KMU MDCAT 2026 paper with answer key (Codes A, B, C, D) and explanations

Answer: C. Bacteriophage — Tadpole virus

A bacteriophage has an icosahedral head and tail, giving the familiar tadpole appearance. Poliovirus is roughly spherical or polyhedral, while tobacco mosaic virus is rod shaped.

The shape pairing in option C is therefore the one that fits.

Answer: C. Weakening of the immune system

HIV damages immune function, particularly by depleting CD4 helper T cells. The body then becomes less able to coordinate a protective response to microbes.

Increased susceptibility follows from this weakened defence. Destruction of red cells or excess digestive enzymes does not explain the defining immune defect.

Answer: C. Pyruvate

In the link reaction, three carbon pyruvate loses one carbon as carbon dioxide. The remaining two carbon acetyl group attaches to coenzyme A.

Acetyl CoA is the product of this conversion, so it is not the starting molecule requested in the question.

Answer: C. 9

Beta oxidation removes two carbon units from an even chain fatty acid, and each unit enters acetyl CoA. Divide the 18 carbons into groups of two.

The number of acetyl CoA molecules is nine. Do not confuse this with the eight beta oxidation cycles needed to cut the chain into nine units.

Working
18 ÷ 2 = 9 acetyl CoA
beta oxidation cycles = 9 − 1 = 8

Answer: B. Carbohydrates, proteins, lipids, nucleic acids

These four groups are major carbon based biological molecules. Their structures and functions make them organic components of living cells.

The competing lists include water, mineral ions, oxygen, or carbon dioxide, which cannot all be classified as organic biomolecules.

Answer: D. Phosphoric acid

A nucleotide contains a phosphate group, sugar, and nitrogenous base. Both DNA and RNA nucleotides have phosphate in their backbone.

Their sugars differ, and thymine is characteristic of DNA rather than RNA. Thus phosphoric acid is their shared component among the options.

Answer: B. Heat of vaporization

When sweat or water on a leaf evaporates, it needs energy to change into vapour. That energy is drawn as heat from the body or leaf surface.

A high heat of vaporization therefore gives evaporative cooling. Heat capacity helps resist temperature changes but is not the main mechanism asked here.

Answer: C. Loss of survival of protoplasm

Water provides the medium for biochemical reactions and supports the structure of protoplasm. Severe dehydration makes normal cellular activity impossible.

At about ten percent water content, the expected result in this question is loss of protoplasm survival, rather than faster growth or more protein synthesis.

Answer: B. Disaccharide

Maltose consists of two glucose units joined by a glycosidic bond. A sugar made from two monosaccharide units is a disaccharide.

It is therefore neither a single sugar unit nor a long polysaccharide chain.

Answer: D. Quaternary

Functional haemoglobin consists of four polypeptide subunits associated in one protein complex. The arrangement of multiple chains defines quaternary structure.

A single subunit can also have its own tertiary folding, but the complete haemoglobin molecule is the intended example of quaternary structure.

Answer: A. Mitochondria / D. Plasma membrane

KMU's official Code A key accepts both A and D.

Plasma membrane glycoproteins are well established and project carbohydrate groups toward the extracellular surface.

The mitochondrial option needs careful wording, because glycosylation of mitochondrial proteins is more specialised than the standard textbook plasma membrane example. Both answers are accepted, exactly as KMU published them.

Answer: D. 1000

Adjacent base pairs in ordinary B form DNA are separated by approximately 0.34 nm along the helical axis. Divide the length by this rise per pair.

The result is 1000 base pairs. A base pair contains two nucleotides, so base pairs and individual nucleotides must not be confused.

Working
Number of base pairs = 340 nm ÷ 0.34 nm per pair
= 1000 base pairs

Answer: B. Lipids; amino acids

Fatty acids and glycerol are the familiar building components of many lipids. Proteins, in turn, are polymers of amino acids.

The two blanks ask for the product class first and the protein building blocks second, which gives lipids followed by amino acids.

Answer: C. Chloroplast

The stroma is the fluid interior of a chloroplast, and thylakoids form its internal membrane system. Their stacked regions are associated with light reactions.

A mitochondrion also has two membranes, but it contains cristae rather than thylakoids, so the internal details identify the chloroplast.

Answer: B. Ribosomes

Both cell types synthesize proteins, so both contain ribosomes. The ribosomes differ in size and composition but share the same basic role.

Mitochondria, endoplasmic reticulum, and Golgi apparatus are membrane bound structures absent from typical prokaryotes.

Answer: D. Mitochondria

Mitochondria carry their own DNA and divide from pre existing mitochondria. This gives them a degree of replication within the cell.

The Golgi apparatus and lysosomes do not have the same independent genetic machinery; mitochondria are the intended choice.

Answer: B. Lipid synthesis and detoxification

Smooth ER is a major site for lipid synthesis and enzymatic detoxification. Removing it directly interferes with both processes.

Protein synthesis on bound ribosomes belongs mainly to rough ER, while aerobic ATP generation belongs mainly to mitochondria.

Answer: B. Holds identical chromatids together

After DNA replication, the two sister chromatids remain joined at the centromeric region until separation. This region also supports kinetochore assembly for spindle attachment.

It is a chromosome structure rather than an ATP producing or protein synthesizing organelle.

Answer: C. Three Na⁺ ions are pumped out and two K⁺ ions are pumped in

For each ATP used, the pump sends three sodium ions out of the neuron and brings two potassium ions in. Both movements run against the respective concentration gradients.

This helps maintain the ion gradients required for resting potential and later nerve impulses; reversing the directions is a common trap.

Answer: B. Dendrites, cell body, axon

Dendrites mainly receive signals, the cell body integrates them, and the axon carries output away. Together they describe a typical neuron's main structural plan.

A ganglion is a collection of neuronal cell bodies, not one part of a single neuron; cilia and cauda equina do not complete this standard list.

Answer: D. Depolarization

A resting membrane near −70 mV becomes less negative as sodium enters. Movement toward a positive value is the depolarizing phase of the action potential.

Repolarization is the later return toward the resting negative value, so it does not name the change described in the stem.

Answer: C. Spinal reflex

The rapid withdrawal is coordinated through a spinal reflex arc before conscious processing becomes the main response. This short path protects tissue from further heat damage.

