UGC NET · ENVIRONMENTAL SCIENCES · SUBJECT CODE 89
40 Important EVS Questions. 10 Units. Explained.
December 2026 revision starts with understanding—not memorising answers. Practise 40 original UGC NET Environmental Science MCQs, solve numericals and learn the common exam traps.
By SWMG Academic Team · Updated 21 October 2026 · Original practice · Approx. 30–40 min
THE PRACTICE BLUEPRINTEVS / 089
40
Concept checks + calculation drills
01020304050607080910
One question at a time. Reveal the explanation only after choosing an answer.
Source clarity: All 40 questions in this article are original syllabus-based practice questions. They are not verbatim official NTA previous-year questions and do not predict December 2026 questions.
UGC NET Environmental Science important questions: where should you start?
Start with the basics of every unit, then move to problems that require equations, interpretation or application. This practice collection covers 40 original questions from all 10 Environmental Sciences units to help you identify gaps before tackling full official previous papers.
QUICK ANSWER
For December 2026 UGC NET EVS Paper 2 preparation, practise across all ten syllabus units, especially your weak conceptual and numerical areas. Use each incorrect answer to trigger a revision action; then solve genuine previous-year papers separately to learn exam framing.
The names follow the official syllabus structure. These 40 questions are a balanced learning selection—not a prediction of unit weightage or a statistically measured list of the most repeated PYQs.
The three-pass method: learn more from fewer questions
Getting the correct option is useful. Knowing why it is correct—and why the other options fail—is what makes a question valuable for revision.
01 · AttemptRead the stem and choose an option without notes. Write your answer on paper.
02 · ExplainOpen the solution, check units and logic, and identify the distractor that almost fooled you.
03 · RevisitRepeat only missed or guessed questions after 48 hours, then try a full mixed set.
How do these relate to PYQs? They test concepts similar to those assessed in syllabus-based multiple-choice examinations, but they are not quoted from the NTA paper. For authentic archived papers see official sources below.
40 unit-wise questions with answer explanations
All questions below use four options. The correct option remains hidden until you choose to reveal it.
01OFFICIAL SYLLABUS UNIT 01
Fundamentals of Environmental Sciences
Use these foundation questions to distinguish environmental systems, flows of matter and energy, and how ecosystems respond to disturbance.
4 original MCQsSystems · biogeochemical cycles · energy transfer · resilience
QUESTION 01UNIT 01 · Concept distinction
Which statement most accurately describes Environmental Sciences as a field?
AIt examines ecosystems only
BIt integrates natural sciences with policy, technology and social dimensions
CIt excludes quantitative modelling
DIt studies pollution without considering humans
Reveal answer & explanation ↗
Correct: B. It integrates natural sciences with policy, technology and social dimensions
Environmental Sciences combines ecology, chemistry, geology, engineering, public policy and human behaviour to study linked environmental problems. An interdisciplinary question may require more than one of these perspectives.
Common trap: Do not treat Environmental Sciences and ecology as identical.
QUESTION 02UNIT 01 · Carbon cycle
Which process directly removes atmospheric carbon dioxide as part of the short-term biological carbon cycle?
ARespiration
BComplete combustion of biomass
CPhotosynthesis
DDecomposition
Reveal answer & explanation ↗
Correct: C. Photosynthesis
Photosynthetic organisms incorporate CO₂ into organic matter using light energy. Respiration, combustion and many decomposition processes generally return CO₂ to the atmosphere.
Common trap: Do not confuse carbon fixation with carbon release.
QUESTION 03UNIT 01 · 2,000 × 0.1 × 0.1 = 20 kJ
If 2,000 kJ is available to producers and 10% is transferred at each successive trophic level, about how much reaches secondary consumers?
