KMUCAT Paper 3 2025

Explanation: A

Explanation: B
The code shifts each letter +3 positions (Caesar cipher).
T → W, R → U, U → X, T → W, H → K (all +3).
Apply to LIE: L → O, I → L, E → H → OLH.

Explanation: D
A book is a tool used to gain knowledge; similarly, a weapon is a tool used for defense. (War and soldier are users/contexts, and “power” is vague—not the primary function.)

Explanation: B
A clause has a subject + verb. If it doesn’t express a complete thought, it’s a dependent (subordinate) clause.
• A phrase lacks a subject–verb pair.
• A sentence is a complete thought.
• A paragraph contains sentences.

Explanation: D
An imperative sentence gives a command, request, or instruction (e.g., “Close the door,” “Please sit down”).
• Statement = declarative, Question = interrogative, Exclamation = exclamatory.

Explanation: B
In standard biology texts, endangered species are those at serious risk of extinction (numbers critically low, “near to extinction”).
• Threatened = likely to become endangered soon.
• Vanished = extinct.
• Wild = simply living in nature.

Explanation: B
A simile compares two unlike things using “like” or “as” (e.g., “as brave as a lion,” “floats like a feather”).
• Metaphor: direct comparison without like/as.
• Personification: human traits to nonhuman things.
• Hyperbole: deliberate exaggeration.

Explanation: C
An exclamatory sentence expresses strong emotion or surprise and typically ends with an exclamation mark (e.g., “What a beautiful day!”).

Explanation: B
Speed v = 90 km/h = 90×1000/3600 = 25 m/s.
Length = distance covered in 20 s = v × t = 25 × 20 = 500 m.

Explanation: C
A metastable state is an excited state with an unusually long lifetime because the transition back to lower energy is quantum-mechanically forbidden or weak, so decay is slow. This is the state exploited in lasers and seen in phosphorescence.

Explanation: B
The Baltimore classification groups viruses by type of nucleic acid (DNA/RNA, ss/ds, +/− sense) and their pathway to mRNA (replication strategy).
• ICTV uses morphology, genetics, and taxonomy ranks.
• Linnaean is general biological taxonomy (kingdom → species).
• Gram stain classifies bacteria, not viruses.

Explanation: C
During intense exercise with insufficient O₂, muscle cells switch to anaerobic glycolysis: pyruvate is reduced to lactic acid (lactate + H⁺) to regenerate NAD⁺. The H⁺ released with lactate enters the bloodstream, where it’s only partly buffered (mainly by the bicarbonate system), so blood pH falls (lactic acidosis).
Why not the others?
A. CO₂: Most CO₂ comes from aerobic metabolism (TCA cycle), not anaerobic glycolysis.
B. Ethanol: Humans don’t produce ethanol in muscle metabolism.
D. Oxygen removal: Low O₂ triggers anaerobic metabolism but doesn’t itself acidify blood; the H⁺ from lactic acid does.

Explanation: B
• A rise in heart rate → higher cardiac output → more O₂ delivery to working muscles and faster removal of CO₂, heat, and lactate.
• With better oxygen supply, muscles can meet more of their ATP demand aerobically, so their reliance on anaerobic glycolysis (which produces lactic acid) is reduced.
Why not the others
A. Increase digestion of carbohydrates: During intense exercise, sympathetic activity suppresses digestion.
B. Increase ventilation: Ventilation rises mainly due to respiratory center control (chemoreceptors, muscle/joint feedback). It’s not an indirect effect of heart rate itself.
D. Reduce need for sweating: Heat production increases in a race; sweating is more needed, not less.
Note: A 200 m sprint is still largely anaerobic, but the rise in heart rate lessens (doesn’t eliminate) the amount of anaerobic contribution required compared with a lower cardiac output.

Explanation: C
Centrioles (within the centrosome) organize microtubules and help form the mitotic spindle that separates chromosomes during cell division.
• Photosynthesis → chloroplasts (plants/algae).
• Detoxification → smooth ER/peroxisomes.
• Protein modification → ER/Golgi apparatus

Explanation: D
Blood from the ileum passes first to the liver via the hepatic portal vein. This “first-pass” route lets the liver detoxify harmful substances (e.g., bacterial toxins, drugs) before they enter the general circulation, protecting tissues.
Why not the others?
A. The liver does deaminate excess amino acids and convert ammonia to urea, but that’s a specific case; the broader protective advantage is detoxification of many absorbed substances.
B. The liver converts excess glucose → glycogen, but glucose is not normally “excreted”; wording is off.
C. Long-chain fat products mostly enter lymph as chylomicrons and bypass the liver initially, so this is incorrect.

Explanation: A
After eating (especially sugars), oral bacteria like Streptococcus mutans ferment the sugars to organic acids (e.g., lactic acid). These acids lower the mouth’s pH. Saliva generally buffers/neutralizes acids (it doesn’t secrete acid), and neurons don’t release acids. Lower pH (≈ below 5.5) promotes enamel demineralization.