It is involuntary and immediate, rather than a learned conditioned response or a deliberate voluntary movement.

Answer: C. Pons

The pons participates in sleep regulation and REM sleep circuitry, so injury there can disturb sleep patterns and dreaming. It also forms part of the brainstem.

The cerebellum chiefly coordinates movement, while the medulla handles vital autonomic functions; the pons best matches this symptom pair.

Answer: A. Midbrain

Dopamine producing neuronal groups such as the substantia nigra and ventral tegmental area are in the midbrain. Injury there can therefore affect dopamine related motor and reward pathways.

The other listed regions have different primary functions and do not locate these classic midbrain dopamine populations.

Answer: B. Limbic system

The amygdala is a limbic structure involved in emotional processing, including fear related responses. It communicates with other brain areas during emotional learning.

It is not a component of the medulla, cerebellum, or reticular formation listed as alternatives.

Answer: C. They remain chemically unchanged

An enzyme binds reactants and lowers the activation energy, then releases products and can be used again. It is not permanently incorporated into those products.

The reaction becomes faster, while the enzyme's overall chemical identity remains unchanged after one catalytic cycle.

Answer: A. Rigid

The classical lock and key model imagines a pre shaped active site complementary to its substrate. In this simplified model, the site is regarded as rigid.

A flexible changing shape is instead emphasised by the induced fit model. The question specifically names lock and key.

Answer: A. 1.5–1.6

Pepsin functions in the acidic environment of the stomach. Its activity is greatest at a very low pH, approximately within the first option's range.

Near neutral or alkaline pH, the active site's charge and shape no longer support the same activity.

Answer: A. Ionization of an active site

Changing pH changes the protonation and charge of amino acid groups at the active site. That can disrupt substrate binding or the catalytic step.

pH is not itself a temperature decrease and does not necessarily remove a prosthetic group; the direct mechanism here is altered ionization.

Answer: A. Competitive inhibitor

A competitive inhibitor resembles the substrate closely enough to bind at the same active site. Substrate and inhibitor compete for access.

A noncompetitive inhibitor acts through a different site or mechanism, so resemblance to the normal substrate identifies the competitive type.

Answer: B. Cannot increase at the same rate

Population can rise more rapidly than food, water, and space. When resources are limited, individuals compete and not every organism survives or reproduces.

The idea does not require resources to stay exactly constant; it requires their increase to lag behind demand.

Answer: C. Inheritance of acquired characteristics

Lamarck proposed that repeated use could lengthen a structure and that the acquired change could pass to offspring. The long giraffe neck was a classic example of his proposed mechanism.

This is a historical explanation, not the accepted modern genetic account of evolution by natural selection.

Answer: B. Natural selection

Fur colours vary, and white rabbits are harder for predators to see against snow. They survive and reproduce more often under those conditions.

Because the environment changes survival by an inherited trait, the example illustrates natural selection rather than deliberate artificial breeding.

Answer: A. It is passed to offspring

Lamarck's theory held that traits acquired through use or disuse during life could be inherited by offspring. Thus a change to a body part would be expected to pass on in that theory.

Modern genetics generally does not support inheritance of an ordinary acquired bodily modification; the question asks specifically what Lamarck proposed.

Answer: D. Corpus luteum

Following ovulation, the remaining follicular tissue reorganizes into the corpus luteum. It secretes progesterone and helps support the endometrium.

The cervix and fimbriae are separate structures, while the endometrium is the uterine lining.

Answer: A. Testosterone secretion

Vasectomy cuts the vas deferens, blocking sperm from entering the ejaculate. The testes remain in place and their hormone producing cells continue to secrete testosterone.

The procedure interferes with sperm transport and fertilization, but it does not directly shut down endocrine function.

Answer: A. Estrogen

Estrogen from developing ovarian follicles drives proliferation and thickening of the uterine lining before ovulation. Deficiency can therefore impair this phase.

Progesterone is more important for the later secretory phase; oxytocin does not cause the proliferative thickening described here.

Answer: B. Syphilis

A primary syphilitic chancre classically presents as a painless ulcer with a firm edge. The symptoms in the stem point to that pattern.

Genital herpes more commonly produces painful grouped vesicles or ulcers, so it is a poorer match.

Answer: C. Floating ribs

Floating ribs, usually pairs 11 and 12, have no anterior attachment to the sternum. That makes option C a literal match to the question.

A note on the wording: false ribs include the floating ribs and other ribs that lack a direct sternal attachment. If “not attached” means no attachment at all, floating ribs are intended; if it means no direct attachment, option B could also be argued. KMU's official key is C.

Answer: D. Cartilage

Cartilage cells lie in a firm extracellular matrix, and cartilage has no blood vessels of its own. Nutrients reach the cells by diffusion.

That matrix resists compression at joints. Bone is vascular, and adipose tissue does not have this cartilage like supporting matrix.

Answer: C. Maintenance of posture and locomotion

Striated multinucleated fibres under voluntary control identify skeletal muscle. Skeletal muscles move bones and help maintain posture.

Heart pumping belongs to cardiac muscle, whereas movement of internal organs belongs mainly to smooth muscle.

Answer: B. Triad

A skeletal muscle triad consists of one transverse T tubule between two terminal cisternae of the sarcoplasmic reticulum. It couples the electrical signal to calcium release.

The sarcomere is a contractile unit, while sarcoplasm and sarcolemma name cytoplasm and membrane, respectively.

Answer: D. Remains constant

The A band corresponds to the length of thick myosin filaments. Sliding filament contraction changes the degree of overlap without shortening the thick filaments.

The I band and H zone can shrink, but the A band's length remains essentially constant.

Answer: D. Cartilaginous joint

Intervertebral discs join vertebral bodies with fibrocartilage, allowing limited movement and absorbing compression. There is no synovial cavity between these vertebral bodies.

The stem describes this cartilaginous joint, rather than a freely movable synovial joint; it also differs from a fibrous suture.

Answer: B. Ball and socket joint

The rounded humeral head fits into the shoulder socket. This arrangement permits movement around several axes, including rotation.

A hinge chiefly bends in one plane, whereas a pivot rotates around one axis; the shoulder is ball and socket.

Answer: D. Smooth movement of joints

Inflammation and damage at a joint increase pain and friction, restricting how smoothly its surfaces move. That explains the main functional effect of arthritis.