A2 kJ
B20 kJ
C200 kJ
D1,800 kJ
Reveal answer & explanation ↗
Correct: B. 20 kJ
Producers → primary consumers: 2,000 × 0.10 = 200 kJ. Primary → secondary consumers: 200 × 0.10 = 20 kJ. The 10% rule is an approximate teaching model, not a fixed value for every ecosystem.
Common trap: Count the number of trophic transfers before calculating.
QUESTION 04UNIT 01 · Ecological resilience
A wetland changes after a flood but later recovers much of its earlier structure. This best illustrates:
AResistance
BResilience
CBioaccumulation
DEutrophication
Reveal answer & explanation ↗
Correct: B. Resilience
Resilience is the capacity to recover after disturbance. Resistance refers to remaining relatively unchanged when a disturbance occurs. Wetlands may display both properties in different circumstances.
Common trap: Recovery after change = resilience; little change during disturbance = resistance.
After Unit 1: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
02OFFICIAL SYLLABUS UNIT 02
Environmental Chemistry
Check quantitative literacy before memorising chemical reactions. pH, equivalents and redox vocabulary transfer into pollution and treatment questions.
4 original MCQspH · alkalinity · redox · atmospheric reactions
QUESTION 05UNIT 02 · pH = −log₁₀(10⁻⁵) = 5
At 25°C, a solution with hydrogen-ion activity approximately 10⁻⁵ has a pH closest to:
A3
B5
C7
D9
Reveal answer & explanation ↗
Correct: B. 5
Using pH = −log₁₀(aH⁺), pH = −log₁₀(10⁻⁵) = 5. In introductory calculations, concentration is often used as an approximation for activity in dilute solutions.
Common trap: The negative sign in the pH equation is essential.
QUESTION 06UNIT 02 · 100 / 50 = 2 meq/L
Water alkalinity of 100 mg/L expressed as CaCO₃ is equivalent to approximately:
A0.2 meq/L
B1 meq/L
C2 meq/L
D5 meq/L
Reveal answer & explanation ↗
Correct: C. 2 meq/L
For alkalinity reported as CaCO₃, 1 meq/L corresponds to 50 mg/L as CaCO₃. Therefore, 100 ÷ 50 = 2 meq/L. This conversion is frequently useful in aqueous chemistry calculations.
Common trap: Do not use the molecular mass 100 g/mol as the equivalent weight; it is 50 g/equiv.
QUESTION 07UNIT 02 · SO₂ + NOₓ
Which statement about acid rain is most accurate?
AUnpolluted rain always has pH exactly 7
BAcid rain is mainly caused by sodium chloride
CSO₂ and NOₓ can produce acids in the atmosphere
DAcid rain occurs only over oceans
Reveal answer & explanation ↗
Correct: C. SO₂ and NOₓ can produce acids in the atmosphere
Sulfur dioxide and nitrogen oxides can be oxidised and converted into sulfuric and nitric acids in the atmosphere. Natural rain is already weakly acidic because dissolved CO₂ forms carbonic acid; acid rain typically refers to rain below that natural background.
Common trap: Avoid assuming neutral pH 7 for unpolluted rain.
QUESTION 08UNIT 02 · Redox fundamentals
Oxidation in an electron-transfer reaction is defined as:
AGain of electrons
BLoss of electrons
CDecrease in oxidation state
DGain of protons only
Reveal answer & explanation ↗
Correct: B. Loss of electrons
Oxidation is loss of electrons and is generally accompanied by an increase in oxidation number. Reduction is gain of electrons. In environmental chemistry, these ideas help explain transformations of metals and contaminants.
Common trap: Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain.
After Unit 2: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
03OFFICIAL SYLLABUS UNIT 03
Environmental Biology
Practise concept selection before the calculation: population models, production, species interactions and pollutant movement through food chains.
In the standard logistic population growth model, absolute population growth rate is maximum when the population is approximately:
AZero
BK/4
CK/2
DK
Reveal answer & explanation ↗
Correct: C. K/2
For dN/dt = rN(1−N/K), growth is maximum at N = K/2 when r and carrying capacity K are constant. Near carrying capacity, net population growth approaches zero.
