CUET PG Disaster Studies 2025 Question Paper (Available): Download Question Paper with Answer Key And Solutions PDF

Step 1: Use the formula for energy of a photon: \[ E = \frac{hc}{\lambda} \]

Step 2: Convert wavelength into meters: \[ \lambda = 200\,nm = 200 \times 10^{-9}\,m \]

Step 3: Substitute the given values: \[ E = \frac{(6.6 \times 10^{-34})(3.0 \times 10^{8})}{200 \times 10^{-9}} \]

Step 4: Simplify: \[ E = \frac{19.8 \times 10^{-26}}{2.0 \times 10^{-7}} = 9.9 \times 10^{-19}\,J \] Quick Tip: Always convert wavelength into meters before substituting in \[ E = \frac{hc}{\lambda}. \] Photon energy is inversely proportional to wavelength.

Step 1: Use Wien’s displacement law: \[ \lambda_{\max} T = b \]
where \( b = 2.9 \times 10^{-3}\,m K \).

Step 2: Substitute the given temperature: \[ \lambda_{\max} = \frac{2.9 \times 10^{-3}}{4830} \]

Step 3: Calculate: \[ \lambda_{\max} \approx 6.0 \times 10^{-7}\,m \]

Step 4: Convert meters to nanometers: \[ 6.0 \times 10^{-7}\,m = 600\,nm \] Quick Tip: For black body radiation problems, directly apply Wien’s law \[ \lambda_{\max} = \frac{2.9 \times 10^{-3}}{T}. \] Higher temperature implies shorter peak wavelength.

Step 1: Coriolis force is a pseudo (apparent) force observed in a rotating frame of reference.
Hence, statement (A) is correct.

Step 2: The magnitude of Coriolis force is zero at the equator and maximum at the poles.
Hence, statement (B) is incorrect.

Step 3: Coriolis force significantly affects large-scale motions such as wind and ocean currents on Earth.
Hence, statement (C) is correct.

Step 4: Therefore, the correct statements are (A) and (C) only. Quick Tip: Coriolis force depends on latitude: - Zero at the equator - Maximum at the poles It strongly influences atmospheric and oceanic circulation.

Step 1: The Richter scale is defined based on the logarithm of the amplitude of seismic waves recorded by seismographs.

Step 2: An increase of 1 unit on the Richter scale corresponds to a tenfold increase in wave amplitude.

Step 3: Hence, the Richter scale follows a logarithmic scale. Quick Tip: On the Richter scale, each whole number increase represents 10 times greater wave amplitude and about 32 times more energy release.

Step 1: Shale is a sedimentary rock and represents the lowest grade before metamorphism begins.

Step 2: With increasing temperature and pressure, shale transforms into slate.

Step 3: Further metamorphism converts slate into phyllite.

Step 4: At the highest grade of metamorphism, phyllite changes into gneiss.

Step 5: Hence, the correct increasing order of metamorphism is: \[ Shale \rightarrow Slate \rightarrow Phyllite \rightarrow Gneiss \] Quick Tip: Increasing grade of metamorphism generally follows: Shale \(\rightarrow\) Slate \(\rightarrow\) Phyllite \(\rightarrow\) Schist \(\rightarrow\) Gneiss.

Step 1: The boundary between the mantle and the outer core is known as the Gutenberg discontinuity.

Step 2: When P-waves enter the outer core from the mantle, they move from a solid to a liquid medium.

Step 3: Due to this change in state, the velocity of P-waves suddenly decreases.

Step 4: Hence, the mantle–outer core boundary is marked by an abrupt decrease in P-wave velocity. Quick Tip: P-waves can travel through both solids and liquids, but their velocity decreases sharply when entering the liquid outer core.

Step 1: Basalt is a mafic igneous rock formed by rapid cooling of lava at the Earth’s surface (extrusive).

Step 2: The plutonic (intrusive) equivalent of an extrusive igneous rock has the same chemical composition but forms deep inside the Earth with slow cooling.

Step 3: Gabbro is a coarse-grained, mafic intrusive igneous rock with the same composition as basalt.

Step 4: Hence, the plutonic equivalent of basalt is gabbro. Quick Tip: Extrusive–Intrusive equivalents: Basalt \(\leftrightarrow\) Gabbro, Rhyolite \(\leftrightarrow\) Granite.

Step 1: Crystal habit refers to the characteristic external shape in which a mineral commonly occurs.

Step 2: Asbestos minerals are known for forming long, thin, thread-like crystals.

Step 3: Such a crystal form is described as fibrous.

Step 4: Hence, the crystal habit of asbestos is fibrous. Quick Tip: Fibrous crystal habit is characterized by elongated, thread-like structures, commonly seen in asbestos minerals.

Step 1: Mohs hardness values of the given minerals are:
Gypsum = 2, Fluorite = 4, Orthoclase = 6, Corundum = 9.

Step 2: Arrange the minerals from lowest to highest hardness.

Step 3: The increasing order is: \[ Gypsum \rightarrow Fluorite \rightarrow Orthoclase \rightarrow Corundum \]

Step 4: Hence, the correct sequence is (D), (B), (A), (C). Quick Tip: Remember key Mohs values: Gypsum (2), Fluorite (4), Orthoclase (6), Corundum (9). Lower number means softer mineral.

Step 1: Muscovite belongs to the mica group.
So, (A) \(\rightarrow\) (III).

Step 2: Actinolite is a member of the amphibole group.
So, (B) \(\rightarrow\) (II).

Step 3: Diopside is a pyroxene mineral.
So, (C) \(\rightarrow\) (IV).

Step 4: Pyrope is a variety of garnet.
So, (D) \(\rightarrow\) (I).

Step 5: Hence, the correct matching is
(A)–(III), (B)–(II), (C)–(IV), (D)–(I). Quick Tip: Common mineral groups: Muscovite \(\rightarrow\) Mica, Actinolite \(\rightarrow\) Amphibole, Diopside \(\rightarrow\) Pyroxene, Pyrope \(\rightarrow\) Garnet.