Gas exchange and red blood cell production occur in different body systems.

Answer: B. 25%

For a single gene cross between heterozygous F1 parents, the genotypes appear in a 1:2:1 ratio. One of four offspring genotypes is homozygous recessive.

The dominant and recessive phenotypes follow 3:1, but the question asks specifically for the homozygous recessive genotype.

Working
Gg × Gg → GG, Gg, Gg, gg
gg = 1 ÷ 4 = 25%

Answer: D. 1:1

Crossing Rr with rr produces Rr and rr offspring in equal proportions. Rr is round and rr is wrinkled.

Therefore the phenotypic ratio of round to wrinkled is 1:1; the 3:1 ratio applies to Rr × Rr instead.

Working
Rr × rr → Rr, rr
round : wrinkled = 1 : 1

Answer: B. No linkage

An unlinked dihybrid makes four types of gametes in equal frequencies. A test cross with a double recessive partner reveals those gametes as four equal phenotype classes.

Complete linkage would favour parental combinations and disrupt the equal 1:1:1:1 result.

Answer: A. Son

A son receives his only X chromosome from his mother and a Y from his father. If that X bears the recessive allele, there is no second normal X to mask it.

A daughter of a carrier mother and unaffected father usually has a normal paternal X, so she is less likely to express this recessive trait.

Answer: C. Colour blindness

Common red green colour blindness is a classic X linked recessive trait. Its frequency is higher in males because males have only one X chromosome.

Vitamin D resistant rickets is usually taught as X linked dominant, while cystic fibrosis is autosomal recessive.

Answer: D. Hypophosphatemic rickets

X linked hypophosphatemic rickets can be expressed when one affected X chromosome is present. That inheritance pattern is dominant.

Haemophilia and common red green colour blindness are recessive X linked examples, helping separate the options.

Answer: B. A recessive allele on the X chromosome

The common inherited haemophilias A and B arise from mutations in clotting factor genes on the X chromosome. The recessive inheritance pattern explains why males are more often affected.

It is not a Y linked or mitochondrial disorder and does not depend on a dominant chromosome 21 allele.

Answer: C. Reduce friction during heart movement

A thin film of fluid separates the layers of the serous pericardium. As the heart beats, this lubrication reduces rubbing between the moving surfaces.

The fluid does not generate the electrical impulse or supply the heart muscle's main nutrients.

Answer: C. Ventricular systole

During ventricular systole, the right ventricle ejects blood toward the lungs and the left ventricle ejects it into the aorta. Both outflows occur as the ventricles contract.

Atrial systole mainly completes ventricular filling; ventricular diastole is the relaxation and filling phase.

Answer: B. Liver

The hepatic portal vein carries nutrient rich blood from the gut to the liver first. The liver can process, store, or regulate absorbed materials before they enter general circulation.

The vena cava receives blood later; it is not the first organ reached by this portal route.

Answer: D. To withstand high pressure

Blood leaves the heart under relatively high pressure, so arteries need a thick muscular and elastic wall. That wall helps them tolerate and smooth pressure changes.

Veins operate at lower pressure and rely more on valves to limit backflow.

Answer: B. Interstitial fluid

Fluid that leaves capillaries enters tissue spaces as interstitial fluid. Once collected into lymphatic vessels, it is called lymph.

It ultimately originates from plasma, but its immediate precursor and closest counterpart is interstitial fluid.

Answer: B. Antibody

Activated B cells can become plasma cells that release antigen specific antibodies. These soluble proteins bind particular targets in an adaptive immune response.

An antigen is the target that stimulates immunity; perforin is associated mainly with cytotoxic immune cells.

Answer: D. Formation of memory cells

Vaccination exposes the adaptive immune system to antigen without the full risk of the natural infection. Memory B and T cells then persist and respond more rapidly on later exposure.

Fever, inflammation, and phagocytosis are useful immediate defences but do not themselves explain the long lasting adaptive memory.

Answer: C. Bronchioles

Trachea and bronchi have cartilage supporting their walls, while smaller bronchioles lack cartilage. Bronchiolar calibre is controlled largely by smooth muscle.

The larynx also contains cartilage, so bronchioles are the structure that meets the “no cartilage” condition.

Answer: C. Filter air and prevent the entry of microorganisms

Mucus traps inhaled particles and microbes. Cilia move the trapped material toward the throat so it can be cleared.

Gas exchange happens mainly across alveoli, not in the nasal cavity; the nasal structures chiefly condition and clean inspired air.

Answer: A. Bicarbonate ion

Most carbon dioxide entering red cells is converted to carbonic acid and then bicarbonate. Bicarbonate moves in the blood and carries the largest share toward the lungs.

Smaller portions travel dissolved in plasma or attached to haemoglobin as carbamino compounds.

Answer: C. Emphysema

Emphysema involves destruction of alveolar walls and loss of elastic recoil. This reduces the effective surface area available for gas exchange.

Chronic bronchitis mainly involves inflamed, mucus producing airways; the alveolar destruction in the question identifies emphysema.

Answer: B. Insulin synthesis

The liver detoxifies many substances, makes bile, and stores vitamin A. Insulin, however, is synthesized and secreted by pancreatic beta cells.

Thus the pancreas rather than the liver is responsible for the activity in option B.

Answer: A. 2–1–3

A swallow begins with the oral phase, continues through the pharyngeal phase, and then passes through the esophagus. Map those names to the numbers in the question.

Oral is 2, pharyngeal is 1, and esophageal is 3; therefore the official answer is the sequence 2–1–3.

Working
Oral (2) → pharyngeal (1) → esophageal (3)

Answer: C. Secretin

Acidic chyme entering the duodenum stimulates secretin release. Secretin promotes bicarbonate rich pancreatic secretion that neutralizes the acid.

Gastrin mainly stimulates gastric activity, while pepsin and trypsin are enzymes rather than the hormone sought here.

Answer: A. Blood capillaries and lacteals

Blood capillaries receive many absorbed sugars and amino acids. A central lacteal transports much of the absorbed lipid in lymph.

Both structures are within the villus and directly carry nutrients away, unlike muscle layers or mucus producing cells.

Answer: B. Move undigested food toward the rectum

Rhythmic contractions push residual intestinal contents along the colon toward the rectum. Water absorption occurs during this transit.