Common trap: Differentiate maximum absolute growth from maximum per-capita growth.
QUESTION 10UNIT 03 · NPP = 1,200 − 400 = 800
An ecosystem has gross primary productivity of 1,200 g C m⁻² yr⁻¹ and autotrophic respiration of 400 g C m⁻² yr⁻¹. Net primary productivity is:
A400
B800
C1,200
D1,600
Reveal answer & explanation ↗
Correct: B. 800
NPP = GPP − autotrophic respiration = 1,200 − 400 = 800 g C m⁻² yr⁻¹. The result represents the carbon fixed by producers after their own respiratory losses.
Common trap: Do not add respiration to GPP; subtract it.
QUESTION 11UNIT 03 · Lichen symbiosis
The association between a lichen-forming fungus and its photosynthetic partner is commonly described as:
AParasitism
BMutualistic association
CAmensalism
DPredation
Reveal answer & explanation ↗
Correct: B. Mutualistic association
Lichens are composite symbiotic associations. The photosynthetic partner supplies organic carbon, while the fungal partner provides structural support and influences water and mineral relations. The interaction is often introduced as mutualism, though its ecology can be more complex.
Common trap: Avoid confusing a symbiotic association with predation.
QUESTION 12UNIT 03 · Biomagnification
Which pollutant class is especially associated with biomagnification in food chains?
ARapidly biodegradable sugars
BPersistent lipophilic pollutants
CDissolved oxygen
DAtmospheric nitrogen
Reveal answer & explanation ↗
Correct: B. Persistent lipophilic pollutants
Persistent, bioaccumulative chemicals such as some organochlorine compounds may rise in concentration at successive trophic levels. Persistence and biological accumulation are key features behind biomagnification.
Common trap: Bioaccumulation occurs within organisms; biomagnification describes increasing concentration along trophic levels.
After Unit 3: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
04OFFICIAL SYLLABUS UNIT 04
Environmental Geosciences
Watch units closely. Hydrological calculations and vegetation indices become easier when you write the equation before substituting numbers.
4 original MCQsgroundwater · porosity · rocks · remote sensing
QUESTION 13UNIT 04 · Q = 2 × 0.05 × 10 = 1 m³/day
For saturated one-dimensional flow through a porous medium, Darcy’s law gives discharge Q = KiA. If K = 2 m/day, i = 0.05 and A = 10 m², Q is:
A0.1 m³/day
B1 m³/day
C10 m³/day
D100 m³/day
Reveal answer & explanation ↗
Correct: B. 1 m³/day
Q = K × i × A = 2 × 0.05 × 10 = 1 m³/day. This simplified application assumes Darcy-flow conditions and compatible units.
A soil sample has a bulk volume of 100 cm³ and a total pore volume of 25 cm³. Its porosity is:
A0.25%
B2.5%
C25%
D75%
Reveal answer & explanation ↗
Correct: C. 25%
Porosity = pore volume / total bulk volume × 100 = 25/100 × 100 = 25%. Porosity is not the same as hydraulic conductivity or permeability.
Common trap: Use bulk volume as denominator, not the volume of solid grains.
QUESTION 15UNIT 04 · Rock cycle
An igneous rock is formed primarily by:
ACompaction of sediment
BCooling and solidification of molten material
CAccumulation of humus
DWeathering of limestone
Reveal answer & explanation ↗
Correct: B. Cooling and solidification of molten material
Igneous rocks form when magma or lava cools and solidifies. Sedimentary rocks commonly develop through deposition and lithification, while metamorphic rocks are transformed by heat and pressure without complete melting.
Common trap: Keep formation mechanisms of three major rock groups distinct.