Step 1: Dolomite is a sedimentary rock formed by chemical precipitation or alteration of limestone.

Step 2: Mudstone is a clastic sedimentary rock formed from compacted mud-sized particles.

Step 3: Hornfels is a metamorphic rock formed by contact metamorphism.
Hence, it is not sedimentary.

Step 4: Slate is also a metamorphic rock derived from shale.

Step 5: Therefore, only dolomite and mudstone are sedimentary rocks. Quick Tip: Sedimentary rocks form by deposition, compaction, or precipitation. Hornfels and slate are metamorphic, not sedimentary.

Step 1: The hydrological (water) cycle involves processes such as evaporation, condensation, precipitation, and runoff.

Step 2: Evaporation of water from oceans, rivers, and lakes requires a continuous supply of energy.

Step 3: This energy is provided by solar radiation from the Sun.

Step 4: Hence, the hydrological cycle on Earth is driven by energy from the Sun. Quick Tip: Solar energy is the main driving force behind evaporation and the entire hydrological cycle on Earth.

Step 1: The Precambrian (including the Neoproterozoic) is the oldest division of geologic time.

Step 2: The Paleozoic era follows the Precambrian, with the Permian as its last period.

Step 3: The Mesozoic era succeeds the Paleozoic, and the Jurassic is a middle period of the Mesozoic.

Step 4: The Cenozoic era is the most recent era, and the Quaternary is its latest period.

Step 5: Hence, the correct chronological order from oldest to most recent is: \[ (A) \rightarrow (D) \rightarrow (C) \rightarrow (B) \] Quick Tip: Geologic time order (oldest to youngest): Precambrian \(\rightarrow\) Paleozoic \(\rightarrow\) Mesozoic \(\rightarrow\) Cenozoic.

Step 1: Frequency of electromagnetic radiation increases as wavelength decreases.

Step 2: Among ultraviolet radiations, UVC has the highest frequency, followed by UVB and then UVA.

Step 3: Infrared radiation has a lower frequency than ultraviolet radiation but higher than microwaves.

Step 4: Microwaves have the lowest frequency among the given radiations.

Step 5: Therefore, the decreasing order of frequency is: \[ UVC \rightarrow UVB \rightarrow UVA \rightarrow Infrared \rightarrow Microwave \] Quick Tip: In the electromagnetic spectrum, frequency decreases from gamma rays to radio waves. Shorter wavelength means higher frequency.

Step 1: Quartz is a naturally occurring crystalline substance and hence a mineral.
So, (A) \(\rightarrow\) (II).

Step 2: Rhyolite is an extrusive (volcanic) igneous rock.
So, (B) \(\rightarrow\) (I).

Step 3: Marble is formed by metamorphism of limestone and is a metamorphic rock.
So, (C) \(\rightarrow\) (IV).

Step 4: Granite forms deep within the Earth by slow cooling of magma and is a plutonic igneous rock.
So, (D) \(\rightarrow\) (III).

Step 5: Hence, the correct matching is
(A)–(II), (B)–(I), (C)–(IV), (D)–(III). Quick Tip: Quartz is a mineral, Rhyolite is volcanic (extrusive), Granite is plutonic (intrusive), Marble is metamorphic.

Step 1: Calculate the molar mass of NaOH: \[ NaOH = 23 + 16 + 1 = 40\,g mol^{-1} \]

Step 2: Calculate the number of moles of NaOH: \[ Moles = \frac{10}{40} = 0.25\,mol \]

Step 3: Convert volume into liters: \[ 500\,mL = 0.5\,L \]

Step 4: Calculate molarity: \[ M = \frac{0.25}{0.5} = 0.50\,M \] Quick Tip: Molarity \(=\dfrac{moles of solute}{volume of solution in liters}\). Always convert volume into liters before calculating molarity.

Step 1: Calcium chloride is commonly used to melt ice on roads.
So, (A) \(\rightarrow\) (III).

Step 2: Sulphuric acid is used as an electrolyte in lead–acid batteries.
So, (B) \(\rightarrow\) (IV).

Step 3: Ethylene glycol is widely used as an antifreeze.
So, (C) \(\rightarrow\) (I).

Step 4: Potassium hydroxide is used as a drain clearing agent.
So, (D) \(\rightarrow\) (II).

Step 5: Hence, the correct matching is
(A)–(III), (B)–(IV), (C)–(I), (D)–(II). Quick Tip: Common applications: Calcium chloride \(\rightarrow\) de-icing, Ethylene glycol \(\rightarrow\) antifreeze, Sulphuric acid \(\rightarrow\) batteries, KOH \(\rightarrow\) drain cleaner.

Step 1: Pauling electronegativity values of the given elements are approximately:
Zn = 1.65, Cd = 1.69, Hg = 2.00.

Step 2: Higher electronegativity means greater tendency to attract electrons.

Step 3: Among the given elements, mercury has the highest electronegativity, followed by cadmium and then zinc.

Step 4: Therefore, the decreasing order of electronegativity is: \[ Hg \rightarrow Cd \rightarrow Zn \] Quick Tip: In group 12 elements, electronegativity generally increases down the group due to relativistic effects, with mercury showing the highest value.

Step 1: Oxygen is the most abundant element in the Earth’s crust, not the least.
Hence, statement (A) is incorrect.

Step 2: The Earth’s core is mainly composed of iron and nickel.
Hence, statement (B) is correct.

Step 3: Oceanic crust is basaltic in nature and is rich in magnesium and iron (SIMA).
Hence, statement (C) is correct.

Step 4: Continental crust is rich in silica and aluminium (SIAL), not magnesium and iron.
Hence, statement (D) is incorrect.

Step 5: Therefore, the correct statements are (B) and (C) only. Quick Tip: Remember Earth composition terms: SIAL \(\rightarrow\) Continental crust (Si + Al), SIMA \(\rightarrow\) Oceanic crust (Si + Mg).