The large intestine is not primarily responsible for carbohydrate digestion or for secreting major digestive enzymes.

Answer: C. Spread of Helicobacter pylori into stomach wall

Helicobacter pylori can colonize the gastric mucosa, disrupt its protective barrier, and promote ulcer formation. That fits chronic discomfort with an endoscopically confirmed ulcer.

The other choices do not give the common direct explanation for gastric mucosal injury asked for here.

Answer: B. Renal cortex

Renal corpuscles contain a glomerulus within Bowman's capsule. They are located in the outer renal cortex.

The medulla contains structures such as loops of Henle and collecting ducts, but not the renal corpuscles requested.

Answer: B. Selective reabsorption

Glucose is filtered at the renal corpuscle and normally reclaimed in the proximal tubule. Finding it in urine can mean reabsorption capacity has been exceeded or the process is impaired.

Thus selective reabsorption is the relevant process among the options; filtration itself is what brings glucose into the filtrate.

Answer: C. Ascending limb

The ascending limb of the loop of Henle transports salts but is effectively impermeable to water. This separation helps establish the medullary osmotic gradient.

The descending limb is water permeable, which is why exchanging the two limbs gives the wrong answer.

Answer: C. ADH release is inhibited, resulting in a large volume of dilute urine

When the body has ample water, ADH release falls. The collecting ducts then reabsorb less water and a larger volume of dilute urine is excreted.

The response prevents excessive water retention; ultrafiltration does not simply stop.

Answer: C. Efferent arteriole

Blood leaves each glomerulus through an efferent arteriole. That vessel branches into peritubular capillaries, and in some nephrons the vasa recta.

The afferent arteriole carries blood into the glomerulus, so it is upstream of this second capillary network.

Answer: D. 10%

The item asks about the level of remaining function at which dialysis or transplantation is usually needed. The official option is below ten percent of normal function.

Clinical decisions also depend on symptoms and laboratory findings, but among the stated thresholds ten percent represents very advanced failure.

Answer: A. Urea

Humans convert much of the highly toxic ammonia produced during amino acid metabolism into urea in the liver. The kidneys then excrete urea.

Uric acid and creatinine are also excreted, but urea is the main nitrogenous product intended by this comparison.

Answer: B. Biotechnology

A DNA probe can recognize a complementary sequence from a microorganism. This molecular identification is a biotechnology method and can be faster than conventional culture.

Antibiotics treat infections and surgery is a procedure; neither names the technology behind the probe.

Answer: C. Genetically altered virus

Engineered viruses can enter target cells and deliver a therapeutic gene. Their natural cell entry machinery makes them useful vectors once modified for safety.

Plasmids can also be vectors in some settings, but the question asks for the most common choice among its listed alternatives.

Answer: B. A group of identical antibodies produced by identical immune cells

A single B cell lineage produces antibodies with the same binding specificity. A monoclonal preparation therefore contains essentially identical antibodies derived from one clone.

A mixed antibody population is polyclonal; radioactive labelling is a possible later modification, not the definition.

Answer: C. Nanoparticles: Efficient drug delivery at target cells and diagnosis of disease

Nanoparticles can be designed to carry medicines to target tissues and support diagnostic imaging or detection. This matches the application in option C.

Biochips are analytical platforms, while mycorrhizae are plant fungus associations; the other pairings assign them unrelated functions.

Answer: C. A + B = B + A, so changing the order does not change the resultant

Vector addition is commutative: reversing the order of two vectors leaves both the magnitude and direction of their resultant unchanged. A head to tail diagram reaches the same endpoint.

The law applies to vectors at any angle; perpendicular vectors are not a special requirement.

Answer: C. Velocity

Linear momentum is the product of a positive scalar mass and the velocity vector. Multiplying by mass changes the vector's size but not its direction.

Force and acceleration can point elsewhere, particularly while an object is slowing or turning.

Working
p⃗ = m v⃗
for m > 0, p⃗ points along v⃗

Answer: C. Velocity changes by equal amount in equal intervals of time

Acceleration measures how rapidly velocity changes. If equal time intervals always produce equal changes in velocity, acceleration stays constant.

Constant velocity would instead mean zero acceleration, a special case; option C directly states the defining test for uniform acceleration.

Working
a⃗ = Δv⃗ ÷ Δt

Answer: B. −4 m/s

Use signed displacement rather than total distance. The particle's position decreases by 20 m over a five second interval.

The negative sign shows motion toward decreasing position; dropping it would turn average velocity into a positive speed.

Working
Δx = 20 − 40 = −20 m
Δt = 15 − 10 = 5 s
v̄ = Δx ÷ Δt = −4 m/s

Answer: B. The 4 kg block has double the acceleration of the 8 kg block

For equal applied force on a frictionless surface, acceleration is inversely proportional to mass. Halving the mass doubles the acceleration.

The 4 kg block therefore has twice the acceleration of the 8 kg block; equal force does not imply equal acceleration.

Working
a₄ = F ÷ 4
a₈ = F ÷ 8
a₄ ÷ a₈ = 2

Answer: A. sin²θ

Only the initial vertical component, u sin θ, determines how high an ideal projectile rises. The height depends on the square of that vertical component.

With launch speed and g fixed, maximum height is directly proportional to sin² θ, not to sin θ alone.

Working
H = (u sin θ)² ÷ (2g)
therefore H ∝ sin² θ

Answer: A. Conserved

An isolated system exchanges no net energy or momentum with its surroundings. Its total energy can change form internally while its total momentum remains fixed.

Individual objects may gain or lose energy or momentum, but the totals for the complete isolated system are conserved.

Answer: C. Zero

Mechanical work on an object requires displacement of that object along the force. The wall does not move during the push.

The person can still expend biological energy, but the mechanical work done on the stationary wall is zero.

Working
W = F s cos θ
s = 0
W = 0 J

Answer: B. 20 W

Power is work divided by elapsed time. Convert two minutes to 120 seconds before dividing.

Using two as though it were seconds would give the misleading 1200 W option.

Working
t = 2 × 60 = 120 s
P = W ÷ t = 2400 ÷ 120 = 20 W

Answer: B. The body's height is much greater than Earth's radius

The expression mgh treats gravitational field strength g as approximately constant over the height change. This approximation works near Earth's surface.