A surface has near-infrared reflectance 0.60 and red reflectance 0.20. Its NDVI is:
A0.20
B0.33
C0.50
D0.80
Reveal answer & explanation ↗
Correct: C. 0.50
NDVI = (NIR − Red)/(NIR + Red) = (0.60 − 0.20)/(0.60 + 0.20) = 0.40/0.80 = 0.50. NDVI is a vegetation index, not a direct measure of biomass in every setting.
Common trap: The denominator is the sum of bands, not the red band alone.
After Unit 4: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
05OFFICIAL SYLLABUS UNIT 05
Energy and Environment
The exam can test both physical laws and simple arithmetic. Power, energy and efficiency are related but different quantities.
4 original MCQswind · solar · electricity · conversion limits
QUESTION 17UNIT 05 · P ∝ v³; 2³ = 8
For fixed air density and swept area, wind power available in the air stream changes by what factor if wind speed doubles?
A2
B4
C8
D16
Reveal answer & explanation ↗
Correct: C. 8
Available wind power P = ½ρAv³. Doubling v multiplies power by 2³ = 8. Actual delivered electricity is lower because the turbine and generator are not perfectly efficient.
Common trap: Wind energy scales with the cube of speed, not the square.
A photovoltaic module of area 1.6 m² receives irradiance 1,000 W/m² and produces 200 W electrical power. Its conversion efficiency is:
A8%
B10%
C12.5%
D20%
Reveal answer & explanation ↗
Correct: C. 12.5%
Incident solar power = 1,000 × 1.6 = 1,600 W. Module efficiency = 200/1,600 × 100 = 12.5% under the stated conditions.
Common trap: Use incident power on the entire module area as denominator.
QUESTION 19UNIT 05 · E = 2 × 3 = 6 kWh
An electrical appliance rated 2 kW operates for 3 hours. Its energy consumption is:
A0.67 kWh
B3 kWh
C6 kWh
D9 kWh
Reveal answer & explanation ↗
Correct: C. 6 kWh
Energy = power × time = 2 kW × 3 h = 6 kWh. A kilowatt is a unit of power; a kilowatt-hour is a unit of energy.
Common trap: Do not treat kW and kWh as interchangeable units.
QUESTION 20UNIT 05 · 16/27 ≈ 59.3%
The Betz limit refers to the maximum ideal fraction of wind-stream power extractable by a simple actuator-disk turbine. Its approximate value is:
A25%
B40%
C59.3%
D100%
Reveal answer & explanation ↗
Correct: C. 59.3%
The theoretical Betz limit is 16/27 ≈ 59.3% under ideal assumptions. Real turbines operate below this ideal value because of mechanical, aerodynamic and electrical losses.
Common trap: The Betz limit is not the overall efficiency of every real wind turbine.
After Unit 5: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
06OFFICIAL SYLLABUS UNIT 06
Environmental Pollution and Control
Pollution problems often require correct baselines: inlet concentration, initial DO, reference intensity and meteorological conditions.
4 original MCQsoxygen demand · treatment · inversion · noise
QUESTION 21UNIT 06 · 8 − 3 = 5 mg/L
A simple undiluted water sample initially has DO of 8 mg/L and after incubation has DO of 3 mg/L. Ignoring seed/dilution corrections, oxygen depletion is:
A3 mg/L
B5 mg/L
C8 mg/L
D11 mg/L
Reveal answer & explanation ↗
Correct: B. 5 mg/L
Oxygen depletion = initial DO − final DO = 8 − 3 = 5 mg/L. Actual standard BOD₅ analysis can require dilution and seed corrections, so this is a simplified concept exercise.
Common trap: Do not call every oxygen-depletion result a fully corrected laboratory BOD₅ value.
QUESTION 22UNIT 06 · (200−50)/200 × 100 = 75%
A wastewater treatment unit reduces a pollutant concentration from 200 mg/L to 50 mg/L. Removal efficiency is:
A25%
B50%
C75%
D150%
Reveal answer & explanation ↗
Correct: C. 75%
Removal efficiency = (C_in − C_out)/C_in × 100 = (200 − 50)/200 × 100 = 75%. This concentration-based estimate assumes comparable flows when interpreting mass removal.