Step 1: Convert ppb to ppm: \[ 310\,ppb = 0.310\,ppm of NO_3^- \]

Step 2: Determine the molar masses: \[ Molar mass of NO_3^- = 14 + (16 \times 3) = 62 \] \[ Molar mass of N = 14 \]

Step 3: Find the fraction of nitrogen in nitrate: \[ \frac{14}{62} \]

Step 4: Calculate concentration of nitrogen: \[ N concentration = 0.310 \times \frac{14}{62} \approx 0.07\,ppm \] Quick Tip: To find elemental concentration from an ion, multiply by the ratio of molar mass of the element to the molar mass of the ion.

Step 1: Write the reduction half-reaction of dichromate ion in acidic medium: \[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \]

Step 2: Balance chromium atoms: \[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \]

Step 3: Balance oxygen atoms by adding water: \[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O \]

Step 4: Balance hydrogen atoms by adding \(H^+\): \[ Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O \]

Step 5: Balance charges by adding electrons: \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

Step 6: Hence, \[ X = 14,\quad Y = 6,\quad Z = 7 \] Quick Tip: In acidic medium, dichromate ion is reduced as: \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \] This reaction is commonly used in redox balancing.

Step 1: Calculate the total mass of the solution: \[ Total mass = 7 + 18 = 25\,g \]

Step 2: Use the formula for mass percent: \[ Mass percent = \frac{Mass of solute}{Mass of solution} \times 100 \]

Step 3: Substitute the values: \[ Mass percent = \frac{7}{25} \times 100 = 28% \] Quick Tip: Mass percent depends only on masses, not on volume: \[ % by mass = \frac{mass of solute}{mass of solution} \times 100. \]

Step 1: The crystallization sequence of minerals from magma is explained by Bowen’s Reaction Series.

Step 2: Olivine crystallizes at the highest temperature and hence forms first.

Step 3: Amphibole crystallizes at a lower temperature than olivine.

Step 4: K-feldspar crystallizes after amphibole at still lower temperatures.

Step 5: Quartz crystallizes at the lowest temperature and hence forms last.

Step 6: Therefore, the correct order of crystallization during cooling is: \[ Olivine \rightarrow Amphibole \rightarrow K-Feldspar \rightarrow Quartz \] Quick Tip: According to Bowen’s Reaction Series, high-temperature mafic minerals crystallize first, while low-temperature felsic minerals crystallize last.

Step 1: The Higher Himalaya is dominated by granitic rocks formed during the Cenozoic era.
So, (A) \(\rightarrow\) (II).

Step 2: The Shivalik Hills consist mainly of young, unconsolidated sedimentary deposits.
So, (B) \(\rightarrow\) (III).

Step 3: The Lesser Himalaya is composed largely of very old Palaeoproterozoic rocks.
So, (C) \(\rightarrow\) (I).

Step 4: The Trans Himalaya includes Phanerozoic sedimentary sequences.
So, (D) \(\rightarrow\) (IV).

Step 5: Hence, the correct matching is
(A)–(II), (B)–(III), (C)–(I), (D)–(IV). Quick Tip: Himalayan divisions: Higher Himalaya \(\rightarrow\) Cenozoic granites, Shivalik \(\rightarrow\) youngest sediments, Lesser Himalaya \(\rightarrow\) very old rocks.

Step 1: The Earth’s crust is composed mainly of oxides and silicate minerals.

Step 2: Oxygen combines with elements such as silicon, aluminium, iron, calcium, and magnesium to form these minerals.

Step 3: By weight percentage, oxygen constitutes about 46% of the Earth’s crust, making it the most abundant element.

Step 4: Hence, the most abundant element in the bulk Earth’s crust is oxygen. Quick Tip: Abundance in Earth’s crust (by weight): Oxygen \(>\) Silicon \(>\) Aluminium \(>\) Iron.

Step 1: Let yellow colour (Y) be dominant over green colour (y), and round seed shape (R) be dominant over wrinkled seed shape (r).

Step 2: The parental cross is: \[ YYRR \times yyrr \]

Step 3: Gametes formed are: \[ YR \quad and \quad yr \]

Step 4: All offspring in the \(F_1\) generation will have the genotype: \[ YyRr \]

Step 5: Since Y and R are dominant alleles, all \(F_1\) plants will show yellow colour and round seed shape. Quick Tip: In a cross between two homozygous parents with contrasting traits, all \(F_1\) offspring are heterozygous and express only the dominant traits.

Step 1: The cell cycle begins with the \(G_1\) phase, during which the cell grows and prepares for DNA synthesis.

Step 2: This is followed by the \(S\) phase, where DNA replication occurs.

Step 3: After DNA synthesis, the cell enters the \(G_2\) phase, involving further growth and preparation for division.

Step 4: Finally, the cell undergoes the \(M\) phase, which includes mitosis and cytokinesis.

Step 5: Hence, the correct sequence is: \[ G_1 \rightarrow S \rightarrow G_2 \rightarrow M \] Quick Tip: Remember the cell cycle order as: \[ G_1 \rightarrow S \rightarrow G_2 \rightarrow M \] Interphase includes \(G_1\), \(S\), and \(G_2\) phases.

Step 1: Natality refers to the number of births occurring in a population during a given period.
So, (A) \(\rightarrow\) (IV).

Step 2: Mortality refers to the number of deaths in a population during a given period.
So, (B) \(\rightarrow\) (II).

Step 3: Immigration is the movement of individuals of the same species into a habitat.
So, (C) \(\rightarrow\) (III).

Step 4: Emigration is the movement of individuals out of a habitat.
So, (D) \(\rightarrow\) (I).

Step 5: Hence, the correct matching is
(A)–(IV), (B)–(II), (C)–(III), (D)–(I). Quick Tip: Population size changes due to: Natality (births) and Immigration (entry) increase population, Mortality (deaths) and Emigration (exit) decrease population.