At heights comparable with or greater than Earth's radius, g changes substantially with distance, so mgh is no longer the appropriate general expression.

Answer: C. Move in a straight line tangent to the circle

The string supplies centripetal force while intact. At the instant it breaks, the ball retains its instantaneous velocity, which points tangent to the circle.

After release, other forces such as gravity can bend its later path; the initial motion is along the tangent.

Answer: B. Radius

Angle in radians equals arc length divided by circle radius. A one radian angle therefore cuts off an arc whose length matches the radius.

Diameter and circumference are different lengths, so they do not define a one radian central angle.

Working
θ = s ÷ r
if s = r, θ = 1 rad

Answer: C. Radian

Angular displacement measures a change in angle. The named SI unit for plane angle is the radian.

Degrees and revolutions are other angle measures, whereas rad/s is an angular speed unit.

Answer: A. Zero

At terminal velocity, upward drag balances downward weight. With equal opposing forces the raindrop no longer accelerates.

The raindrop is still moving, but constant velocity requires zero net force rather than zero individual forces.

Working
F_net = mg − F_drag = 0

Answer: C. Reduced

Just after a parachute opens, a large upward drag slows the skydiver. As downward speed falls, the drag also falls until it balances weight at the new terminal speed.

The question asks what happens during the slowing phase: resistance reduces from its immediate post opening value. At the new terminal velocity it is steady and equals weight.

Answer: A. 0.025 m²

For approximately incompressible steady flow, the product of area and speed stays constant. Doubling speed requires half the original cross sectional area.

Half of 0.05 m² is 0.025 m², the value in option A.

Working
A₁v₁ = A₂v₂
A₂ = 0.05 × 2 ÷ 4 = 0.025 m²

Answer: B. Maintain the same blood flow rate

A narrower vessel resists blood flow more strongly. To sustain a given flow rate, the heart must provide a greater pressure difference.

The higher pressure compensates for the extra resistance; it does not itself widen the artery.

Working
Q = ΔP ÷ R
if R increases and Q is maintained, ΔP must increase

Answer: B. One wavelength

A travelling wave advances one wavelength in the time taken for one complete oscillation. This is the defining relation among speed, wavelength, and period.

Half a wavelength corresponds to half a period at the same wave speed.

Working
v = λ ÷ T
distance in T = vT = λ

Answer: A. Perpendicular; parallel

In a transverse wave, medium particles oscillate perpendicular to the propagation direction. In a longitudinal wave, their motion is parallel to propagation.

Think of a vibrating string for transverse motion and compressions in air for longitudinal motion.

Answer: C. 320 m

The wave speed is frequency times wavelength. Multiply that speed by eight seconds to obtain the distance the disturbance travels.

It is easy to stop after calculating speed; the question asks for distance over a specified duration.

Working
v = fλ = 25 × 1.6 = 40 m/s
d = vt = 40 × 8 = 320 m

Answer: B. 0.6 m/s

Near ordinary temperatures in air, sound speed increases by approximately 0.6 m/s for each 1 °C rise. This is a standard near room temperature approximation.

The relationship is approximate because the exact speed depends on gas composition and absolute temperature.

Answer: C. No heat is exchanged with the surroundings

Compressions and rarefactions during a sound wave occur quickly. Laplace treated these changes as approximately adiabatic.

An adiabatic process exchanges negligible heat with its surroundings, unlike the slower isothermal assumption.

Answer: A. 1.0 × 10³ J

Use the first law with work done by the gas taken as positive: change in internal energy equals heat received minus work performed. The gas receives 1500 J and performs 500 J of work.

Subtracting gives a positive 1000 J increase; adding heat and work would use the wrong sign convention.

Working
Q = 1.5 × 10³ J
W = 5.0 × 10² J
ΔU = Q − W = 1.0 × 10³ J

Answer: C. 586 J

The supplied constant volume molar heat capacity is 21 J mol⁻¹ K⁻¹. Add the gas constant to obtain the constant pressure molar heat capacity, then multiply by one mole and 20 K.

The result rounds to 586 J, matching option C; using C_V directly would give 420 J.

Working
C_P = C_V + R = 21 + 8.315 = 29.315 J mol⁻¹ K⁻¹
Q = nC_PΔT = 1 × 29.315 × 20 = 586.3 J

Answer: A. Decrease

A rapidly expanding gas can do work with little time to receive heat from its surroundings. Its internal energy then falls.

For an ideal gas, internal energy is linked to temperature, so the temperature decreases.

Answer: A. One-fourth of its original value

Coulomb's law gives an inverse square dependence on the distance between point charges. Replacing r with 2r multiplies the denominator by four.

Charge magnitudes are unchanged, so the force becomes one quarter of its starting value.

Working
F ∝ 1 ÷ r²
F_new ÷ F_old = r² ÷ (2r)² = 1 ÷ 4

Answer: C. 6.0 × 10⁴ J

Voltage is energy transferred per unit charge. Multiply the battery potential difference by the total charge.

The result is 6.0 × 10⁴ J, the value shown in option C.

Working
E = VQ = 12 × 5000 = 60000 J = 6.0 × 10⁴ J

Answer: B. 0.5 × 10⁶ N C⁻¹

For one isolated infinite plane sheet, the electric field on either side has magnitude surface charge density divided by twice the vacuum permittivity. The field is equal in magnitude on the two sides.

Using σ/ε₀ would give 10⁶ N C⁻¹ and miss the factor of two for a single sheet.

Working
E = σ ÷ (2ε₀)
= (8.85 × 10⁻⁶) ÷ (2 × 8.85 × 10⁻¹²)
= 0.5 × 10⁶ N C⁻¹

Answer: A. Directed flow of free electrons

Conduction electrons already move randomly because of thermal motion. An applied electric field adds a small net drift, creating steady current.

Random motion alone averages to zero current; the directed component is the key distinction.

Answer: A. N A⁻¹ m⁻¹

A current carrying wire of length L feels magnetic force F = BIL when it is perpendicular to the field. Rearranging gives B in newtons per ampere metre.

This is equivalent to the tesla and to Wb m⁻²; the option's inverse powers of ampere and metre matter.

Working
B = F ÷ (IL)
[B] = N A⁻¹ m⁻¹

Answer: C. Doubled

Faraday's law makes induced emf proportional to the number of turns times the rate of change of magnetic flux through each turn. The stated flux change rate stays fixed.

