Previous Year Paper

UGC NET Paper 2: Environmental Science 21st Nov 2021 (Previous Year Paper)

Held in 2021

100 questions120 minutes
With answers
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Question 1

The place of origin of the National River Ganga is :

  1. AOption (a)

    Bhagirath Kharak

  2. BOption (b)

    Chaukhamba Peak

  3. COption (c)

    Gomukh

  4. DOption (d)

    Pindari Glacier

View answer & explanation

Correct answer

Answer: (c)C.

Gomukh

Why this answer is correct

The correct answer is Gomukh.

Key Points

  • River Ganga
  • It originates in the Himalaya Mountains at Gomukh, the terminus of the Gongotri Glacier. Here, it is known as the Bhagirathi.
  • It cuts through the Central and the Lesser Himalayas in narrow gorges.
  • At Devprayag, the Bhagirathi meets the Alaknanda; hereafter, it is known as the Ganga.
  • The river has a length of 2,525 km.
  • It is shared by Uttarakhand (110 km) and Uttar Pradesh (1,450 km), Bihar (445 km) and West Bengal (520 km).
  • The river finally discharges itself into the Bay of Bengal near the Sagar Island.
  • Left tributaries: Gomti, Ghaghara, Gandak, Burhi Gandak, Koshi, Mahananda, Ramganga, Garra.
  • Right tributaries: Yamuna, Sone, Punpun, Damodar.

Question 2

The difference in values of theoretical oxygen demand and actual oxygen demand is observed because of :

(A) difference in rate of reaeration and deoxygenation

(B) addition of dead cell biomass of bacteria as waste

(C) accumulation of waste in bacterial tissue

(D) formation of humus which resists degradation

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (C) Only

  2. BOption (b)

    (B) and (D) Only

  3. COption (c)

    (A) and (D) Only

  4. DOption (d)

    (B) and (C) Only

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (D) Only

Why this answer is correct

Explanation:

Theoretical Oxygen Demand:

It is a method of calculating the amount of oxygen required by various constituents of the organic matter present in wastewater. Therefore, D is correct.

Actual Oxygen Demand:

It depends on the biodegradability of the compound and the specific organism metabolizing the compound and is measured experimentally by Biological Oxygen Demand (BOD). BOD final is calculated after 5 days at the temperature of 200C. Therefore, B is correct.

Reaeration and deoxygenation are important factors as it is of biological activity where the uptake of oxygen by microorganisms in performing the decomposition process. Therefore, A is not correct.

Hence, option 2 is the correct answer.

Additional Information 

Chemical Oxygen Demand: It is the amount of required oxygen to decompose both the organic and inorganic matter present in the wastewater.

Thus, ThOD > COD > Initial BOD > Final BOD.

Question 3

Water remains in liquid state over a wide range of temperature because of its :

(A) Specific heat

(B) Heat of vaporisation

(C) Density at 0°C

(D) Bipolar nature

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (C) Only

  2. BOption (b)

    (A) and (D) Only 

  3. COption (c)

    (A) and (B) Only

  4. DOption (d)

    (B) and (D) Only

View answer & explanation

Correct answer

Answer: (c)C.

(A) and (B) Only

Why this answer is correct

Explanation

Properties of water: 

  • Specific heat: It is the amount of heat energy that is required to increase the temperature of a unit mass of a substance by one degree. Water has a high specific heat and a lot of energy must be absorbed by it before its temperature changes. Therefore, A is correct.
  • The heat of vaporization: It is the energy required to change a given quantity of a substance into a gas. The strength of hydrogen bonds between water molecules are strong, water has to be heated to 100 ℃ before it changes phase. The heat of vaporization for water is high at about 40.65 kJ/mol. Therefore, B is correct.
  • Less dense solid phase: Water in its solid phase (ice) is less dense than its liquid phase. For instance, if we put ice into a glass of water, it doesn't sink but floats. This phenomenon is also related to the hydrogen bonds between water molecules.
  • Bipolar nature: It is the presence of positive and negative charges in water where the molecule has a slightly positive charge on one side (where the hydrogen atoms are attached) and slightly negative on the other (just oxygen).
  • Absorption of infrared radiation: Water can absorb infrared radiation (heat) from the sun. As a result of this, the oceans and other water bodies act as heat reservoirs and can help moderate the Earth's climate.
  • Melting point (0℃) and Boiling point (100 ℃): This large difference between the melting and boiling point is very important because it means that water can exist as a liquid over a large range of temperatures.

Hence, option 3 is the correct answer.

Question 4

The crystal structure of a compound/material can be determined using:

(A) Scanning Electron Microscope

(B) X‐Ray Fluorescence

(C) Transmission Electron Microscopy

(D) X‐Ray Diffraction

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (D) Only

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (A) and (C) Only

  4. DOption (d)

    (C) and (D) Only

View answer & explanation

Correct answer

Answer: (d)D.

(C) and (D) Only

Why this answer is correct

Explanation

Scanning Electron Microscope(SEM):

  • The Scanning electron microscope uses electrons instead of light to form an image.
  • It is widely used in materials science for research, quality control, and failure analysis.
  • It plays a major role in nanotubes, nanofibers, superconductors, and alloy strength investigation.

X-Ray Fluorescence:

  • It is a non-destructive analytical technique used to determine the elemental composition of materials.
  • A primary X-ray initially generated in an X-ray tube hits an inner shell electron of the atom and ejects the electron from the atom.
  • The open position is filled by an electron from a further outer shell and fluorescence radiation is emitted.

Transmission electron microscope:

  • This microscope uses a particle beam of electrons to visualize specimens and generate a highly-magnified image.
  • It magnifies objects up to 2 million times.
  • This technique is widely used to observe the features of very small specimens.
  • It uses an accelerated beam of electrons, which passes through a very thin specimen to observe the features including structure and morphology.

X-Ray Diffraction:

  • It is an experimental science determining the atomic and molecular structure of a crystal, in which the crystalline structure allows the beam of incident X-rays to diffract into many specific directions.
  • It is used to obtain a three-dimensional molecular structure from a crystal.
  • A purified sample at high concentration is crystallized and the crystals are exposed to an x-ray beam.

​Hence, Transmission Electron Microscopy & X‐Ray Diffraction are used to determine the crystal structure of a compound/material. So, option 4 is the correct answer.

Question 5

Which of the following are living fossils?

(A) Ginkgo biloba

(B) Panthera tigris

(C) Elephas Maximas

(D) Latimavia

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (D) Only

  2. BOption (b)

    (A) and (C) Only

  3. COption (c)

    (B) and (C) Only

  4. DOption (d)

    (A) and (B) Only

View answer & explanation

Correct answer

Answer: (a)A.

(A) and (D) Only

Why this answer is correct

Explanation

Living fossil

  • An organism that remained essentially unchanged from earlier geologic times and whose close relatives are usually extinct.

Example:

  • Ginkgo biloba also called the maidenhair tree, which is the only remaining representative of the Ginkgoaceae family.
  • Latimeria is a rare genus of fish that includes two extant species- Latimeria chalumnae and Latimeria menadoensis.

Hence, option 1 is the correct answer.

Question 6

Consider the following statements :

(A) Earth is farthest from the sun at the time of perehelion

(B) When perehelion occurs, it is winter season in the northern hemisphere

(C) Seasons on earth occur due to inclination in earth’s axis of rotation

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (B) Only

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (A) and (C) Only

  4. DOption (d)

    (A), (B) and (C)  

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (C) Only

Why this answer is correct

Explanation:

Perihelion:

  • It is the phenomenon where the earth is closest to the sun.
  • This occurs about two weeks after the December solstice.

Aphelion:

  • It is the phenomenon where the earth is farthest from the sun.
  • This occurs two weeks after the June solstice.

The earth’s spin axis is titled with respect to its orbital plane which causes seasons.

Hence, option 2 is the correct answer.

Question 7

Ferromagnesian minerals are :

(A) a group of silicates

(B) black, brown or green colour minerals

(C) a group of bicarbonates

(D) highly resistant to weathering and erosional process

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A), (B) and (D) Only

  2. BOption (b)

    (B), (C) and (D) Only

  3. COption (c)

    (A) and (B) Only

  4. DOption (d)

    (C) and (D) Only

View answer & explanation

Correct answer

Answer: (c)C.

(A) and (B) Only

Why this answer is correct

Explanation:

Ferromagnesian minerals:

  • These are silicate minerals.
  • These minerals contain a higher proportion of iron and magnesium. These minerals are dark in color.
  • They tend to look metallic in their lustre.
  • They have relatively high density. They also have magnetic properties.

      

Hence, option 3 is the correct answer.

Question 8

Consider the following statements regarding clay minerals in soil :

(A) Some clay particles are technically not colloids yet they have colloid like properties

(B) There are four major types of colloids present in soil

(C) All types of clays are crystalline in nature

(D) Soil humus are not considered as colloids

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (C) Only

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (A) and (B) Only

  4. DOption (d)

    (B) and (D) Only

View answer & explanation

Correct answer

Answer: (c)C.

(A) and (B) Only

Why this answer is correct

Explanation:

Clay minerals in soil:

  • There are four major types of colloids present in the soil.
  • Layer silicate clays
  • Iron and aluminum oxide clays (sesquioxide clays)
  • Allophane and associated amorphous clays
  • Humus.
  • Layer silicate clays, iron and aluminum oxide clays, allophane, and associated amorphous clays are inorganic colloids while humus is an organic colloid.
  • Some clay particles are technically not colloids yet they have colloid-like properties
  • The humus is often referred to as an organic colloid and consists of various chains and loops of linked carbon atoms.
  • The humus colloids are not crystalline.
  • They are composed basically of carbon, hydrogen, and oxygen rather than of silicon, aluminum, iron, oxygen, and hydroxyl groups.

Question 9

Consider the following statements:

(A) Richter magnitude is based upon the amplitude of the largest seismic wave produced during an earthquake

(B) An earthquake with Richter magnitude 7 producers 10 times larger displacement on the seismogram than does a magnitude 6

(C) S‐waves travel faster than P waves through solid materials

Choose the most appropriate answer from the options given below:

  1. AOption (a)

    (A) and (B) Only

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (A) and (C) Only

  4. DOption (d)

    (A), (B) and (C)

View answer & explanation

Correct answer

Answer: (a)A.

(A) and (B) Only

Why this answer is correct

The correct answer is A and B only.

Key Points

  • Ritcher Scale: It is used to determine the magnitude of an earthquake**.**
  • It is an open-ended logarithmic scale** i.e, **the logarithm of the amplitude of the largest seismic wave is recorded by a seismograph.
  • The vibration in an earthquake with a magnitude of 7 is 10 times greater in magnitude than that of an earthquake with a magnitude of 6 considering it is a base-10 logarithm scale.
  • The order of velocity of seismic waves is:

           P-waves > S-waves > R-waves > L-waves

Therefore, only statements A and B are correct. 

Additional Information

Earthquake waves (seismic waves) are basically of two types: 1. Body waves 2. Surface waves

  • Body waves are of two types: Primary or P - waves and Secondary or S - waves.
  • Surface waves are of two types: Rayleigh or R - waves and Love or L - waves.
Seismic WavesProperties
Primary or P-wavesThey are analogous to sound waves where particles move to and fro in the propagation of rays. They travel fastest in solid material.
Secondary or S-wavesThey are analogous to water ripples or light waves where particles move at the right angle to rays.
Rayleigh or R-wavesThey are dispersive like sound waves.
Love or L-wavesThey are transverse waves that vibrate the ground in the horizontal direction perpendicular to the direction that waves are traveling in.

Question 10

Consider the following statements regarding lines of regression between two variables X and Y:

(A) The correlation between X and Y is the geometric mean of the regression coefficients of Y on X and X on Y.

(B) If the variables are uncorrelated, the two lines of regression (i.e. Y on X and X on Y) are perpendicular to each other

(C) If the regression co-efficient Y on X is greater than one, the co-efficient of X on Y must also be greater than one

Choose the most appropriate answer from the options given below:

  1. AOption (a)

    (A) and (B) Only 

  2. BOption (b)

    (B) and (C) Only 

  3. COption (c)

    (A) and (C) Only 

  4. DOption (d)

    (A), (B) and (C)

View answer & explanation

Correct answer

Answer: (a)A.

(A) and (B) Only 

Why this answer is correct

The correct answer is (A) and (B) Only.

 Key Points

  • The regression coefficient is a statistical measure that represents the average functional relationship between two or more variables.
  • In regression analysis, one variable is regarded as dependent, while the other is regarded as independent.
  • As a result, it assesses the degree to which one variable is dependent on another.
  • The geometric mean of the two regression coefficients equals the correlation coefficient.
  • Properties of the regression line and correlation coefficient:
  • ​If the correlation coefficient rxy = 0, then the two lines of regression are perpendicular to each other.
  • ​If the correlation coefficient rxy = +1, then the two lines of regression are parallel to each other.
  • If one regression coefficient is greater than unity, then the other regression coefficient must be lesser than unity.

Therefore, statement C is incorrect.

Question 11

Match List I with List II :

List – I (Waste type)List – II (Toxic elements)
(A)Wood preservatives(i)Cadmium
(B)Food cans(ii)Mercury
(C)Colored printing inks(iii)Arsenic
(D)Fluorescent lamps(iv)Lead

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (iii), (B) ‐ (i), (C) ‐ (ii), (D) ‐ (iv)

  2. BOption (b)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

  3. COption (c)

    (A) ‐ (iii), (B) ‐ (ii), (C) ‐ (iv), (D) ‐ (i)

  4. DOption (d)

    (A) ‐ (iv), (B) ‐ (ii), (C) ‐ (iii), (D) ‐ (i)

View answer & explanation

Correct answer

Answer: (b)B.

(A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

Why this answer is correct

Concept-

Toxic elements-

  • Organisms require some of the trace elements as essential micronutrients, including copper, iron, molybdenum, zinc, and in some cases aluminum, nickel, and selenium. Under certain conditions, however, these same elements can accumulate to high concentrations in organisms and cause ecological damage.
  • Trace elements that are most often associated with environmental toxicity are the heavy metals cadmium, chromium, cobalt, copper, iron, lead, mercury, nickel, silver, tin, and zinc, as well as the lighter elements aluminum, arsenic, and selenium.
  • Human activities have caused additional examples of pollution by toxic elements, particularly in the vicinity of industrial sources such as smelters.
  • In addition, emissions of mercury and lead from power plants and automobiles have caused widespread contamination of remote environments, although it is not yet certain that this is causing ecological damage.

Wood preservative-

  • Wood preservative products are those that control wood degradation problems due to fungal rot or decay, sapstain, molds, or wood-destroying insects.
  • Wood preservatives containing chromated arsenicals include preservatives containing chromium, copper, and arsenic.

Food cans - It contains lead.

Colored printing inks - It contains cadmium.

Fluorescent lamps - It contains mercury.

So the correct option is option 2.

Question 12

Match List I with List II :

List – I (Wild Animal / Plant)List – II (Schedule of Wildlife Protection Act)
(A)Musk Deer(i)IV
(B)Himalayan Black Bear(ii)VI
(C)Pitcher Plant(iii)I
(D)House rat(iv)II

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (ii), (D) ‐ (i)

  2. BOption (b)

    (A) ‐ (i), (B) ‐ (ii), (C) ‐ (iii), (D) ‐ (iv)

  3. COption (c)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (i)

  4. DOption (d)

    (A) ‐ (ii), (B) ‐ (iii), (C) ‐ (iv), (D) ‐ (i)

View answer & explanation

Correct answer

Answer: (a)A.

(A) ‐ (iii), (B) ‐ (iv), (C) ‐ (ii), (D) ‐ (i)

Why this answer is correct

The correct answer is (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (ii), (D) ‐ (i)

Confusion PointsAn incorrect pair has been given in the PYP as Musk deer and Bison both are in the Schedule I. We have replaced Bison with House Rat to provide correct matching of the pairs. 

  • Musk Deer - Schedule-I
  • Himalayan Black Bear - Schedule II
  • Pitcher Plant - Schedule VI
  • Bison (Gaur) - Schedule-I
  • House rat - Schedule IV

 

Key Points 

Schedules of Wildlife Protection ActExplanationExamples
I- It includes endangered species that require strict protection. The species are protected from poaching, killing, trading, and other forms of exploitation. - Under this Schedule, a person is subject to the harshest penalties for breaking the law. - Species listed in this Schedule are prohibited from being hunted throughout India unless there is a threat to human life or a disease that is incurable.Bengal tiger, Snow leopard, Himalayan bear, Kashmiri stag, Musk deer, Rhinocerous, Golden langur etc.
II- Animals on this list are also given special protection, with trade prohibited. - They cannot be hunted unless there is a threat to human life or they are suffering from a disease/disorder that is incurable.Assamese Macaque, Pig Tailed Macaque, Stump Tailed Macaque, Bengal Hanuman langur, Himalayan Black Bear, Jackal etc.