Explanation: C
The ulna has the olecranon process (the bony tip of the elbow), which is the insertion of the triceps brachii—the main extensor of the forearm.
A: The elbow hinge is mainly humerus–ulna (trochlea ↔ trochlear notch); the radius–ulna articulation at the proximal end is a pivot joint (not the elbow hinge).
B: The ulna does not form a joint at the shoulder; that’s the humerus–scapula (ball-and-socket).
D: The ulna doesn’t link the radius to the scapula; the radius and ulna form the forearm, while the scapula is part of the shoulder girdle.

Explanation: C
ABO blood groups are controlled by three alleles: I^A, I^B, i.
• Phenotype A = I^AI^A or I^Ai
• Phenotype B = I^BI^B or I^Bi
To have one child with A and another with B in the same family, the parents must be able to pass on I^A to one child and I^B to another. This is possible only if at least one parent carries two different alleles (is heterozygous).
Examples that work:
• AB × O (I^AI^B × ii) → children: I^Ai (A) or I^Bi (B)
• A (I^Ai) × B (I^Bi) → children can be A, B, AB, or O
Counterexamples:
• Both parents homozygous (e.g., I^AI^A × I^BI^B, or I^AI^A × ii, or I^BI^B × ii) cannot produce both A and B children.
So, the only conclusion that must be true is: at least one parent is heterozygous.

Explanation: A
The cerebellum coordinates posture, balance, and the timing/precision of movements. Damage causes ataxia, tremors, and unsteady gait (poor coordination).
B. (temperature control): hypothalamus.
C. (heartbeat): medulla/ANS centers.
D. (visual recognition): occipital/association cortex.

Explanation: C
Pinocytosis is “cell drinking,” where the cell membrane invaginates to take in extracellular fluid and dissolved solutes in small vesicles.
• Phagocytosis = “cell eating” (solid particles).
• Exocytosis = vesicles fuse outwards to export substances.
• Chloroplast = organelle for photosynthesis (not a transport process).

Explanation: D
• Glycogenesis = synthesis of glycogen from glucose (storage form in liver & muscle).
• Gluconeogenesis (A) = making glucose from non-carbohydrate sources.
• Glycogenolysis (B) = breakdown of glycogen to glucose.
• Glycolysis (C) = breakdown of glucose → pyruvate for ATP.

Explanation: D
• Maltose, lactose, sucrose are all disaccharides (two monosaccharides linked by a glycosidic bond).
• Fructose is a monosaccharide (a ketohexose), not a disaccharide.

Explanation: A
Glucose and fructose join via an O-glycosidic linkage to form sucrose—specifically an α(1 → 2) glycosidic bond between C1 of α-D-glucose and C2 of β-D-fructofuranose. Peptide and ester bonds are for proteins and lipids, respectively; hydrogen bonds are noncovalent.

Explanation: D
• Apoenzyme = the protein part alone, lacking its required cofactor → inactive.
• Holoenzyme = apoenzyme + cofactor (active form).
• Coenzyme = organic cofactor (e.g., NAD⁺).
• Prosthetic group = tightly bound cofactor.
• Active site = region where substrate binds (not the whole enzyme).

Explanation: A
Coenzymes are organic cofactors, and most are derived from vitamins (especially B-complex)—e.g., NAD⁺ from niacin (B3), FAD from riboflavin (B2), CoA from pantothenic acid (B5), TPP from thiamine (B1), PLP from pyridoxine (B6), biotin, cobalamin (B12).
Minerals (metal ions like Zn²⁺, Mg²⁺) are inorganic cofactors, not coenzymes. Hormones are signaling molecules, not cofactors.

Explanation: C
• mRNA carries the codon sequence copied from DNA to the ribosome, where it’s translated into a polypeptide.
• tRNA (not mRNA) brings amino acids to ribosomes (A).
• DNA replication (B) is done by DNA polymerases, not mRNA.
• rRNA (not mRNA) forms core ribosomal structure (D).

Explanation: C
The dark reaction (Calvin–Benson cycle) fixes CO₂ via RuBP and the first stable product formed is 3-phosphoglycerate (3-PGA), a 3-carbon compound.
(Note: In C₄ plants, the first product of the Hatch–Slack pathway is a 4-carbon oxaloacetate, but the Calvin cycle itself still yields 3-PGA.).

Explanation: C
Tobacco mosaic virus (TMV) has a rigid, rod-like helical capsid made of repeating protein subunits wrapped around its RNA.
A Spherical → many animal viruses (e.g., orthomyxoviruses)
B Tadpole-like → typical of some bacteriophages (head + tail)
D Polyhedral → e.g., adenoviruses (icosahedral)

Explanation: D
• SARS — caused by a coronavirus (virus).
• FIV — Feline Immunodeficiency Virus (retrovirus).
• SV — Simian virus (e.g., SV40), a virus.
• Mycoplasma — bacteria (wall-less prokaryotes), not viruses.