Doubling the turns therefore doubles the magnitude of the induced emf; its sign depends on the direction convention.

Working
Magnitude of ε = N × magnitude of ΔΦ/Δt
2N gives 2 × magnitude of ε

Answer: B. 90°

A sinusoidal AC quantity written as I = I₀ sin θ reaches its maximum positive value when sin θ equals one. That occurs at 90 degrees in the first cycle.

At 0 and 180 degrees the sine is zero, and at 270 degrees it is maximally negative.

Working
sin 90° = 1
I = +I₀

Answer: A. Almost stops the current

Reverse bias increases the depletion region's barrier to majority carriers. An ordinary diode therefore passes only a very small reverse leakage current before breakdown.

It does not become a better conductor in the reverse direction; “almost stops” accounts for the small leakage.

Answer: C. Neither emits nor absorbs energy

A stationary quantum state has a definite allowed energy. An atom remaining in that same state does not continuously radiate energy.

Emission or absorption accompanies a transition between different energy levels, with photon energy equal to their difference.

Answer: A. Line emission

Excited atoms can emit photons with particular energies as electrons transition to lower levels. Those discrete energies appear as bright lines at specific frequencies.

A line absorption spectrum instead has dark lines against a continuous background.

Answer: A. Constant

For a given radioactive isotope under ordinary conditions, each nucleus has a fixed probability of decay per unit time. The half life consequently stays constant as the sample shrinks.

The activity falls with time, but the time required for the remaining nuclei to halve does not depend on the initial quantity.

Answer: D. 100 g

Add the molar masses of one calcium atom, one carbon atom, and three oxygen atoms in CaCO₃. One mole has a mass numerically equal to this formula mass in grams.

The sum is 100 g per mole, so a one mole sample has mass 100 g.

Working
M(CaCO₃) = 40 + 12 + 3(16) = 100 g mol⁻¹
mass of 1 mol = 100 g

Answer: B. Produces minimum number of moles of product

The limiting reagent runs out first in the reaction according to the balanced mole ratio. It sets the maximum product that can actually be formed.

Compare the product amounts predicted from each reactant: the smaller prediction identifies the limiting reagent.

Answer: C. Chemists use percentage yield to express the efficiency of a chemical reaction

Percentage yield compares what a reaction actually produced with its theoretical maximum. Chemists use it to express the efficiency of obtaining a product.

Actual yield comes from experiment; theoretical yield comes from the balanced equation. Reversing these definitions makes the other statements wrong.

Working
Percent yield = (actual yield ÷ theoretical yield) × 100%

Answer: A. Azimuthal

The azimuthal or angular momentum quantum number l distinguishes s, p, d, and f orbital shapes. For example, s orbitals are spherical and p orbitals have two lobes.

The principal number chiefly describes shell and energy, while the magnetic number describes orientation.

Answer: C. Four

No two electrons in one atom can have an identical set of all four quantum numbers. Even two electrons in the same orbital must have opposite spin quantum numbers.

The full set is n, l, m_l, and m_s; matching only some of them is allowed.

Working
(n, l, m_l, m_s): all four cannot match for two electrons

Answer: B. [Ar] 4s²

Calcium has 20 electrons. Argon accounts for 18, leaving two electrons to occupy the 4s orbital.

The correct shorthand therefore has the argon core followed by 4s², as in option B.

Answer: C. Absolute temperature

For an ideal gas, mean translational kinetic energy depends only on temperature measured in kelvins. It rises when absolute temperature rises.

Pressure and volume may change along with temperature, but they are not the direct proportional variable in this relation.

Working
Mean E_k = (3/2) k_B T

Answer: B. Halve

Boyle's law gives an inverse relationship between pressure and volume when temperature and amount of gas stay fixed. Doubling pressure requires half the original volume.

A direct doubling of volume would contradict the constant product P V.

Working
P₁V₁ = P₂V₂
if P₂ = 2P₁, V₂ = V₁ ÷ 2

Answer: B. Double

Charles's law states that volume is proportional to kelvin temperature at constant pressure and gas amount. A factor of two in absolute temperature gives a factor of two in volume.

The word “absolute” matters: doubling a Celsius reading is not generally the same operation.

Working
V₁/T₁ = V₂/T₂
T₂ = 2T₁ → V₂ = 2V₁

Answer: C. −273

Zero kelvin is the lower limit of thermodynamic temperature. Converting kelvins to Celsius subtracts 273.15.

Rounded to the options supplied, absolute zero is −273 °C.

Working
t(°C) = T(K) − 273.15
at T = 0 K, t ≈ −273 °C

Answer: C. Greater

In the usual introductory comparison at comparable conditions, liquid particles have more freedom of motion and greater average kinetic energy than particles in the solid state. This is the intended ordering.

Strictly, at the same absolute temperature, mean translational kinetic energy cannot be inferred from phase alone, and the question does not specify the temperature. KMU's official key is C.

Answer: A. Intermolecular

While a substance boils at constant pressure, added energy separates molecules into the vapour phase rather than increasing their average kinetic energy. The temperature stays approximately constant.

The heat overcomes intermolecular attractions; it does not break the covalent bonds inside individual molecules.

Answer: A. Electronegative

A hydrogen bonded to nitrogen, oxygen, or fluorine can interact strongly with a lone pair on another electronegative atom. High electronegativity produces the strong bond polarity required.

Electropositive or nonpolar atoms do not provide the same typical hydrogen bond attraction.

Answer: C. Opposite

The NaCl lattice alternates positive sodium and negative chloride ions. Electrostatic attraction between unlike charges stabilizes the ordered crystal.

A given ion is nearest to ions of opposite charge, rather than to a shell of identical charges.

Answer: B. Molten

In the solid lattice, ions are held in fixed positions and cannot carry charge through the bulk. Melting frees the ions to move.

The mobile ions make molten ionic substances conduct electricity; an aqueous solution would also conduct but is not an option.

Answer: C. 6

A unit cell is specified by three edge lengths and three angles between the edges. These six values describe its geometry.

Its symmetry may make some values equal, but the standard complete parameter list still contains six.

Working
a, b, c and α, β, γ = six parameters

Answer: B. Released

Bringing oppositely charged gaseous ions together lowers electrostatic potential energy as a crystal forms. Energy is released to the surroundings.