Common trap: Always divide by the inlet value.
QUESTION 23UNIT 06 · Stable boundary layer
A temperature inversion near the ground commonly leads to:
AStronger vertical mixing
BReduced pollutant accumulation
CSuppressed upward mixing and possible pollutant buildup
DImmediate elimination of particulate matter
Reveal answer & explanation ↗
Correct: C. Suppressed upward mixing and possible pollutant buildup
An inversion means temperature increases with height over a layer, inhibiting vertical convection. Pollutants released below the inversion can be trapped near the surface under suitable meteorological conditions.
Common trap: An inversion is a stability condition, not a pollution-control device.
QUESTION 24UNIT 06 · 10 log₁₀ 2 ≈ 3 dB
If sound intensity doubles while reference intensity is unchanged, the sound-intensity level increases by approximately:
A1 dB
B3 dB
C10 dB
D20 dB
Reveal answer & explanation ↗
Correct: B. 3 dB
Difference = 10 log₁₀(I₂/I₁) = 10 log₁₀ 2 ≈ 3.01 dB. A tenfold intensity increase corresponds to 10 dB.
Common trap: For intensity use 10 log, not 20 log (which applies to pressure amplitudes under specified conditions).
After Unit 6: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
07OFFICIAL SYLLABUS UNIT 07
Solid and Hazardous Waste Management
Pair definitions with operational logic. Waste prevention, compost conditions, landfill contamination and daily quantities should connect.
4 original MCQshierarchy · leachate · compost · mass balance
QUESTION 25UNIT 07 · Waste hierarchy
Which action is generally placed highest in a conventional waste-management hierarchy?
ALandfilling
BIncineration
CWaste prevention or reduction
DDisposal after mixing
Reveal answer & explanation ↗
Correct: C. Waste prevention or reduction
Preventing waste at source avoids material use, collection and disposal burdens. Reuse and recycling follow in common hierarchy frameworks; disposal is typically the least preferred option.
Common trap: Waste-to-energy is not the same as waste prevention.
QUESTION 26UNIT 07 · Landfill management
Leachate from a landfill is best defined as:
APurified groundwater
BLiquid that has percolated through waste and may carry dissolved or suspended contaminants
CA solid compost fraction
DGas produced during anaerobic digestion
Reveal answer & explanation ↗
Correct: B. Liquid that has percolated through waste and may carry dissolved or suspended contaminants
Rainwater and other liquids passing through refuse can dissolve or carry contaminants, forming leachate. Its management is a major design issue for sanitary landfills.
Common trap: Distinguish leachate (liquid) from landfill gas.
QUESTION 27UNIT 07 · Aerobic composting
An actively aerated composting process depends primarily on:
AOxygen-consuming microbial decomposition
BComplete exclusion of oxygen
CDirect combustion of all organic matter
DUse of only inorganic waste
Reveal answer & explanation ↗
Correct: A. Oxygen-consuming microbial decomposition
Aerobic composting uses oxygen-dependent microbial activity to transform biodegradable organic materials into more stable products. Aeration, moisture and carbon-to-nitrogen balance affect performance.
Common trap: Anaerobic digestion is a different process with different conditions and outputs.
QUESTION 28UNIT 07 · 500 × 0.6 = 300 kg/day
A city produces 500 kg/day of solid waste, of which 60% is biodegradable. How much biodegradable waste is generated daily?
A200 kg
B250 kg
C300 kg
D400 kg
Reveal answer & explanation ↗
Correct: C. 300 kg
Biodegradable fraction = 500 × 0.60 = 300 kg/day. This calculation does not imply all 300 kg will be recovered or composted; collection and processing losses matter.
Common trap: Convert percentage to decimal before multiplication.
