1. Which of the following statement(s) concerning natural regions of the

Which of the following statement(s) concerning natural regions of the world is/are correct ?

  • 1. Equatorial climatic regions have less urbanization
  • 2. Mediterranean climatic regions have high level of urbanization
  • 3. Equatorial climatic regions have high concentration of human population
  • 4. Mediterranean climatic regions practice plantation farming

Select the correct answer using the code given below :

1 and 2 only
1, 2 and 3
2 and 4
1 only
This question was previously asked in
UPSC NDA-2 – 2016
Statement 1: Equatorial rainforest regions, characterized by dense vegetation, high humidity, and challenging environment, historically and in many areas today, have lower levels of urbanization compared to other major climatic zones. While significant cities exist, large parts of the region remain less developed and urbanized. This statement is correct.
Statement 2: Mediterranean climatic regions, known for their pleasant climate (warm, dry summers; mild, wet winters), fertile land in coastal areas, and accessibility, have historically been centers of civilization and continue to attract high levels of settlement and urbanization. Many major cities worldwide are located in these regions. This statement is correct.
Statement 3: Equatorial climatic regions are generally not known for high concentrations of human population across the entire region. While some specific areas like river deltas, coasts, or specific resource-rich zones might have high density, vast areas, particularly deep rainforests, are sparsely populated. Regions with high population concentration are typically found in monsoon climates (South Asia, East Asia) or parts of temperate zones. This statement is incorrect.
Statement 4: Plantation farming, involving large-scale cultivation of single cash crops, is more characteristic of tropical and subtropical regions (e.g., coffee, tea, rubber, sugarcane, oil palm). Mediterranean agriculture is more known for specialized farming of crops like grapes, olives, citrus fruits, figs, wheat, and vegetables, often on smaller scales or in a more intensive manner rather than extensive plantation systems common in the tropics. This statement is incorrect.
Different natural regions have distinct characteristics that influence human activities, settlement patterns, and economic development. Climate, vegetation, and resources play crucial roles. Mediterranean climates are highly attractive for human habitation and agriculture, leading to high urbanization, while dense equatorial rainforests pose significant challenges for large-scale settlement and development.
Mediterranean agriculture is often referred to as ‘horticultural farming’ focusing on fruits, vegetables, and high-value crops, along with some cereal cultivation. The challenges in equatorial regions include dense vegetation, heavy rainfall, high temperatures, diseases, and difficult terrain, although resource exploitation and specific agricultural practices (like shifting cultivation or specialized crops) do occur.

2. Which of the following statements concerning temperature is/are correc

Which of the following statements concerning temperature is/are correct ?

  • 1. In winter season isotherms in Northern Hemisphere are more or less circular on continents
  • 2. Isotherms are parallel to latitude in Southern Hemisphere in oceans
  • 3. Minimum temperature during winter is recorded in Northern Canada

Select the correct answer using the code given below :

1 and 2 only
1 and 3 only
1, 2 and 3
2 only
This question was previously asked in
UPSC NDA-2 – 2016
Statement 1: In winter season, continents in the Northern Hemisphere become significantly colder than the surrounding oceans due to differential heating and cooling rates. This thermal contrast leads to isobars bending equatorward over land and poleward over oceans. While not perfectly circular, the isotherms tend to form closed patterns reflecting cold centers over the large landmasses, deviating from the zonal pattern. Compared to the Southern Hemisphere, the influence of land is pronounced, making the continental isotherms less parallel to latitudes and somewhat ‘circular’ around cold pools. This statement is generally considered correct in depicting the characteristic pattern.
Statement 2: The Southern Hemisphere is dominated by oceans. Oceans have a more uniform temperature distribution and heat/cool more slowly than land. As a result, isotherms in the Southern Hemisphere, particularly over vast oceanic areas and away from significant landmasses like Antarctica or the southern tips of continents, tend to be more regular and run broadly parallel to the lines of latitude. This statement is correct.
Statement 3: While Northern Canada experiences extremely low temperatures in winter, the absolute minimum temperatures on Earth during winter are typically recorded in places like Siberia (Russia) or Antarctica (Vostok Station holds the world record). Northern Canada is among the coldest regions, but it doesn’t necessarily record *the* minimum temperature globally. This statement is incorrect.
Isotherms are lines connecting points of equal temperature. Their patterns are influenced by latitude, altitude, distribution of land and water, ocean currents, and winds. Landmasses cause greater temperature extremes (hot summers, cold winters) compared to oceans, leading to bending of isotherms. The Southern Hemisphere’s oceanic dominance results in more zonal (latitude-parallel) temperature distribution.
The bending of isotherms over landmasses in winter (equatorward bend) and summer (poleward bend) is a direct consequence of the continentality effect. The Southern Hemisphere’s temperature patterns are simpler due to the lack of large east-west oriented landmasses at middle and high latitudes.