III & IV- Schedules III and IV contain species that are not endangered. - This includes protected species that are not hunted, but the penalty for any violation is less severe than in the first two schedules.Chital (spotted deer), Bharal (blue sheep), Hyena, Nilgai, Sambhar (deer), Sponges, House rat, etc.
V- This list includes animals that are considered vermin (small wild animals that carry disease and destroy plants and food). These animals are huntable.Common Crows, Fruit Bats, Rats, Mice
VI- It regulates the cultivation of a specific plant and limits its possession, sale, and transportation.
Plant cultivation and trade are only permitted with the prior permission of a competent authority.
Beddomes’ cycad (Native to India) Blue Vanda (Blue Orchid) Red Vanda (Red Orchid) Kuth (Saussurea lappa) Slipper orchids (Paphiopedilum spp.) Pitcher plant (Nepenthes khasiana)

Additional Information

  • Wildlife (Protection) Act, 1972: The Act was enacted for the protection of plants and animal species. 
  • It extends to the whole of India except the State of Jammu and Kashmir.
  • The Central Government appoints the Director of Wildlife Preservation.
  • The State Governments appoint a Chief Wildlife Warden (CWLW) who heads the Wildlife Wing of the department and exercises complete administrative control over Protected Areas (PAs) within a state.
  • Bodies Constituted under the Act:
  • National Board for Wildlife
  • State Board for Wildlife
  • Central Zoo Authority
  • National Tiger Conservation Authority
  • Wildlife Crime Control Bureau

Question 13

Match List I with List II :

List – I (Pollutant)List – II (Health Effects)
(A)Cadmium(i)Severe skin condition ‘Chloracne’
(B)Dioxin(ii)Cancer
(C)Carbofuran(iii)Weakening of bones
(D)Benzo (a) Pyrene(iv)Impeded neuro‐transmission

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (i), (D) ‐ (ii)

  2. BOption (b)

    (A) ‐ (iv), (B) ‐ (i), (C) ‐ (ii), (D) ‐ (iii)

  3. COption (c)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii) 

  4. DOption (d)

    (A) ‐ (iii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (ii)

View answer & explanation

Correct answer

Answer: (d)D.

(A) ‐ (iii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (ii)

Why this answer is correct

Explanation:

PollutantSourcesHealth Effects
CadmiumBatteries, pigments, metal coatings, and plastics.Weakening of bones, muscle pain, kidney, liver, and lung disease
DioxinCombustion, chemical manufacturing units, refineries, and photochemical processes.Reproductive and developmental problems, immune system damage and causes Severe skin condition ‘Chloracne’ which is a skin eruption of blackheads and cysts appearance
CarbofuranBroad spectrum carbamate pesticides that kill insects, mites, and pest control in agricultureWeakness, Sweating, Nausea, Vomiting, blurred vision, loss of coordination and Impeded neuro‐transmission
Benzo (a) PyreneWood burning, coal tar, automobile exhaust, oil and gas products.Skin rashes, burning sensation, and bronchitis cause mutation and leads to Cancer

Question 14

Identify the environmental system with lowest entropy 

  1. AOption (a)

    Land in monoculture agriculture sustained by high energy input 

  2. BOption (b)

    Soils containing a balanced community of living organisms

  3. COption (c)

    Coal reduced to CO2

  4. DOption (d)

    Polluted rivers 

View answer & explanation

Correct answer

Answer: (b)B.

Soils containing a balanced community of living organisms

Why this answer is correct

The correct answer is Soils containing a balanced community of living organisms.

Key Points

  • Entropy is a term central to the field of thermodynamics. It is a thermodynamic property that measures the degree of disorder or randomness. Hence, entropy can serve as a dynamic crucial concept to characterize the organization of the unique thermodynamic states of life and earth's habitable environments.
  • The second law states "For isolated systems that do not exchange energy or mass with their surroundings, the entropy of that system increases".
  • Soils that contain a balanced community of living organisms are in a constant exchange of energy through the processes like nutrient cycling, homeostasis, energy conversions, decomposition, etc. Therefore, this system has the lowest entropy.

Therefore, soils containing a balanced community of living organisms have the lowest entropy.

Question 15

Which chemical form of mercury is completely water soluble?  

  1. AOption (a)

    Elemental mercury in liquid state 

  2. BOption (b)

    Monomethyl mercury 

  3. COption (c)

    Elemental mercury in vapour state 

  4. DOption (d)

    Dimethyl mercury 

View answer & explanation

Correct answer

Answer: (d)D.

Dimethyl mercury 

Why this answer is correct

The correct answer is Dimethyl mercury.

Key Points

  • Elemental mercury is inert, stable, and insoluble in water but it dissolves in lipids.
  • Monomethyl mercury is soluble in most organic solvents but insoluble in water (inorganic solvent).
  • The solubility of mercury vapour in water has been measured by means of atomic absorption spectrophotometry over the temperature range of 5 - 60°C under atmospheric pressure. The aqueous solubility obeys Henry's law at each temperature.
  • The heat of the solution of mercury vapour is found to be -5.3 KCal/mol.
  • Dimethylmercury is capable of permeating many materials, including plastic and rubber compounds.

The water solubilities of different compounds of mercury are:

  • Hg (mercury) water solubility is 0.000056 g/L
  • CH3HgCl (methyl mercury chloride) water solubility is 0.1 g/L
  • (CH3)2Hg (dimethyl mercury) water solubility is 1 g/L

​Therefore, Dimethyl mercury is completely soluble in water.

Question 16

A gas-fired, combined heat and power plant produces 5000 MWh of electricity and heat of 30,000 GJ. If the input energy required to run the power plant is 60,000 GJ, the efficiency of the plant is : 

  1. AOption (a)

    35%

  2. BOption (b)

    40%

  3. COption (c)

    50%

  4. DOption (d)

    80%

View answer & explanation

Correct answer

Answer: (d)D.

80%

Why this answer is correct

Explanation:

Given:

Power produced = 5000 MWh

The heat produced (in the form of energy) = 30,000 GJ

Input energy = 60,000 GJ

To get the given values in the same units, we have to convert 5000 MWh in MJ

That is,  Heat InputkW hr electricity\frac{Heat \ Input}{kW\ hr\ electricity} = 5000MWhr ×1Mj/sec1 Mw ×3600 sec/hr5000 MW hr\ \times \frac{1 Mj/sec}{1\ Mw}\ \times 3600\ sec/hr

= 18000000 MJ

1 MJ = 0.001 GJ

Therefore, Total input energy = 18000 GJ + 30,000 GJ = 48000 GJ

We know, Efficiency=Output EnergyInput EnergyEfficiency = {Output\ Energy{} \over Input\ Energy}

Therefore,   Efficiency=4800060000Efficiency = {48000{} \over 60000} x 100 = 80%

Question 17

Which of the following compounds has the least Ozone Depleting Potential?

  1. AOption (a)

    CFC‐11

  2. BOption (b)

    Halons

  3. COption (c)

    HCFCs

  4. DOption (d)

    Methyl Bromide

View answer & explanation

Correct answer

Answer: (c)C.

HCFCs

Why this answer is correct

Concept:

Ozone depletion Potential (ODP):

  • It is a measure of a chemical compound that is the relative amount of degradation to the ozone layer it can cause with trichlorofluoromethane (CFC-11) being fixed at an ODP of 1.0.
  • ODP can be estimated from the molecular structure of a given substance.

Ozone-depleting substances include:

  • Chlorofluorocarbons (CFCs)
  • Hydrobromofluorocarbons (HBFCs)
  • Halons
  • Methyl Bromide- ODP
  • Carbon tetrachloride
  • Methyl Chloroform
  • Hydrochlorofluorocarbons(HCFCs)- least ODP ranging from 0.01-0.1. Hence,  option 3 is the correct answer.

Ammonia and Carbon dioxide do not cause depletion of Ozone, so they have zero ODP.

Question 18

Which one among the following is not an example of detrital sedimentary rocks? 

  1. AOption (a)

    Sandstone

  2. BOption (b)

    Chalk

  3. COption (c)

    Shale

  4. DOption (d)

    Conglomerate

View answer & explanation

Correct answer

Answer: (b)B.

Chalk

Why this answer is correct

Concept:

  • Processes of weathering produce sediment. Particles continue to be broken down as they are transported and sorted (By the processes of erosion).
  • Following deposition, sediments become sedimentary rocks through the processes of lithification.

There are two types of sedimentary rocks.

1) Detrital sedimentary rocks:  Rocks that are formed from transported solid material. 

Detritus - Latin for "worn down." 

Types of detrital sedimentary rocks:

  • Conglomerate - Forms from cemented GRAVEL (a mixture of cobbles, pebbles, and coarse sand), such as from a stream gravel bar. The gravel particles are "rounded" by erosion processes.
  • Breccia- Forms from cemented "gravel-sized" fragments of rock, but the rock fragments are very "angular" in shape. Broken rock fragments are common in fault zones and in volcanic areas.
  • Sandstone - A sedimentary rock formed by cemented SAND. (Sand particles are between 1/16-2mm; sandstone usually displays a high degree of sorting due to wind and water transport processes
  • SIltstone- Forms from silt (particle sizes: 1/16-1/256mm).
  • Shale - Forms from clay(particles: <1/256mm); shale displays laminae (layering).

2) Chemical sedimentary rocks:

  • Formed when dissolved substances precipitate from water to form sediments and eventually rock; include salts such as rock salt, Gypsum, and Limestone, and biochemically mediated rocks such as peat, and reef rock (organic limestone).

So chalk is not an example of detritus sedimentary rocks.

Question 19

According to the U.S. Department of Agriculture system of classification, the size of silt particles in soil ranges between :

  1. AOption (a)

    0.02 to 0.2 mm

  2. BOption (b)

    0.05 to 0.5 mm

  3. COption (c)

    0.002 to 0.05 mm

  4. DOption (d)

    0.0002 to 0.002 mm

View answer & explanation

Correct answer

Answer: (c)C.

0.002 to 0.05 mm

Why this answer is correct

Explanation:

  • Soil particles are classified to determine the texture of the soil and grouped with similar attributes. The size of the particle majorly influences the water-holding capacity and the porosity of the soil.
  • Majorly two classification systems viz. International Union/Society of Soil Sciences (ISSS/IUSS) and United Nations Department of Agriculture (USDA) are as follows,
ISSS/IUSSUSDA
CategorySize range (mm)CategorySize range (mm)
Gravel>2.0Gravel>2.0
Coarse sand0.2 – 2.0Very coarse sand1.0 – 2.0
Fine Sand0.02 – 0.2Coarse sand0.50 – 1.0
Silt0.002 – 0.02Medium sand0.25 – 0.50
Clay<0.002Fine sand0.10 – 0.25
Very fine sand0.05- 0.10
Silt0.002-0.05
Clay<0.002

Hence, Option 3 is the correct answer.**

Question 20

Which one among the following has the lowest hydraulic conductivity (m/day)?

  1. AOption (a)

    Gravel

  2. BOption (b)

    Sandstone

  3. COption (c)

    Sand

  4. DOption (d)

    Clay

View answer & explanation

Correct answer

Answer: (d)D.

Clay

Why this answer is correct

Concept-

  • A coefficient describing the relative ease with which groundwater can move through a permeable layer of rock or soil.
  • Soils are generally divided on the basis of their particle size into coarse-grained and fine-grained soils.
  • Coarse-grained soils are of particle size greater than 0.075 mm or 75 microns and particles smaller than this make up fine-grained soils.
  • Further coarse grains are divided into gravel and sand, and fine grains are divided into silt and clay sized.

  • Gravel-sized particles are larger than 4.75mm and sand particles are between 4.75mm to 75 microns.
  • Silt-sized particles are of size 75 microns to 2 microns, and clay-sized particles are of size less than 2 microns.
  • The order of hydraulic conductivity will be Gravel > Sandstone > sand > Clay.

Question 21

‘‘Slash and burn’’ is a term that is usually associated with the practice of :

  1. AOption (a)

    Illicit felling and encroachment

  2. BOption (b)

    Grazing and lopping

  3. COption (c)

    Incineration of waste

  4. DOption (d)

    Shifting cultivation

View answer & explanation

Correct answer

Answer: (d)D.

Shifting cultivation

Why this answer is correct

Explanation

Shifting Cultivation

  • It is a form of agriculture where the farmers cultivate the land for two or three seasons temporarily and shifts to another area.
  • In this farming, the forest land is cleared by burning and is allowed for the cultivation of crops. This burnt biomass increases the soil nutrients. Therefore, this system of cultivation is called slash-and-burn farming.
  • This farming is also known as primitive subsistence farming.

Different names for shifting cultivation

  • Jhuming/Bewar – Northeastern states of India
  • Podu- Andhra Pradesh
  • Ladang – Malaysia, Indonesia
  • Milpa – Mexico
  • Roca – Brazil
  • Ray- Vietnam

Hence option 4 is the correct answer.

Question 22

A species that has profound effect on the ecological processes of an ecosystem and whose removal or loss would result in disturbing the community structure is known as :

  1. AOption (a)

    Dominant species

  2. BOption (b)

    Keystone species

  3. COption (c)

    Indicator species

  4. DOption (d)

    Flagship species

View answer & explanation

Correct answer

Answer: (b)B.

Keystone species

Why this answer is correct

Explanation:

Dominant species

  • These are species that have a relatively high abundance to other species in a community and ultimately have proportionate effects on the functions of the ecosystem.
  • Example: Conifers are the dominant species in the temperate forest.

Keystone species

  • These species have a greater impact on the population of a particular ecosystem and maintain biodiversity and play a critical role in the food web.
  • Without these species, an entire ecosystem would radically change or cease to exist altogether.
  • Example: Bees as pollinator supports the reproduction of flowering plants in higher proportion.

Hence, option 2 is the correct answer.

Indicator species

  • These species reflect the environmental condition by their presence, absence, or in abundance.
  • They are the signals to diagnose the health condition of a particular ecosystem.
  • Example: Mosses often indicate acid soil.

Flagship species

  • These species are the icon that needs to be conserved in a social context. Thus, inducing people to support the conservation of these species.
  • Example: The Bengal Tiger is a flagship species by World Wildlife Fund.

Question 23

The flagship species of Kedarnath Wildlife sanctuary is : 

  1. AOption (a)

    Elephant

  2. BOption (b)

    Musk Deer

  3. COption (c)

    Tiger

  4. DOption (d)

    Wild Boar

View answer & explanation

Correct answer

Answer: (b)B.

Musk Deer

Why this answer is correct

The correct answer is Musk Deer.

Key Points

  • Kedarnath Wildlife Sanctuary 
  • It is situated in Chamoli, Uttrakhand.
  • It is the largest protected area in the region of the Western Himalayas.
  • It is best known for the endangered musk deer and is thus called the Kedarnath Musk Deer Sanctuary.
  • The sanctuary takes its name from the famous Hindu temple of Kedarnath which is just outside its northern border.
  • The sanctuary is rich in fauna and one can find animals like snow leopards, Indian leopards, jackals, Himalayan black bears, yellow-throated martens, red fox, etc.
  • Some of the popular avifauna are snow partridge, Himalayan monal, Koklass pheasant, little-pied flycatcher, grey-cheeked warbler, West Himalayan bush warbler, etc.
  • The management of the wildlife sanctuary is done by the Uttarakhand Forest department.
  • The Divisional Forest Officer(DFO) Kedarnath Wildlife Division is responsible for scientific management of the wildlife sanctuary as per the approved management plan.

Question 24

The maximum permissible limit of BOD (mg/l) in the treated waste water that can be discharged into garden for watering is :

  1. AOption (a)

    20

  2. BOption (b)

    30

  3. COption (c)

    40

  4. DOption (d)

    10

View answer & explanation

Correct answer

Answer: (b)B.

30

Why this answer is correct

Concept:

  • BOD represents the amount of biodegradable organic matter present in water.
  • It also indicates the amount of treatment for purifying. If the BOD is high the extent of treatment required will be high.
  • The permissible limit declared by CPCB for various parameters is given below so that treated wastewater can be discharged into the garden for watering-
ParametersPermissible limit by CPCB(mg/l)
BOD30
COD250
TSS100
Oil and grease10
Total Hardness180
pH7.5

Question 25

At STP, 1 ppb sulphur dioxide is equivalent to :

  1. AOption (a)

    ∼ 7.31 μg m-3

  2. BOption (b)

    ∼ 3.16 μg m-3

  3. COption (c)

    ∼ 4.52 μg m-3

  4. DOption (d)

    ∼ 2.86 μg m-3

View answer & explanation

Correct answer

Answer: (d)D.

∼ 2.86 μg m-3

Why this answer is correct

Explanation

  • The mass units used to measure the air pollutants include milligrams (one-thousandth of a gram), micrograms (one-millionth of a gram), nanograms, and pictograms calculated per volume of air (cubic meters or cubic feet).
  • However, these concentrations can also be expressed as parts per million (ppm) or parts per billion (ppb) by volume. This can be done based on the molecular weight of the chemical.

The formula for conversion between µg/m3 (micrograms per cubic meter) and ppb (parts per billion) is as follows:

Converting from ppb to µg/m3 is:

Concentration (µg/m3) = molecular weight x concentration (ppb) / 24.45 ………..… (I)

Calculations:

  • STP refers to the standard temperature at 25°C and the pressure is at 1 atmosphere (1 atm = 1.01325 bar).
  • The number 24.45 in the equation is the volume (liters) of a mole (gram molecular weight) of a gas at STP.