Step 1: Sickle cell anemia is caused by a single amino acid substitution (glutamic acid replaced by valine) in the beta globin chain.

Step 2: Hemoglobin has two beta chains, so this single amino acid change is present in both beta chains.
Hence, statement (A) is correct.

Step 3: There is only one amino acid change, not two, and it does not occur in a single beta chain alone.
Hence, statement (B) is incorrect.

Step 4: Sickle cell anemia is an inherited genetic disorder affecting hemoglobin in red blood cells.
Hence, statement (C) is correct.

Step 5: Therefore, the correct statements are (A) and (C) only. Quick Tip: Sickle cell anemia is caused by a point mutation in the beta globin gene, leading to abnormal hemoglobin (HbS) and sickle-shaped red blood cells.

Step 1: Induction refers to the process of switching on a gene by an inducing molecule.
So, (A) \(\rightarrow\) (IV).

Step 2: Repression involves switching off gene expression using a repressor or other agents.
So, (B) \(\rightarrow\) (I).

Step 3: Transfection is the introduction of foreign DNA (plasmid) into eukaryotic cells.
So, (C) \(\rightarrow\) (II).

Step 4: Transformation is the uptake of plasmid DNA by prokaryotic cells such as bacteria.
So, (D) \(\rightarrow\) (III).

Step 5: Hence, the correct matching is
(A)–(IV), (B)–(I), (C)–(II), (D)–(III). Quick Tip: Remember: Induction \(\rightarrow\) gene ON, Repression \(\rightarrow\) gene OFF, Transformation \(\rightarrow\) prokaryotes, Transfection \(\rightarrow\) eukaryotes.

Step 1: Bioaerosol sampling involves techniques that collect airborne biological particles based on size, inertia, or electrical charge.

Step 2: Multiple hole impacters collect particles by inertial impaction and are commonly used in bioaerosol sampling.

Step 3: Centrifugal impacters use centrifugal force to concentrate and collect airborne particles.

Step 4: Electrostatic precipitation collects bioaerosols by charging particles and attracting them to oppositely charged surfaces.

Step 5: Magnetic transformation is not used as a standard method for concentrating or sampling bioaerosols. Quick Tip: Bioaerosol sampling methods rely on inertial, centrifugal, or electrostatic principles. Magnetic techniques are not used for this purpose.

Step 1: Covaxin is developed using the whole SARS-CoV-2 virus.

Step 2: The virus is rendered non-infectious by chemical/heat inactivation so that it cannot cause disease.

Step 3: Although inactivated, the virus retains its antigenic properties and can stimulate an immune response.

Step 4: Hence, Covaxin contains a heat-inactivated (killed) form of the SARS-CoV-2 virus. Quick Tip: Vaccines can be based on: live attenuated virus, inactivated virus, subunit proteins, or nucleic acids (DNA/mRNA). Covaxin is an \textbf{inactivated whole-virus vaccine}.

Step 1: DNA replication occurs simultaneously on both strands (leading and lagging strands), not on one strand at a time.
Hence, statement (A) is incorrect.

Step 2: DNA replication is semi-conservative, meaning each daughter DNA molecule contains one parental and one newly synthesized strand.
Hence, statement (B) is correct.

Step 3: In eukaryotic cells, DNA replication begins at multiple origins along the chromosome.
Hence, statement (C) is correct.

Step 4: Eukaryotic cells possess efficient DNA repair mechanisms; mutations do not occur in every cycle due to absence of repair.
Hence, statement (D) is incorrect.

Step 5: Therefore, the incorrect statements are (A) and (D) only. Quick Tip: DNA replication facts: - It is semi-conservative - Occurs on both strands simultaneously - Starts at multiple origins in eukaryotes - DNA repair mechanisms maintain fidelity

Step 1: The isoelectric point (\(pI\)) is the pH at which an amino acid has no net charge.

Step 2: Amino acids with acidic side chains (–COOH) have low isoelectric points.

Step 3: Aspartic acid contains an extra carboxyl group in its side chain, making it acidic.

Step 4: Due to this extra acidic group, the \(pI\) of aspartic acid lies below 4.

Step 5: Hence, the amino acid with isoelectric point at \(pH < 4\) is aspartic acid. Quick Tip: Isoelectric point trends: Acidic amino acids \(\rightarrow\) low \(pI\), Basic amino acids \(\rightarrow\) high \(pI\), Neutral amino acids \(\rightarrow\) intermediate \(pI\).

Step 1: In most plants, starch is the primary storage product of photosynthesis.

Step 2: However, many temperate grasses and some dicot plants store carbohydrates differently.

Step 3: In these plants, sucrose is converted into polymers of fructose.

Step 4: These fructose polymers are collectively known as fructans.

Step 5: Hence, the correct answer is fructans. Quick Tip: Temperate grasses often store food as fructans instead of starch, which helps them tolerate cold and drought conditions.

Step 1: Glycolysis occurs in the cytoplasm and does not require oxygen.

Step 2: The Krebs cycle depends indirectly on oxygen because it requires NAD\(^+\) and FAD, which are regenerated only when the ETS operates.

Step 3: The electron transport system (ETS) requires oxygen as the final electron acceptor.

Step 4: Hence, among the three steps of respiration, only glycolysis can occur in the absence of oxygen. Quick Tip: Glycolysis is an anaerobic process, whereas Krebs cycle and ETS are aerobic processes.

Step 1: Plants of the Compositae (Asteraceae) family show a special type of inflorescence.

Step 2: In this inflorescence, numerous small flowers (florets) are densely packed on a common flat receptacle.

Step 3: This head-like arrangement gives the appearance of a single flower.

Step 4: Such an inflorescence is specifically called a capitulum. Quick Tip: Capitulum is characteristic of the Compositae (Asteraceae) family, e.g., sunflower, marigold.

Step 1: Eudicots are characterized by the presence of two cotyledons.
So, (A) \(\rightarrow\) (II).

Step 2: Ovary is the basal part of the carpel or fused carpels that contains ovules.
So, (B) \(\rightarrow\) (I).