3. 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
(Geomorphic Agent)
List II
(Feature)
A. Ground water 1. Plunge pool
B. Running water 2. Horns
C. Glacier 3. Playas
D. Wind 4. Lapies

Code :

4 1 2 3
3 2 1 4
3 1 2 4
4 2 1 3
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is A) 4 1 2 3. This matches the features to their primary geomorphic agents:
– Ground water (A) is a significant agent in karst landscapes, forming features like sinkholes, caves, and Lapies (also known as Karren), which are solutional features on limestone or dolomite surfaces. So, A-4.
– Running water (B), particularly rivers, is responsible for erosional features like valleys, canyons, and waterfalls. Plunge pools are common erosional features found at the base of waterfalls, carved by the impact and turbulence of falling water. So, B-1.
– Glacier (C) is a powerful erosional and depositional agent in high latitudes and altitudes. Glacial erosion shapes mountains, creating features like cirques, arêtes, and sharp, pyramidal peaks called Horns (or pyramidal peaks) formed when several cirques meet. So, C-2.
– Wind (D) is a dominant geomorphic agent in arid and semi-arid regions. Wind can erode, transport, and deposit sediment, forming features like sand dunes, loess deposits, and erosional landforms. Playas are flat, dry lakebeds found in desert basins, which are depositional features associated with ephemeral lakes and wind processes (deflation and deposition). So, D-3.
– Groundwater creates dissolution features in soluble rocks (karst).
– Running water (rivers) creates features like valleys and erosional landforms at waterfalls (plunge pools).
– Glaciers carve distinctive landscapes in mountainous and polar regions (horns, cirques, U-shaped valleys).
– Wind shapes arid landscapes through erosion, transport, and deposition (playas, dunes).
– Lapies are small-scale solution features that can form on rock surfaces exposed to rainfall or covered by a thin soil layer, often part of larger karst systems shaped by groundwater.
– Horns are characteristic features of glaciated mountain ranges, resulting from the headward erosion of multiple cirques around a peak.
– Playas are temporary lakes; when they dry, the flat surface is often subject to wind deflation and deposition of fine sediments.

4. Which one of the following is the correct descending order of countrie

Which one of the following is the correct descending order of countries in terms of per capita availability of arable land ?

India – China – Brazil
China – Brazil – India
Brazil – China – India
Brazil – India – China
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is D) Brazil – India – China. Per capita availability of arable land is calculated by dividing the total arable land area by the total population. Using approximate figures (as of recent years, trends were similar earlier):
– Brazil: Large land area, relatively smaller population compared to India and China. High per capita arable land.
– India: Significant arable land area, but very high population. Moderate per capita arable land.
– China: Significant arable land area, but also very high population (similar to India’s), and slightly less arable land area than India. Low per capita arable land compared to India and Brazil.

Therefore, the descending order of per capita availability of arable land is Brazil > India > China.

– Per capita arable land = Arable land area / Population.
– Brazil has a large land area and lower population density compared to India and China.
– India and China have very large populations, which significantly reduces their per capita figures despite having large absolute areas of arable land.
– Factors affecting arable land availability include total land area, climate, topography, soil quality, land use patterns, and population size.
– China’s arable land area has been under pressure due to urbanization, industrialization, and desertification.
– India faces challenges related to land degradation, fragmentation, and high population density.

5. Which of the following statements with regard to coverage rate of Aadh

Which of the following statements with regard to coverage rate of Aadhar cards during 2015 is/are correct ?