Molecular weight of Sulphur dioxide (SO2) = 32.065 (S) + 2(15.999) O2 = 64.06 g/mol

Now, substituting the above values in equation (I) give,

Concentration (µg/m3) = 64.06 × 124.45\frac{64.06 \ \times\ 1}{24.45} = 2.62 µg/m3.

The nearest value is in Option 4. Hence, it is the correct answer.

Question 26

Which of the following organisms is the most tolerant to sewage pollution?

  1. AOption (a)

    Scenedesmus

  2. BOption (b)

    Daphnia

  3. COption (c)

    Chironomus

  4. DOption (d)

    Chlorella

View answer & explanation

Correct answer

Answer: (c)C.

Chironomus

Why this answer is correct

The correct answer is Chironomus.

Key Points

  • The Chironomidae are very important constituents of the benthic macro-invertebrate community at the stations downstream of the effluent.
  • Out of all the microbes enlisted, Chironomus riparius is the most tolerant species.
  • They contain the red blood pigment 'haemoglobin', which allows them to absorb oxygen from the water more readily.
  • As a result, they can survive in habitats with very low levels of oxygen, including heavily polluted areas.

Therefore, the most tolerant organism to sewage pollution is Chironomus.

Question 27

The process of illuviation in soil indicates :

  1. AOption (a)

    Removal of particles from the top soil

  2. BOption (b)

    Removal of particles from the C horizon

  3. COption (c)

    Percolation of soluble salts into deeper soil layers

  4. DOption (d)

    Accumulation of particles in the B horizon

View answer & explanation

Correct answer

Answer: (d)D.

Accumulation of particles in the B horizon

Why this answer is correct

Explanation

illuviation:

  • It is the process of deposition of illuvium. illuvium is the material that got removed from one horizon (a process called eluviation) of the soil to another.
  • It includes organic matter, silicate clay, and hydrous oxides of iron and aluminum.
  • This removal is usually by the action of rainwater.
  • This transport of material may be either mechanical or chemical.
  • This accumulation of these materials occurs in the subsoil horizon or B horizon. This zone is also known as zones of illuviation.

Hence, option 4 is the correct answer.

Question 28

Carbonmonooxide is harmful to human beings, as it :

  1. AOption (a)

    causes cancer

  2. BOption (b)

    causes eye irritation

  3. COption (c)

    depletes ozone layer

  4. DOption (d)

    has high affinity with haemoglobin as compared to oxygen

View answer & explanation

Correct answer

Answer: (d)D.

has high affinity with haemoglobin as compared to oxygen

Why this answer is correct

Concept:

  • Blood is a connective tissue, that has an iron-containing pigment called Haemoglobin.
  • Haemoglobin gives a red colour to the blood and is responsible for the transport of O<sub>2 </sub>molecules throughout the body.
  • When haemoglobin combines with oxygen from the lungs it forms oxyhaemoglobin.
  • Oxyhaemoglobin transports oxygen to the cells of the body.
  • Oxygen gets attached to Haemoglobin because it shows an affinity towards it, but other gases such as Carbon monoxide (CO), Carbon dioxide (CO2) also show an affinity towards it.

Explanation:

  • When the outside environment is rich in CO, and it is being inhaled, it starts to displace oxygen bind on Haemoglobin.
  • The displacement occurs because CO shows several hundred times stronger affinities towards haemoglobin than oxygen, and it forms carboxyhaemoglobin.
  • Carbon monoxide is harmful to human beings, as it has a high affinity with haemoglobin as compared to oxygen.
  • This leads to the scarcity of oxygen in the body, and sufficient oxygen is not available for organs.
  • Concentrations as low as 20 or 30 PPM can be harmful if exposed for several hours.

Question 29

In a class of 13 students, 3 students failed in a test. The marks of 10 students who passed were 6, 9, 4, 8, 7, 6, 5, 6, 7, 5. The median value of marks of all the 13 students is :

  1. AOption (a)

    5

  2. BOption (b)

    5.5

  3. COption (c)

    6

  4. DOption (d)

    6.5

View answer & explanation

Correct answer

Answer: (c)C.

6

Why this answer is correct

Solution:

Ascending order is 4, 5, 5, 6, 6, 6, 7, 7, 8, 9

and all the failed students have marks less than 4.

data in ascending order is x, y, z, 4, 5, 5, 6, 6, 6, 7, 7, 8, 9

number of observations is 13

Median is middle-term i.e., 7

Hence, the median is 6.

Question 30

The probability that a standard normal variate lies between -∞ and +1 is:

  1. AOption (a)

    ∼0.84

  2. BOption (b)

    ∼0.78

  3. COption (c)

    ∼0.72

  4. DOption (d)

    ∼0.68

View answer & explanation

Correct answer

Answer: (a)A.

∼0.84

Why this answer is correct

The correct answer is ∼0.84.

Key Points

  • Normal distribution or continuous probability distribution presents a bell-shaped symmetrical curve called a normal distribution curve.
  • In the normal curve, all distribution is bilaterally symmetrical about the mean.

For every normal curve,

  1. 68.26% of data points fall within 1 standard deviation of mean.
  2. 95.44% of data points fall within 2 standard deviations of mean.
  3. 99.73% of data points fall within 3 standard deviations of mean.​

 

 

 

 

 

 

 

 

 

 

 

 

          Fig. Area under Standard Normal Curve

Distance from mean ordinate

  • μ to +1σ = 34.13% = 0.3413
  • μ to -1σ = 34.13% = 0.3413
  • -1σ to -2σ = 13.59% = 0.1359
  • -2σ to -3σ = 2.14% = 0.0214
  • -3σ to -∞ = 0.14% = 0.0014

          ​=> 0.3413+0.3413+0.1359+0.0214+0.0014 = 0.8413 = ∼0.84

Where μ = Mean

​Therefore, the correct answer is ∼0.84.

Additional Information

Distance from the mean ordinate in terms of ± σArea under the curve
Z = ±0.674550% = 0.50
Z = ±1.068.27% = 0.6827
Z = ±1.9695% = 0.95
Z = ±2.0095.45% = 0.9545
Z = ±2.5899% = 0.99
Z = ±3.099.73% = 0.9973

Question 31

Population of a city doubled in a span of 20 years. Assuming exponential growth at a constant rate over the given period, the growth rate would have been :

  1. AOption (a)

    -3.5%

  2. BOption (b)

    -3.0%

  3. COption (c)

    -2.5%

  4. DOption (d)

    -2.0%

View answer & explanation

Correct answer

Answer: (a)A.

-3.5%

Why this answer is correct

Concept:

 Let the exponential growth equation of population can be modeled as

N = N0e−λt{N_0}{e^{ - λ t}} 

Where N = Population at any time t

λ = Growth rate constant, t = time

Calculation:

The population of the city doubled in a span of 20 years

N = 2N0, t = 20 yrs

2N0 = N0e-λt 

λ  = 0.69320\frac{0.693}{20} = - 3.5 %

Question 32

Among all the unbiased Estimators Of the population regression coefficients which are linear in the dependent variable (say Yi), the least squares estimators have the smallest variance.

This statement is known as:

  1. AOption (a)

    Bayes' theorem

  2. BOption (b)

    Central limit theorem

  3. COption (c)

    Chebyshev's theorem

  4. DOption (d)

    Gauss-Markov theorem

View answer & explanation

Correct answer

Answer: (d)D.

Gauss-Markov theorem

Why this answer is correct

Explanation

Bayes’ theorem

  • It states that the conditional probability of an event, based on the occurrence of another event, is equal to the likelihood of the second event given the first event multiplied by the probability of the first event.

Central Limit theorem

  • It explains that the sampling distribution of the mean will always be normally distributed, as long as the sample size is large enough.

Chebyshev’s theorem

  • This estimates the minimum proportion of observations that fall within a specified number of standard deviations from the mean.

Gauss- Markov theorem

  • It states that when the linear regression model satisfies the first six classical assumptions, then ordinary least squares OLS regression produces unbiased estimates that have the smallest variance of all possible linear estimators.

​Hence, option 4 is the correct answer.

Question 33

Consider the following statements regarding organic carbon estimation following Walkley and Black method :

(A) Organic carbon estimated by this method excludes most of the less active organic matter

(B) 100% of the total oxidizable organic carbon is oxidized in this method

(C) Soil organic matter percentage can be calculated by multiplying estimated organic carbon percentage by 1.724

(D) K2SO4 and H2SO4 are used as reagents

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (B) and (D) Only

  2. BOption (b)

    (A) and (B) Only

  3. COption (c)

    (A) and (C) Only

  4. DOption (d)

    (B) and (C) Only

View answer & explanation

Correct answer

Answer: (c)C.

(A) and (C) Only

Why this answer is correct

Concept-

Walkley and Black method

  • The determination of soil organic carbon is based on the Walkley & Black chromic acid wet oxidation method.
  • Oxidizable organic carbon in the soil is oxidized by 0.167 M potassium dichromate (K2Cr2O7) solution in concentrated sulfuric acid.
  • The heat of the reaction raises the temperature which is sufficient to induce substantial oxidation.
  • Organic carbon estimated by this method excludes most of the less active organic matter.
  • Recoveries of the total Soil Organic Carbon by this method can typically be between 75 – 90 % in surface soils and will vary with soil type and depth.
  • Walkley & Black found that on average about 77% of the organic C was recovered by the heat of the dilution procedure, and they proposed that a correction factor of 1.3 be used to account for unrecovered organic C.
  • Soil organic matter percentage can be calculated by multiplying the estimated organic carbon percentage by 1.724
  • For soils that are very high in organic carbon content, the Walkley & Black method may result in low test results, due to the incomplete oxidation of the organic carbon in the sample.
  • So 100% of the total oxidizable organic carbon is not oxidized in this method.
  • Smaller sample weights should be used for samples with very high carbon content.
  • This method is for the determination of organic carbon in soils. It is not applicable to soils containing significant amounts of carbonized materials.

Question 34

Match List I with List II :

List – I Mineral TypeList – II Geometric Shape
(A)Helicitic(i)Arranged like a star
(B)Micaceous(ii)Consists of small ellipsoid or spheroids
(C)Oolitic(iii)Consists of thin plate like lamelae
(D)Stellate(iv)Twisted and curved, like coral branching

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (i), (D) ‐ (ii)

  2. BOption (b)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (i)

  3. COption (c)

    (A) ‐ (i), (B) ‐ (ii), (C) ‐ (iii), (D) ‐ (iv)

  4. DOption (d)

    (A) ‐ (i), (B) ‐ (ii), (C) ‐ (iv), (D) ‐ (iii)

View answer & explanation

Correct answer

Answer: (b)B.

(A) ‐ (iv), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (i)

Why this answer is correct

The correct answer is (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (i).

Key Points

  • Helicitic mineral is a type of recrystallization structure where the poikiloblasts overgrow the pre-existing foliation appearing in twists and curves. This is akin to coral branching in terms of shape. This texture, therefore, indicates post-tectonic crystallization of poikiloblasts.
  • Micaceous minerals form very thin layers that peel off or cleave very easily. They characteristically split into flexible sheets much like lamelae.
  • Oolitic minerals occur in crystalline aggregates that are rounded and less than 4mm in size. Oolite is a sedimentary rock composed of ooids (ooliths), which are spheroidal grains with a nucleus and mineral cortex accreted around them, the sphericity of which increases with distance from the nucleus. For example oolitic hematite, chamosite etc.
  • Stellate minerals are long thin crystals radiating outwards from a centre producing a star-like shape, such as astrophyllite, wavellite etc.

Question 35

Match List I with List II :

List – I Air Quality IndexList – II Categorization
(A)0-50(i)Poor
(B)51-100(ii)Severe
(C)201-300(iii)Good
(D)401-500(iv)Satisfactory

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (ii), (D) ‐ (i)

  2. BOption (b)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

  3. COption (c)

    (A) ‐ (i), (B) ‐ (ii), (C) ‐ (iv), (D) ‐ (iii)

  4. DOption (d)

    (A) ‐ (i), (B) ‐ (ii), (C) ‐ (iii), (D) ‐ (iv)

View answer & explanation

Correct answer

Answer: (b)B.

(A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

Why this answer is correct

Explanation:

Air Quality Index:

  • AQI is an overall scheme that transforms individual air pollutant (e.g. SO2, CO, PM10) levels into a single number, a simple and lucid description of air quality for the citizens.
  • AQI relates to health impacts and citizens can avoid unnecessary exposure to air pollutants;
  • AQI indicates compliance with National Air Quality Standards;
  • AQI prompts local authorities to take quick actions to improve air quality;
  • AQI guides policymakers in take broad decisions; and
  • AQI encourages citizens to participate in air quality management.

AQI categories and range:

CategorizationAir Quality Index
0 - 50Good
51 - 100Satisfactory
101 - 200Moderate
201 - 300Poor
301 - 400Very Poor
> 401Severe

So the correct answer is option 2.

Question 36

Match List I with List II :

List – I (Hydropower Projects)List – II (River)
(A)Sardar Sarovar project(i)Bhagirathi River
(B)Tehri Dam project(ii)Narmada River
(C)Pong Dam(iii)Mahanadi River
(D)Hirakund Dam(iv)Beas River

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (i), (B) ‐ (ii), (C) ‐ (iv), (D) ‐ (iii)

  2. BOption (b)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (i)

  3. COption (c)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

  4. DOption (d)

    (A) ‐ (ii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (iii)

View answer & explanation

Correct answer

Answer: (d)D.

(A) ‐ (ii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (iii)

Why this answer is correct

The correct answer is (A) ‐ (ii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (iii).

Key Points

Hydropower ProjectsRiver
Sardar Sarovar projectNarmada River
Tehri Dam projectBhagirathi River
Pong DamBeas River
Hirakund DamMahanadi River

 

 

 

 

Additional Information

  • Hydropower projects are classified as large and small hydro projects based on their sizes.
  • India is the world's seventh-largest producer of hydroelectric energy, with a total capacity of 47,057 MW.
  • The Shivanasamudra hydroelectric power station was the first hydroelectric power facility.
  • Tehri Dam is India's tallest structure, and Tehri Hydro Electric Power Plant is the country's tallest hydroelectric power facility.
  • The Koyna Hydroelectric Project is India's largest completed hydroelectric power plant. It has a capacity of 1960 MW.

Question 37

Identify the correct sequence of Life Time (years) of the green house gases in increasing order :

(A) CH4 (Methane)

(B) N2O (Nitrous Oxide)

(C) CFC‐12

(D) O3 (Tropospheric)

Choose the correct answer from the options given below :

  1. AOption (a)

    A, D, B, C

  2. BOption (b)

    D, A, C, B

  3. COption (c)

    B, A, D, C

  4. DOption (d)

    B, D, A, C

View answer & explanation

Correct answer

Answer: (b)B.

D, A, C, B

Why this answer is correct

The correct answer is D, A, C, and B.

Key Points

  • The greenhouse effect is the process by which solar radiation is absorbed by greenhouse gases rather than reflected back into space. This insulates the earth's surface and keeps it from freezing.
  • A greenhouse gas (GHG) is a gas that absorbs and emits radiant energy in the thermal infrared spectrum, resulting in the greenhouse effect.
  • Different GHGs are Water vapour, Carbon dioxide (CO2), Methane (CH4), Nitrous oxide (N2O), CFCs, HFCs, Perfluorocarbons and Tropospheric ozone.
  • These GHGs have a different lifetime for which they can remain in the atmosphere.
Greenhouse gases (GHGs)Average lifetime (years)
Carbon dioxide50-200
Methane12
Nitrous oxide109
CFC-1152
CFC-12100
HCFC-2212
HFC-325
HFC-134a14
Tetrafluoromethane50000
Hexafluoroethane10000
Sulfur hexafluoride3200
Tropospheric OzoneHours-weeks

Question 38

Identify the correct sequence of Net Heating Value (MJ/Kg) in increasing order for the following fuels.

(A) Charcoal

(B) Bituminous coal

(C) Regular gasoline

(D) Natural gas

(E) Oak wood

Choose the correct answer from the options given below :

  1. AOption (a)

    A, E, B, C, D

  2. BOption (b)

    E, B, A, C, D

  3. COption (c)

    B, A, E, D, C

  4. DOption (d)

    E, A, B, C, D

View answer & explanation

Correct answer

Answer: (d)D.

E, A, B, C, D

Why this answer is correct

The correct answer is E, A, B, C, D.

Key Points

  • Net Heating Value (NVH) is the potential energy available in a fuel sample that is the heat generated by the complete combustion of a unit volume of gas, excluding heat recoveries by condensing the water vapour formed.
FuelsNet Heating Value (MJ/kg)
Charcoal29.6
Bituminous coal33
Regular gasoline46
Natural gas55
Oak wood10 - 11

 

 

 

 

 

Additional Information

  • High Heating Value (HHV) refers to the energy released in combustion when condensing the water vapour resulting from the combustion, causing latent heat evaporation as a result.

Question 39

Given below are two statements :

Statement I : Mean deviation is a better measure of dispersion than the quartile deviation.

Statement II : Mean deviation is obtained by dividing the algebraic sum of deviations about mean by total number of observations.