Step 3: Apomixis is the formation of seeds and fruits without fertilization.
So, (C) \(\rightarrow\) (IV).

Step 4: Orchid roots possess a special spongy tissue called velamen.
So, (D) \(\rightarrow\) (III).

Step 5: Hence, the correct matching is
(A)–(II), (B)–(I), (C)–(IV), (D)–(III). Quick Tip: Key associations: Eudicots \(\rightarrow\) two cotyledons, Apomixis \(\rightarrow\) asexual seed formation, Orchid roots \(\rightarrow\) velamen tissue.

Step 1: Floating green microorganisms like \textit{Volvox and \textit{Spirogyra are phytoplanktons, not zooplanktons.
Hence, statement (A) is incorrect.

Step 2: \textit{Cyclops and \textit{Cypris are animals and belong to zooplanktons, not phytoplanktons.
Hence, statement (B) is incorrect.

Step 3: Benthic animals live at the bottom of water bodies; molluscs are common examples.
Hence, statement (C) is correct.

Step 4: The euphotic zone is the upper layer of water that receives sufficient sunlight for photosynthesis.
Hence, statement (D) is incorrect.

Step 5: Nektons are actively swimming aquatic animals such as fishes.
Hence, statement (E) is correct.

Step 6: Therefore, the correct statements are (C) and (E) only. Quick Tip: Aquatic groups: Phytoplankton \(\rightarrow\) photosynthetic drifters, Zooplankton \(\rightarrow\) drifting animals, Benthos \(\rightarrow\) bottom dwellers, Nekton \(\rightarrow\) active swimmers.

Step 1: Cadmium is a toxic heavy metal that can accumulate in the human body, particularly affecting bones and kidneys.

Step 2: Chronic exposure to cadmium leads to severe bone pain, bone softening, and kidney damage.

Step 3: The disease caused by cadmium poisoning was first reported in Japan and is known as Itai-Itai disease.

Step 4: Hence, cadmium poisoning causes Itai-Itai disease. Quick Tip: Heavy metal poisoning examples: Mercury \(\rightarrow\) Minamata disease, Cadmium \(\rightarrow\) Itai-Itai disease.

Step 1: Above-ground living biomass refers to the total mass of living plant material present above the soil surface.

Step 2: Sequoia (giant sequoia/redwood) trees attain enormous height, girth, and longevity compared to other tree species.

Step 3: Due to their massive size and long life span, Sequoia trees accumulate the highest above-ground biomass and nutrient content at maturity.

Step 4: Hence, among the given options, Sequoia has the highest above-ground living biomass. Quick Tip: Giant Sequoia trees are among the largest living organisms on Earth, known for their exceptional biomass and carbon storage capacity.

Step 1: Basal metabolic rate (BMR) refers to the minimum amount of energy required to maintain vital physiological functions.

Step 2: It is measured when the organism is at rest, in a post-absorptive state, and under neutral temperature conditions.

Step 3: BMR represents the lowest rate of metabolism necessary for survival.

Step 4: Hence, the energy expended under these conditions is called basal metabolic rate. Quick Tip: Basal metabolic rate is measured under strict resting conditions and reflects the minimum energy needed for basic life processes.

Step 1: Catalytic destruction of ozone is most prominent over Antarctica due to extreme cold and the presence of polar stratospheric clouds.
Hence, statement (A) is correct.

Step 2: The thickness of the ozone layer is measured in Dobson Units (DU), not in angstrom (\AA) units.
Hence, statement (B) is incorrect.

Step 3: Ozone thinning also occurs over the Arctic, though it is less severe than over Antarctica.
Hence, statement (C) is correct.

Step 4: Polar stratospheric clouds provide surfaces for reactions that release active chlorine, which destroys ozone.
Hence, statement (D) is correct. Quick Tip: Ozone layer thickness is always expressed in \textbf{Dobson Units (DU)}, not in length units like angstrom or nanometer.

Step 1: Radioactive decay follows first-order kinetics because the rate of decay is proportional to the number of undecayed nuclei present.

Step 2: For a first-order reaction, the half-life is given by: \[ T_{1/2} = \frac{0.693}{k} \]

Step 3: The half-life expression does not contain the initial concentration term.

Step 4: Hence, radioactive decay is a first-order process and its half-life is independent of the initial concentration. Quick Tip: All radioactive decay processes follow first-order kinetics, and their half-life remains constant regardless of the amount of substance present.

Step 1: Victor Ernest Shelford proposed the Law of Tolerance, which explains limits of survival for organisms.
So, (A) \(\rightarrow\) (II).

Step 2: Justus von Liebig gave the Law of Minimum, stating that growth is controlled by the scarcest resource.
So, (B) \(\rightarrow\) (I).

Step 3: David Lack proposed the Theory of Maximum Reproduction, explaining population regulation in birds.
So, (C) \(\rightarrow\) (IV).

Step 4: Charles Darwin and Alfred Russel Wallace jointly proposed the theory of Natural Selection.
So, (D) \(\rightarrow\) (III).

Step 5: Hence, the correct matching is
(A)–(II), (B)–(I), (C)–(IV), (D)–(III). Quick Tip: Key ecological laws: Liebig \(\rightarrow\) Law of Minimum, Shelford \(\rightarrow\) Law of Tolerance, Darwin & Wallace \(\rightarrow\) Natural Selection.

Step 1: Biological classification follows a hierarchical system from broader to more specific categories.

Step 2: The correct sequence is: \[ Kingdom \rightarrow Phylum \rightarrow Class \rightarrow Order \rightarrow Family \]

Step 3: Substituting the given symbols: \[ (B) \rightarrow (C) \rightarrow (A) \rightarrow (D) \rightarrow (E) \]

Step 4: Hence, the correct hierarchy is (B), (C), (A), (D), (E). Quick Tip: Remember the taxonomic hierarchy as: \textbf{Kingdom → Phylum → Class → Order → Family → Genus → Species}.

