v = u + at, s = ut + (1/2)at², and v² = u² + 2as. Any of these can be used depending on the given and required variables.

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36. Deserts, fertile plains and moderately forested mountains are the char

Deserts, fertile plains and moderately forested mountains are the characteristic features of which one of the following regions of India ?

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South-Western border along Arabian Sea
Coromandel Coast
North-Eastern Frontier
North-Western India
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The region characterized by deserts, fertile plains, and moderately forested mountains is North-Western India.
North-Western India includes the states of Rajasthan (home to the Thar Desert), Punjab and Haryana (part of the fertile Indo-Gangetic plains), and parts of the Himalayan and Aravalli ranges (moderately forested mountains). This combination of diverse geographical features makes North-Western India unique in having all three characteristics.
Other options do not fit this description: The South-Western border along the Arabian Sea includes coastal plains and the Western Ghats (dense forests and hills) but no deserts. The Coromandel Coast includes coastal plains, fertile delta regions, and the Eastern Ghats (hills/mountains) but no deserts. The North-Eastern Frontier is predominantly mountainous and hilly with dense forests and fertile valleys, but no deserts.

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37. Areas which are engines for economic growth supported by quality infra

Areas which are engines for economic growth supported by quality infrastructure and complemented by an attractive fiscal package are known as

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Export Processing Zones
Duty Free Tariff Zones
Special Economic Zones
Technology Parks
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The correct answer is Special Economic Zones (SEZs).
Special Economic Zones (SEZs) are specifically designated areas in a country with special economic regulations that differ from the rest of the country. They are designed to promote trade, investment, job creation, and effective administration. SEZs are typically supported by quality infrastructure and often include attractive fiscal packages, such as tax incentives, customs duty exemptions, and streamlined procedures, to attract domestic and foreign businesses.
Export Processing Zones (EPZs) are an earlier form of SEZs, primarily focused on promoting exports through tariff exemptions and simplified customs procedures. Duty Free Tariff Zones (DFTFs) are areas where goods can be imported and exported without customs duties, often used for trading and re-processing. Technology Parks are areas focused on promoting technology and IT-related industries, often with specialized infrastructure but not necessarily the broad fiscal and regulatory benefits of a full SEZ. SEZs encompass a broader mandate for economic growth across various sectors, supported by comprehensive policies and infrastructure.

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38. Consider the following statements: 1. All Echinoderms are not marine

Consider the following statements:

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  • 1. All Echinoderms are not marine.
  • 2. Sponges are exclusively marine.
  • 3. Insects are found in all kinds of habitats.
  • 4. Many primates are arboreal.

Which of the statements given above is/are correct ?

1, 3 and 4 only
3 and 4 only
2 and 4 only
3 only
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The correct answer is B) 3 and 4 only.
Let’s examine each statement:
1. All Echinoderms are not marine: This statement is incorrect. Echinodermata (which includes starfish, sea urchins, sea cucumbers, brittle stars, and crinoids) is a phylum consisting *entirely* of marine animals. There are no freshwater or terrestrial echinoderms.
2. Sponges are exclusively marine: This statement is incorrect. While the vast majority of sponges (Phylum Porifera) are marine, there are a number of species found in freshwater habitats (e.g., the family Spongillidae).
3. Insects are found in all kinds of habitats: This statement is correct. Insects (Class Insecta) are the most diverse group of animals and have colonized almost every habitat on Earth, including terrestrial, freshwater, and some marine environments (coastal areas, surface films, etc.). They are absent from the open ocean deep sea.
4. Many primates are arboreal: This statement is correct. Arboreal animals live in trees. Many species of primates (like most monkeys and some apes) are well-adapted for life in trees and spend most of their time in forest canopies.
Based on the analysis, statements 3 and 4 are correct.
Echinoderms are known for their radial symmetry (usually pentaradial in adults) and unique water vascular system. Sponges are simple multicellular animals, filter feeders, lacking true tissues and organs. Insects are characterized by a chitinous exoskeleton, a three-part body (head, thorax, and abdomen), three pairs of jointed legs, compound eyes, and typically two pairs of wings. Primates are characterized by traits like relatively large brains, grasping hands and feet, and forward-facing eyes.

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39. Which of the following pairs of vector and disease is/are correctly ma

Which of the following pairs of vector and disease is/are correctly matched?

  • 1. Anopheles : Malaria
  • 2. Aedes aegypti : Chikungunya
  • 3. Tsetse fly : Filariasis
  • 4. Bed bugs : Sleeping sickness

Select the correct answer using the code given below :

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1, 2 and 3 only
1 and 2 only
1 and 4 only
2 only
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The correct answer is B) 1 and 2 only.
Vectors are organisms that transmit pathogens from one host to another. Let’s evaluate each pair:
1. Anopheles : Malaria – Correct. Female Anopheles mosquitoes are the vectors for the malaria parasite (Plasmodium).
2. Aedes aegypti : Chikungunya – Correct. Aedes aegypti mosquitoes are known vectors for Chikungunya virus, Dengue virus, Zika virus, and Yellow Fever virus.
3. Tsetse fly : Filariasis – Incorrect. Tsetse flies are vectors for African Trypanosomiasis (Sleeping Sickness). Filariasis (caused by parasitic worms) is typically transmitted by mosquitoes (Culex, Anopheles, or Aedes species) or black flies, depending on the type of filariasis.
4. Bed bugs : Sleeping sickness – Incorrect. Bed bugs feed on blood but are not primary vectors for major human diseases. Sleeping sickness is transmitted by the Tsetse fly.
Disease vectors play a crucial role in the spread of many infectious diseases. Vector control is an important strategy for preventing outbreaks and reducing the burden of vector-borne diseases worldwide. Examples include controlling mosquito populations to prevent malaria and dengue, or using insecticides to target tsetse flies.

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40. Which one of the following elements will not react with dilute HCl to

Which one of the following elements will not react with dilute HCl to produce H2 ?

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Hg
Al
Mg
Fe
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The correct answer is A) Hg.
Dilute HCl is a non-oxidizing acid. Metals that are more reactive than hydrogen will react with such acids to displace hydrogen gas (H₂). The reactivity of metals can be determined from the electrochemical series (or reactivity series). Metals above hydrogen in the series are more reactive and will displace hydrogen from dilute acids. Metals below hydrogen will not. The relevant part of the reactivity series is (from more reactive to less reactive): … Al > Zn > Fe > Pb > H > Cu > Hg > Ag > Au …
– Al, Mg, and Fe are all above hydrogen in the series, so they will react with dilute HCl to produce H₂ gas.
– Hg (Mercury) is below hydrogen in the series, so it will not react with dilute HCl to produce H₂ gas.
The reaction of a reactive metal (M) with dilute HCl is typically represented as: M(s) + nHCl(aq) → MClₙ(aq) + n/2 H₂(g), where ‘n’ is the valency of the metal in the chloride. For example, Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Metals below hydrogen can react with oxidizing acids (like concentrated nitric acid or hot concentrated sulfuric acid) but produce different products (e.g., oxides of nitrogen or sulfur dioxide) instead of hydrogen gas.

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