UPSC NDA-2 Archives

11. White Phosphorus glows in the dark due to:

White Phosphorus glows in the dark due to:

[amp_mcq option1=”amorphous character” option2=”slow oxidation” option3=”high ignition temperature” option4=”good conducting property of electricity” correct=”option2″]

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UPSC NDA-2 – 2015

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The correct answer is B) slow oxidation.

White phosphorus exhibits phosphorescence, which is a form of chemiluminescence resulting from its slow oxidation when exposed to air. This process releases energy in the form of light, causing it to glow faintly in the dark. This is often referred to as ‘cold light’.

Amorphous character, high ignition temperature (white phosphorus has a very low ignition temperature, around 30°C), and good electrical conductivity are not the reasons for its glow. Its low ignition temperature makes it highly reactive and dangerous.

12. The main constituent of Vinegar is:

The main constituent of Vinegar is:

[amp_mcq option1=”Acetic acid” option2=”Ascorbic acid” option3=”Citric acid” option4=”Tartaric acid” correct=”option1″]

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UPSC NDA-2 – 2015

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The correct answer is A) Acetic acid.

Vinegar is primarily an aqueous solution of acetic acid. Typically, it contains 5-20% acetic acid by volume. Acetic acid is responsible for vinegar’s characteristic sour taste and pungent smell.

Other options are incorrect. Ascorbic acid is Vitamin C. Citric acid is found in citrus fruits. Tartaric acid is found in grapes and tamarinds. While small amounts of other acids may be present depending on the production method, acetic acid is the main constituent.

13. 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:

List I (Forest Conservation Type)	List II (Place)
A. National Park	1. Dudhwa
B. Sanctuary	2. Bhitarkanika
C. Biosphere Reserve	3. Chilika
D. Tiger Reserve	4. Nokrek

[amp_mcq option1=”2 3 4 1″ option2=”1 4 3 2″ option3=”2 4 3 1″ option4=”1 3 4 2″ correct=”option4″]

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UPSC NDA-2 – 2015

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Let’s match the conservation types with the correct places:
A. National Park: Dudhwa National Park is located in Uttar Pradesh.
B. Sanctuary: Chilika Lake area includes the Nalabana Bird Sanctuary in Odisha.
C. Biosphere Reserve: Nokrek National Park is part of the Nokrek Biosphere Reserve in Meghalaya.
D. Tiger Reserve: Dudhwa National Park is also a designated Tiger Reserve. Bhitarkanika is primarily known as a National Park and Wildlife Sanctuary, famous for its mangrove ecosystem and saltwater crocodiles, and is not a major Tiger Reserve.

Mapping based on common or primary classifications:
National Park – Dudhwa (1)
Sanctuary – Chilika (3)
Biosphere Reserve – Nokrek (4)
Tiger Reserve – Dudhwa (1)

Let’s examine the options given the list format (A, B, C, D matched with 1, 2, 3, 4):
Option A: A-2 (Bhitarkanika), B-3 (Chilika), C-4 (Nokrek), D-1 (Dudhwa).
Option B: A-1 (Dudhwa), B-4 (Nokrek), C-3 (Chilika), D-2 (Bhitarkanika).
Option C: A-2 (Bhitarkanika), B-4 (Nokrek), C-3 (Chilika), D-1 (Dudhwa).
Option D: A-1 (Dudhwa), B-3 (Chilika), C-4 (Nokrek), D-2 (Bhitarkanika).

Based on the common designations:
A. National Park: Dudhwa (1)
B. Sanctuary: Chilika (3)
C. Biosphere Reserve: Nokrek (4)
D. Tiger Reserve: Dudhwa (1)

Let’s re-evaluate the options against these probable matches.
Option D suggests:
A. National Park – Dudhwa (1) – Correct
B. Sanctuary – Chilika (3) – Correct (Nalabana Bird Sanctuary within Chilika)
C. Biosphere Reserve – Nokrek (4) – Correct (Nokrek Biosphere Reserve)
D. Tiger Reserve – Bhitarkanika (2). This is incorrect. Bhitarkanika is a National Park/Sanctuary, not a Tiger Reserve.