In the light of the above statements, choose the correct answer from the options given below :

  1. AOption (a)

    Both Statement I and Statement II are true

  2. BOption (b)

    Both Statement I and Statement II are false

  3. COption (c)

    Statement I is true but Statement II is false

  4. DOption (d)

    Statement I is false but Statement II is true

View answer & explanation

Correct answer

Answer: (c)C.

Statement I is true but Statement II is false

Why this answer is correct

The correct answer is Statement I is true but Statement II is false.

Key Points

Properties of a good measure of dispersion:

  • It should be simple to understand.
  • It should be easy to compute.
  • It should be rigidly defined.
  • It should be based on each and every point of distribution.
  • It should have sampling stability.
  • It should not be unduly affected by extreme items.

Quartile deviation:

  • Quartile deviation is an absolute measure of dispersion.
  • Quartile deviation gives the average amount by which the two quartiles differ from the median. The median ± quartile deviation covers exactly 50% of the observations.
  • Quartile deviation ignores 50% of items i.e., the first 25% and the last 25%. As the value of quartile deviation does not depend upon every item of the series, therefore, it cannot be regarded as a good method of measuring dispersion.

Mean deviation:

  • ​The mean deviation is also known as the average deviation. It is the average difference between the items in a distribution and the median or mean of that series.
  • It is based on each and every item of data and is less affected by the value of extreme items. Therefore, is comparatively a better measure of deviation than range and quartile deviation.

 Therefore, statement I is true but statement II is false.

Question 40

Given below are two statements :

Statement I: The log‐normal distribution is positively skewed.

Statement II: A negatively skewed distribution has its long tail to the left.

In light of the above statements, choose the correct answer from the options given below :

  1. AOption (a)

    Both Statement I and Statement II are true

  2. BOption (b)

     Both Statement I and Statement II are false

  3. COption (c)

    Statement I is true but Statement II is false

  4. DOption (d)

    Statement I is false but Statement II is true

View answer & explanation

Correct answer

Answer: (a)A.

Both Statement I and Statement II are true

Why this answer is correct

The correct answer is Both Statement I and Statement II are true.

Key Points

  • A distribution that is non-symmetrical is called skewed distribution. The curve stretches to one side of the centre.

          The distribution can be of two types: 1. Positively Skewed distribution 2. Negatively Skewed distribution

  1. Positively Skewed distribution: Here greater range of values spread from the right side of the curve. Therefore, the long tail is towards the right side. Log-normal distributions are positively skewed due to low mean values and high variances in the random variables.
  2. Negatively Skewed distribution: Here greater range of values spread from the left side of the curve. Therefore, the long tail is towards the left side.

Therefore, both statements are true.

Question 41

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : In the hybrid car technology, cars run on both petrol and battery based electrical engines.

Reason R : Battery of the hybrid car is externally charged.

In the light of the above statements, choose the correct answer from the options given below :

  1. AOption (a)

    Both A and R are true and R is the correct explanation of A

  2. BOption (b)

    Both A and R are true but R is NOT the correct explanation of A 

  3. COption (c)

    A is true but R is false

  4. DOption (d)

    A is false but R is true

View answer & explanation

Correct answer

Answer: (c)C.

A is true but R is false

Why this answer is correct

The correct answer is A is true but R is false.

Key Points

  • Hybrid cars have both an Internal Combustion Engine (ICE) and an onboard electric power system made up of a generator, electric motor and battery storage.
  • ICE runs on gasoline or diesel or other alternative fuels like natural gas, propane, biodiesel and ethanol.
  • When electric power is put into a motor/generator, it acts as an electric motor delivering mechanical shaft power; when shaft power is put in, it acts as a generator delivering electricity.
  • For slow-speed driving, the car can run entirely on its batteries.
  • When cruising along, the ICE and motor/generator provide power, with perhaps some shaft power driving a small additional generator to recharge the batteries.
  • When brakes are applied, the electric motor becomes an electric generator that recharges the batteries.

​Therefore, assertion (A) is true but reason (R) is false.

Question 42

The Biological Diversity Act 2002 provides the following provisions :

(A) Regulation of access to biodiversity

(B) Does not allow local communities for using their biological resources

(C) Establishment of state biodiversity boards

(D) Equitable sharing of benefits of biological resources with local communities

Choose the correct answer from the options given below :

  1. AOption (a)

    (A), (B) and (C) Only

  2. BOption (b)

    (B), (C) and (D) Only

  3. COption (c)

    (A), (B) and (D) Only

  4. DOption (d)

    (A), (C) and (D) Only

View answer & explanation

Correct answer

Answer: (d)D.

(A), (C) and (D) Only

Why this answer is correct

The correct answer is (A), (C) and (D) Only.

Key Points

  • Biological Diversity Act which was passed in 2002, aims to conserve biological resources, manage their sustainable use, and enable fair and equitable sharing of benefits arising from the use and knowledge of biological resources with local communities.

The act called for a three-tiered structure to govern:

  • The National Biodiversity Authority
  • The State Biodiversity Boards
  • The Biodiversity Management Committees​

The National Biodiversity Authority (NBA) was established in 2003 by the Central Government to implement India’s Biological Diversity Act (2002).

  • It is a Statutory body.
  • The NBA has its Headquarters in Chennai, Tamil Nadu, India.
  • The State Biodiversity Boards are established by the State Governments in accordance with Section 22 of the Act.

Therefore, only statements A, C, and D are correct.

Question 43

The Government of India has introduced a scheme of eco-labelling of consumer products as “Ecomark”.

(A) This scheme was introduced in 1986

(B) This scheme was introduced in 1991

(C) ‘ISI was given the mark for this scheme

(D) An ‘earthen pitcher’ was made the symbol for this scheme

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (D) Only

  2. BOption (b)

    (A) and (C) Only

  3. COption (c)

    (B) and (D) Only

  4. DOption (d)

    (C) and (D) Only  

View answer & explanation

Correct answer

Answer: (c)C.

(B) and (D) Only

Why this answer is correct

The correct is (B) and (D) Only.

Key Points

  • Eco mark is a certification mark issued by the Bureau of Indian Standards (BIS) for products that are environmentally safe and meet the BIS's standards.
  • It was issued by the Ministry of Environment and Forest in 1991.
  • In India, the Ecomark symbol is an earthen pot or 'Matka'.

Additional Information

  • In India, the Ecomark label is given to environmentally friendly products. It is a certification mark that is available for products in approximately 16 categories.
  • Soaps and detergents, paints, food items, lubricating oils, packing/packaging materials, architectural paints and powder coatings, batteries, electrical and electronic goods, food additives, wood substitutes, cosmetics, aerosols and propellants, plastic products, textiles, fire-extinguisher, leather, and coir & coir products are among the product categories covered by the scheme.

Therefore, only statements B and D are correct.

Question 44

Consider the following statements :

(A) Arithmetic Mean is greater than the geometric mean for two non-equal positive real numbers

(B) Correlation coefficient is a confirmatory tool to establish a causal relationship between two variables

(C) Regression coefficient is independent of change of origin

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) and (B) Only

  2. BOption (b)

    (B) and (C) Only  

  3. COption (c)

    (A) and (C) Only

  4. DOption (d)

    (A), (B) and (C) 

View answer & explanation

Correct answer

Answer: (c)C.

(A) and (C) Only

Why this answer is correct

The correct answer is (A) and (C) Only.

Key Points

Measure of Central Tendency:

In any distribution, when the original item differs in size the value of arithmetic mean (AM), geometric mean (GM), and harmonic mean (HM) would also differ and will be in the following order:

  • ​AM ≥ GM ≥​ HM
  • The equality sign holds only if all the numbers x1, x2.....xn are identical.

Regression:

  • The coefficient of X in the line of regression of Y on X is called the regression coefficient of Y on X. It represents a change in the value of a dependent variable (Y) corresponding to a unit change in the value of an independent variable (X).
  • The regression coefficient is independent of the change of the origin. This implies that there will be no effect on the regression coefficients if any constant is subtracted from the value of x and y.

Correlation:

  • Correlation is a statistical measure that describes the size and direction of a relationship between two or more variables.
  • A correlation between variables, however, does not automatically mean that the change in one variable is the cause of a change in the values of other variables. Therefore, a correlation between two variables does not imply causation.

Therefore, only statements A and C are correct.

Direction: Read the passage given below. Based on it answer the question.

According to a study published recently in the journal Nature Climate Change, while black carbon has a large effect on snow darkening and resultant melting of snow, dust particles transported from as far as Saudi Arabia that get deposited in the Western Himalayan Region (WHR) have a large role to play in melting of snow, particularly at higher elevations. Dust transported as elevated, aerosol layers get deposited at 1 – 5 km elevations, black carbon emission is mostly a surface phenomenon and influences melting of snow from surface to about 3 km elevation. 

Based on remote sensing data of spatial distribution of dust aerosol concentration over the Indian subcontinent and dust‐ induced snow albedo reduction over Himalayas during the period 2011‐2016 and simulations, it has been inferred that the relative impacts of dust and black carbon vary with surface elevation of snow pack. This is in addition to snow‐melt caused by warming due to climate change. Earlier studies have shown that the magnitude of snow mass decrease is about 1mm per year at 1 km elevation, about 5 mm per year at 4.5 km elevation and about 3 mm per year at 6 km elevation. Though black carbon has a larger snow albedo darkening effect than dust due to a larger mass absorption efficiency, the study found that radiative effects of dust deposited on snow are comparable to black carbon in the WHR at higher elevations. This is mainly because the deposition of the dust by mass is 100‐1000 times more than black carbon. As the elevation increases, the influence of dust comes greater than black carbon and this coincides with maximum intensity of snow melt reduction seen at 3‐5 km elevation. Between these two black carbon mainly contributes to snow melt at lower elevation while dust is the major contributor for snow melt at higher elevation.  

Westerlies transport dust particles as elevated aerosol layers at maximum intensities mostly during the pre‐monsoon period and this gets deposited at higher elevations in the WHR.  

Due to global warming, snow cover at lower elevations in the Himalayas will occur less frequently or totally disappear compared with snow cover at higher elevations. The annual contribution of dust to snow melt will therefore likely increase in future as the black carbon effect at lower elevation weakness with dwindling snow pack.

Question 45

Based on Passage: View directions & passage

Snow melt in Himalayas is on account of :

(A) Dust deposition

(B) Black carbon deposition

(C) Increase in albedo

(D) Climate change

  1. AOption (a)

    (A) and (B) only

  2. BOption (b)

    (A), (B), (C) and (D)

  3. COption (c)

    (A), (B) and (C) only

  4. DOption (d)

    (A), (B) and (D) only  

View answer & explanation

Correct answer

Answer: (d)D.

(A), (B) and (D) only  

Why this answer is correct

The correct answer is (A), (B) and (D) only. 

Key Points

Consider the first sentence of the first paragraph:

  • ​According to a study published recently in the journal Nature Climate Change, while black carbon has a significant effect on snow darkening and the resultant melting of snow, dust particles transported from as far as Saudi Arabia that gets deposited in the Western Himalayan Region (WHR) have a significant role to play in melting of snow, particularly at higher elevations.

​Consider the second sentence of the second paragraph:

  • ​This is in addition to snow‐melt caused by warming due to climate change.
  • ​Thus it can be concluded that the reasons for the melting of snow in the Himalayas are black carbon deposition, deposition of dust particles, and climate change.

​Therefore, the correct answer is (A), (B), and (D) only. ​

Question 46

Based on Passage: View directions & passage

Maximum rate of snow mass reduction per annum is at elevation of :

  1. AOption (a)

    1 km

  2. BOption (b)

    3‐5 km

  3. COption (c)

    6 km

  4. DOption (d)

     2 km

View answer & explanation

Correct answer

Answer: (b)B.

3‐5 km

Why this answer is correct

The correct answer is 3‐5 km.

Key Points

Take a look at the second last sentence of the second paragraph:

  • ​As the elevation increases, the influence of dust comes greater than black carbon and this coincides with the maximum intensity of snow melt reduction seen at 3‐5 km elevation.

So maximum snow melting rate occurs at an elevation of 3-5 km.

Therefore, the correct answer is 3‐5 km.

Additional Information

  • Dust particles and black carbon deposition have a cumulative effect on the melting of snow i.e., both increase the rate of snow melting in the Himalayas.

Question 47

Based on Passage: View directions & passage

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :

Assertion A : Reduction in snow albedo is more in case of black carbon than dust

Reason R : Black carbon has higher mass absorption efficiency compared to dust.

In the light of the above statements, choose the correct answer from the options given below :

  1. AOption (a)

    Both A and R are true and R is the correct explanation of A

  2. BOption (b)

    Both A and R are true but R is NOT the correct explanation of A

  3. COption (c)

    A is true but R is false

  4. DOption (d)

    A is false but R is true

View answer & explanation

Correct answer

Answer: (a)A.

Both A and R are true and R is the correct explanation of A

Why this answer is correct

The correct answer is Both A and R are true and R is the correct explanation of A.

Key Points

  • Examine the third sentence of the second paragraph:
  • Though black carbon has a larger snow albedo darkening effect than dust due to a larger mass absorption efficiency, the study found that the radiative effects of dust deposited on snow are comparable to black carbon in the WHR at higher elevations.
  • ​This proves our assertion is true which states that the reduction in snow albedo is more due to black carbon than dust.
  • Examine the fourth sentence of the second paragraph:
  • ​This is mainly because the deposition of the dust by mass is 100‐1000 times more than black carbon.
  • ​This means that although more dust gets deposited in the Himalayas than black carbon, due to its higher mass absorption efficiency i.e., absorbing more solar radiation per unit mass than that of dust particles, black carbon causes more melting of snow than dust.
  • This proves our reason is true and also proves that the reason is the correct explanation of the assertion.

Therefore, the correct answer is Both A and R are true and R is the correct explanation of A.

Question 48

Based on Passage: View directions & passage

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :

Assertion A : Radiative effects of dust deposited on snow are comparable to black carbon in the WHR at higher elevations.

Reason R : Both black carbon and dust darken the snow cover.

In the light of the above statements, choose the correct answer from the options given below :

  1. AOption (a)

    Both A and R are true and R is the correct explanation of A

  2. BOption (b)

    Both A and R are true but R is NOT the correct explanation of A

  3. COption (c)

    A is true but R is false

  4. DOption (d)

    A is false but R is true

View answer & explanation

Correct answer

Answer: (b)B.

Both A and R are true but R is NOT the correct explanation of A

Why this answer is correct

The correct answer is Both A and R are true but R is NOT the correct explanation of A.

Key Points

  • Analyze the third sentence of the second paragraph:
  • Though black carbon has a larger snow albedo darkening effect than dust due to a larger mass absorption efficiency, the study found that the radiative effects of dust deposited on snow are comparable to black carbon in the WHR at higher elevations.
  • This proves our assertion is true which states that the radiative effects of dust deposited on snow are comparable to black carbon in the WHR at higher elevations.
  • By analysing the same sentence again, we can say that both black carbon and dust have a darkening effect on the snow which proves our reason is true.
  • But the reason does not correctly explain the assertion as their radiative effect is comparable only because more amount of dust gets deposited in the west Himalayan region i.e., almost 100-1000 times the mass of black carbon and not because black carbon and dust both darken the snow causing its melting.

Therefore, the correct answer is Both A and R are true but R is NOT the correct explanation of A.

Question 49

Based on Passage: View directions & passage

The passage clearly brings out the finding that :  

  1. AOption (a)

    Black carbon causes global warming

  2. BOption (b)

    Dust causes global warming

  3. COption (c)

    Dust modifies the radiative properties of snow cover to the same extent as black carbon

  4. DOption (d)

    Black carbon and dust are competing agents in melting of snow packs

View answer & explanation

Correct answer

Answer: (d)D.

Black carbon and dust are competing agents in melting of snow packs

Why this answer is correct

The correct answer is Black carbon and dust are competing agents in the melting of snow packs.

Key Points

  • Take a look at the first sentence of the first paragraph:
  • According to a study published recently in the journal Nature Climate Change, while black carbon has a large effect on snow darkening and the resultant melting of snow, dust particles transported from as far as Saudi Arabia that gets deposited in the Western Himalayan Region (WHR) have a large role to play in melting of snow, particularly at higher elevations.
  • ​Take a look at the fourth sentence of the second paragraph:
  • ​Though black carbon has a larger snow albedo darkening effect than dust due to a larger mass absorption efficiency, the study found that the radiative effects of dust deposited on snow are comparable to black carbon in the WHR at higher elevations.
  • ​​Take a look at the fifth sentence of the second paragraph:
  • ​This is mainly because the deposition of the dust by mass is 100‐1000 times more than black carbon.
  • ​Take a look at the last sentence of the second paragraph:
  • ​Between these two black carbon mainly contributes to snow melt at lower elevations while dust is the major contributor to snow melt at higher elevations.
  • ​By analyzing these sentences and the passage as a whole, we can see that it is both black carbon and dust which is transported from far-off places like Saudi Arabia and get deposited in the Western Himalayan Region and act together in the melting of snow. While the black carbon melts the snow at the lower elevations usually, the dust gets deposited at higher altitudes and causes the melting of snow there.
  • It can also be concluded that although more amount of dust gets deposited than black carbon in the Himalayas, due to the larger mass absorption efficiency of black carbon, it causes more melting than the dust.

