UGC NTA NET/JRF Exam, Environmental Sciences, June-2025

Total Questions: 99

51. Which of the following is NOT an element and component of hazard and risk assessment of waste?

Correct Answer: 3. Reduction in use of hazardous substances
Solution:

Why this is NOT an element of hazard/risk assessment: “Reduction in use of hazardous substances” is a risk-reduction/control strategy (source reduction/inherent safety) applied after assessment. The assessment phase comprises:

• Process knowledge & documentation (flowsheets, materials, SDSs, waste inventories) — needed to identify hazards.

• Process & equipment integrity (design, inspection, maintenance) — informs failure likelihood and exposure scenarios.

• Standards, codes, laws & practices — provide criteria/benchmarks for evaluation and compliance.

• Assessment proper = hazard identification → exposure assessment → toxicity/dose-response → risk characterization. Control (like substitution/reduction) belongs to mitigation/management, not assessment.

52. Match the LIST-I with LIST-II.

LIST-I (Class)LIST-II (Disease)
A. Water borneI. Trachoma
B. Water basedII. Yellow fever
C. Water washedIII. Typhoid
D. Water relatedIV. Schistosomiasis

Choose the correct answer from the options given below:

Correct Answer: 3. A-III, B-IV, C-I, D-II
Solution:

• Water-borne (A → III Typhoid): Ingested contaminated water (e.g., Salmonella typhi).

• Water-based (B → IV Schistosomiasis): Pathogen has an aquatic intermediate host (snails); humans infected by water contact.

• Water-washed (C → I Trachoma): Due to poor hygiene/insufficient water; transmission reduced by face-washing.

• Water-related (D → II Yellow fever): Vector-borne near water (Aedes mosquitoes breed in/near water containers).

53. Consider the following statements about HPLC:

A. In normal phase HPLC, the stationary phase is non-polar and the mobile phase is polar.
B. In normal phase HPLC, the least polar compound eludes first and increasing polarity of mobile phase decreases elution time.
C. In reverse phase HPLC, the most polar compound eludes first and increasing polarity of mobile phase increases elution time.
D. In the isocratic mode of HPLC analysis, the polarity of mobile phase remains unchanged throughout the run.
E. The resolution of analysis depends on proper balance of intermolecular forces among the analyte, the mobile phase and the stationary phase.
Choose the correct answer from the options given below: 

Correct Answer: 3. B, C, D and E only
Solution:

A. False: Normal-phase HPLC uses polar stationary phase (e.g., silica) and non-polar mobile phase (e.g., hexane), not the reverse.

B. True: In normal phase, the least polar analyte elutes first; increasing mobile-phase polarity competes with analyte-stationary interactions →shorter elution time.

C. True: In reverse-phase (non-polar stationary; polar mobile), most polar analytes elute first; making the mobile phase more polar (more aqueous) generally increases retention times (especially for hydrophobes).

D. True: Isocratic run = mobile-phase composition (hence polarity) constant throughout.

E. True: Resolution depends on intermolecular forces among analyte, mobile phase, stationary phase (controls retention/selectivity/ efficiency).

54. Match the LIST-I with LIST-II.

LIST-I (National Park)LIST-II (River)
A. DudhawaI. Jhelum
B. KalesarII. Sharda
C. KazinagIII. Teesta
D. KhangchendzongaIV. Yamuna

Choose the correct answer from the options given below: 

Correct Answer: 2. A-II, B-IV, C-I, D-III
Solution:

• Dudhwa (A) → Sharda (II): Park along the Sharda/Mahakali river (UP-Nepal border).
• Kalesar (B) → Yamuna (IV): Shivalik park in Haryana, drained by Yamuna system.
• Kazing (C) → Jhelum (I): Kashmir park in Baramulla region tied to Jhelum basin.
• Khangchendzonga (D) → Teesta (III): Sikkim park feeding the Teesta river system.

55. Match the LIST-I with LIST-II.

LIST-I (Ways of expressing probabilities)LIST-II (Situations expressing probabilities)
A. Decimal NotationsI. Random variables
B. Scientific NotationsII. Conversions
C. PercentageIII. Computations
D. Pictorial FormIV. For low probabilities

Choose the correct answer from the options given below:

Correct Answer: 2. A-III, B-IV, C-II, D-I
Solution:

56. Which of the following statements regarding statistical tool to assess various aspects of a relationship is/are true?

A. For assessing agreement between two quantitative variables, Cohen's Карра is used.
B. The residuals in a linear relationship should follow normal distribution.
C. For assessing strength of relationship between exposure and outcome (yes Vs no), relative risk is used.
D. When dependent variable is quantitative and one independent variable is nonlinear, we do Curvi linear regression.
E. There are three parameters for expressing linear relationship between one dependent and one independent variable (both quantitative).
Choose the most appropriate answer from the options given below:

Correct Answer: 2. B and C only
Solution:

A. False: Cohen's kappa measures agreement for categorical ratings (nominal/ordinal). For two quantitative measures, use Intraclass Correlation Coefficient (ICC) or BlandAltman analysis (and Lin's concordance correlation) to assess agreement.