Explanation: A
• Hepatitis A virus (HAV) spreads primarily by the fecal–oral route: ingestion of food or water contaminated with infected feces (e.g., poor hand hygiene, contaminated shellfish, unsafe water). Outbreaks often occur where sanitation is inadequate.
• Hepatitis B (HBV): transmitted via blood and body fluids (unprotected sex, shared needles, perinatal from mother to baby, needlestick injuries). Not fecal–oral.
• Hepatitis C (HCV): mainly blood-borne (shared needles, transfusions before screening, needlestick); sexual/perinatal transmission is less common than HBV.
• (For completeness: Hepatitis E (HEV)—not listed here—also spreads fecal–orally, often via contaminated water, especially in areas with poor sanitation.)
Quick contrasts
• Incubation: HAV ~2–6 weeks; HBV ~6 weeks–6 months; HCV ~6–10 weeks.
• Chronicity: HAV does not become chronic; HBV and HCV can become chronic (especially HCV).
• Vaccines: Vaccines available for HAV and HBV; no widely used vaccine for HCV.
So, the only option transmitted by the oro-fecal (fecal–oral) route is Hepatitis A.

Explanation: D
Tuberculosis & leprosy are caused by Mycobacterium (acid-fast actinomycetes).
Tuberculosisis caused by Mycobacterium tuberculosisand leprosy by Mycobacterium leprae.
• Streptomyces (A) are soil bacteria best known for producing antibiotics, not TB or leprosy.
• Actinomycetes (B) is a broad group (e.g., Actinomyces, Streptomyces); they don’t include the disease agents above in standard school classification.
• Mycoplasma (C) are wall-less bacteria causing atypical pneumonia, not TB or leprosy.
Since Mycobacterium is not an option, the correct choice is D. None of these.

Explanation: D
• Pseudomonas → rod-shaped (bacillus)
• Clostridium → rod-shaped (bacillus)
• Salmonella → rod-shaped (bacillus)
• Streptococcus → spherical bacteria in chains (cocci), not bacilli.

Explanation: D
Tobacco smoke triggers inflammation and enzyme activity that destroys elastic fibres in alveolar walls. Loss of elasticity causes air trapping, reduced surface area for gas exchange, and breathlessness—hallmarks of emphysema (a form of COPD). Bronchitis = airway inflammation/mucus; pneumonia = alveoli filled with fluid; lung cancer = uncontrolled cell growth.

Explanation: C
The primary function of the respiratory system is gas exchange—bringing O₂ into the body and removing CO₂. Smell involves olfactory receptors (sense function, not the core “basic ability”), electrolytes are transported via blood/kidneys, and creatinine is excreted by the kidneys, not the lungs.

Explanation: C
An efficient respiratory surface is thin, moist, large in area, and richly supplied with blood to maintain steep diffusion gradients—blood delivers CO₂ to the surface and carries away O₂. Lymph isn’t used for gas transport, and “supply of gases” is achieved by ventilation, not by a vascular supply.

Explanation: D
Airway pathway: trachea → bronchi → bronchioles → terminal/respiratory bronchioles → alveolar ducts → alveolar sacs → alveoli.
Bronchioles ultimately open into alveolar sacs, which are clusters composed of many alveoli (the actual gas-exchange units). Hence both alveoli and alveolar sacs are correct endpoints in this context.

Explanation: C
A nerve impulse (action potential) is the wave of electrochemical change—rapid depolarization and repolarization—propagating along a neuron’s membrane.
• Reflex arc: the pathway (receptor → sensory → CNS → motor → effector).
• Reflex action: the automatic response produced via the reflex arc.
• Influx: inward flow (e.g., Na⁺ influx), not the whole traveling signal.

Explanation: A
In non-myelinated neurons, the action potential is regenerated at every point along the axon membrane, producing continuous conduction.
• Saltatory conduction happens only in myelinated axons, jumping node-to-node.
• Calling it simply “electrical” is incomplete (it’s electrochemical).
• So “All of them” is incorrect.

Explanation: A
Testosterone, estrogen, and progesterone are all steroid hormones, synthesized from cholesterol. They’re lipid-soluble, act via intracellular receptors, and modulate gene transcription. (Amino acid/protein/polypeptide options refer to peptide or amine hormones like insulin, ADH, or adrenaline.)

Explanation: B
In the pancreas, the islets of Langerhans contain:
• β (beta) cells → insulin (lowers blood glucose)
• α (alpha) cells → glucagon
• δ (delta) cells → somatostatin
• Acinar cells (exocrine pancreas) → digestive enzymes, not hormones.

Explanation: B
Modern insulin is produced by recombinant DNA: the human insulin gene is inserted into a plasmid and introduced into bacteria (usually E. coli), which then synthesize human insulin in large fermenters. 
Note: Some products also use yeast—fungi—but the classic and most common answer is bacteria.