​Therefore, the finding of the passage is that black carbon and dust are competing agents in the melting of snow packs.

Question 50

The water is classified as ‘‘aggressive’’ if : 

  1. AOption (a)

    The Ca2+ concentration is much less than CaCO3 saturation

  2. BOption (b)

    The Ca2+ concentration is equal to the CaCO3 saturation

  3. COption (c)

    The Ca2+ concentration is more than the CaCO3 saturation

  4. DOption (d)

    The Ca2+ concentration is double of the CaCO3 saturation

View answer & explanation

Correct answer

Answer: (a)A.

The Ca2+ concentration is much less than CaCO3 saturation

Why this answer is correct

Explanation:

Aggressive/Scaling Water:

  • A particular feature of calcium bicarbonate only exists in the dissolved form. When it precipitates, it does so in the calcium carbonate form, one that is only slightly soluble and it can only remain in the water when it is balanced out by available CO2.

Ca [HCO3]2

CaCO3 + CO2 + H2O

  • As the CO2 content has an influence on pH, there will also be a pH value for the equilibrium between this water and the calcium carbonate (Langelier pH or saturation pH written as pHs).

Therefore, we can "classify" water on this basis as follows,

  • If pH > pHs, water will tend to deposit CaCO3 which is then termed scale forming.
  • If pH < pHs, water will tend to dissolve CaCO3 and is then termed aggressive.

Hence, option 1 is the correct answer.

Question 51

The minimum time needed for the formation of one inch of the top soil is :

  1. AOption (a)

    Less than 10 years

  2. BOption (b)

    10‐50 years

  3. COption (c)

    100‐400 years

  4. DOption (d)

    1,000‐2,000 years

View answer & explanation

Correct answer

Answer: (c)C.

100‐400 years

Why this answer is correct

Explanation:

Topsoil:

  • It is the uppermost layer of the soil. It has a high level of organic matter and microbial activity.
  • The depth of the topsoil usually extends up to 5-10 inches. This occupies O or A Horizon.
  • Formation of topsoil typically takes around more than 100-500 years which is an incredibly slow process. Hence option 4 is the correct answer.

Soil Formation:

  • It is a complex natural process and formed through different series of development.
  • This weathering of rock into the soil is called the process of pedogenesis.
  • There are five factors that influence the formation of soil: parent material, climate, biota, topography, and time.
  • The various soil-forming processes such as soil enrichment (organic and inorganic matter added to the soil), removal (leaching), translocation (upward and downward movement within the soil), and transformation (one form of compound to another) are involved.

Question 52

Which zone of a pond/lake ecosystem does not receive light?

  1. AOption (a)

    Littoral

  2. BOption (b)

    Limnetic

  3. COption (c)

    Profundal

  4. DOption (d)

    Epilimnion

View answer & explanation

Correct answer

Answer: (c)C.

Profundal

Why this answer is correct

Concept-

Different factors such as distance from the shore, penetration of light, depth of water, plant and animal species, etc. determine the following zones found in the pond ecosystem:

  • Littoral zone: It is the zone closer to the shore. It contains shallow water and allows easy penetration of light. Rooted plant species occupy it. Animal species include reeds, crawfish, snails, insects, etc.
  • Limnetic zone: The limnetic zone refers to the open water of the pond with an effective penetration of light. This zone is dominated by phytoplankton. Animal species mainly include small fishes and insects.
  • Profundal zone: The region of a pond below the limnetic zone is called a profound zone with no effective light penetration. Some amphibians and small turtles occupy it.
  • Benthic zone: The bottom zone of a pond is benthic and is occupied by a community of decomposers. The decomposers are called benthos.

Question 53

Which of the following mixture of gases can be found in biogas?

  1. AOption (a)

    CO, CH4, NO, H2S, H2O vapor

  2. BOption (b)

    CO2, CH4, NH3, H2S, H2O vapor

  3. COption (c)

    CO2, CH4, N2O, NH3, H2O vapor

  4. DOption (d)

    CO2, CH4, NO, NH3

View answer & explanation

Correct answer

Answer: (b)B.

CO2, CH4, NH3, H2S, H2O vapor

Why this answer is correct

Concept:

Biogas:

  • The byproducts of anaerobic digestion of organic materials are commonly referred to as Biogas because of the biological nature of the gas production.
  • Biogas technology refers to the production of a combustible gas (called biogas) and a value-added fertilizer (called slurry or sludge) by the anaerobic fermentation of organic material under certain controlled conditions.
  • Biogas is produced by microbial activities and can be used only at the place where it is produced.
  • The main constituents of biogas are - 
  • Methane (CH4) (55-65 %)
  • carbon dioxide (CO2) ( 30-45 %)
  • Traces of Hydrogen Sulfide (H2S)
  • Gases like methane, hydrogen, and carbon monoxide may all be burned or oxidized.
  • The use of biogas as fuel is made possible by this energy release.
  • It is used as a fuel, for instance, in cooking for heating purposes, and employed in a gas engine.

The correct answer is option 2.

Question 54

India became a party to CITES in the year :

  1. AOption (a)

    1972

  2. BOption (b)

    1976

  3. COption (c)

    1974

  4. DOption (d)

    1986

View answer & explanation

Correct answer

Answer: (b)B.

1976

Why this answer is correct

The correct answer is 1976.

Key Points

  • India became a party to CITES in the year 1976. 
  • ​A resolution was adopted in 1963 at the IUCN meeting and CITES came into force on 1 July 1975.
  • CITES (Convention on International Trade in Endangered Species of Wild Fauna and Flora)
  • It is an international agreement between governments.
  • Its aim is to ensure that international trade in specimens of wild animals and plants does not threaten the survival of the species.
  • It provides the framework for the countries and the parties have to prepare domestic laws for implementation.
  • It prepares a list of region wise species to be protected.
  • CITES works by subjecting international trade in specimens of selected species to certain controls.
  • All import, export, re-export and introduction from the sea of species covered by the Convention has to be authorized through a licensing system.
  • Headquarters: Geneva, Switzerland.
  • Members: 184
  • Secretary-General: Ivonne Higuero

Question 55

Time clocked by an athlete in a 100 m race is an example of which type of variable?

  1. AOption (a)

    discrete

  2. BOption (b)

    continuous

  3. COption (c)

     nominal

  4. DOption (d)

    oridnal

View answer & explanation

Correct answer

Answer: (a)A.

discrete

Why this answer is correct

Concept Used:

A continuous variable is a variable whose value is obtained by measuring, i.e., one which can take on an uncountable set of values.

In contrast, a variable is a discrete variable if and only if there exists a one-to-one correspondence between this variable and {\displaystyle \mathbb {N} }

\mathbb {N}

,

Since time clocked is only measured at .01 seconds interval.

Hence, the time clocked by an athlete in a 100 m race is an example of a discrete variable.

Question 56

For a regression model, the following information is given :

Unexplained sum of squares = 25

Explained sum of squares = 75

The value of coefficient of determination for this model is

  1. AOption (a)

    0.33

  2. BOption (b)

    3

  3. COption (c)

    0.67

  4. DOption (d)

    0.75

View answer & explanation

Correct answer

Answer: (d)D.

0.75

Why this answer is correct

Explanation

The coefficient of determination R2 is the ratio of the explained variation to the total variation.

Given values:

Explained sum of squares or Explained variation = 75

Unexplained sum of squares or Unexplained variation =25

The total variation = 75 + 25 = 100.

Therefore, the value of R2 = 75/100 = 0.75.

Hence, option 4 is the correct answer.

Question 57

Consider the following statements regarding incineration of urban waste :

(A) Incineration of urban waste is a clean process

(B) Under ideal condition, incineration may reduce the volume of waste by 75–95%

(C) Ash is generated as by‐product of incineration

(D) This process of pollutant abatement is inexpensive

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (B) Only

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (B), (C) and (D) Only

  4. DOption (d)

    (A), (B) and (D) Only

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (C) Only

Why this answer is correct

Explanation:

Incineration:

It is a method of thermal treatment of waste materials. It converts waste materials into ash, flue gas, and heat. Hence, C is correct.

Advantages

Reduced Quantity of Waste: It decreases the quantity of waste by as much as 95% and reduces the solid quantity of the original waste by as much as 80-85%. Hence, B is correct.

  • Efficient Waste Management: Incineration can burn up to 90% of the total waste generated and sometimes even more.
  • Production of Heat and Power: Incineration plants generate energy from waste that can be used to generate electricity or heat.
  • Reduced Reliance on Transportation: waste does not have to be driven long distances for dumping.
  • Better control over Odour and Noise: waste gets burnt inside a facility where the byproducts of the incineration process can be controlled.
  • Prevents the Production of Methane Gas.
  • Eliminates Harmful Germs and Chemicals.

Disadvantages

  • It requires high operating costs and is expensive. Hence, D is not correct.
  • It leads to significant levels of pollution. The smoke produced during the burning process may include acid, dioxin, particulates, heavy metals, etc. These gases are poisonous to the environment and cause health effects. Therefore, it is not considered a clean process. Hence, A is not correct.
  • Incineration does not encourage recycling processes.

Therefore, option 2 is the correct answer

Question 58

Match List I with List II :

List – I Water Quality ParameterList – II Chemical/Reagent used in Analysis
(A)Iron(i)Barium Chloride
(B)Dissolved Oxygen(ii)Phenanthroline
(C)Calcium(iii)Murexide
(D)Sulphate(iv)Manganous Sulphate

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (i), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (iv)

  2. BOption (b)

    (A) ‐ (ii), (B) ‐ (iv), (C) ‐ (iii), (D) ‐ (i)

  3. COption (c)

    (A) ‐ (ii), (B) ‐ (iii), (C) ‐ (i), (D) ‐ (iv)

  4. DOption (d)

    (A) ‐ (iv), (B) ‐ (ii), (C) ‐ (iii), (D) ‐ (i)

View answer & explanation

Correct answer

Answer: (b)B.

(A) ‐ (ii), (B) ‐ (iv), (C) ‐ (iii), (D) ‐ (i)

Why this answer is correct

Concept:

Iron:

  • A commonly used method for the determination of trace amounts of iron involves the complexation of Fe2+ with 1,10-phenanthroline to produce an intensely red-orange colored complex and further test procedures are carried out to measure the amount of Iron.

Dissolved Oxygen-

  • It is measured using Winkler's method.
  • Manganous Sulphate is used as a reagent here.

Calcium-

  • Murexide, also known as ammonium purpurate or MX, is the ammonium salt of purpuric acid.
  • This metallochromic indicator undergoes a color change in the presence of calcium, so it is used to measure Calcium in water.

Sulphate-

  • Test of sulphate is based on the reaction of soluble sulphate with barium chloride in presence of dilute hydrochloric acid to form barium sulphate which appears as solid particles (turbidity) in the solution.

Question 59

Identify the correct sequence of the following data products of Indian satellites according to their spatial resolutions :

(A) Cartosat ‐ 2 PAN

(B) Oceansat ‐ 2 OCM

(C) Resourcesat ‐ 2 AWIFS

(D) IRS ‐ ID LISS ‐ 3

Choose the correct answer from the options given below :

  1. AOption (a)

    C > A > B > D

  2. BOption (b)

    B > D > C > A

  3. COption (c)

    C > B > D > A

  4. DOption (d)

    B > C > D > A

View answer & explanation

Correct answer

Answer: (d)D.

B > C > D > A

Why this answer is correct

The correct answer is B > C > D > A.

Key Points

  • Spatial resolution refers to the distance represented by a pixel in an image.
SatellitesSpatial Resolution (in meters)Applications
Oceansat - 2 OCM360mIdentification of potential fishing zones, sea state forecasting, coastal zone studies and inputs for weather forecasting and climatic studies.
Resourcesat -2 AWIFS56mUrban planning, mapping, vegetation discrimination, land mapping, natural resources management and disaster management.
IRS - ID LISS - 323.5mRemotely sensed data for applications in the areas of agriculture, hydrology, geology, drought and flood monitoring, marine studies, snow studies and land use.
Cartosat - 2 PAN0.45 - 0.8mUseful for cartographic applications, urban and rural infrastructure development and management, and various other Land Information System (LIS) and Geographical Information System (GIS) applications.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Therefore, the decreasing order of spatial resolution of some Indian satellites is Oceansat - 2 OCM > Resourcesat -2 AWIFS > IRS - ID LISS - 3 > Cartosat - 2 PAN.

Question 60

Identify the correct sequence of natural surfaces according to their albedo:

(A) Ocean

(B) Fresh snow

(C) Grass

(D) Desert Sand

Choose the correct answer from the options given below :

  1. AOption (a)

    A < B < C < D

  2. BOption (b)

    C < A < D < B

  3. COption (c)

    C < A < B < D

  4. DOption (d)

    A < C < D < B 

View answer & explanation

Correct answer

Answer: (d)D.

A < C < D < B 

Why this answer is correct

The correct answer is A < C < D < B.

Key Points

  • Albedo is the measurement of the reflectivity of sunlight by any object.
  • Objects reflecting more sunlight have high albedo while those reflecting less have a low albedo.
SurfacesAlbedo (%)
Fresh snow80
Desert sand40
Grass25
Ocean6

 

 

 

 

Therefore, The increasing order of albedo of natural surfaces is Oceans < Grass < Desert < sand < Fresh snow.

Additional Information

  • Objects with higher albedo are cooler in temperature and while those with lower albedo are hotter in temperature.
  • The albedo of a given region depends on soil composition and moisture, vegetation types, level of urbanization, etc.
  • Earth’s overall albedo is measured to be 30%.

Question 61

Identify the correct sequence in ascending order of the grade of metamorphism :

(A) Gneiss

(B) Slate

(C) Schist

(D) Shale

Choose the correct answer from the options given below :

  1. AOption (a)

    (A), (B), (C), (D)

  2. BOption (b)

    (D), (C), (B), (A)

  3. COption (c)

    (B), (D), (C), (A)

  4. DOption (d)

    (D), (B), (C), (A)

View answer & explanation

Correct answer

Answer: (d)D.

(D), (B), (C), (A)

Why this answer is correct

The correct answer is: (D), (B), (C), (A)

Key Points

Metamorphic rocks:

  • Derived from the Greek word, ‘meta’- change and ‘morphos’ meaning form.
  • These rocks are formed by changing the mineral in the presence of high temperature and pressure.
  • Both igneous and sedimentary rocks can become metamorphic rocks.

Metamorphic grade:

  • It refers to the range of metamorphic changes the rock undergoes.
  • It progresses from little metamorphic change (low grade) to significant metamorphic change (high grade).
  • Low-grade metamorphism begins at temperatures and pressures just above sedimentary rock conditions. In the above option, ns Shale is the sedimentary rock and starts with low-grade and gradually high-grade metamorphism.

Therefore, the ascending order sequence of metamorphism grade would be **shale **→ slate → phyllite → schist → gneiss.

Hence, option 4 is the correct answer.

Additional Information

Metamorphic  RockOrigin Rock
MarbleLimestone (Carbonate Sedimentary rock)
SlateShale (Clastic Sedimentary rock)
GneissGranite (Intrusive igneous rock)
QuartziteSandstone (Clastic Sedimentary rock)

Question 62

According to the Solid Waste Management Rules 2016, arrange the following metals in decreasing order of their permissible concentration (mg/kg) in the prepared compost:

(A) Chromium

(B) Lead

(C) Cadmium

(D) Copper

Choose the correct answer from the options given below :

  1. AOption (a)

    A, B, C, D

  2. BOption (b)

    B, D, A, C

  3. COption (c)

    D, B, A, C

  4. DOption (d)

    D, A, B, C

View answer & explanation

Correct answer

Answer: (c)C.

D, B, A, C

Why this answer is correct

Concept-

  • The compost which is to be used as fertiliser for food crops should abide by the more stringent FCO, 2009, while compost used as a soil conditioner and for other purposes should at least meet the requirements of SWM Rules, 2016.
  • Phosphatic fertiliser is in short supply in the country. The use of phosphate-rich organic manure (PROM) can reduce the use of phosphatic fertilisers to some extent; PROM is formed by the mixing of rock phosphate with MSW-derived compost.

The FCO, 2013 specified quality standards for PROM, while retaining the standards specified in FCO, 2009 for organic compost.

ParametersOrganic Compost
Cadmium (mg/kg)5.00
Chromium (mg/kg)50.00
Lead (mg/kg)100.00
Copper (mg/kg)300.00
C/N ratio< 20
Nickel (mg/kg50

So the decreasing order is Copper > Lead > Chromium > Cadmium.

Question 63

Which article in Indian constitution recognizes the “Fundamental Duties” for the protection of environment and wildlife? 

  1. AOption (a)

    Article 42 

  2. BOption (b)

    Article 21 

  3. COption (c)

    Article 51A(g) 

  4. DOption (d)

    Article 484

View answer & explanation

Correct answer

Answer: (c)C.

Article 51A(g) 

Why this answer is correct

The correct answer is Article 51A(g).