The reverse process, pulling an ionic crystal completely apart into gaseous ions, requires energy input. Always check which direction the question states.

Answer: A. Endothermic

Increasing temperature adds heat to an equilibrium system. The equilibrium shifts toward the direction that absorbs this added heat.

That direction is endothermic; the exact chemical products cannot be named without the reaction equation.

Answer: B. Ionization

Sodium acetate supplies acetate ions, already produced when acetic acid dissociates. Adding this common ion shifts the acid equilibrium toward undissociated acetic acid.

This is the common ion effect, so the fraction of acetic acid ionizing decreases.

Working
CH₃COOH ⇌ H⁺ + CH₃COO⁻
added CH₃COO⁻ shifts left

Answer: B. Formation of ammonia

The forward synthesis of ammonia releases heat. Lowering temperature favours the heat producing, exothermic direction at equilibrium.

This concerns equilibrium yield; a lower temperature can also make the reaction slower, which is why real industry chooses a compromise.

Answer: C. Initiate a chemical reaction

Reacting particles must reach a transition state before products can form. Activation energy is the minimum energy barrier for this reaction route.

It does not necessarily mean enough energy to break every bond in every reactant molecule or to increase the final yield.

Answer: A. Faster

A smaller activation energy barrier lets a larger fraction of collisions succeed at the same temperature. The reaction proceeds more rapidly.

Catalysts exploit this by giving an alternate route with lower activation energy without changing the overall equilibrium position.

Answer: B. Positive

Under the usual chemistry sign convention, heat entering the system is q > 0. Heat leaving it is q < 0.

The statement says heat is absorbed from the surroundings, so q is positive regardless of whether work also occurs.

Working
q_system > 0 for heat absorbed

Answer: C. Initial and final states

Enthalpy is a state function: its overall change depends on the start and end states. A reaction may be divided into different intermediate steps without changing the sum.

That is why enthalpy changes for steps can be added to find an otherwise difficult overall reaction enthalpy.

Answer: B. +2 → 0

Reduction is a decrease in oxidation number, corresponding to gain of electrons in a simple ion. Moving from +2 to 0 is a two unit decrease.

The other listed transitions increase oxidation number and therefore describe oxidation.

Working
+2 → 0: gain of 2 e⁻
oxidation number falls by 2

Answer: C. Electron pairs

Valence shell electron pairs repel each other and arrange as far apart as practicable around the central atom. Bonding and lone pairs together influence shape.

Electron mass or volume is not the organizing rule of VSEPR; the geometry arises from electron pair repulsions.

Answer: D. 4

Promoting one carbon 2s electron to an empty 2p orbital gives four singly occupied valence orbitals. This simple excited state account explains carbon's four bonding possibilities.

One 2s and three 2p orbitals then each hold a single electron.

Working
Ground valence: 2s² 2p²
excited valence: 2s¹ 2p³
singly occupied orbitals = 4

Answer: C. Ionic

A bond dipole arises from unequal sharing of electrons. Comparing the measured dipole with the ideal fully separated charge estimate can indicate the bond's percentage ionic character.

It does not by itself measure whether a bond is acidic, basic, or organic.

Answer: A. CCl₄

Carbon tetrachloride does not hydrolyze readily at room temperature. Carbon lacks suitable low lying vacant orbitals for the usual hydrolysis route available to heavier group 14 tetrachlorides.

Silicon tetrachloride, for example, reacts readily with water, which helps identify CCl₄ among the choices.

Answer: B. Filled

Copper adopts [Ar] 3d¹⁰ 4s¹ rather than the simple predicted [Ar] 3d⁹ 4s². The completely filled 3d subshell is relatively stable.

The relevant word is “filled,” not merely “half filled,” which is associated with other common configuration exceptions.

Working
Cu: [Ar] 3d¹⁰ 4s¹

Answer: A. Carbon tetrachloride

Radical substitution can replace each of methane's four hydrogen atoms step by step. With excess chlorine, substitution proceeds toward the fully chlorinated product.

The limiting or shorter exposure conditions would favour mixtures containing chloromethane, dichloromethane, and chloroform.

Working
CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄

Answer: C. Longest with maximum number of double bonds

The selected parent must contain the multiple bond and, when applicable, the maximum possible number of double bonds. Length is then considered within the relevant selection rules.

A highly branched chain or a chain chosen for triple bonds alone would miss the alkene features being named.

Answer: A. Dehydration

Removing H and OH from neighbouring positions of an alcohol eliminates a molecule of water and forms a carbon carbon double bond. This process is dehydration.

Dehydrohalogenation instead removes a hydrogen halide from a haloalkane, so the starting material distinguishes the methods.

Answer: A. Planar molecule with six sp²-hybridized carbon atoms

Each carbon in benzene is sp² hybridized, leaving a p orbital for the delocalized π system. The six carbons form a planar ring.

The printed alternative mentioning a nonplanar ring is wrong even though it includes alternating bonds; planarity is essential to the intended description.

Answer: D. Substitution

Addition across benzene's ring would disrupt its aromatic π system. Substitution replaces a ring hydrogen while allowing aromaticity to be restored.

This is why electrophilic substitution, rather than the ordinary alkene addition pattern, is typical of benzene.

Answer: A. Acetaldehyde

Catalysed hydration of ethyne first gives the enol vinyl alcohol. That unstable enol tautomerizes to the more stable carbonyl compound acetaldehyde.

The question asks for the ultimate product, so stopping at the initial enol would miss the rearrangement.

Working
HC≡CH + H₂O → CH₂=CHOH → CH₃CHO

Answer: A. Alphabetical order

Different substituent names are listed alphabetically when constructing an IUPAC name, once the parent and locants have been chosen. For example, bromo precedes chloro.

Their atomic masses or the amount of branching do not decide the order in the final name.

Answer: C. C–F

Fluorine is small, so the carbon fluorine bond is short and strong. Among the listed carbon halogen single bonds, it has the greatest bond dissociation energy.

Carbon iodine is the longest and generally weakest of the listed series; do not equate a heavy halogen with a stronger bond.

Answer: C. Tertiary

Lucas reagent replaces the OH group with chloride, producing an insoluble alkyl chloride and visible cloudiness. Tertiary alcohols react rapidly because a relatively stable tertiary carbocation can form.