Step 1: In amensalism, one species is harmed while the other remains unaffected.
Hence, statement (A) is correct.

Step 2: Mutualism is an interaction in which both species benefit, not just one.
Hence, statement (B) is incorrect.

Step 3: Commensalism is an interaction where one species benefits and the other is neither harmed nor benefited.
Hence, statement (C) is incorrect.

Step 4: In competition, both interacting species suffer due to limited resources.
Hence, statement (D) is correct.

Step 5: Therefore, the correct statements are (A) and (D) only. Quick Tip: Key ecological interactions: Mutualism \((+,+)\), Commensalism \((+,0)\), Amensalism \((-,0)\), Competition \((-,-)\).

Step 1: The human forearm consists of two long bones: radius and ulna.

Step 2: The radius is located on the lateral (thumb) side of the forearm.

Step 3: Sternum is a flat bone in the chest, tibia and fibula are bones of the lower leg.

Step 4: Hence, the bone present in the forearm is the radius. Quick Tip: Forearm bones: Radius (thumb side) and Ulna (little finger side).

Step 1: Genetic disorders can be inherited as autosomal dominant or autosomal recessive traits.

Step 2: Huntington's disease is caused by an autosomal dominant mutation, meaning a single copy of the mutant gene is sufficient to cause the disorder.

Step 3: Cystic fibrosis, Tay–Sachs disease, and sickle cell anemia are autosomal recessive disorders.

Step 4: Hence, the genetic disorder caused by a dominant gene is Huntington's disease. Quick Tip: Autosomal dominant disorder example: Huntington's disease. Autosomal recessive disorders: Cystic fibrosis, Tay–Sachs disease, Sickle cell anemia.

Step 1: Fossils are remains or impressions of organisms preserved in rocks.

Step 2: During the Precambrian era, life forms were mostly simple, soft-bodied, and microscopic.

Step 3: Soft-bodied organisms have very low chances of fossilization.

Step 4: Hence, rocks of the Precambrian era contain the least number of fossils. Quick Tip: Precambrian era represents the earliest history of Earth, with primitive life forms and very scarce fossil records.

Step 1: The Brundtland Commission was formally known as the World Commission on Environment and Development (WCED).

Step 2: In 1987, the commission published a report defining the concept of sustainable development.

Step 3: The report was titled Our Common Future.

Step 4: Hence, the Brundtland Commission Report is referred to as \textit{Our Common Future. Quick Tip: Sustainable development was formally defined in the 1987 Brundtland Report titled \textit{Our Common Future.

Step 1: Keoladeo National Park is located in Rajasthan.
So, (A) \(\rightarrow\) (II).

Step 2: Loktak Lake is situated in Manipur.
So, (B) \(\rightarrow\) (III).

Step 3: Pong Dam Lake is located in Himachal Pradesh.
So, (C) \(\rightarrow\) (IV).

Step 4: Wular Lake is located in Jammu \& Kashmir.
So, (D) \(\rightarrow\) (I).

Step 5: Hence, the correct matching is
(A)–(II), (B)–(III), (C)–(IV), (D)–(I). Quick Tip: Important Indian wetlands: Keoladeo \(\rightarrow\) Rajasthan, Loktak \(\rightarrow\) Manipur, Pong \(\rightarrow\) Himachal Pradesh, Wular \(\rightarrow\) Jammu \& Kashmir.

Step 1: The Red List of Threatened Species is an authoritative inventory of the conservation status of biological species.

Step 2: It is published and maintained by the International Union for Conservation of Nature (IUCN).

Step 3: The Red List assesses the extinction risk of species worldwide using standardized criteria.

Step 4: Hence, the institution that publishes the Red List is IUCN. Quick Tip: The IUCN Red List categorizes species as Extinct, Endangered, Vulnerable, Near Threatened, etc., based on their risk of extinction.

Step 1: Ramsar sites are wetlands of international importance designated under the Ramsar Convention.

Step 2: According to updated state-wise data, Tamil Nadu has the highest number of Ramsar sites in India, with more than any other state. :contentReference[oaicite:0]{index=0

Step 3: The other states listed (Maharashtra, Andhra Pradesh, Gujarat) have fewer Ramsar sites in comparison. :contentReference[oaicite:1]{index=1

Step 4: Therefore, the state with the highest number of Ramsar sites as of February 2025 is Tamil Nadu. Quick Tip: Ramsar sites are distributed unevenly across Indian states, with Tamil Nadu leading in count due to numerous important wetlands designated under the Ramsar Convention. :contentReference[oaicite:2]{index=2}

Step 1: Nitrous oxide (\(N_2O\)) is a potent greenhouse gas.
Hence, (A) is correct.

Step 2: Carbon tetrachloride (\(CCl_4\)) is a halogenated compound that absorbs infrared radiation and acts as a greenhouse gas.
Hence, (B) is correct.

Step 3: Ozone (\(O_3\)) present in the troposphere acts as a greenhouse gas.
Hence, (C) is correct.

Step 4: Water vapour is the most abundant natural greenhouse gas in the atmosphere.
Hence, (D) is correct.

Step 5: Oxygen (\(O_2\)) does not absorb infrared radiation and is not a greenhouse gas.
Hence, (E) is incorrect.

Step 6: Therefore, the correct greenhouse gases are (A), (B), (C) and (D). Quick Tip: Greenhouse gases absorb infrared radiation and trap heat. Major examples include \(CO_2\), \(CH_4\), \(N_2O\), water vapour, and ozone.

Step 1: Smog produced from coal combustion releases sulphur dioxide (\(SO_2\)) into the atmosphere.

Step 2: In the presence of water vapour, \(SO_2\) forms sulphuric acid aerosols and particulate matter.

Step 3: This type of smog is rich in reducing agents and is commonly called reducing smog or London smog.

Step 4: Photochemical smog, on the other hand, is formed from nitrogen oxides and hydrocarbons under sunlight.