Key Points

  • Article 51A(g) of the Indian Constitution
  • **“It shall be duty of every citizen of India to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures.”**​
  • The Directive principles under the Indian constitution directed towards ideals of building welfare state. Healthy environment is also one of the elements of welfare state.
  • Article 42 ​
  • The State shall make provision for securing just and humane conditions of work and for maternity relief.
  • Article 21
  • No person shall be deprived of his life or personal liberty except according to procedure established by law.

Important Points

  • Articles 36-51 under Part-IV of the Indian Constitution deals with Directive Principles of State Policy (DPSP).
  • They are borrowed from the Constitution of Ireland, which had copied it from the Spanish Constitution.
  • DPSP are non-binding in nature which means they are not enforceable by the courts for their violation.
  • These principles emphasize that the State shall try to promote the welfare of the people by providing them basic facilities like shelter, food and clothing.
  • It denotes the ideals that the State should keep in mind while formulating policies and enacting laws.

Question 64

Identify the correct sequence of the given soil texture classes in decreasing order of stability of wet clods :

(A) Loam

(B) Clay

(C) Sandy Loam

(D) Silt Loam

(E) Clay Loam

Choose the correct answer from the options given below :

  1. AOption (a)

    A, B, D, E, C

  2. BOption (b)

    B, E, D, A, C

  3. COption (c)

    B, D, A, C, E

  4. DOption (d)

    B, E, A, C, D

View answer & explanation

Correct answer

Answer: (b)B.

B, E, D, A, C

Why this answer is correct

The correct answer is B, E, D, A, and C.

Key Points

  • The texture of the soil is one of the primary deciding features of the porosity of soils. The amount of organic matter present in the soil is directly proportional to its water-holding capacity.
  • Clay has the highest proportion of organic matter. It has the highest porosity and lowest permeability. Therefore, the wet clods formed are highly stable.
  • Primary soil particles are held together by cohesive forces, and secondary particles and organic matter form soil aggregates. Aggregate stability is the measure of the ability of soil aggregates to bind together and resist degradation or breaking apart.
  • Organic matter increases aggregate stability. Since clay has the highest amount of organic matter, therefore, it has the highest aggregate stability.
  • Since sandy loam has the lowest percentage of organic matter, therefore, its aggregate stability is lowest i.e., the stability of micro-aggregates is lowest.

The amount of organic matter and porosity vary in the order of:

  • Clay > Clay loam > Silt loam > Loam > Sandy loam

Therefore, the stability of wet clods of given texture classes are in the order of:

  • Clay > Clay loam > Silt loam > Loam > Sandy loam

Question 65

Match List I with List II :

List – IList – II
(A)Animal Anatomical Waste(i)White
(B)Waste sharp(ii)Blue
(C)Discarded catheters(iii)Yellow
(D)Disposable glasswares(iv)Red

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (ii), (B) ‐ (iii), (C) ‐ (iv), (D) ‐ (i)

  2. BOption (b)

    (A) ‐ (iii), (B) ‐ (ii), (C) ‐ (iv), (D) ‐ (i)

  3. COption (c)

    (A) ‐ (iv), (B) ‐ (i), (C) ‐ (ii), (D) ‐ (iii)

  4. DOption (d)

    (A) ‐ (iii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (ii)

View answer & explanation

Correct answer

Answer: (d)D.

(A) ‐ (iii), (B) ‐ (i), (C) ‐ (iv), (D) ‐ (ii)

Why this answer is correct

Explanation:

Bio-medical waste means any waste, which is generated during the diagnosis, treatment, or immunization of human beings or animals or research activities pertaining thereto or in the production or testing of biological or health camps.

Hence, Option 4 is the correct answer.

Question 66

The pioneer community in human evolution is :

  1. AOption (a)

    Pastoralists (Nomadic) 

  2. BOption (b)

    Agrosociety 

  3. COption (c)

    Hunting and Gathering Society 

  4. DOption (d)

    Industrial Society 

View answer & explanation

Correct answer

Answer: (c)C.

Hunting and Gathering Society 

Why this answer is correct

Explanation:

Human Evolution:

  • It is the process by which human beings occupied the earth through various stages of living.
  • Homo sapiens, first evolved 315,000 years ago.
  • During the Miocene Epoch which is 23 million to 5.3 million years ago, we were the probable beginners of life and developed into modern humans only recently in the Pleistocene Epoch is 2.6 million to 11,700 years ago.
  • Pastoralist societies are people who lived in the countryside during the period of 8500-6500 B.C.
  • Humans started growing cultivation and evolved as an agrarian society ty during the Neolithic period around 10,000 -8,000 years ago.
  • For survival during the earlier times of 12,000 years ago, humans were occupied with hunting and gathering which is a subsistence lifestyle for food.
  • Industrial society developed as a modernized era of human society during recent times where machines were playing the important role in the production process.

Hence, option 3 is the correct answer.

Question 67

Name the person, who launched the ‘‘Green Belt Movement’’ in Kenya, and got the Nobel Peace Prize : 

  1. AOption (a)

    Wangari Maathai 

  2. BOption (b)

    Gehard Ertl 

  3. COption (c)

    Koichi Tanaka 

  4. DOption (d)

    Ryojii Noyori 

View answer & explanation

Correct answer

Answer: (a)A.

Wangari Maathai 

Why this answer is correct

Explanation

Green belt movement:

  • It was founded in the year 1977 under the auspices of the National Council of Women of Kenya (NCWK).
  • It is an environmental organization that empowers communities, particularly women, to conserve the environment and improve livelihoods.
  • Wangari Maathai was the founder of this movement and movement and he got the Nobel Peace prize in the year 2004. She authored four books namely The Green Belt Movement, Unbowed: A Memoir, The Challenge for Africa; and Replenishing the Earth.
  • The well-known work of this organization is the planting of trees and now addressing the problem of global deforestation.
  • This movement aims at organizing women in rural Kenya to plant trees, combat deforestation, restore their main sources of fuel for cooking, generate income and stop soil erosion.

Hence, option 1 is the correct answer.

Question 68

At locations characterized by latitude ϕ = 30°. pressure is found to vary by 1.0 mb over a distance of 100 km. If the density of air mass is ~1.25 kg/m3, the geostrophic wind velocity is

  1. AOption (a)

    ~11 m/s

  2. BOption (b)

    ~15 m/s

  3. COption (c)

    ~26 m/s

  4. DOption (d)

    ~7 m/s

View answer & explanation

Correct answer

Answer: (a)A.

~11 m/s

Why this answer is correct

Explanation:

Given values are,

  • Latitude- 300 (Sin 300 = 0.5)
  • Pressure- 1.0 mb (1 mb x 100 = 100 Pa)
  • Density of Air = 1.25 kg/m3
  • Distance = 100 km (100 km x 1000 = 100000 m)

Also,

​​The angular velocity of Earth= 7.3 x 10-5 rad/s

Geostrophic wind velocity (Vg)  = Pressure(Pa)(2 ×Angular velocity of Earth)× Density of Air ×Latitude × Distance(m)\frac{Pressure(Pa )}{ (2 \ \times Angular\ velocity\ of \ Earth) \times\ Density\ of \ Air \ \times Latitude \ \times \ Distance(m)}

Vg = 100(2×7.3×10−5)× 1.25 ×0.5 × 100000\frac{100}{ (2 \times 7.3 \times 10^{-5}) \times\ 1.25\ \times 0.5\ \times \ 100000}

Vg = 10.95 = ~11 m

Hence, option 1 is the correct answer.

Question 69

The internal relationship between α, β and γ diversity can be presented as :

  1. AOption (a)

    β=γα\rm \beta=\frac{\gamma}{\alpha}

  2. BOption (b)

    β = γ × α

  3. COption (c)

    α = γ × β

  4. DOption (d)

    γ = β + α

View answer & explanation

Correct answer

Answer: (a)A.

β=γα\rm \beta=\frac{\gamma}{\alpha}

Why this answer is correct

Explanation

Alpha Diversity (α)

  • It refers to the species diversity within a community.
  • This generally describes the size or a defined unit of an ecosystem.
  • For example Species present in a patch of a grassland ecosystem.

Beta Diversity (β)

  • It refers to the species diversity between two communities.
  • It is the ratio between regional and local species diversity.
  • A geographical area will act like a barrier between two like a river or mountain.
  • For example Diversity between woodland and hedgerow habitats.

Gamma Diversity (ϒ)

  • It refers to the species diversity of two or more ecosystems.
  • It is a large-scale study of a biome or landscape.
  • For example Diversity in the entire mountain slope

The interrelationship among them is represented as,

β = γα\frac{\gamma }{\alpha} (or) ϒ= β x α,  

Hence, option 1 and 2 is the correct answer.

Question 70

Which of the following is not a molecular method for identification of microbial population form soil or water? 

  1. AOption (a)

    MPN

  2. BOption (b)

    rRNA sequencing

  3. COption (c)

    RT–PCR

  4. DOption (d)

    PCR

View answer & explanation

Correct answer

Answer: (a)A.

MPN

Why this answer is correct

Explanation:

Bacterial Identification:

  • The isolated bacteria are further processed through various methods to identify their number and type.
  • These methods include the Biochemical method, Serological tests, Molecular techniques, etc.
  • Molecular Methods include DNA-DNA hybridization, DNA or RNA bases sequencing, Polymerase chain reaction PCR, real-time PCR, etc.

MPN:

  • The most probable number (MPN) is a technique for estimating the viable numbers of bacteria by multiplying the inoculation broth in a series of dilutions.
  • This test is completed in three steps. These include the presumptive test, the confirmed test, and the completed test.

Hence, option 1 is the correct answer.

Question 71

The rate of increase in temperature with depth in the earth also known as geothermal gradient, is equal to :

  1. AOption (a)

    6.8°C/Km

  2. BOption (b)

    20°C/Km

  3. COption (c)

    25°C/Km

  4. DOption (d)

    37.4°C/Km

View answer & explanation

Correct answer

Answer: (c)C.

25°C/Km

Why this answer is correct

Concept-

  • The geothermal gradient is defined as the increase in temperature with depth in the Earth.
  • In normal continental crust, a typical geothermal gradient within the first 3 to 5 kilometers (2 or 3 miles) of Earth’s surface is about 25°C/km.
  • This gradient, however, is not sustained but decreases to no more than about 16°C/km at a depth of 40 km.

  • The temperature profile of the Earth's layers. Temperature increases with depth
  • The interior of the Earth is extremely hot and reaches temperatures over 5000°C near the core, which is not much colder than the surface of the Sun.

Question 72

Carbon content in a certain type of coal is 0.81. Assume that the combustion efficiency of coal is ~100%, then how much CO2 will be produced on combustion of 1 kg of coal?

  1. AOption (a)

    -3 kg

  2. BOption (b)

    -5.4 kg

  3. COption (c)

    -1.8 kg

  4. DOption (d)

    -4.8 kg

View answer & explanation

Correct answer

Answer: (a)A.

-3 kg

Why this answer is correct

Concept-

The carbon content in coal is 0.81.

Carbon content in coal describes the amount of carbon in 1 kg of coal.

Here in 1 kg of coal, carbon content is 810 gm.

C + O2 → CO2

12 gm of carbon reacts with 32 gm oxygen to form 44 gm of CO2.

So 810 gm of carbon will form  = 44 × 81012\frac{44\ \times \ 810}{12} = 2970 gm of CO2.

So the amount of CO2 produced will be almost -3 kg.

Question 73

The ash content in coal used in stand‐alone thermal power plants in India should not be more than

  1. AOption (a)

    20%

  2. BOption (b)

    27%

  3. COption (c)

    34%

  4. DOption (d)

    40%

View answer & explanation

Correct answer

Answer: (c)C.

34%

Why this answer is correct

Concept-

  • The ministry of environment, forests and climate change (MoEFCC) has taken a decision that will prove disastrous for people living near power plants, according to environmentalists.
  • It has done away with the condition for power plants to use.
  • The Ash content of coal produced in the country is generally 25 to 45 % whereas the average ash content of imported coal varies from 10 to 20 %.
  • Indian Coal has comparatively higher ash content than imported coal due to the drift theory of the formation of coal deposits in India.
  • Coal seams formed due to drift theory contain higher ash as compared to the in-situ theory of formation.
  • As per the Ministry of Environment and Forest & Climate Change (MOEF&CC) stipulation, coal-based thermal power plant (situated beyond 500 km from the pit-head) be supplied with, and shall use, raw or blended or beneficiated coal with ash content not exceeding thirty-four percent, on a quarterly average basis.
  • Coal Companies are supplying coal to their customers maintaining an ash percentage below 34% on a quarterly average basis as per the aforesaid MOEF&CC stipulation.

Question 74

Energy crops :

(A) are fast-growing plants

(B) can be grown on degraded, waste, saline, or water-logged lands

(C) can not be used to produce liquid fuels

(D) usually have high calorific values

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (B) Only

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (A), (B) and (C) Only

  4. DOption (d)

    (A), (B) and (D) Only

View answer & explanation

Correct answer

Answer: (d)D.

(A), (B) and (D) Only

Why this answer is correct

The correct answer is (A), (B) and (D) Only.

Key Points Characteristics of energy crops:

  1. ​Energy crops are one of the primary sources of biomass-based feedstocks studied for biofuel production.
  2. They are low-cost, low-maintenance crops grown solely for the production of energy through combustion (not for food).
  3. They grow at a very fast rate.
  4. These crops are transformed into solid, liquid, or gaseous fuels like pellets, bioethanol, or biogas.
  5. Energy crops require low establishment inputs, and low fossil fuel inputs, be adaptable to marginal lands and provide high biomass and energy yields that are expected to help reduce global warming and combat climate change (GCC).
  6. Highly erodible, flood-prone, compacted, saline, acidic, contaminated, or sandy soils, reclaimed mine soils, urban marginal sites, and abandoned or degraded croplands are examples of marginal or degraded lands suitable for energy crops.
  7. Energy crops with a higher proportion of fibrous material (lignocellulose) have the highest calorific value, making it advantageous to maximise the yield of this plant fraction for energy and fuel production.
  8. Fuel wood (dedicated energy crop) generates usable heat for the residential, commercial, and electric utility sectors.
  9. Examples: Maize or sweet sorghum and herbaceous and woody plants, such as miscanthus (Miscanthus giganteus), switchgrass (Panicum virgatum), kenaf (Hibiscus cannabinus), reed canary grass (Phalaris arundinacea) or giant reed (Arundo donax), and eucalyptus (Eucalyptus globulus).

Therefore, statements A, B, and D are correct.

Question 75

In Isokinetic air sampling :

(A) The inlet is faced against the prevailing wind

(B) Sampling flow rate is chosen so that the inlet speed is the same as the wind speed

(C) There should be the least possibility of distortion of particles sampled

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A), (B) and (C)

  2. BOption (b)

    (B) and (C) Only

  3. COption (c)

    (A) and (C) Only

  4. DOption (d)

    (A) and (B) Only

View answer & explanation

Correct answer

Answer: (a)A.

(A), (B) and (C)

Why this answer is correct

The correct answer is (A), (B) and (C).

Key Points

  • Isokinetic air sampling is a method of collecting airborne particulate matter in which the collector is designed in such a way that the airstream entering it has the same velocity as the air passing around and outside the collector.

 Principles of isokinetic air sampling:

  • The aim is to capture particles that pass through a defined area for a specified amount of time without interfering with their paths. That is why the inlet of the air sampler is kept against the prevailing winds.
  • The velocity of the air entering the sampling probe nozzle equals the velocity of undisturbed air at that point.
  • The air is disturbed as little as possible so that the same particles (no more and no less) enter the nozzle as would have passed through the area of the nozzle if it did not exist.

Therefore, all the statements about isokinetic air sampling are correct.

Question 76

Match List I with List II

List – IList – II
(A)Stenothermal(i)Daphnia
(B)Eurythermal(ii)Heteroneries
(C)Cyclomorphosis(iii)Toads
(D)Circalunar rhythms(iv)Fishes

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (i), (D) ‐ (ii)

  2. BOption (b)

    (A) ‐ (ii), (B) ‐ (iv), (C) ‐ (iii), (D) ‐ (i)

  3. COption (c)

    (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (i)

  4. DOption (d)

    (A) ‐ (iii), (B) ‐ (ii), (C) ‐ (i), (D) ‐ (iv)

View answer & explanation

Correct answer

Answer: (a)A.

(A) ‐ (iv), (B) ‐ (iii), (C) ‐ (i), (D) ‐ (ii)

Why this answer is correct

The correct answer is (A) ‐ (iv), (B) ‐ (iii), (C) ‐ (i), (D) ‐ (ii)

Key Points

ExplanationExamples
Stenothermal organismsThe organisms that can survive under a narrow range of temperatures.Insects, Fish, Penguins, Pythons, Crustaceans, Crocodiles, etc.
Eurythermal organismsThe organisms that can tolerate a wide range of temperatures.Humans, Cows, Monkeys, Sheep, Toads, Goats, etc.
Cyclomorphosis- The occurrence of seasonal changes in an organism's phenotype over successive generations. - It occurs in small aquatic invertebrates that reproduce through parthenogenesis and produce several generations per year.Rotifers, Cladocerans, Daphnia, etc.
Circalunar rhythms- A biorhythm associated with the lunar cycle (approximately 29.5 days). - Many organisms' reproductive cycles, particularly those of marine organisms, are linked to changing levels of moonlight and the tidal cycle, both of which are governed by lunar phases.Corals, Annelids (Heteronereis), Insects, Echinoderms, Teleosts, etc.