After Unit 7: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
08OFFICIAL SYLLABUS UNIT 08
Environmental Assessment, Management and Legislation
Memorise the role of each EIA stage and international agreement, then test your understanding with realistic situations rather than isolated acronyms.
4 original MCQsEIA stages · mitigation · EMS · conventions
QUESTION 29UNIT 08 · EIA workflow
During environmental impact assessment (EIA), the scoping stage is intended mainly to:
AIssue a final project clearance automatically
BIdentify important impacts and define the study’s terms of reference
CReplace all field studies
DCalculate a project’s profit only
Reveal answer & explanation ↗
Correct: B. Identify important impacts and define the study’s terms of reference
Scoping identifies the significant environmental issues that need assessment and helps frame study boundaries and terms of reference. Screening, impact prediction, consultation and appraisal serve different functions.
Common trap: Do not confuse screening (whether EIA is required) with scoping (what EIA should study).
In the usual mitigation hierarchy, which action is considered before minimising or compensating for harm?
AOffsetting impacts
BAvoiding impacts
CMonitoring after closure only
DIncreasing project capacity
Reveal answer & explanation ↗
Correct: B. Avoiding impacts
Avoidance is preferred because an impact not created does not need repair. Minimisation, restoration and offsets are considered later according to context and applicable requirements.
Common trap: Offsets are not a first-choice substitute for preventing damage.
QUESTION 31UNIT 08 · Environmental management systems
ISO 14001 is principally associated with:
AA mandatory global emission limit for every industry
BEnvironmental management systems
CRadiocarbon dating
DThe definition of endangered species
Reveal answer & explanation ↗
Correct: B. Environmental management systems
ISO 14001 specifies requirements for an environmental management system that an organisation can use to manage environmental responsibilities. It does not itself impose one universal emission limit for all sectors.
Common trap: A management-system standard is not the same thing as a national emission standard.
QUESTION 32UNIT 08 · Environmental conventions
Which international agreement primarily addresses the phase-out of ozone-depleting substances?
AMontreal Protocol
BParis Agreement
CConvention on Biological Diversity
DRamsar Convention
Reveal answer & explanation ↗
Correct: A. Montreal Protocol
The Montreal Protocol controls production and consumption of many ozone-depleting substances. The Paris Agreement concerns climate change; CBD focuses on biodiversity; Ramsar concerns wetlands.
Common trap: Do not confuse stratospheric ozone protection with climate-change treaty goals.
After Unit 8: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
09OFFICIAL SYLLABUS UNIT 09
Statistical Approaches and Modelling
Practice calculations and interpretation separately. Statistical significance, association and variability are not interchangeable concepts.
4 original MCQsmean · dispersion · correlation · hypothesis tests
QUESTION 33UNIT 09 · 20 / 4 = 5
The arithmetic mean of observations 2, 4, 6 and 8 is:
A4
B5
C6
D8
Reveal answer & explanation ↗
Correct: B. 5
Mean = sum of observations divided by their count = (2+4+6+8)/4 = 20/4 = 5. The mean is sensitive to extreme values, unlike the median in many datasets.
Common trap: Divide by the number of observations, not by the largest observation.
QUESTION 34UNIT 09 · r = −0.90
A Pearson correlation coefficient of −0.90 suggests:
AA strong negative linear association
BNo linear association
CA strong positive linear association
DProof of causation
Reveal answer & explanation ↗
Correct: A. A strong negative linear association
A value close to −1 indicates a strong negative linear association between two measured variables. Correlation alone does not prove that one variable causes the other.
Common trap: Strong correlation does not automatically establish causation.
QUESTION 35UNIT 09 · 0.03 < 0.05
A hypothesis test produces p = 0.03, with a preselected significance level α = 0.05. Under the usual decision rule, the result is:
AFail to reject H₀
BReject H₀
CProve H₁ with certainty
DIgnore the p-value
Reveal answer & explanation ↗
Correct: B. Reject H₀
Because p < α, reject the null hypothesis under the specified testing assumptions. Statistical significance is not equivalent to practical importance or a 97% probability that the alternative hypothesis is true.