B. True: In simple/Multiple linear regression, statistical inference assumes the errors/residuals are approximately normal, with independence and constant variance (homoscedasticity). Normality pertains to residuals, not necessarily the raw variables.

C. True: With a binary outcome (yes/no) in cohort or cross-sectional analysis, Relative Risk (risk ratio) is a standard measure of strength of association between exposure and outcome RR = [a/(a+b)/(c/c+d)]. (In case-control designs, odds ratio is typically used.)

D. False: Variables are not "linear" or "nonlinear"; the relationship can be. If the Y-X relation is curved, we fit polynomial/ transform models (still linear in parameters, estimated by OLS) or nonlinear regression. The statement's wording ("one independent variable is non-linear") is incorrect as a criterion.

E. False: A straight line is fully specified by two parameters-slope and intercept. (Error variance is part of the stochastic model, not a third line parameter.)

57. Consider the following statements and choose the INCORRECT statement.

Correct Answer: 5. (*)
Solution:

1. Chromate ions are soluble in water - True. The chromate ion (CrO₄²⁻) readily exists in aqueous solution; many chromate salts of alkali metals and ammonium are watersoluble (e.g., Na₂ CrO₄, K₂CrO₄). Insolubility arises with specific cations (e.g., PbCrO₄), not from the ion itself.

2. Arsenate ion are insoluble in water - False. The arsenate ion (AsO₄³⁻) is an aqueous oxyanion; alkali arsenates (e.g., Na₃AsO₄, K₃AsO₄) are water-soluble. Some heavymetal arsenates are sparingly soluble, but the statement that "arsenate ion [is] insoluble" is incorrect as a general claim.

3. Arsenite is more toxic than arsenate - True. As(III) (arsenite) typically shows greater acute toxicity than As(V) (arsenate) because As(III) binds strongly to thiol (-SH) groups in enzymes and proteins, disrupting cellular function.

4. Chromite is more toxic than chromate - False. Chromite (FeCr₂O₄) is a mineral with Cr(III) bound in a solid lattice and exhibits low bioavailability/toxicity relative to soluble Cr(VI) chromate, which is highly toxic and carcinogenic.

Therefore, both statement 2 and statement 4 are incorrect.

58. In an absorbed dose of 0.2 Gy of slow neutron radiation produces the same biological effect as 2 Gy of gamma radiation, then the relative biological effectiveness for slow neutron would be ________ .

Correct Answer: 4. 10
Solution:

Relative Biological Effectiveness (RBE) = (dose of reference radiation producing effect) ÷ (dose of test radiation producing the same effect).

59. How deep into Earth, plate recycling occurs?

Correct Answer: 4. Mantle-core boundary
Solution:

Plate recycling by subduction carries cold, dense lithosphere deep into the mantle. Seismic tomography images show slab graveyards near the core-mantle boundary (~2900 km), indicating that recycling extends through the lower mantle to the CMB (though some slabs stagnate near the 660-km discontinuit

60. Choose the CORRECT statements about biological degradation of waste:

A. End product of biological degradation is always less toxic than parent compound.
B. pH of soils play vital role and at low pH fungi dominates and likely to form epoxide.
C. Persistence is a 'conditional' property of biodegradable compounds.
D. Mineralization is the complete conversion of an organic compound to the end products of CO₂ and water.
E. Recalcitrant compounds cannot be biodegraded under any circumstances.
Choose the correct answer from the options given below: 

Correct Answer: 2. В, С and D only
Solution:

B. pH controls microbial guilds; fungi dominate at low pH and can form epoxides - Correct. Soil pH strongly shapes community structure. In acidic soils, fungi often outcompete bacteria and drive key oxidative steps.

Fungal monooxygenases (e.g., P450s) can epoxidize alkenes/aromatics, producing epoxide intermediates during biodegradation.

C. Persistence is 'conditional' - Correct. A compound's persistence depends on environment: temperature, moisture, redox, nutrients, sorption, and presence of competent microbes/enzymes. The same chemical may degrade quickly in one setting yet persist in another.

D. Mineralization → CO₂ + H₂O (aerobic) - Correct. Mineralization is complete microbial conversion of organic carbon to inorganics. In aerobic systems this means CO₂, + H₂O (and mineral salts). (Note: anaerobic mineralization yields CH₂ + CO₃ ; the statement's aerobic framing is standard in many exam contexts.)

A. End products are always less toxic - Incorrect. Biodegradation can generate more toxic intermediates (e.g., epoxides, aldehydes) before ultimate mineralization; therefore "always less toxic" is not true.

E. Recalcitrant compounds can never biodegrade - Incorrect. "Recalcitrant" means resistant, not impossible. With adapted consortia, cometabolism, or altered conditions (e.g., bioavailability, redox), many such compounds do degrade, albeit slowly.