Additional Information

  • Circannual rhythms are self-sustaining endogenously driven biological oscillations that last for about a year in the absence of environmental cycles.
  • Examples: population rhythm in carpet beetles, the urge to migrate in birds, hibernation cycles in ground squirrels and cycles of reproductive activity and moulting in ungulates etc.
  • Circadian rhythms are 24-hour cycles of physical, mental, and behavioural changes.
  • Examples: the sleep-wake cycle, the body temperature cycle, the cycle in the secretion of hormones etc.

Question 77

Given below are two statements :

Statement I : Gasoline mixed with biofuels (ethanol) has generally more energy content than pure gasoline.

Statement II : Energy content per unit volume of ethanol is more than that of gasoline.

In the light of the above statements, choose the correct answer from the options given below :

  1. AOption (a)

    Both Statement I and Statement II are true

  2. BOption (b)

    Both Statement I and Statement II are false

  3. COption (c)

    Statement I is true but Statement II is false

  4. DOption (d)

    Statement I is false but Statement II is true

View answer & explanation

Correct answer

Answer: (b)B.

Both Statement I and Statement II are false

Why this answer is correct

The correct answer is Both Statement I and Statement II are false.

Key Points

  • Ethanol (CH3CH2OH) is also known as grain alcohol. It is widely used as an oxygenate to reduce CO emissions, in which case the gasoline/ethanol mixture is usually referred to as gasohol.
  • Ethanol contains less energy per gallon than gasoline. Blending ethanol and gasoline at a ratio of 85% to 15% (E85), the resultant fuel is nearly 30% less powerful than pure gasoline. The energy density of E85 is 22.8 MJ/L.
  • There is a huge difference between the energy density of pure gasoline and ethanol. The energy density of ethanol is 21 MJ/L while that of pure gasoline is 32.2 MJ/L.

Therefore, both statements I and II are false.

Question 78

Which of the following are correct in the context of the Elephant Project?

(A) It was launched in February. 1992

(B) It was launched in April, 1973

(C) It is being implemented in 16 states

(D) Project is being executed in 50 reserves

(E) Project is being executed in 32 reserves

Choose the correct answer from the options given below :

  1. AOption (a)

    (B), (C) and (D) Only 

  2. BOption (b)

    (A), (C) and (D) Only

  3. COption (c)

    (A), (C) and (E) Only 

  4. DOption (d)

    (B), (C), (D) and (E) Only

View answer & explanation

Correct answer

Answer: (c)C.

(A), (C) and (E) Only 

Why this answer is correct

The correct answer is (A), (C), and (E) Only.

 Key Points

  • Project Elephant was launched in 1992 by the Ministry of Environment and Forests, Govt. of India for free-ranging populations of wild Asian Elephants.
  • The project is mainly implemented in 16 states, including Arunachal Pradesh, Assam, Andhra Pradesh, Chhattisgarh Jharkhand, Kerala, Karnataka, Meghalaya, Maharashtra, Nagaland, Orissa, Tamil Nadu, Uttaranchal, Uttar Pradesh, and West Bengal as per 2019 and currently updated to 22 States.
  • There are around 32 Elephant Reserves across the country, the first one being the Singhbhum Elephant Reserve of Jharkhand.
  • Project Elephant has been formally implementing the MIKE (Monitoring of Illegal Killing of Elephants) program of CITES in 10 Elephant Reserves since 2004.
  • There are 88 Elephant corridors throughout India majorly spread across Southern India, Central India, and North-Eastern states.
  • The Ministry of Environment and Forests, in collaboration with the Wildlife Trust of India, has launched the Hathi Mere Sathi campaign The campaign aimed to raise public awareness and foster friendships between elephants and the local population.

Therefore, statements A, C, and E are correct.

Question 79

Given below are two statements :

Consider the following two statements with reference to Gaussian Plume Model.

Statement I: Turbulent diffusion causes mixing of pollutants at a rate faster than molecular diffusion in atmosphere.

Statement II: The distance from the stack, where maximum ground level concentration is observed, is relatively large when the atmosphere is more turbulent.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. AOption (a)

    Both Statement I and Statement II are correct 

  2. BOption (b)

    Both Statement I and Statement II are incorrect 

  3. COption (c)

    Statement I is correct but Statement II is incorrect 

  4. DOption (d)

    Statement I is incorrect but Statement II is correct  

View answer & explanation

Correct answer

Answer: (c)C.

Statement I is correct but Statement II is incorrect 

Why this answer is correct

The correct answer is Statement I is correct but Statement II is incorrect. 

Key Points

  • The Gaussian plume model is used to estimate the pollution concentration at ground level at a distance from its source.
  • The highest concentration of pollution would be along the central line with decreasing concentrations as the distance from this central line increases.

This model follows certain assumptions:

  1. The rate of emissions from the source is constant.
  2. The windspeed is constant both in time and with elevation.
  3. The pollutant is conservative; that is, it is not lost by decay, chemical reaction, or deposition.
  4. The terrain is relatively flat, open country.
  • The Gaussian point-source dispersion equation relates average, steady-state pollutant concentrations to the source strengths, windspeed, effective stack height and atmospheric conditions.
  • The highest peak downwind concentration occurs when the atmosphere is very unstable (turbulent) rather than stable.
  • The turbulence in an unstable atmosphere brings the looping plume to earth very quickly, resulting in high peak values near the stack (source).

Therefore, statement I is correct but statement II is incorrect.

Direction: Read the following passage carefully and answer the question :

Ecologists and Economists have joined forces to estimate the value of the services that the World’s ecosystems provide. There are many direct goods such as foods and indirect services, such as the assimilation of waste that ecological systems provide on a renewable basis. Many ecosystems provide more than one service. For example, swamps are important in flood control, water supply and waste treatment and as habitat. In turn, most services are provided by more than one ecosystem; for instance, many systems are involved in nutrient cycling. Robert Constanza and his colleagues estimated that the World’s ecosystems provide atleast US 33trillionworthofservicesannually.Thisstaggeringfigureismorethanthetotalgrossnationalproduct,whichisaroundUS 33 trillion worth of services annually. This staggering figure is more than the total gross national product, which is around US 19 trillion per year. The majority of the services are currently outside the market system and include atmospheric gas regulation, waste treatment and nutrient cycling, which at US 17trillion,wasbyfarthemostexpensiveservicesperformed,butevenifweeliminatethis,thetotalannualvaluewouldstillbeawhoppingUS 17 trillion, was by far the most expensive services performed, but even if we eliminate this, the total annual value would still be a whopping US 16 trillion. About 63 percent of the estimated value of natural ecosystems is contributed by marine system; with most of the value coming from coastal systems (US 10.6trillionperyear).About37percentoftheestimatedvaluecamesfromterrestrialsystems,mainlyforests(US 10.6 trillion per year). About 37 percent of the estimated value cames from terrestrial systems, mainly forests (US 4.7 trillion) and wetlands (US 4.9trillion).Ifecosystemserviceswereactuallypaidfor,theglobalpricesystemwouldbeverydifferentfromwhatitistoday.Thepriceofcommoditieswouldskyrocket.However,becauseecosystemservicesarelargelyoutsidethemarket,theyareusuallyignoredorgrosslyundervalued 4.9 trillion). If ecosystem services were actually paid for, the global price system would be very different from what it is today. The price of commodities would skyrocket. However, because ecosystem services are largely outside the market, they are usually ignored or grossly undervalued.

Question 80

Based on Passage: View directions & passage

Which one of the following ecosystems provide maximum estimated value of ecosystem services?

  1. AOption (a)

    Forest ecosystem

  2. BOption (b)

    Coastal ecosystem

  3. COption (c)

    Grassland ecosystem

  4. DOption (d)

    Desert ecosystem

View answer & explanation

Correct answer

Answer: (b)B.

Coastal ecosystem

Why this answer is correct

Explanation:

  • Among the various natural ecosystems, the marine system provides around 63 percent of the estimated value and the terrestrial ecosystem is around 37 percent.
  • The value is about 10.6 trillion US dollars per year coming from the coastal systems.

Hence, option 2 is the correct answer.

Question 81

Based on Passage: View directions & passage

Swamp ecosystems provide the following ecosystem services.

(A) Assimilation of waste

(B) Provide habitat for organisms

(C) Terrestrial fruits

(D) Flood control

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A) and (C) only

  2. BOption (b)

    (B), (C) and (D) only

  3. COption (c)

    (A), (B) and (D) only

  4. DOption (d)

    (A), (B) and (C) only

View answer & explanation

Correct answer

Answer: (c)C.

(A), (B) and (D) only

Why this answer is correct

Explanation:

  • It is to be noted that many ecosystems provide more than one service.
  • This includes a swampy ecosystem.
  • Swamps are important in flood control, water supply, waste treatment, and as habitat.

Hence, option 3 is the correct answer.

Question 82

Based on Passage: View directions & passage

If the ecosystem services are paid for, the price of commodities will :

  1. AOption (a)

    Increase marginally

  2. BOption (b)

    become very high

  3. COption (c)

    remain unaffected 

  4. DOption (d)

    decrease marginally

View answer & explanation

Correct answer

Answer: (b)B.

become very high

Why this answer is correct

Explanation:

  • It was evident from the passage that, even after eliminating the most expensive services it was found that nearly 16 trillion US dollars would be the annual estimated value of the services provided by the ecosystem.
  • But majority of the services are outside the market system currently.
  • Therefore, if the ecosystem services were paid for, the price of commodities will become very high.

Hence, option 2 is the correct answer.

Question 83

Based on Passage: View directions & passage

What was the estimated value per year for the global ecosystem services outside the market system?

  1. AOption (a)

    US 19trillion 19 trillion

  2. BOption (b)

    US 33trillion 33 trillion

  3. COption (c)

    US 16trillion 16 trillion

  4. DOption (d)

    US 17trillion 17 trillion

View answer & explanation

Correct answer

Answer: (d)D.

US 17trillion 17 trillion

Why this answer is correct

Explanation:

  • From the passage, it was evident that the majority of the services are currently outside the market system.
  • This was estimated at 17 trillion US dollar per year.

Hence, option 4 is the correct answer.

Question 84

Based on Passage: View directions & passage

Ecosystem services which have no market value are :

(A) Providing food

(B) Nutrient cycling

(C) Atmospheric gas regulation

(D) Water supply

Choose the most appropriate answer from the options given below :

  1. AOption (a)

     (A) and (B) only

  2. BOption (b)

    (B) and (C) only

  3. COption (c)

    (A) and (D) only

  4. DOption (d)

    (C) and (D) only

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (C) only

Why this answer is correct

Explanation

  • Ecosystems provide trillion worth of services per year.
  • But, the majority of the services are outside the market system.
  • These services include atmospheric gas regulation, waste treatment, and nutrient cycling.

Hence, option 2 is the correct answer.

Question 85

Which method is used to separate proteins on the basis of their sizes? 

  1. AOption (a)

    Ion exchange chromatography

  2. BOption (b)

    Thin layer chromatography

  3. COption (c)

    Adsorption chromatography

  4. DOption (d)

    Gel filtration chromatography

View answer & explanation

Correct answer

Answer: (d)D.

Gel filtration chromatography

Why this answer is correct

Explanation:

Chromatography Techniques: 

  • It is the biophysical technique that is used to separate, identify and purify the components of a mixture for both qualitative and quantitative analysis.
  • This method involves the analyte that combines within a liquid or gaseous mobile phase pumped through the stationary phase.

Gel filtration chromatography:

  • It separates the proteins based on molecular size or weight differences.
  • The microporous packing material made of gel is used to separate the molecules pumped through it where the small-sized molecules access the interior of the pores partly or wholly.

Ion exchange chromatography:

  • It separates the molecules based on their affinity to ion exchangers between the target ions present in the sample solution. Two types include cationic and anionic exchangers can be used.

Thin layer chromatography:

  • The liquid mobile phase is run through the flat surface of the finely divided solid/liquid stationary phase to separate the individual components in the mixture.
  • This technique is mostly used for non-volatile or low-volatility compounds.

Adsorption/Column chromatography:

  • The separation is done through the solid stationary phase (Silica gel or Alumina) and the liquid mobile phase single or mixture solvent.
  • The separation is mainly due to the adsorption force that existed to hold the molecules based on the interaction with the adsorbate at different rates.

Hence, option 4 is the correct answer.

Question 86

Among the following effects of noise on people, identify the Auditory effects :

(A) Annoyance

(B) Speech interference

(C) Acoustical privacy

(D) Sleep interference

(E) Hearing loss

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A), (B) (C) and (E) Only

  2. BOption (b)

    (B) and (E) Only

  3. COption (c)

    (B), (C) and (E) Only

  4. DOption (d)

    (A), (C), (D) and (E) Only

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (E) Only

Why this answer is correct

Concept-

  • Noise is often defined as ‘unwanted sound’ and occupational noise is the most frequently studied type of noise exposure.
  • The auditory effects of noise refer to noise-induced hearing loss and problems due to the masking of sounds by noise.
  • Non-auditory effects refer to annoyance, sleep disturbance, stress, hypertension, cardiovascular disease, and effects on performance.
  • Some effects, such as the effects of noise on accidents, may have both an auditory and non-auditory component.
  • The main auditory effects include:
  • Acoustic trauma
  • Temporary hearing loss
  • Permanent hearing loss
  • Tinnitus

Question 87

Match List I with List II :

List – I (Method)List – II (Estimation of)
(A)X‐Ray Fluorescence meter(i)Functional groups
(B)Quartz Crystal Microbalance(ii)Inorganic compounds
(C)Fourier Transform Infrared Spectrometer(iii)Multi‐element analysis
(D)Laser Microprobe Mass Spectrometry(iv)Particulate matter

Choose the correct answer from the options given below :

  1. AOption (a)

    (A) ‐ (i), (B) ‐ (iii), (C) ‐ (iv), (D) ‐ (ii)

  2. BOption (b)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (ii), (D) ‐ (i)

  3. COption (c)

    (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

  4. DOption (d)

    (A) ‐ (i), (B) ‐ (iii), (C) ‐ (ii), (D) ‐ (iv)

View answer & explanation

Correct answer

Answer: (c)C.

(A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii)

Why this answer is correct

The correct answer is (A) ‐ (iii), (B) ‐ (iv), (C) ‐ (i), (D) ‐ (ii).

Key Points 

Analytical TechniquesApplications
X‐Ray Fluorescence meter- The elemental composition of materials can be determined using XRF (X-ray fluorescence), a non-destructive analytical technique. - XRF analyzers measure the fluorescent (or secondary) X-ray emitted by a sample when it is excited by a primary X-ray source to determine the chemistry of the sample.
Quartz Crystal Microbalance- QCM is a gravimetric sensor. - QCM is capable to detect very slight mass changes, around a nanogram in real time. - It is commonly used to determine airborne particulate matter mass concentration.
Fourier Transform Infrared Spectrometer- Fourier transform infrared spectroscopy (FTIR) spectroscopy detects functional groups such as vibrational bands like N-H, O-H, C-H, and C = O (ester, amine, ketone, aldehyde). - The vibrational modes of plant pigment, especially chlorophyll is monitored using FTIR.
Laser Microprobe Mass Spectrometry- Laser microprobe mass spectrometry (LMMS) provides spot analysis with a lateral resolution of 1-5 microns. - It yields information on elements, molecular identification of inorganic substances, and structural characterisation of organic molecules of typically less than 2-3 kDa.

Question 88

What is the correct order of abundance of following heavy metals in the earth’s crust?

(A) Fe

(B) Mg

(C) Al

(D) Cu

Choose the correct answer from the options given below :

  1. AOption (a)

    A > C > B > D

  2. BOption (b)

    A > C > D > B

  3. COption (c)

    C > A > B > D

  4. DOption (d)

    C > A > D > B

View answer & explanation

Correct answer

Answer: (c)C.

C > A > B > D

Why this answer is correct

The correct answer is C > A > B > D.

Key Points

  • Heavy metals are metallic elements with a relatively high density relative to water and are toxic at low concentrations.
  • Heavy metals are naturally occurring elements in the Earth's crust.
  • The abundance of different heavy metals in the earth's crust are:
Heavy metalsAbundance in Earth's crust (%)
Al8.23
Fe5.63
Mg2.33
Cu0.006
Ni0.0084
Zn0.007
Au0.000004
Ag0.0000075
Pt0.0000005

 

 

 

 

 

 

 

 

Therefore, the correct order of abundance of heavy metals in the earth’s crust is Al > Fe > Mg > Cu.

Question 89

In the context of an urban environment, arrange the following in order of increasing residence time:

(A) Formaldehyde

(B) Carbon‐mono‐oxide

(C) Peroxyacetyl nitrate

(D) Hydroxyl radical

Choose the correct answer from the options given below:

  1. AOption (a)

    A < D < B < C

  2. BOption (b)

    A < B < D < C

  3. COption (c)

    D < A < C < B

  4. DOption (d)

    B < C < A < D

View answer & explanation

Correct answer

Answer: (c)C.