Common trap: A p-value is not the probability that the null hypothesis is true.
QUESTION 36UNIT 09 · σ = √[(4+4)/2] = 2
For values 10 and 14 treated as the complete population, the population standard deviation is:
A1
B2
C4
D8
Reveal answer & explanation ↗
Correct: B. 2
Population mean = 12. Squared deviations are (10−12)² = 4 and (14−12)² = 4. Population variance = (4+4)/2 = 4, so population SD = √4 = 2.
Common trap: Population SD divides squared deviations by N; sample SD uses N−1.
After Unit 9: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
10OFFICIAL SYLLABUS UNIT 10
Contemporary Environmental Issues
Explain each issue in plain language first, then connect mechanism, environmental effect and the correct policy or management response.
The global warming potential (GWP) of a greenhouse gas compares its integrated warming effect with CO₂ over:
ANo specified interval
BA specified time horizon
COne second only
DThe duration of a solar eclipse
Reveal answer & explanation ↗
Correct: B. A specified time horizon
GWP compares the radiative influence of a unit mass emission against CO₂ over a defined time horizon, commonly 100 years. The selected period matters because gases have different atmospheric lifetimes.
Common trap: A GWP value should not be quoted without its time horizon and assessment context.
QUESTION 38UNIT 10 · CO₂ + H₂O ⇌ H₂CO₃
Ocean acidification is primarily linked to:
AAbsorption of atmospheric CO₂ and formation of carbonic acid
BConversion of all seawater to pure water
CIncreasing pH from a loss of hydrogen ions
DRemoval of all dissolved inorganic carbon
Reveal answer & explanation ↗
Correct: A. Absorption of atmospheric CO₂ and formation of carbonic acid
As seawater absorbs CO₂, carbonate equilibria shift and hydrogen-ion concentration increases, lowering pH. This can affect carbonate availability and calcifying organisms. Ocean acidification does not mean that average seawater has become pH < 7.
Common trap: A falling pH is acidification even when the water remains alkaline.
QUESTION 39UNIT 10 · Good up high, harmful nearby
Why is stratospheric ozone beneficial while elevated ground-level ozone is a pollutant?
AOzone has different chemistry at different heights
BStratospheric ozone absorbs harmful UV, while surface ozone can harm health and vegetation
CGround-level ozone is always harmless
DOnly ground-level ozone absorbs UV
Reveal answer & explanation ↗
Correct: B. Stratospheric ozone absorbs harmful UV, while surface ozone can harm health and vegetation
The stratospheric ozone layer helps screen ultraviolet radiation. Near the surface, ozone is a photochemical oxidant that can damage respiratory tissues and vegetation. The molecule is the same; its environmental role depends strongly on location.
Common trap: Location and exposure determine whether ozone is protective or harmful.
QUESTION 40UNIT 10 · Resource circularity
Which strategy is most consistent with a circular-economy approach?
ADesigning products for easy repair, reuse and material recovery
BReplacing all durable products after one use
CUsing landfills as the first choice
DMaximising virgin-material use
Reveal answer & explanation ↗
Correct: A. Designing products for easy repair, reuse and material recovery
Circular-economy strategies seek to keep products and materials in use and reduce waste through durable design, maintenance, reuse, repair and appropriate recycling. These approaches should still be evaluated for their overall environmental impacts.
Common trap: Recycling alone is not the entire circular economy; product design and reuse matter.
After Unit 10: Try its questions again without viewing the solutions. Add each incorrect concept to an error log and revisit it after two days. Jump to unit index ↑
Numerical revision: seven formulas to remember
Learn the variables and units instead of remembering the answer to just one worked question.