Explanation: B
• Hypothalamus (preoptic area) = the body’s thermostat. It receives input from skin (peripheral) and core (central) thermoreceptors and compares to the set-point (~37°C). It then triggers appropriate responses.
• Blood vessels near the skin = major effectors for heat exchange.
• Vasodilation → increases blood flow to skin → more heat lost by radiation, convection, and conduction.
• Vasoconstriction → reduces skin blood flow → conserves heat.
• Skeletal muscles = effectors for shivering thermogenesis. Rapid, involuntary contractions generate heat when core temperature falls.
Note: Sweat glands are also important effectors (evaporative cooling), but the option set that includes them (A) replaces the hypothalamus with the cerebellum, which makes it incorrect.
Why the other options are wrong
A. … cerebellum …
The cerebellum coordinates movement and balance; it does not regulate temperature. (Sweat glands are correct, but the cerebellum makes the whole set wrong.)
C. Kidneys … cerebellum …
Kidneys regulate fluid/electrolyte balance and acid–base status, not moment-to-moment thermoregulation; cerebellum also not involved.
D. Kidneys …
Again, kidneys aren’t part of the immediate thermoregulatory control loop, so the set isn’t “all involved.”
Summary: Thermoregulation = hypothalamic control (integrator) + skin blood vessels and skeletal muscles (effectors), with sweat glands as additional effectors for heat loss. Hence B.

Explanation: C
• All viruses are obligate intracellular parasites: they must enter living host cells to replicate because they lack their own ribosomes, ATP-generating systems, and metabolic machinery.
Why the others are wrong
A. Most of the cell is cytoplasm — Viruses are acellular: no cytoplasm, just a nucleic acid core inside a protein capsid (± lipid envelope).
B. Their outer layer is cellulosic — Viral coats are protein capsids (some with a lipid envelope); cellulose is in plant cell walls, not viruses. Bacteria have peptidoglycan; fungi have chitin/glucans.
D. They have a nucleus containing DNA or RNA — Viruses have no nucleus. They possess either DNA or RNA (never both) enclosed in a capsid. Bacteria lack a true nucleus; fungi do have one, but the statement isn’t true for viruses.

Explanation: D
Per one glucose in aerobic respiration:
• Glycolysis: 0 CO₂
• Link reaction (pyruvate → acetyl-CoA): 2 CO₂ (1 per pyruvate × 2)
• Krebs cycle: 4 CO₂ (2 per turn × 2 turns)
Total = 6 CO₂ per glucose.
For two glucose molecules: 6 × 2 = 12 CO₂.

Explanation: B
• A chromosome is made of one very long, continuous DNA double helix plus associated proteins (histones).
• Along this DNA, discrete functional segments called genes occur in sequence. So: chromosome → long DNA → genes.
Why the others are wrong
A: Mixes up terms—genes are sections of DNA, not the other way around.
C: Says genes are divided into DNA molecules—backwards; DNA contains genes.
D: Says genes include chromosomes—again inverted; chromosomes contain genes.

Explanation: B
• Tt means the organism is heterozygous at the height gene: it carries two different alleles—T and t—on a pair of homologous chromosomes.
• In diploids, an individual has two alleles of the same gene (one from each parent). Tt shows that the gene for height has ≥2 allelic forms in the population (at least T and t), though the individual has only two.
Why the others are wrong
A. Reverses terms: alleles are forms of a gene, not genes of an allele.
C. Implies two different genes for height; here T and t are alleles of the same gene, not separate genes.
D. Says “one allele … with two forms,” which is self-contradictory; alleles are the forms of a gene.

Explanation: B
At the arterial end of capillaries, capillary hydrostatic pressure (blood pressure) exceeds the opposing oncotic pressure, causing bulk filtration of water and small solutes (glucose, ions, etc.) out of the capillary into the tissue fluid (interstitial fluid). Larger proteins and cells remain in the blood. At the venous end, plasma oncotic (osmotic) pressure predominates and pulls fluid back into capillaries.
Why others are incorrect
A. Active transport: Not the main mechanism for bulk movement from blood to tissue fluid; endothelial active transport is limited (e.g., vesicular transport).
C. Capillarity: Refers to movement in narrow tubes by surface tension—not the driver of capillary-tissue exchange.
D. Osmosis: Primarily drives reabsorption into capillaries at the venous end due to plasma proteins, not the outward movement to form tissue fluid.

Explanation: C
• The right atrium only has to push blood a short distance into the right ventricle → low pressure task → thin wall.
• The right ventricle must pump blood through the pulmonary arteries to the lungs → needs higher pressure/force → thicker muscular wall.
Typical pressures: right atrium ~0–8 mmHg vs right ventricle systolic ~15–30 mmHg — reflecting the greater work the ventricle does.
Why the others are wrong
A. Different oxygen concentrations — Myocardial thickness depends on work/pressure load, not O₂ levels.
B. Right atrium pumps blood around the body — False; the left ventricle does systemic pumping.
D. Right ventricle receives more blood than the right atrium — In steady state, flow volumes are equal through all chambers (no net gain/loss).

Explanation: A
A dialysis machine uses a semipermeable membrane that lets small solutes (e.g., urea, salts/ions) and water cross, but retains large molecules and cells. Plasma proteins (like albumin, ~66 kDa) are too large to pass through the membrane pores, so they do not diffuse into the dialysate.
Check on the other options
• Salts (B) and Urea (C): small, freely diffusible; move down their concentration gradients into dialysate.
• Water (D): crosses the membrane (mainly via ultrafiltration/osmosis) to remove excess fluid.
Hence, proteins stay in the blood; the rest can move out.