Step 5: Hence, the smog described is reducing smog. Quick Tip: Reducing (London) smog is associated with coal burning, high \(SO_2\) levels, fog, and cool humid conditions, whereas photochemical smog occurs in sunny, warm climates.

Step 1: Environmental ethics deals with moral relationships between humans and the natural environment.

Step 2: Ecocentrism places intrinsic value on all components of ecosystems, including living and non-living entities.

Step 3: Unlike anthropocentrism, ecocentrism does not prioritize human interests over nature.

Step 4: Hence, the philosophy that grants direct moral consideration to the environment is ecocentrism. Quick Tip: Anthropocentrism is human-centered, while ecocentrism is nature-centered and values ecosystems intrinsically.

Step 1: Landfill sites contain large amounts of organic waste that undergo anaerobic decomposition.

Step 2: Anaerobic decomposition produces methane as a major gas.
Hence, methane is emitted from landfills.

Step 3: Decomposition of nitrogen- and sulphur-containing organic matter releases ammonia and hydrogen sulphide.
Hence, \(NH_3\) and \(H_2S\) are emitted.

Step 4: Ozone is not directly emitted from landfill sites; it is a secondary pollutant formed in the atmosphere by photochemical reactions.

Step 5: Therefore, ozone is not emitted from landfill sites. Quick Tip: Landfill gases mainly include methane, carbon dioxide, ammonia, and hydrogen sulphide. Ozone is formed photochemically in the atmosphere, not emitted directly.

Step 1: The reference sound pressure in air is: \[ p_0 = 20\,\muPa = 2 \times 10^{-5}\,Pa \]

Step 2: Use the sound pressure level (SPL) formula: \[ SPL = 20 \log_{10}\!\left(\frac{p}{p_0}\right) \]

Step 3: Substitute the given values: \[ SPL = 20 \log_{10}\!\left(\frac{0.002}{2 \times 10^{-5}}\right) = 20 \log_{10}(100) \]

Step 4: Evaluate: \[ \log_{10}(100) = 2 \] \[ SPL = 20 \times 2 = 40\,dB \] Quick Tip: Sound pressure level in air is calculated using \[ SPL = 20 \log_{10}\!\left(\frac{p}{20\,\muPa}\right). \] An increase by a factor of 10 in pressure corresponds to a \(20\,dB\) increase.

Step 1: Continuous inhalation of coal dust in coal mines leads to deposition of dust particles in the lungs.

Step 2: This causes chronic lung disease characterized by inflammation and fibrosis.

Step 3: The disease is medically termed coal workers’ pneumoconiosis, also known as anthracosis.

Step 4: Hence, black lung disease is pneumoconiosis or anthracosis. Quick Tip: Occupational lung diseases: Coal dust \(\rightarrow\) Pneumoconiosis (Anthracosis), Silica dust \(\rightarrow\) Silicosis, Asbestos \(\rightarrow\) Asbestosis.

Step 1: The littoral zone is the shallow region of a lake near the shore.

Step 2: This zone receives sufficient sunlight that reaches the bottom.

Step 3: Because light reaches the lake bottom, rooted aquatic plants can grow here.

Step 4: Limnetic and profundal zones are deeper and do not support rooted vegetation.

Step 5: Hence, rooted vegetation is supported in the littoral zone. Quick Tip: Lake zones: Littoral \(\rightarrow\) shallow, light reaches bottom, rooted plants present; Limnetic \(\rightarrow\) open surface water; Profundal \(\rightarrow\) deep, no light.

Step 1: Metallic minerals are those from which metals can be extracted economically.

Step 2: Bauxite is the principal ore of aluminium, a metallic element.

Step 3: Dolomite, gypsum, and kyanite are non-metallic minerals.

Step 4: Hence, the metallic mineral among the given options is bauxite. Quick Tip: Metallic minerals yield metals on processing (e.g., bauxite \(\rightarrow\) aluminium), whereas non-metallic minerals do not.

Step 1: Dissolved oxygen (DO) in water depends strongly on temperature.

Step 2: Solubility of oxygen in water increases as temperature decreases.

Step 3: In winter, colder water can hold more dissolved oxygen than warmer summer water.

Step 4: Therefore, dissolved oxygen remains higher in cold winter compared to summer. Quick Tip: Cold water holds more dissolved gases than warm water. Hence, dissolved oxygen levels are higher in winter than in summer.

Step 1: Average salinity of ocean water is commonly expressed as \(35\,ppt\) (parts per thousand).

Step 2: Converting parts per thousand to percentage: \[ 35\,ppt = \frac{35}{1000} \times 100 = 3.5% \]

Step 3: Hence, the average salinity of ocean water is \(3.5%\). Quick Tip: Ocean water salinity is typically expressed as \(35\,ppt\), which is numerically equal to \(3.5%\).

Step 1: Soil development is governed by five classic soil-forming factors, commonly remembered as CLORPT.

Step 2: Climate controls temperature and rainfall, influencing weathering and leaching.

Step 3: Organisms (plants, animals, microbes) contribute organic matter and affect soil structure.

Step 4: Relief (topography) influences drainage, erosion, and soil depth.

Step 5: Parent material determines the mineral composition of the soil.

Step 6: Time allows soil-forming processes to act and horizons to develop.

Step 7: Hence, all the given factors influence soil development. Quick Tip: Soil-forming factors = \textbf{CLORPT}: Climate, Organisms, Relief, Parent material, Time.

Step 1: The Water (Prevention and Control of Pollution) Act was enacted in 1974.
So, it is the oldest among the given Acts.

Step 2: The Air (Prevention and Control of Pollution) Act was enacted in 1981.

Step 3: The Environment (Protection) Act was enacted in 1986, after the Bhopal gas tragedy.

Step 4: The Biodiversity Act was enacted in 2002, making it the most recent among the given Acts.