Let’s re-examine Option A:
A. National Park – Bhitarkanika (2) – Correct (Bhitarkanika is a National Park)
B. Sanctuary – Chilika (3) – Correct (Nalabana Bird Sanctuary within Chilika)
C. Biosphere Reserve – Nokrek (4) – Correct (Nokrek Biosphere Reserve)
D. Tiger Reserve – Dudhwa (1) – Correct (Dudhwa is a Tiger Reserve)

Option A provides the correct mapping for all four entries. There was a mix-up in my initial assessment of which option matched which set of numbers.

Matching specific protected areas in India with their designated conservation status (National Park, Sanctuary, Biosphere Reserve, Tiger Reserve).

Different conservation statuses offer varying levels of protection. National Parks have the highest level of protection. Wildlife Sanctuaries offer protection to specific species or habitats. Biosphere Reserves are large areas designed for conservation, research, and sustainable development, often including core zones (like National Parks/Sanctuaries), buffer zones, and transition zones. Tiger Reserves are areas specifically designated for the conservation of tigers.

14. Which one among the following places is not an iron-ore mining area?

Which one among the following places is not an iron-ore mining area?

[amp_mcq option1=”Badampahar” option2=”Zawar” option3=”Bailadila” option4=”Anantpur” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2015

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Zawar is a historically significant mining area located in Rajasthan, primarily known for its lead and zinc mines, particularly the Rampura Agucha mine which is a major zinc deposit. While other minerals may be present, Zawar is not primarily recognized as an iron-ore mining area. Badampahar in Odisha and Bailadila in Chhattisgarh are major iron-ore mining regions in India. Anantapur district in Andhra Pradesh also has known iron ore deposits.

Knowledge of the principal mineral resources associated with specific mining locations in India is required.

Bailadila is one of the largest iron ore deposits in the world, known for high-grade haematite ore. Badampahar was one of the first iron ore mines in India. Zawar has been mined for zinc since ancient times.

15. The following figure represents the velocity-time graph of a moving ca

The following figure represents the velocity-time graph of a moving car on a road:
Which segment of the graph represents the retardation?

[amp_mcq option1=”AB” option2=”BC” option3=”CD” option4=”None” correct=”option3″]

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In a velocity-time (v-t) graph, acceleration is represented by the slope of the graph. Retardation (or deceleration) occurs when the acceleration is in the opposite direction to the velocity, causing the object to slow down. This corresponds to a negative slope in the v-t graph, assuming positive velocity.
– Segment AB has a positive slope, indicating positive acceleration (velocity increasing).
– Segment BC is horizontal, indicating zero slope and thus zero acceleration (constant velocity).
– Segment CD has a negative slope, indicating negative acceleration. Since the velocity is positive during this segment, a negative acceleration means retardation or deceleration (velocity decreasing).

The slope of a velocity-time graph represents acceleration. A negative slope indicates deceleration or retardation if the velocity is in the positive direction.

If the velocity were negative and the slope were positive, it would also represent slowing down (velocity increasing towards zero from a negative value). In this graph, velocity is positive throughout the plotted segments.

16. Three equal resistances when combined in series are equivalent to 90 o

Three equal resistances when combined in series are equivalent to 90 ohm. Their equivalent resistance when combined in parallel will be:

[amp_mcq option1=”10 ohm” option2=”30 ohm” option3=”270 ohm” option4=”810 ohm” correct=”option1″]

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Let R be the resistance of each of the three equal resistors. When connected in series, the equivalent resistance (Rs) is the sum of individual resistances: Rs = R + R + R = 3R. Given Rs = 90 ohm, we have 3R = 90 ohm, so R = 30 ohm. When the three equal resistors are connected in parallel, the equivalent resistance (Rp) is given by the formula 1/Rp = 1/R + 1/R + 1/R = 3/R. Thus, Rp = R/3. Substituting R = 30 ohm, we get Rp = 30 ohm / 3 = 10 ohm.

Calculation of equivalent resistance for resistors connected in series and parallel.

For N equal resistors of resistance R, the equivalent resistance in series is NR, and the equivalent resistance in parallel is R/N.

17. Who among the following discovered the antibiotic producing fungus fro

Who among the following discovered the antibiotic producing fungus from Penicillium genus?

[amp_mcq option1=”Louis Pasteur” option2=”Sir Alexander Fleming” option3=”Stanley Prusiner” option4=”Robert Hooke” correct=”option2″]

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Sir Alexander Fleming discovered penicillin in 1928, observing that a mould of the genus *Penicillium* contaminated a staphylococcus culture plate and killed the bacteria. This serendipitous discovery led to the identification of penicillin as the first widely used antibiotic.