Explanation: A
Syphilis is a sexually transmitted infection caused by the spirochete bacterium Treponema pallidum. It’s a tightly coiled, motile bacterium detected by dark-field microscopy or serologic tests (e.g., VDRL/RPR with treponemal confirmatory tests).
Why the others are wrong
B. Mycoplasma pneumoniae — causes atypical (“walking”) pneumonia.
C. Plasmodium — protozoa causing malaria.
D. Vibrio cholerae — bacterium causing cholera.

Explanation: B
After vaccination, memory B cells undergo class switching and affinity maturation. On re-exposure (or sustained response), they predominantly produce high-affinity IgG, which has a long serum half-life (~21 days) and persists for years via ongoing memory responses—hence long-term immunity.
Why the others are not correct
IgA: Main mucosal antibody (tears, saliva, gut, breast milk); not the primary driver of systemic long-term post-vaccine protection.
IgM: First antibody made in primary response; short-lived, lower affinity (pentamer).
IgE: Involved in allergy and defense against helminths; not responsible for vaccine-induced long-term protection

Explanation: C
• Males are XY: if they inherit one mutant allele on the X (from their mother), they express the disease (no second X to mask it).
• Females are XX: they would need two mutant copies (one on each X) to be affected; with one mutant copy they’re typically carriers and usually asymptomatic.
• Classic hemophilias: A (factor VIII deficiency) and B (factor IX deficiency)—both X-linked recessive.
Why the others are wrong
A. Dominant autosomal — Would affect males and females similarly across generations.
B. Y-linked — Would pass father → son only; hemophilia does not follow this pattern.
D. Acquired later due to liver disease — Acquired coagulopathies exist but are not the reason hemophilia is more common in males; the sex difference is due to X-linked inheritance.
Quick inheritance note: A carrier mother has a 50% chance to pass the mutant X to each child → 50% of sons affected, 50% of daughters carriers.

Explanation: D
Right after ovulation, the ruptured follicle becomes the corpus luteum, which starts secreting large amounts of progesterone. By about 3 days post-ovulation, progesterone is clearly rising toward its mid-luteal peak (≈day 21 in a 28-day cycle), preparing and maintaining the endometrium.
Why not the others
• FSH (A) and LH (B): both fall after the pre-ovulatory LH/FSH surge.
• Oestrogen (C): it drops right after ovulation (though there’s a smaller secondary rise later), so it’s not higher than in the few days before ovulation, when it was near its peak.

Explanation: B
Endometriosis is the presence of endometrial-like tissue outside the uterus (e.g., ovaries, pelvic peritoneum). It’s a female-only condition and a well-known cause of infertility. It can impair fertility by:
• Pelvic adhesions/distorted anatomy → blocked or tethered fallopian tubes; impaired ovum pickup.
• Ovarian involvement (endometriomas) → reduced ovarian reserve/ovulation issues.
• Inflammation & cytokines in peritoneal fluid → toxic to gametes/embryos; impaired fertilization and implantation.
• Luteal/implantation defects due to altered endometrial receptivity.
It does not occur in males, so options A and C are incorrect; D is wrong because it can indeed cause infertility in affected females.

Explanation: C
The cerebellum coordinates and fine-tunes voluntary movements and maintains posture and balance (equilibrium). Damage causes ataxia, intention tremor, and balance problems.
Why not the others
• Medulla oblongata (A): autonomic functions (breathing, heart rate), reflexes.
• Cerebrum (B): initiates/plans voluntary movement (motor cortex) but is not the primary center for balance/coordination.
• Hypothalamus (D): homeostasis (temperature, hunger), endocrine control—not movement/balance coordination.

Explanation: B
Work has the same dimensions as energy:
Work = Force × distance  ⇒  [MLT⁻²] [L]=[ML²T⁻²]
Torque is also force × perpendicular distance, so it has the same dimensions: [ML²T⁻²].
Why the others are wrong
• Impulse (A): Force × time → [MLT⁻¹].
• Power (C): Work ÷ time → [ML²T⁻³].
• Angular momentum (D): r × p → [ML²T⁻¹].
Note: Work (a scalar, joule = N·m) and torque (an axial vector, N·m) share dimensions but are not interchangeable quantities.

Explanation: D
Magnetic flux Φ has unit weber (Wb), and 1 Wb = 1V.
So the time rate of change dΦ/dt has units Wb/s = V, i.e., potential difference. By Faraday’s law, the induced emf (a voltage) is E=-dΦ/dt.
Dimensional check (base units):
• [Φ] = Wb = V ⇒ [Φ] = ML² T^–2 I^–1
• [dΦ/dt] = ML²T^–3 I^–1 =[Volt]
Why the others are wrong
• Current (A): [I]
• Resistance (Ω): [V/A]
• Magnetic induction (Tesla, T ): [Wb/m²]
Only potential difference matches  dΦ/dt.