D < A < C < B

Why this answer is correct

The correct answer is D < A < C < B.

Key Points

Formaldehyde: 

  • Also called methanal, is a colorless chemical commonly used in many manufacturing processes.
  • It easily becomes a gas at room temperature making it a member of a larger group of chemicals known as VOCs.
  • It also causes Sick Building Syndrome.
  • Sources: Panelling, plyboard, carpets, ceiling tiles.
  • Residence time in the atmosphere has been estimated at between 0.1 to 1.2 days.

Carbon monoxide:​

  • ​It is colorless, odorless, flammable gas which is lighter than air.
  • CO is an asphyxiant and is fetal at 1000 ppm.
  • Residence time is less than 3 years.

Peroxyacetyl nitrate:

  • ​Acetaldehyde reacts with hydroxyl radical and molecular oxygen to form acetyl peroxy radical which combines with NO2 to form peroxyacetyl nitrates.
  • It is a potent eye irritant.
  • PANs can remain in the atmosphere for about 3 months after formation under cold conditions. However, in warmer areas, PANs persist only for a few hours.

Hydroxyl radical:

  • ​It is the single most reactive intermediate species in atmospheric chemical processes.
  • Residence time is less than 1 second because of their instability and high reactivity.​

​Therefore, the correct order of increasing residence time is Hydroxyl radical, Formaldehyde, Peroxyacetyl nitrate, and Carbon monoxide.

Question 90

Identify the correct sequence of moisture content (%) in ascending order in the following categories of wastes :

(A) Animal solids and organic waste

(B) Trash

(C) Rubbish

(D) Garbage

(E) Refuse

Choose the correct answer from the options given below :

  1. AOption (a)

    E, D, B, C, A

  2. BOption (b)

    C, B, D, E, A

  3. COption (c)

    A, B, C, E, D

  4. DOption (d)

    B, C, E, D, A

View answer & explanation

Correct answer

Answer: (d)D.

B, C, E, D, A

Why this answer is correct

Concept:

Moisture content: 

  • Moisture content is defined as the ratio of the weight of water (wet weight - dry weight) to the total weight of the wet waste.
  • Moisture increases the weight of solid wastes, and thereby, the cost of collection and transport.

In addition, the moisture content is a critical determinant in the economic feasibility of waste treatment by incineration, because wet waste consumes energy for the evaporation of water and raising the temperature of water vapor.

  • Garbage-
  • This refers to animal and vegetable wastes resulting from the handling, sale, storage, preparation, cooking, and serving of food.
  • Garbage comprising these wastes contains putrescible (rotting) organic matter, which produces an obnoxious odor and attracts rats and other vermin.
  • Trash-
  • Trash Is broken furniture, old tires, boxes of stuff collecting dust, etc.
  • Refuse-
  • Refuse, also called municipal solid waste, is nonhazardous solid waste that requires collection and transport to a processing or disposal site.
  • Refuse includes garbage and rubbish.
  • Rubbish-
  • Rubbish is mostly dry material such as glass, paper, cloth, or wood that does not readily decompose.
  • Some forms of rubbish are recyclable,
  • Animal solids and organic wastes will have the maximum moisture content.
  • So the correct order of sequence of moisture content will be Trash < Rubbish < Refuse < Garbage < Animal solids and organic waste.

Question 91

What is the concentration of H+ in a solution of 0.1 N NaOH at 25°C? 

  1. AOption (a)

    1013 M

  2. BOption (b)

    10-13 M

  3. COption (c)

    10-14 M

  4. DOption (d)

    1014 M

View answer & explanation

Correct answer

Answer: (b)B.

10-13 M

Why this answer is correct

Concept-

Normality:

  • Normality refers to the gram equivalent of the substance being dissolved in one liter of the solution.
  • Gram equivalent weight can be defined as the molecule’s reactive capacity. It is measured as ‘N,’ eq/L or meq/L where ‘eq’ stands for equivalents while ‘meq’ stands for milliequivalent.
  • Normality = GrameqofsoluteVolumeofsolutioninlitre\frac{Grameq of solute}{Volume of solution in litre}

Molarity:

  • Molarity is the most commonly used measure of concentration in a solution.
  • Molarity may also be referred to as molar concentration.
  • It can be defined as the number of moles of a solute dissolved per litre of solution.

 

[H+][OH-] = 10-14

[H+], [OH-] = Molar concentration of H+ and OH- ion respectively.

Relation between Normality and Molarity-

Normality = Molarity × MolarmassEquivalentmass\frac{Molar mass}{Equivalent mass}

Calculation:

Solution is given as 0.1 N NaOH.

NaOH → Na+ + OH-

Valence of OH- ion is 1.

So normality = Molarity

Molarity = 0.1 

So the molar concentration of OH- = 0.1

[H+] = )\(\frac{{{{10}^{ - 14}}}}{{0.1}} = 10-13 M

Question 92

Following information is given regarding a one-way ANOVA test :

There are ‘k’ treatments. each having sample size n1, n2, n3, ......., nk such that the total sample size N = n1 + n2 + n3 + ... + nk.

The degrees of freedom for the mean error sum of squares in the test will be

  1. AOption (a)

    k

  2. BOption (b)

    k - 1

  3. COption (c)

    N - 1

  4. DOption (d)

    N - k

View answer & explanation

Correct answer

Answer: (d)D.

N - k

Why this answer is correct

Explanation

Degrees of Freedom

k Σi = 1

ni Σj = 1  

(yij – mean yij) = 0

Degrees of freedom for Total sum of squares (TSS) = N - 1

kΣi = 1 ni  (yij – mean yij) =0

Degrees of freedom for Sum of squares total (SST) = k - 1

D.f of TSS = D.f of SST + D.f of SSE

Therefore,

Degrees of freedom for error sum of square (SSE) = D.f of TSS - D.f of SST

D.f of SSE = N - 1 - (k - 1) = N - k.

Hence, option 4 is the correct answer.

Question 93

Consider the following statements regarding the application of vermicompost :

(A) Vermicompost is an excellent base for the establishment of beneficial microorganisms

(B) Presence of earthworms helps in aerating the soil

(C) Plants respond slowly compared to compost or farm yard manure

(D) Application of vermicompost augments the growth of N‐fixing microbes

(E) Earthworm casts do not harbor vesicular arbuscular micorrhizal (VAM) propagules

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A), (B) and (E) Only

  2. BOption (b)

    (A), (D) and (E) Only

  3. COption (c)

    (A), (B) and (D) Only

  4. DOption (d)

    (B), (C) and (E) Only

View answer & explanation

Correct answer

Answer: (c)C.

(A), (B) and (D) Only

Why this answer is correct

Concept:

Vermicomposting:

  • Vermicomposting is an organic and biological process in which earthworm species are primarily used to convert organic matter or biodegradable wastes into manure.
  • The produced vermicomposts are rich in nutrition and thus, they are widely used as biofertilizers in organic farming and sewage treatment plants.
  • Compost mainly refers to the end products formed by the biodegradation of organic matter. Vermicompost refers to the end product developed by the degradation of organic matter.
  • Vermicompost can be produced in lesser spaces, in smaller containers, and both indoors and outdoors.
  • The extensive channeling and burrowing by earthworms loosens and aerates the soil and improves soil drainage. Soils with earthworms drain up to 10 times faster than soils without earthworms.
  • Vermicompost enhances soil biodiversity by promoting the beneficial microbes which in turn enhances plant growth directly by production of plant growth-regulating hormones and enzymes and indirectly by controlling plant pathogens, nematodes, and other pests, thereby enhancing plant health and minimizing yield loss.
  • The application of vermicompost augments the growth of N‐fixing microbes.

Question 94

Which of the following are included in the list of hazardous wastes as per Hazardous Waste Rules, 2016?

(A) Fly ash

(B) Empty barrels contaminated with hazardous waste

(C) Red mud

(D) Chemical-containing residue arising from decontamination

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (A), (B) and (C) Only 

  2. BOption (b)

    (B) and (D) Only

  3. COption (c)

    (B) and (C) Only

  4. DOption (d)

    (A), (B) and (D) Only

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (D) Only

Why this answer is correct

The correct answer is (B) and (D) Only.

Key Points

The Hazardous Waste Management Rules, 2016, are designed to guarantee the safe generation, processing, treatment, packaging, storage, transportation, use, reprocessing, collecting, conversion, and sale, destruction, and disposal of hazardous waste.

Different hazardous wastes as per the Hazardous Waste Rules, 2016 are:

  1. ​Sludge and filter press cake arising out of the production of zinc sulphate and other zinc compounds. ​
  2. Residues containing cadmium and arsenic.
  3. Acidic and alkaline residues.
  4. Cyanide, nitrate, or nitrite-containing sludge.
  5. Mercury-bearing sludge is generated from the mercury cell process.
  6. Chromium-bearing residue and sludge.
  7. Empty barrels/containers/liners contaminated with hazardous chemicals /wastes.
  8. Chemical-containing residue arising from decontamination.
  9. Sludge from wet scrubbers.
  10. Exhaust air or gas cleaning residue.

Therefore, only B and D are correct.

Question 95

Which of the following deposition mechanisms dominate(s) for particles in the size range 3-10 μ m?

(A) Diffusion

(B) Impaction

(C) Sedimentation

Choose the most appropriate answer from the options given below :

  1. AOption (a)

    (B) only

  2. BOption (b)

    (B) and (C) only

  3. COption (c)

    (A) only

  4. DOption (d)

    (A) and (C) only

View answer & explanation

Correct answer

Answer: (b)B.

(B) and (C) only

Why this answer is correct

Explanation:

Inhalation is the most important route of exposure to air pollutants. These particles will enter the respiratory system's air passages and get deposited based on the particulate material's size, shape, and density (particles, dust, mist, or fume that have become suspended in the surrounding air).

Deposition of these particles in the lungs occurs in one of the following ways:

Interception:

  • A particle is intercepted or deposited when it travels so close to the surface of the airway passages that an edge of the particle touches the surface. Eg. Asbestos.
  • Fibers with a diameter of 1 µm and a length of 200 µm would be deposited in the bronchial tree.

Diffusion:

  • When particles are in random motion, they deposit on the lung walls mostly by chance.
  • This movement is also known as the "Brownian motion". The smaller the particle size, the more vigorous the movement.
  • Diffusion is the most important mechanism for deposition in the small airways and alveoli with a particle size of less than 0.5 µm. Therefore, A is not correct.

Impaction:

  • Many particles do not turn with the air but rather impact or stick to a surface in the particles' original path if there would be a bend in the airway system.
  • Typically, most particles of 3 - 10 µm (aerodynamic diameter) are deposited in the nose or throat and cannot penetrate the lower tissues of the respiratory tract. Therefore, B is correct.

Sedimentation:

  • As particles travel through the air, gravitational forces and air resistance eventually overcome their buoyancy (the tendency for the particle to stay up). The result is that the particles will settle on the surface of the lung. This occurs when the aerodynamic **diameter of the particle is greater than 3 µm. Therefore, C is correct.**​

Hence, option 2 is correct.

Question 96

The portion of UV spectrum which readily destroys DNA and causes skin cancer is : 

  1. AOption (a)

    320 mm to 400 mm 

  2. BOption (b)

    280 mm to 320 mm 

  3. COption (c)

    100 mm to 280 mm 

  4. DOption (d)

    400 mm to 700 mm 

View answer & explanation

Correct answer

Answer: (b)B.

280 mm to 320 mm 

Why this answer is correct

Explanation:

Ultraviolet Spectrum

There are 3 types of UV radiation

Ultraviolet A

  • Wavelength ranges from 315- 399 nm.
  • Not absorbed by the ozone layer.
  • This radiation causes skin tanning and eye damage.

Ultraviolet B

  • Wavelength ranges from 280-314 nm.
  • Mostly absorbed by the ozone layer, but sometimes reach the Earth’s surface.
  • This radiation is very harmful, destroys DNA, and causes skin cancer.

Hence, option 2 is the correct answer.

Ultraviolet C

  • Wavelength ranges from 100-279 nm.
  • Completely absorbed by the ozone layer and atmosphere.

Question 97

Free energy (ΔG) value of zero indicates :

  1. AOption (a)

    Spontaneous reaction in forward direction

  2. BOption (b)

    The state of equilibrium

  3. COption (c)

    Spontaneous reaction in backward direction

  4. DOption (d)

    No reaction at all

View answer & explanation

Correct answer

Answer: (b)B.

The state of equilibrium

Why this answer is correct

Concept-

  • Enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy.
  • The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states.
  • This set of conditions required for determining ΔG° is rarely, if ever, encountered when carrying out chemical reactions, so the literal meaning of ΔG° is of questionable value.
  • However, the sign and magnitude of ΔG° are extremely helpful in describing whether the reaction is product-favored or reactant-favored when it has reached equilibrium, which is sometimes known as the extent of a chemical reaction.
  • ΔG° < 0 means that the reaction is product-favored at equilibrium; the odds are in favor of making a significant amount of product at equilibrium.
  • ΔG° > 0 means that the reaction is reactant-favored at equilibrium; the odds are against making a significant amount of product at equilibrium.
  • ΔG° = 0 means that the reaction is neither product-favored nor reactant-favored at equilibrium; there are 50:50 odds that a significant amount of product will be made at equilibrium.

Question 98

For wind power generation, the percentage of maximum extractable power contained in the wind is :  

  1. AOption (a)

    -45%

  2. BOption (b)

    -59%

  3. COption (c)

    -62%

  4. DOption (d)

    -37%

View answer & explanation

Correct answer

Answer: (b)B.

-59%

Why this answer is correct

Explanation:

  • Wind turbines work by slowing down passing wind in order to extract energy.
  • The kinetic energy from the wind can be used to spin the turbine.
  • This is helpful in the generation of electricity.
  • The theoretical maximum efficiency for a wind turbine is calculated by the Betz limit. This was introduced in 1919 by Albert Betz.
  • It was estimated that 59.35% is the maximum efficiency of the turbine. But in common, the efficiencies usually range between 35-45%.

Hence, option 2 is the correct answer.

Question 99

Pyrolysis of solid wastes refers to :

  1. AOption (a)

    High temperature burning in presence of excess air

  2. BOption (b)

    High temperature anaerobic distillation

  3. COption (c)

     Low temperature combustion in pressure of limited air

  4. DOption (d)

    Ambient anaerobic distillation

View answer & explanation

Correct answer

Answer: (b)B.

High temperature anaerobic distillation

Why this answer is correct

Concept:

Pyrolysis:

  • It is the chemical decomposition of organic (carbon-based) materials through heat application.
  • It is a thermochemical treatment, which can be applied to any organic product.
  • In this treatment process, materials are exposed to a very high temperature, and in the absence of oxygen, It goes through a chemical and physical separation into different molecules.
  • The rate of pyrolysis increases with an increase in temperature.

Additional Information

CombustionGasificationPyrolysis
Thermal processing of solid waste by chemical oxidation with stoichiometric or excess amounts of airProcess of partial combustion of solid waste in which air is supplied less than stoichiometric air.Thermal processing of waste in the complete absence of air.
End products-hot gases, water vapor (flue gas), and noncombustible residue (ash)End products-Flue gases (carbon monoxide, hydrogen, carbon dioxide, hydrocarbons (methane), condensable liquids, solid residue (char).End products-solids (char), liquids (tar/oil), and gases (hydrogen. methane, carbon monoxide, carbon dioxide, etc.)
Energy can be recovered by heat exchange from the hot combustion gases.Energy efficient technique for reducing the volume of solid waste and energy recovery.Endothermic process, an external source of heat is required.

Question 100

Identify the correct sequence of conversion of a fertile nuclear material to a fissile material involving the following nuclei

(A) Np146239\rm Np_{146}^{239}

(B) U147239\rm U_{147}^{239}

(C) Pu145239\rm Pu_{145}^{239}

(D) U146238\rm U_{146}^{238}

Choose the correct answer from the options given below :

  1. AOption (a)

    A, B, C, D

  2. BOption (b)

    A, B, D, C

  3. COption (c)

    B, D, A, C

  4. DOption (d)

    D, B, A, C

View answer & explanation

Correct answer

Answer: (d)D.

D, B, A, C

Why this answer is correct

The correct answer is D, B, A, and C.

Key Points

The binding energy released by U-238 absorbing a thermal neutron is less than the critical energy. So neutrons must possess additional energy for fission to be possible. Consequently, U-238 is fissionable but not fissile.

  • U146238\rm U_{146}^{238} + 0n1 → 146U239
  • Np146239\rm Np_{146}^{239}  →Pu145239\rm Pu_{145}^{239}  + -1e0 

​Explanation:

Neptunium-239 was the first trans-uranium element produced synthetically**.** This isotope has a beta decay which forms the daughter product Plutonium-239.

Therefore, the correct sequence of conversion is U146238\rm U_{146}^{238}→U147239\rm U_{147}^{239}→Np146239\rm Np_{146}^{239}→Pu145239\rm Pu_{145}^{239}​

100 questions in the full paper