Concept
Relation
Watch out for
pH
−log₁₀(aH⁺)
Activity vs concentration approximation
Alkalinity
mg/L as CaCO₃ ÷ 50 = meq/L
Equivalent weight
Net primary productivity
NPP = GPP − Ra
Autotrophic respiration only
Groundwater discharge
Q = K i A
Compatible distance/time units
NDVI
(NIR − Red)/(NIR + Red)
Correct bands in denominator
Wind-stream power
P = ½ρAv³
Cube of wind speed
Pollutant removal
(Cin−Cout)/Cin × 100
Use inlet as baseline
Other frequently useful calculations include energy in kWh, porosity, oxygen depletion, mean, standard deviation and simple statistics decisions. Recheck units in every step.
Where to find genuine UGC NET Environmental Science previous-year questions
The most reliable approach is to keep official past-paper practice separate from original concept exercises. Use authentic papers to study question wording, distractors and time management. This article provides original explanations for practice only.
For every question you miss, record four things: the unit, the underlying concept, the wrong assumption and the revision action. After five questions, look for patterns. Repeated errors in units and conversions require a different fix from errors in factual recall.
Already working with a schedule? Read SWMG’s Environmental Science preparation strategy for a connected system of concept learning, authentic PYQs, numericals and mock tests.
Use these 40 questions in a seven-day revision sprint
Do not spend the whole week rereading answer explanations. Make the work visible in your error log.
One week. Four rounds of learning.
This is an editorial practice routine, not an official exam timetable.
Days 1–2Attempt Units 1–5 and list weak concepts.
Days 3–4Attempt Units 6–10 and solve calculations again.
Day 5Attempt authentic PYQs for the weak units.
Day 6Review your error log and revisit guessed answers.
Day 7Take a mixed mock and revise Paper 1 too.
RepeatTrack accuracy, reasoning and calculation mistakes.
Suggested tracking: A correct answer without confidence is worth revisiting. Mark each question confident, guessed or incorrect, and re-test the last two groups. This habit often reveals more than a raw score alone.
WHEN A STRUCTURED PREPARATION PATH HELPS
Need a clear sequence for your weaker EVS units?
If these practice questions expose conceptual gaps or difficulty with calculations, you may benefit from a unit-wise learning sequence with PYQ discussions, tests and revision. SWMG offers Environmental Science Paper 2 learning options with Dr. Mukesh Goyal. Review the active curriculum and inclusions before choosing a plan.
Course format, fees, validity and included resources may change; the current course pages are the final reference.
Frequently asked questions
Are these 40 questions genuine UGC NET previous-year questions?
No. They are 40 original, syllabus-based PYQ-style practice questions created for revision. Genuine previous-year papers should be consulted from the UGC previous-question-paper archive and relevant official NTA resources. Do not cite these original questions as NTA examination questions.
How many units are included in UGC NET Environmental Sciences Paper 2?
The subject is listed as Environmental Sciences, code 89, and has ten official syllabus units. This article includes four original practice questions from each of the ten units.
Which units should I practise first for the December 2026 cycle?
Start with a short diagnostic across all ten units. Prioritise units in which you miss foundational questions, then add calculation practice in chemistry, geosciences, energy, pollution and statistics. Do not treat these selected exercises as a verified unit-wise question-weightage forecast.
Does completing these 40 questions guarantee NET or JRF?
No. These questions are a short diagnostic and revision set, not a complete syllabus replacement or a guarantee of qualification. Use the official syllabus, authentic previous papers, mocks, mistake analysis and Paper 1 preparation.
Where can I find real Environmental Science PYQs?
Start with the University Grants Commission previous-question-paper archive. Some older papers use a different exam format; use the latest NTA bulletin to understand current rules and check official current-cycle resources.
Is Paper 1 preparation still needed for Environmental Science aspirants?
Yes. UGC NET includes the common Paper 1 and the subject-specific Paper 2. A Paper 2-only question bank should complement, not replace, your Paper 1 preparation.