Explanation: C
The Boltzmann constant k_B links temperature to energy (e.g., average kinetic energy = 3/2 k_BT) and has the SI-defined exact value k_B=1.380649 × 10^–23 J, K^–1 (option C matches to two/three sig figs).
Why others are wrong
A. 6.022 × 10²³ is Avogadro’s number (unitless), not k_B.
B. 1.8 × 10^–23 has the right units but wrong value.
D. 8.314 J. K^–1 is the gas constant R and should be 8.314 J.  K^–1; not k_B.

Explanation: B
Uniform acceleration means constant acceleration. For a body of fixed mass m, Newton’s second law gives a = F_net/m. Thus:
• If F_net​ is constant and nonzero, a is constant and nonzero → uniform acceleration.
• If F_net = 0, then a = 0 → no acceleration (uniform velocity, not uniform acceleration).
• If F_net​ increases uniformly with time, then a increases with time → non-uniform acceleration.

Explanation: D

Explanation: C
Two waves of the same frequency and amplitude traveling in opposite directions superpose to form a standing wave. The displacement becomes
y(x ,t )=2A cos(kx) sin(ωt),
showing fixed nodes (where cos(kx) = 0, displacement always zero) and antinodes (max oscillation). Energy oscillates locally; there’s no net energy transport along the medium.
Why not the others
A (Transverse) and B (Longitudinal): Those describe the type of the medium’s oscillation; standing waves can be either transverse (on a string) or longitudinal (in air columns).
D (Electromagnetic waves): Not implied; the phenomenon applies to any wave type meeting the given conditions.

Explanation: C
In a standing wave, adjacent nodes are separated by λ/2. Antinodes lie midway between nodes. Therefore, the distance from a node to the nearest antinode is (λ/2)/2 = λ/4.

Explanation: B
On a horizontal surface with friction, the frictional force does negative work (opposes motion). For the body to move against friction, an applied force in the direction of motion must do positive work on the surroundings (energy goes into overcoming friction and becomes heat).
• If speed is constant, the network is zero, but the work done by the body (applied force) is still positive, equal in magnitude to the negative work of friction.
• If the body speeds up, the applied work exceeds the frictional loss (still positive).

Explanation: D

Explanation: B
When a wave passes from one medium to another, the source still determines the oscillation rate, so frequency stays the same. The speed changes to the value in the new medium, so the wavelength adjusts via v = fλ. Intensity usually changes due to reflection/absorption at the boundary.

Explanation: B
Polarization is a defining property of transverse waves: their oscillations are perpendicular to the direction of propagation, so you can specify (or filter) the direction of vibration—that’s polarization.
Why the others are wrong
A. Not all transverse waves involve oscillating atoms (e.g., electromagnetic waves have oscillating fields, not atoms).
C. Not all transverse waves travel in a vacuum; only EM waves do—mechanical transverse waves (like on a string) require a medium.
D. Since A is false, “Both A & C” is false.

Explanation: C

Explanation: D
Irreversible (spontaneous) processes produce entropy. By the Clausius inequality, dS > δQ_rev/T for an irreversible change, so entropy increases. In an isolated system (δQ = 0), this gives ΔS > 0. (For reversible processes, ΔS_univ = 0.)
Eliminate others
• A. Decreases: violates the second law (for an isolated system).
• B. Remain constant: true only for reversible processes.
• C. Zero: again only for reversible processes (net change of the universe).

Explanation: D
The thermodynamic (Kelvin) scale is defined from fundamental laws (Carnot/second law and, in SI, the fixed value of k_B​), so it’s independent of the working substance used to measure temperature.
• Celsius/Centigrade (A/C) and Fahrenheit (B) are empirical scales tied to properties of particular substances (e.g., water), hence not universally substance-independent.

Explanation: A
Carnot efficiency depends only on the reservoir temperatures (in kelvin):
η = 1 − T_c/T_h.​​
If the source (hot reservoir) temperature T_h​ increases while the sink T_c​ stays the same, the ratio T_c​/T_h​ gets smaller, so η increases. (Conversely, raising Tc would lower efficiency.).

Explanation: B
The electric field is the negative gradient of electric potential:
This points in the direction of greatest decrease of potential. Units match:
V/m = N/C.
• A/C are energies, not fields.
• D (electron volt) is a unit of energy, not a field.

Explanation: A
• If two conductors have the same resistivity ρ (at a given reference temperature), they also have the same conductivity σ = 1/ρ → D is not helpful.
• Resistance R = ρL/A and conductance G = 1/R depend on the dimensions L,A, not just the material → B and C can’t uniquely differentiate materials.
• The temperature coefficient of resistivity α is a material property describing how ρ changes with temperature; different metals can share the same ρ at one temperature but have different α. Hence A

Explanation: B
In a bipolar junction transistor (BJT) operating in its active region (the usual mode for amplification), the emitter–base junction is forward biased so carriers are injected from the emitter into the base. The collector–base junction is simultaneously reverse biased to sweep those carriers into the collector, enabling current amplification.