The discovery of penicillin by Alexander Fleming revolutionized medicine by providing an effective treatment for bacterial infections.

Louis Pasteur was a pioneering microbiologist known for pasteurization and the germ theory of disease. Stanley Prusiner discovered prions, which are infectious proteins. Robert Hooke was a 17th-century scientist known for his microscopic observations and coining the term “cell”.

18. Boric acid is an acid because its molecule:

Boric acid is an acid because its molecule:

[amp_mcq option1=”accepts OH^– from water” option2=”combines with proton from water molecule” option3=”contains replaceable H⁺ ion” option4=”gives up a proton” correct=”option1″]

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Boric acid (H₃BO₃) is a weak monoprotic Lewis acid. It does not directly donate a proton (H⁺) from its own molecule in aqueous solution. Instead, it accepts a hydroxide ion (OH⁻) from a water molecule, releasing a proton from the water molecule in the process. The reaction is: H₃BO₃ + H₂O ⇌ [B(OH)₄]⁻ + H⁺. Therefore, it acts as an acid by accepting an OH⁻ ion from water.

Boric acid’s acidity is explained by its behavior as a Lewis acid, accepting an electron pair from the hydroxide ion of water.

Although boric acid contains three hydrogen atoms, they are not easily ionizable as protons in the way that typical Brønsted-Lowry acids like HCl or H₂SO₄ are. Its acidic nature arises from its ability to form a coordinate covalent bond with the hydroxide ion.

19. The IST meridian 82½°E passes through a number of States in India. Whi

The IST meridian 82½°E passes through a number of States in India. Which one of the following sets of States is correct in this respect?

[amp_mcq option1=”Uttarakhand, Uttar Pradesh, Chhattisgarh and Andhra Pradesh” option2=”Uttar Pradesh, Jharkhand, Chhattisgarh and Odisha” option3=”Uttarakhand, Uttar Pradesh, Madhya Pradesh and Chhattisgarh” option4=”Uttar Pradesh, Odisha, Andhra Pradesh and Chhattisgarh” correct=”option4″]

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The Indian Standard Time (IST) meridian, 82°30′ East longitude, passes through five states in India: Uttar Pradesh, Madhya Pradesh, Chhattisgarh, Odisha, and Andhra Pradesh. Option D lists four of these states: Uttar Pradesh, Odisha, Andhra Pradesh, and Chhattisgarh. The other options include states that the meridian does not pass through (Uttarakhand, Jharkhand). Among the given choices, Option D is the correct set of states through which the IST meridian passes.

Knowledge of the geographical path of the IST meridian through Indian states is required.

The city of Mirzapur in Uttar Pradesh is often cited as the location near which the IST meridian passes. The meridian serves as the reference for timekeeping across India, which is 5 hours and 30 minutes ahead of Coordinated Universal Time (UTC+5:30).

20. Which one of the following statements is not correct?

Which one of the following statements is not correct?

[amp_mcq option1=”If the velocity and acceleration have opposite sign, the object is slowing down” option2=”If the velocity is zero at an instant, the acceleration should also be zero at that instant” option3=”If the velocity is zero for a time interval; the acceleration is zero at any instant within the time interval” option4=”If the position and velocity have opposite sign, the object is moving towards the origin” correct=”option2″]

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Statement B is not correct. If the velocity of an object is zero at a particular instant, its acceleration at that instant does not necessarily have to be zero. A classic example is an object thrown vertically upwards: at the peak of its trajectory, its instantaneous velocity is zero, but the acceleration due to gravity is still acting on it and is non-zero (approximately 9.8 m/s² downwards).

Acceleration is the rate of change of velocity. Zero velocity at an instant only means the object is momentarily stopped; it does not imply that the velocity is not changing at that instant.

Statement A is correct: if velocity and acceleration have opposite signs, the object is slowing down (e.g., positive velocity, negative acceleration). Statement C is correct: if velocity is zero for a time interval, the object is stationary, meaning its velocity is constant (zero), and thus its acceleration is zero. Statement D is correct: if position and velocity have opposite signs (e.g., positive position, negative velocity means moving towards origin from the positive side; negative position, positive velocity means moving towards origin from the negative side), the object is moving towards the origin (assuming origin is at position=0).