Secondary alcohols react more slowly, while primary alcohols generally do not cloud immediately at room temperature.

Answer: B. Alcohol

Lithium aluminium hydride is a strong reducing agent that converts an ester to alcohol products. The acyl portion becomes a primary alcohol, and the alkoxy portion is released as an alcohol after workup.

Stopping at an aldehyde would require different conditions or a milder reagent; the strong reduction continues.

Working
RCOOR′ → RCH₂OH + R′OH after workup

Answer: A. Aldehyde

An aldehyde carbonyl carbon is generally less sterically hindered than a ketone carbonyl carbon. It also receives less electron donation from alkyl groups.

These effects make nucleophilic attack easier on a typical aldehyde than on a comparable ketone.

Answer: D. Tollens'

Tollens reagent contains a silver complex that aldehydes can reduce to metallic silver. The deposited silver produces the characteristic mirror.

Ketones generally do not give the same test. Fehling reagent gives a different coloured precipitate rather than a silver mirror.

Answer: C. Lipoproteins — Lipids

Lipoproteins contain a protein component associated with lipids, so that pairing fits. Glycoproteins instead have carbohydrate groups attached.

Chromoproteins contain coloured groups, and nucleoproteins contain nucleic acids, ruling out the other printed combinations.

Answer: B. Hot-melt

Hot melt adhesive is applied while heated and fluid. It solidifies as it cools, bonding surfaces in packaging and bookbinding.

A pressure sensitive adhesive stays tacky under pressure, so the setting mechanism in the stem identifies hot melt.

Answer: A. Nylon-6,6

Nylon 6,6 forms from a diamine and a dicarboxylic acid, creating amide links while eliminating small molecules. That is condensation polymerization.

Polyethene, PVC, and polystyrene are commonly formed by addition polymerization of alkene monomers.

Answer: D. Contrast

“I like cake” and “my friend likes chocolate” present different preferences. The transition “however” signals that the second clause contrasts with the first.

It does not add a matching example, give a sequence, or state a special exception to a rule.

Answer: C. They reduce air pollution

The passage connects their popularity with pollution concerns and says electric cars produce no emissions during operation. The matching answer is reduced air pollution.

The passage does not claim that they are faster or cheaper to manufacture, and it says they use rechargeable batteries rather than petrol.

Answer: B. Singing is my favorite hobby

“Singing” is an -ing form used as a noun and occupies the subject position before “is.” That makes the second sentence the requested example.

In “I love singing,” it is an object; “is singing” is part of a verb phrase, and “singing bird” is descriptive.

Answer: C. Under

“Under the bed” expresses a position lower than and covered by the bed. It fits the ordinary spatial meaning of the sentence.

“Inside” would require enclosure, while “to” and “from” express different relationships.

Answer: C. At

“Turn left at the traffic signal” treats the signal as the point where the turn happens. The sentence uses a location marker, so “at” is idiomatic.

“Into” suggests entering something, and “between” needs two reference points.

Answer: C. The light went out while I was reading

The ongoing background action takes past continuous “was reading.” The shorter event that interrupts it takes simple past “went out.”

The other options mix present and past time in ways that do not fit the intended narrative.

Working
Ongoing: I was reading
interrupting event: the light went out

Answer: A. The number of students in Class 9 is twenty

The head of the subject is singular “number” in the phrase “the number of students.” The verb therefore takes singular “is.”

“Students” is plural but belongs to the modifying phrase; it does not control the agreement here.

Answer: C. Late

“Stay out late” means remain away until a late hour. “Late” works directly as the adverb modifying the action in this expression.

“Lately” means recently, while “later” compares a time with an earlier one.

Answer: C. Wearing red clothes causes intelligence

Clothing colour has no credible causal connection to intelligence. The assertion confuses an arbitrary feature with a cause.

The other examples describe familiar plausible effects, such as rain wetting roads and exercise improving health.

Answer: D. 24

There are four choices for the first chair, three for the second, two for the third, and one for the last. Multiply the choices because chair positions are distinct.

This is the permutation of four people in four seats, not a selection of an unordered group.

Working
4 × 3 × 2 × 1 = 4! = 24

Answer: D. 36

The successive differences are 2, 3, 4, 5, 6, and 7. The next increase is therefore eight.

Adding eight to 28 gives 36, which is the listed option D.

Working
Differences: +2, +3, +4, +5, +6, +7, +8
28 + 8 = 36

Answer: A. HR

IT meets last, so it takes 2 pm. Finance must meet before HR, while HR cannot take the 9 am slot.

The only consistent arrangement is Finance at 9 am and HR at 11 am.

Working
9 am Finance
11 am HR
2 pm IT

Answer: B. Granddaughter

Salman is Amna's father, and Amna is Aimen's mother. Following those two parent child links places Salman one generation above Amna and two above Aimen.

From Salman's viewpoint, Aimen is his granddaughter; Sarah's relation is extra information and does not change the answer.

Working
Salman → Amna → Aimen

Answer: C. 150

The hospital performs the same count on each of six days in the week. Multiply the daily count by the number of working days.

The result is 150 surgeries, matching option C rather than a seven day total.

Working
25 surgeries/day × 6 days = 150 surgeries

Answer: C. Cooperate with the officer and comply with traffic regulations

Parking in the prohibited area has already occurred, so the appropriate response is to cooperate with the officer and follow traffic rules. That addresses the situation responsibly.

Arguing, fleeing, or leaving without permission avoids the officer's instructions instead of resolving the violation.

Answer: B. To improve and adapt

Monitoring reveals whether a decision is producing its intended result. Feedback allows adjustments when circumstances or evidence change.

The purpose is to improve the action, not to obstruct progress, silence feedback, or assign blame automatically.

Answer: A. Human rights are only one of many factors influencing economic growth

The argument observes improved labour rights but low growth and assumes rights therefore have no helpful effect. Economic growth depends on many other conditions as well.

One factor can support growth without being sufficient to guarantee it, so the observed outcome does not disprove the claimed contribution.

Answer: A. Grapefruit is more expensive than grapes

Oranges are cheaper than grapefruit, and oranges are more expensive than grapes. Put those two comparisons in one order.

Grapefruit is at the top and grapes at the bottom; therefore grapefruit costs more than grapes.

Working
Grapefruit > oranges > grapes