Explanation: D
Standard resistivity of copper at ∼20° C is about 1.68 × 10^–8 Ω ⋅ m (often rounded to 1.7 × 10^–8).
• 1.52 × 10^–8 and 1.54 × 10^–8are closer to silver (∼1.59 × 10^–8 Ω⋅ m).
• 5.00 × 10^–8 is far too high for copper (more like poorer conductors).
Since none of the given numbers matches copper’s accepted value, D is correct.

Explanation: B
• The magnetic part of the Lorentz force is the cross product of velocity and magnetic field, scaled by charge: F_m = qv × B.
• Direction: given by the right-hand rule (perpendicular to both v and B ).
• Magnitude: |F_m | = qv B sinθ (zero if v ‖B ).
Why others are wrong
A. Writes  q × v × Bⁿ – a  cross product with a scalar q is not defined.
C. Uses v(q × B) – again, q × B is meaningless.
D. qE + q(v × B) is the full Lorentz force (electric + magnetic); the question asked only for the magnetic force.

Explanation: D
• Avogadro’s constant N_A = 6.022 × 10²³ mol^–1 is the number of entities in 1 mole.
• Therefore one atom corresponds to 1/N_A moles ( ≈ 1.66 × 10^–24 mol ).

Eliminate others
A. (23 g): that’s the molar mass of Na (mass of 1 mole), not one atom.
B. (N_A moles): would be an astronomically huge amount of atoms.
C. (positively charged): a sodium atom is neutral; Na⁺ is the ion.

Explanation: D
• A gram atom of an element = its atomic mass in grams = 1 mole of its atoms.
• Chlorine’s atomic mass ≈ 35.5 g (average of its isotopes), so 1 gram atom of Cl = 35.5 g of chlorine atoms = 1 mole of Cl atoms → both A and B are true.
• 71 g (C) corresponds to 1 mole of chlorine molecules (Cl₂) (a gram molecule), not a gram atom.

Explanation: B
The elementary charge ee is exactly 1.602 176 634 × 10⁻¹⁹ C by SI definition. An electron carries this charge with a negative sign, so −1.6 × 10⁻¹⁹C (to 2 – 3 sig figs).

Explanation: C

Explanation: A
In the Bohr model for hydrogen-like atoms, electron speed scales as
v_n ∝ 1/n
(precisely v_n = Zαc/n). So, the lowest principal quantum number (n = 1) has the highest electron velocity. Even in multi-electron atoms, inner-shell (lower-
n) electrons move faster due to stronger effective nuclear attraction.

Explanation: B

Explanation: B

Explanation: C
For nonpolar molecules, van der Waals (London dispersion) forces strengthen with greater molar mass and polarizability (more electrons, larger electron cloud). Among the choices:
• H₂ (2 g/mol) < N₂ (28) < O₂ (32) < Cl₂ (≈71)
Thus Cl₂ has the largest, most polarizable electron cloud and the strongest dispersion forces between its molecules.

Explanation: D

Explanation: C
Both methylamine (CH₃NH₂) and hydrogen peroxide (H₂O₂) have X–H bonds (N–H and O–H, respectively) and lone pairs on electronegative atoms, so they can donate and accept hydrogen bonds. These relatively strong intermolecular attractions greatly raise their boiling points compared with similar-mass molecules that lack H-bonding.
• Methylamine: N–H···N hydrogen bonds between molecules.
• Hydrogen peroxide: O–H···O hydrogen-bond network.
(Dipole–dipole and dispersion forces are present too, but hydrogen bonding is the dominant reason for their condensed (liquid) behavior.).

Explanation: D
• Liquid crystals are materials that show an intermediate (mesomorphic) phase with some order like solids but flow like liquids.
• Soap solutions form lyotropic liquid-crystal phases.
• DNA solutions can exhibit cholesteric/nematic liquid-crystal ordering.
• Cholesteryl formate (spelled “format” here) is a classic cholesteric liquid crystal.
• Engine oil is an ordinary isotropic liquid (a hydrocarbon mixture) and does not exhibit liquid-crystal phases.

Explanation: D
In metallic crystals, lattice points are occupied by metal cations immersed in a “sea” of delocalized electrons (metallic bonding).
Why not the others
• Ionic crystals: lattice points are alternately occupied by both cations (often metals) and anions (non-metals), not metal ions alone.
• Covalent crystals: lattice points are atoms linked in a giant covalent network (e.g., diamond, Si).
• Molecular crystals: lattice points are neutral molecules held by intermolecular forces (e.g., ice, CO₂(s), I₂).

Explanation: C
• SiO₂ (Quartz): giant 3D covalent network of Si–O–Si bonds.
• Diamond: 3D network of sp³ C–C bonds.
• Graphite: 2D covalent network sheets of sp² carbon (strong within layers; weak van der Waals between layers).
• Selenium (Se): exists as Se₈ rings or helical chains; solids are held together between molecules/chains mainly by van der Waals interactions → not a 3D covalent network.

Explanation: B
A plane of symmetry (mirror plane) splits a crystal so that each point on one side is the mirror image of a point on the other.
Why the others are wrong
• Axis of rotation: line about which the crystal can be rotated to coincide with itself.
• Center of symmetry (inversion center): every point x, y, z has an equivalent at −x,−y,−z.
• Axis of inversion: not a standard term; inversion is about a center, not an axis.

