UPSC NDA-1 Archives

31. Consider the following diagram : In which one among the following lett

Consider the following diagram :
In which one among the following lettered areas of the diagram would erosion most likely change the shapes of the riverbed ?

[amp_mcq option1=”A” option2=”B” option3=”C” option4=”D” correct=”option3″]

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UPSC NDA-1 – 2015

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The diagram shows a meandering river. In a meandering river bend, water flows faster on the outer bank (due to centrifugal force and deeper water) and slower on the inner bank (where sediment is deposited, forming a point bar). Erosion is most active on the outer bank of the bend due to the faster current. Area C is located on the outer bank of the river bend where erosion is dominant, leading to the cutting away of the bank and changing the shape of the riverbed and bend over time. Areas A and D are on the inner bank where deposition occurs. Area B is likely in a straighter section or near the transition, but erosion is most pronounced at the apex of the outer bend, which is represented by C.

Erosion in a meandering river is concentrated on the outer bends where water flows fastest, while deposition occurs on the inner bends where water is slower.

Over time, continuous erosion on the outer bends and deposition on the inner bends cause the meanders to migrate laterally across the floodplain and can eventually lead to the formation of oxbow lakes when a meander neck is cut off.

32. 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 (Island)	List II (Location)
A. Continental island	1. Mauritius
B. Coral island	2. Madagascar
C. Volcanic island	3. Andaman and Nicobar islands
D. Mountain island	4. Maldives

[amp_mcq option1=”A-2, B-4, C-1, D-3″ option2=”A-2, B-1, C-4, D-3″ option3=”A-3, B-1, C-4, D-2″ option4=”A-3, B-4, C-1, D-2″ correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2015

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Let’s match the types of islands with their examples:
A. Continental island: These are islands that were formerly part of a continent and became detached due to changes in sea level, geological uplift, or separation of landmasses. Madagascar is a classic example of a continental island that broke away from Africa. So, A matches with 2.
B. Coral island: These islands are formed from the accumulation of coral skeletons and associated organic material. Maldives is an archipelago of coral atolls, making it a prime example of a coral island chain. So, B matches with 4.
C. Volcanic island: These islands are formed by volcanic activity, either from underwater volcanoes reaching the surface or continental volcanic activity. Mauritius was formed by volcanic activity. So, C matches with 1.
D. Mountain island: This term can refer to islands that are essentially the peaks of submerged mountain ranges or fold mountains. The Andaman and Nicobar Islands are believed to be the northern extension of the Arakan Yoma mountain range of Myanmar, making them mountain islands. So, D matches with 3.
Putting it together: A-2, B-4, C-1, D-3. This matches option A.

Islands are classified based on their formation process: Continental islands separated from continents, Coral islands built by coral polyps, Volcanic islands formed by volcanoes, and Mountain islands representing submerged mountain peaks.

Other examples: Great Britain is a continental island. Lakshadweep in India is also a coral island group (atoll). Hawaii is a famous example of a volcanic island chain. Indonesia and Japan are archipelagos largely made up of volcanic islands.

33. Which one of the following weather conditions indicates a sudden fall

Which one of the following weather conditions indicates a sudden fall in barometer reading ?

[amp_mcq option1=”Stormy weather” option2=”Calm weather” option3=”Cold and dry weather” option4=”Hot and sunny weather” correct=”option1″]

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UPSC NDA-1 – 2015

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A barometer measures atmospheric pressure. A sudden fall in barometer reading indicates a rapid decrease in atmospheric pressure. Falling atmospheric pressure is often associated with approaching low-pressure systems. Low-pressure systems are typically accompanied by unstable atmospheric conditions, rising air, cloud formation, precipitation, and often, stormy weather. Conversely, rising pressure indicates stable conditions (usually associated with clear or fair weather), and steady high pressure indicates calm, dry, and often clear weather. A sudden fall specifically points towards a rapidly developing or approaching weather disturbance, which is characteristic of stormy weather.

A sudden drop in atmospheric pressure (barometer reading) indicates unstable weather conditions, often preceding or accompanying storms.

Forecasters use changes in atmospheric pressure to help predict weather changes. A steady fall may indicate rain, while a rapid fall often indicates a strong storm or depression approaching. High pressure generally leads to clear, calm, and stable weather.

34. Which of the following statements with regard to the western coastal p

Which of the following statements with regard to the western coastal plain of India are correct ?

1. It is a narrow belt.
2. It is an example of submerged coastal plain.
3. It provides natural conditions for development of ports.
4. It has well developed deltas.