  • 1. The coverage rate of Aadhar cards in western India is quite high and it is above 90 per cent in Gujarat
  • 2. The coverage rate of Aadhar cards is the highest in southern India and it is almost 100 per cent in the states of Telangana and Andhra Pradesh
  • 3. The penetration of Aadhar cards has been very low in north-eastern part of India and it is the lowest in Assam and Meghalaya

Select the correct answer using the code given below :

2 only
1 and 3 only
2 and 3 only
1, 2 and 3
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is D) 1, 2 and 3. Based on the Aadhar enrolment data and trends reported around 2015-16:
1. The coverage rate in Western India, including Gujarat, was indeed quite high, generally above 90% for the eligible population.
2. Southern states consistently showed high coverage rates. Telangana and Andhra Pradesh were among the leading states with very high enrolment figures, often cited as approaching universal coverage for the eligible population by 2015-16.
3. States in the North-Eastern part of India, particularly Assam and Meghalaya, faced significant challenges and delays in Aadhar enrolment due to various factors, resulting in very low coverage rates compared to the national average and other states during this period.
– Aadhar enrolment progress varied significantly across Indian states.
– States in the South and West generally achieved higher coverage rates earlier.
– States in the North-East, especially Assam and Meghalaya, had considerably lower coverage due to unique challenges.
– By the end of 2015 and early 2016, the Aadhar project had achieved substantial coverage nationally, but regional disparities existed.
– The low coverage in Assam was partly linked to the ongoing process of updating the National Register of Citizens (NRC).
– These trends were widely reported in government documents and media coverage regarding the Aadhar rollout.

6. If one mixes up ashes with animal fat, the substance received in the c

If one mixes up ashes with animal fat, the substance received in the crude form is called

Pheromone
Soap
Cement
Concrete
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is B) Soap. Historically, soap was produced by boiling animal fat with lye, which is an alkaline solution obtained from leaching ashes (often wood ashes containing potassium carbonate, which reacts with water to form potassium hydroxide, or using lime to convert sodium carbonate from plant ashes into sodium hydroxide). The process, called saponification, involves the hydrolysis of fats or oils (triglycerides) by an alkali to produce glycerol and fatty acid salts, which are soap. Mixing ashes (source of alkali) with animal fat is a crude method of making soap.
– Soap is made through saponification, a reaction between fats/oils and an alkali.
– Animal fat provides triglycerides.
– Ashes, particularly from wood, can provide alkali (potassium carbonate/hydroxide) when mixed with water.
– This historical process yields crude soap.
– Pheromones are chemical signals released by organisms.
– Cement is a binder substance used in construction.
– Concrete is a composite material made from cement, aggregate, water, and sometimes admixtures.

7. Emulsion is known as a

Emulsion is known as a

colloidal solution of substances having different physical states
true solution
distillation mixture for making alcohols
colloidal solution of two liquids
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is D) colloidal solution of two liquids. An emulsion is a specific type of colloid where one liquid is dispersed throughout another liquid in the form of tiny droplets. The two liquids are typically immiscible or poorly miscible.
– Emulsions are colloidal systems.
– The dispersed phase and the dispersion medium are both liquids.
– Examples include milk (fat droplets dispersed in water) and mayonnaise (oil dispersed in vinegar).
– Colloidal solutions can involve different physical states (e.g., solid in gas for smoke, liquid in gas for fog), but an emulsion is specifically liquid in liquid.
– A true solution is a homogeneous mixture where the solute is dissolved at the molecular or ionic level, unlike a colloid which has larger dispersed particles.
– Distillation is a separation technique, not a type of solution or mixture in this context.

8. The setting time of cement is lowered by adding

The setting time of cement is lowered by adding

oxides of aluminium
gypsum
oxides of magnesium
silica
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is A) oxides of aluminium. The question asks what is added to lower the setting time of cement, which means making it set faster. Gypsum (option B) is added to *increase* the setting time (retard setting). While standard accelerators are typically calcium chloride or alkali aluminates, reactive forms of aluminium oxides or phases rich in alumina like Calcium Aluminate Cements (CAC) or specific additives containing activated alumina can act as accelerators, causing rapid setting. Therefore, among the given options, oxides of aluminium (in a suitable form or context as present in cement phases like C₃A) are the most plausible candidates for contributing to or accelerating the setting process compared to the other options which are either retarders (gypsum) or primary components/cause of expansion (magnesium oxide).
– Setting time refers to the time required for cement paste to lose its plasticity.
– Additives are used to control setting time: accelerators decrease setting time, retarders increase it.
– Gypsum is a common retarder, preventing flash set by reacting with C₃A.
– Oxides of aluminium are present in cement clinker primarily as tricalcium aluminate (C₃A), which hydrates rapidly and contributes to early setting. Certain forms of reactive alumina can act as accelerators.
– Calcium chloride (CaCl₂) is a very common and effective accelerator used in cement.
– Oxides of magnesium (MgO), if present in excess in portland cement, can cause delayed expansion and soundness issues, not primarily affecting initial setting time in a controlled manner.
– Silica (SiO₂) is a main component of cement (silicates C₃S and C₂S), which are responsible for strength development, but adding pure silica is not a method to control setting time.
– The phrasing “lowered by adding” implies acceleration. While gypsum is a crucial additive related to setting time, its effect is the opposite (increasing/retarding). This points towards A as the intended answer, assuming a context where an aluminium oxide based additive is used for acceleration.