Explanation: D
For H₂ + I₂ ⇌ 2HI at constant T,
• Forward rate: r_f = k_f [H₂ ][I₂ ] – as time passes, [H₂ ]  and [I₂ ]  drop  ⇒ r_f  decreases.
• Reverse rate: r_r = k_r [HI]²  – as HI forms, [HI] rises ⇒r_r  increases.
They become equal at equilibrium (dynamic equilibrium).
Why others are wrong
A: Forward rate does not increase; it falls as reactants are consumed.
B: Both rates are affected until equilibrium.
C: Reverse rate initially increases, not decreases.

Explanation: B
A buffer solution contains a weak acid–conjugate base pair (e.g., CH₃COOH/CH₃COO^–) or a weak base–conjugate acid pair (e.g., NH₃/NH₄⁺).  By Le Chatelier’s principle, it neutralizes small added amounts of H⁺ or OH^–, so its pH changes only slightly—not perfectly constant.
Why the others are wrong
A. Has constant pH: No real solution has perfectly constant pH; buffers only minimize pH change.
C. Strong acid + strong base: This gives a neutral salt and no buffering (no weak/conjugate pair).
D. Salts of strong acids and strong bases: Such salts are typically neutral in water and do not buffer.

Explanation: D
“Non-aqueous medium” means a solvent other than water. Options A–C are aqueous (water-based) solutions, while kerosene is a non-aqueous organic solvent. Equilibrium constant expressions (in terms of activities or, in dilute cases, concentrations) are perfectly applicable in non-aqueous solvents like kerosene for reactions occurring in that medium.

Explanation: C
For a reaction whose rate depends on A as r = k[A]ⁿ, if [A] = 1 mol. pL⁻¹ r = k ⋅ 1ⁿ. (This holds for any order n, including zero order.)
Note: The units of k depend on the reaction order, but numerically the rate equals k when [A] = 1.

Explanation: B
In a multi-step (mechanistic) reaction, the rate-determining step (RDS) is the slowest elementary step—the bottleneck. It limits the overall rate because subsequent fast steps can’t proceed faster than this slow step. Often, the observed rate law reflects the molecularity/concentrations involved in the RDS (with possible modifications if fast pre-equilibria are present).
Why the others are wrong
A. Fastest step: Never limits the overall rate.
C. “Intermediate step” (unspecified): Any step could form/use intermediates; only the slowest one controls the rate.
D. First step: It could be fast or slow; being first doesn’t guarantee it’s rate-determining.

Explanation: B
In a zero-order reaction, the rate law is r = k[A]⁰ = k. The rate is independent of reactant concentration.
Examples: enzyme-catalyzed reactions at substrate saturation (Michaelis–Menten zero-order region), or certain surface/photochemical reactions where the surface or light intensity—not [A]—limits the rate.

Explanation: C
Molar mass M(CaCO₃) = 40.08(Ca) + 12.01(C) + 3 × 16.00(O) ≈ 100.09 g∖mol^–1
Fraction of Ca = 40.08/100.09 ≈ 0.40 ⇒ ≈ 40% by mass.

Explanation: B
Potassium (Z = 19) fills up to the argon core ( [Ar], 18 electrons), then places the 19th electron in the 4s subshell: [Ar] 4s¹.
Eliminate others
• A[Ar]4s² → that’s calcium (Z = 20).
• C[Kr]4s¹→ krypton core is too large ( 36e^–).
• D [He] 4s¹→ helium core is too small ( 2e^–); skips the needed inner shells.

Explanation: C
Number of atoms = (moles) × (atoms per molecule).
• H₂O: 0.5 mol × 3 = 1.5 N_A atoms
• CO₂: 1.0 mol × 3 = 3 N_A atoms
• CH₄: 1.0 mol × 5 = 5 N_A atoms ← most
• HCl: 2.0 mol × 2 = 4 N_A atoms
So, methane sample has the greatest total atoms.

Explanation: C
1. Assume 100 g sample → C = 40.0 g, H = 6.7 g, O = 53.3 g.
2. Convert to moles:
• C: 40.0/12 ≈ 3.3340.0/12 ≈ 3.33
• H: 6.7/1 ≈ 6.76.7/1 ≈ 6.7
• O: 53.3/16≈3.3353.3/16≈3.33
3. Divide by the smallest (≈3.33): C : H : O ≈ 1 : 2 : 1 → empirical formula = CH₂O (empirical mass = 30).
4. Given Mr is between 55 and 65, the closest whole multiple of 30 is 2 × 30 = 60 → molecular formula = (CH₂O)₂ = C₂H₄O₂.

Explanation: B
• p orbitals (px, py, pz) have two lobes on opposite sides of the nucleus → the classic dumbbell shape.
• s: spherical.
• d: mostly four-lobed “cloverleaf” (except dz₂²​ which is dumbbell + torus).
• f: more complex, multi-lobed shapes.