Select the correct answer using the code given below :

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

This question was previously asked in

UPSC NDA-1 – 2015

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Let’s evaluate each statement:
1. It is a narrow belt: The western coastal plain of India is generally narrow, especially compared to the eastern coastal plain. This statement is correct.
2. It is an example of submerged coastal plain: The western coastal plain is believed to be largely a submerged coastal plain, which is why it is relatively narrow and has less developed deltas. This statement is correct.
3. It provides natural conditions for development of ports: Due to its submerged nature, the coast is often indented and has natural harbours, making it suitable for port development (e.g., Mumbai, Kochi, Kandla, etc.). This statement is correct.
4. It has well developed deltas: The rivers flowing westwards into the Arabian Sea are generally short and swift and do not form significant deltas; instead, they often form estuaries. Well-developed deltas (like the Ganga, Mahanadi, Godavari, Krishna, Cauvery) are characteristic of the eastern coastal plain where rivers are longer and gentler. This statement is incorrect.
Based on the evaluation, statements 1, 2, and 3 are correct.

The Western Coastal Plain of India is narrow, largely submerged, has numerous natural harbours suitable for ports, and lacks well-developed deltas, unlike the Eastern Coastal Plain.

The western coastal plain extends from the Rann of Kachchh in the north to Kanyakumari in the south. It is divided into different sections: the Konkan coast (Maharashtra and Goa), the Kanara coast (Karnataka), and the Malabar coast (Kerala). The absence of large deltas is due to the steeper gradient and shorter courses of the west-flowing rivers, which mostly originate in the Western Ghats.

35. 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 (Region)	List II (Vegetation)
A. Selvas	1. Conifers
B. Savannas	2. Mosses and Lichens
C. Taiga	3. Epiphytes
D. Tundra	4. Grasses and trees

Code :

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

This question was previously asked in

UPSC NDA-1 – 2015

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Let’s match the regions with their characteristic vegetation:
A. Selvas: This term is often used for the equatorial rainforests, particularly in the Amazon basin. These forests are dense and rich in biodiversity, including many epiphytic plants that grow on other plants for support. So, Selvas match with Epiphytes (3).
B. Savannas: Savannas are grasslands with scattered trees, found in tropical and subtropical regions with distinct wet and dry seasons. So, Savannas match with Grasses and trees (4).
C. Taiga: Taiga is the boreal forest, a biome characterized by coniferous forests, found in cold climates. So, Taiga match with Conifers (1).
D. Tundra: Tundra is a treeless polar or alpine biome characterized by permafrost, low temperatures, and short growing seasons. Vegetation is typically limited to mosses, lichens, grasses, sedges, and dwarf shrubs. So, Tundra match with Mosses and Lichens (2).
Putting it together: A-3, B-4, C-1, D-2. This matches option C.

Distinct biomes are characterized by specific climate conditions and resulting vegetation types. Selvas (tropical rainforests) feature epiphytes, Savannas are grasslands with scattered trees, Taiga (boreal forests) are dominated by conifers, and Tundra has low vegetation like mosses and lichens.

Selvas is a term specifically associated with the Amazon rainforest. Epiphytes are plants that grow harmlessly on another plant and get moisture and nutrients from the air, rain, or accumulated debris, common in humid tropical forests. Savannas cover large areas of Africa, Australia, and South America. Taiga is the largest terrestrial biome, located south of the Arctic tundra. Tundra is found in high latitudes and high altitudes.

36. A spring can be used to determine the mass m of an object in two ways

A spring can be used to determine the mass m of an object in two ways : (i) by measuring the extension in the spring due to the object; and (ii) by measuring the oscillation period for the given mass. Which of these methods can be used in a space-station orbiting Earth ?

[amp_mcq option1=”Both” option2=”Only the extension method” option3=”Only the oscillation method” option4=”Neither” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

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Method (i) involves measuring the extension of a spring caused by the weight of the object (F = mg, where F is the force causing extension, m is mass, and g is acceleration due to gravity). In a space station orbiting Earth, objects are in a state of apparent weightlessness because they are in free fall. The effect of gravity is effectively absent for this type of measurement, so there would be no significant extension of the spring due to the object’s mass in the absence of gravitational force acting downwards. Method (ii) involves measuring the oscillation period of a mass attached to a spring. The period of oscillation (T) of a simple harmonic oscillator like a mass-spring system is given by T = 2π√(m/k), where m is the mass and k is the spring constant. This formula depends on mass (m) and spring constant (k), but not on gravity (g). Therefore, the oscillation method can be used to determine mass in a weightless environment like an orbiting space station.

Measuring extension relies on weight (force due to gravity), which is effectively zero in orbit. Measuring oscillation period relies on inertial mass and spring constant, which are unaffected by gravity.