9. Two balls, A and B, are thrown simultaneously. A vertically upward wit

Two balls, A and B, are thrown simultaneously. A vertically upward with a speed of 20 m/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9.8 m/s²?

The balls will collide after 3s at a height of 30·2 m from the ground
The balls will collide after 2s at a height of 20·1 m from the ground
The balls will collide after 1s at a height of 15·1 m from the ground
The balls will collide after 5s at a height of 20 m from the ground
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is C) The balls will collide after 1s at a height of 15·1 m from the ground.
Let the origin be the ground level, with the upward direction as positive.
For ball A (thrown upward from ground):
Initial position, y₀_A = 0
Initial velocity, u_A = +20 m/s
Equation of motion: y_A(t) = y₀_A + u_A*t – (1/2)gt² = 0 + 20t – (1/2)(9.8)t² = 20t – 4.9t²

For ball B (thrown downward from 40 m):
Initial position, y₀_B = 40 m
Initial velocity, u_B = -20 m/s
Equation of motion: y_B(t) = y₀_B + u_B*t – (1/2)gt² = 40 – 20t – (1/2)(9.8)t² = 40 – 20t – 4.9t²

Collision occurs when y_A(t) = y_B(t):
20t – 4.9t² = 40 – 20t – 4.9t²
Add 20t and 4.9t² to both sides:
40t = 40
t = 1 second

Substitute t = 1s into either equation to find the height:
y_A(1) = 20(1) – 4.9(1)² = 20 – 4.9 = 15.1 m
y_B(1) = 40 – 20(1) – 4.9(1)² = 40 – 20 – 4.9 = 15.1 m

The balls collide after 1 second at a height of 15.1 m from the ground.

– Use equations of motion under constant acceleration (gravity).
– Define a consistent coordinate system (origin and positive direction).
– Set the positions of the two objects equal to find the time of collision.
– Use the time of collision to find the position (height) of collision.
– The acceleration due to gravity (g) is taken as 9.8 m/s² downwards.
– The relative velocity approach could also be used for the time of collision: v_rel = v_A – v_B. Initial relative velocity = 20 – (-20) = 40 m/s. The relative acceleration is g – g = 0. The initial separation is 40m. Time to collide = separation / relative velocity = 40m / 40m/s = 1s. This confirms the time calculation.

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

Which one of the following statements is not correct ?

The SI unit of charge is ampere-second
Debye is the unit of dipole moment
Resistivity of a wire of length l and area of cross-section a depends upon both l and a
The kinetic energy of an electron of mass m kg and charge e coulomb, when accelerated through a potential difference of V volt, is eV joule
This question was previously asked in
UPSC NDA-2 – 2016
The correct answer is C) Resistivity of a wire of length l and area of cross-section a depends upon both l and a. This statement is incorrect. Resistivity ($\rho$) is an intrinsic property of the material of the wire and depends only on the material composition and temperature, not on its dimensions (length ‘l’ or area of cross-section ‘a’). The resistance (R) of the wire depends on resistivity, length, and area (R = $\rho$ * l/a).
– Resistivity is a material property.
– Resistance depends on material, length, and area.
– Charge unit is Coulomb (C) or Ampere-second (A·s).
– Debye is a unit for electric dipole moment.
– Kinetic energy gained by a charge ‘e’ accelerated by potential ‘V’ is eV.
– Statement A is correct: The SI unit of charge is Coulomb (C), and 1 Coulomb = 1 Ampere * 1 second (C = A·s).
– Statement B is correct: Debye (D) is a common unit used to measure the electric dipole moment of molecules.
– Statement D is correct: When a charge ‘e’ is accelerated through a potential difference ‘V’, the work done on the charge is W = V * e. By the work-energy theorem, this work is converted into kinetic energy, so Kinetic Energy = eV.