Step 5: Hence, the correct chronological order (old to recent) is: \[ (D) \rightarrow (C) \rightarrow (A) \rightarrow (B) \] Quick Tip: Important environmental Acts in India (chronological): Water Act (1974) \(\rightarrow\) Air Act (1981) \(\rightarrow\) Environment Protection Act (1986) \(\rightarrow\) Biodiversity Act (2002).

Step 1: Use Beer–Lambert law: \[ A = \varepsilon \, l \, c \]
where \(A\) = absorbance, \(\varepsilon\) = molar extinction coefficient, \(l\) = path length (standard cuvette \(= 1\,cm\)), \(c\) = concentration.

Step 2: Substitute the given values: \[ 0.24 = 96000 \times 1 \times c \]

Step 3: Solve for concentration: \[ c = \frac{0.24}{96000} = 2.5 \times 10^{-6}\,M \]

Step 4: Convert molarity to micromolar: \[ 2.5 \times 10^{-6}\,M = 2.5\,\muM \] Quick Tip: For spectrophotometry problems, remember: \[ A = \varepsilon l c \] For a standard cuvette, \(l = 1\,cm\), which simplifies calculations.

Step 1: In some lower animals, light does not determine the direction of movement but affects the speed of movement.

Step 2: Such non-directional response to light, where locomotory speed changes with light intensity, is called kinesis.

Step 3: When this response is specifically due to light, it is termed photokinesis.

Step 4: Hence, regulation of speed of locomotion by light is known as photokinesis. Quick Tip: Phototaxis \(\rightarrow\) directional movement toward/away from light. Photokinesis \(\rightarrow\) change in speed due to light intensity.

Step 1: Apply Beer–Lambert law: \[ A = \varepsilon \, l \, c \]

Step 2: Substitute the given values: \[ 0.2 = 10000 \times 1 \times c \]

Step 3: Calculate concentration: \[ c = \frac{0.2}{10000} = 2 \times 10^{-5}\,M \]

Step 4: Convert molarity to micromolar: \[ 2 \times 10^{-5}\,M = 20\,\muM \] Quick Tip: At \(280\,nm\), protein absorbance mainly arises from aromatic amino acids (tryptophan and tyrosine). For a \(1\,cm\) cuvette, concentration is simply \(c = A/\varepsilon\).

Step 1: The true value of the standard lies in the range: \[ 15 \pm 0.6 \Rightarrow 14.4 to 15.6\,ppb \]

Step 2: Most of the reported values lie close to \(15\,ppb\) and within or very near the acceptable range, indicating good accuracy.

Step 3: The measurements are closely clustered with small variation among themselves, indicating good precision.

Step 4: Since the values are both close to the true value and close to each other, the data is both accurate and precise. Quick Tip: Accuracy refers to closeness to the true value, while precision refers to closeness among repeated measurements. Good analytical data should ideally be both accurate and precise.

Step 1: Bragg’s equation is given by: \[ n\lambda = 2d \sin \theta \]
where \(d\) is the interplanar spacing of crystal planes and \(\theta\) is the angle of incidence.

Step 2: This equation explains the diffraction of X-rays by crystal lattices.

Step 3: X-ray diffractometers utilize Bragg’s law to determine crystal structure and interplanar spacing.

Step 4: The other instruments listed operate on absorption or emission principles, not diffraction. Quick Tip: Bragg’s law is exclusively associated with \textbf{X-ray diffraction (XRD)} and is used to study crystal structures.

Step 1: Biological Oxygen Demand (BOD) measures the amount of oxygen used by microorganisms to oxidize organic matter.
So, (A) \(\rightarrow\) (III).

Step 2: Chemical Oxygen Demand (COD) measures the total oxygen required to oxidize organic and inorganic substances in water.
So, (B) \(\rightarrow\) (IV).

Step 3: Eutrophication refers to nutrient enrichment leading to excessive biological growth.
So, (C) \(\rightarrow\) (I).

Step 4: Turbidity indicates the clarity of water and is measured by light transmission.
So, (D) \(\rightarrow\) (II). Quick Tip: BOD \(\rightarrow\) biological oxidation of organic matter, COD \(\rightarrow\) chemical oxidation (organic + inorganic), Eutrophication \(\rightarrow\) nutrient enrichment, Turbidity \(\rightarrow\) water clarity.

Step 1: Acceleration is the time derivative of velocity: \[ a(t) = \frac{dv}{dt} \]

Step 2: Differentiate the given velocity function: \[ v(t) = a_0 t^3 + v_0 \] \[ a(t) = \frac{d}{dt}(a_0 t^3 + v_0) = 3a_0 t^2 \]

Step 3: Substitute \(t = 3\,s\): \[ a(3) = 3a_0 (3)^2 = 3a_0 \times 9 = 27a_0 \] Quick Tip: Acceleration is obtained by differentiating velocity with respect to time. Constants like \(v_0\) vanish on differentiation.

Step 1: The Bardenpho process is an advanced biological wastewater treatment process.

Step 2: It consists of alternating anaerobic, anoxic, and aerobic zones.

Step 3: These conditions allow effective biological removal of nitrogen (via nitrification and denitrification).

Step 4: The process also promotes enhanced biological phosphorus removal (EBPR).

Step 5: Hence, along with nitrogen, phosphorus is also removed in the Bardenpho process. Quick Tip: The Bardenpho process is designed for \textbf{simultaneous removal of nitrogen and phosphorus} from wastewater using biological treatment stages.

Step 1: Butterfly feeds on plant material and acts as a primary consumer.

Step 2: Dragon fly is an insect predator that feeds on smaller insects like butterflies.

Step 3: Frog feeds on insects such as dragon flies.

Step 4: Snake feeds on frogs.

Step 5: Hawks are top predators and feed on snakes.

Step 6: Therefore, the correct food chain sequence is: \[ Butterfly \rightarrow Dragon fly \rightarrow Frog \rightarrow Snake \rightarrow Hawks \] Quick Tip: In food chains, energy flows from lower trophic levels (herbivores) to higher trophic levels (top carnivores).