The oscillation method is actually used on the International Space Station (ISS) with a device called the Body Mass Measurement Device (BMMD) to measure the mass of astronauts. The BMMD is essentially a chair attached to springs that oscillates, and the oscillation period is used to calculate the mass. This demonstrates the practicality of the oscillation method for mass determination in a microgravity environment.

37. When heat rays are reflected from Earth, gases like Carbon dioxide, Ni

When heat rays are reflected from Earth, gases like Carbon dioxide, Nitrous oxide do not allow them to escape back to the space causing our planet to heat up. These gases are known as

[amp_mcq option1=”Noble gas” option2=”Green-house gas” option3=”Hot gas” option4=”Blue gas” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2015

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The question describes the phenomenon of the greenhouse effect, where certain gases in the atmosphere trap outgoing infrared radiation (heat rays) reflected from the Earth’s surface, causing the planet to warm up. Gases like carbon dioxide, nitrous oxide, methane, and water vapour are known for this property. These gases are specifically called greenhouse gases. Therefore, option B is the correct answer.

Greenhouse gases trap outgoing infrared radiation, causing the greenhouse effect and warming the planet.

Noble gases (like Helium, Neon, Argon) are inert and do not significantly interact with radiation in this way. “Hot gas” and “Blue gas” are not scientific terms used to describe these atmospheric gases and their effect on temperature. The greenhouse effect is a natural process essential for keeping the Earth warm enough to support life, but increased concentrations of these gases due to human activities are causing enhanced global warming.

38. When the sun is 30° above the horizon, shadow of one tree is 17·3 m lo

When the sun is 30° above the horizon, shadow of one tree is 17·3 m long. What is the height of this tree ?

[amp_mcq option1=”20 m” option2=”17·30 m” option3=”10 m” option4=”1·73 m” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

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This problem can be solved using trigonometry, specifically the tangent function in a right-angled triangle.

The tree stands vertically, forming a right angle with the ground. The sun’s rays create a shadow on the ground. The angle of elevation of the sun (30°) is the angle between the ground (shadow) and the line from the end of the shadow to the top of the tree. Let H be the height of the tree (opposite side) and L be the length of the shadow (adjacent side). We have the relationship: tan(angle) = Opposite / Adjacent. So, tan(30°) = H / L.
Given L = 17.3 m and tan(30°) = 1/$\sqrt{3}$.
H = L * tan(30°) = 17.3 m * (1/$\sqrt{3}$).

The value of $\sqrt{3}$ is approximately 1.732.
H $\approx$ 17.3 / 1.732.
Notice that 17.3 is very close to 10 * 1.73. If we assume the shadow length is precisely $10\sqrt{3}$ meters (which is approximately 17.32 m), then:
H = $(10\sqrt{3}) \times (1/\sqrt{3}) = 10$ meters.
Given the options, 10 m is the most likely intended answer, implying the shadow length 17.3 m was an approximation for $10\sqrt{3}$ m.

39. We use CFL to save electrical energy and to provide sufficient light.

We use CFL to save electrical energy and to provide sufficient light. The full form of CFL is

[amp_mcq option1=”Condensed filament light” option2=”Compact filament lamp” option3=”Condensed fluorescent lamp” option4=”Compact fluorescent lamp” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

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CFL is a type of fluorescent lamp designed to replace incandescent lamps.

The acronym CFL stands for Compact Fluorescent Lamp. These lamps are designed to be smaller and more energy-efficient than traditional linear fluorescent lamps, allowing them to fit into the same fixtures as incandescent bulbs.

CFLs work by passing an electric current through a tube containing argon and a small amount of mercury vapor. This generates invisible ultraviolet light, which excites a fluorescent coating on the inside of the tube, causing it to emit visible light. They use significantly less energy and last much longer than incandescent bulbs, although they contain mercury and require special disposal.

40. The product of conductivity and resistivity of a conductor

The product of conductivity and resistivity of a conductor

[amp_mcq option1=”depends on pressure applied” option2=”depends on current flowing through conductor” option3=”is the same for all conductors” option4=”varies from conductor to conductor” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

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Resistivity ($\rho$) and conductivity ($\sigma$) are inverse properties of a material concerning its ability to conduct electricity.

By definition, conductivity is the reciprocal of resistivity ($\sigma = 1/\rho$). Therefore, the product of conductivity and resistivity is always equal to 1 ($\sigma \times \rho = (1/\rho) \times \rho = 1$). This relationship holds true for all materials, regardless of whether they are conductors, insulators, or semiconductors, and it is independent of the physical conditions like pressure or the current flowing through the conductor (assuming the material obeys Ohm’s Law).

Resistivity is a fundamental property of the material itself, like density or melting point. Conductivity is simply another way of expressing this property. Their product being 1 is a mathematical consequence of their definition relative to each other.