MCQ #1 of 150
Biology
NUMS 2021
[NUMS 2021]
If fertilization occurs, the young embryo implants into which layer of the uterus?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The human uterus consists of three anatomical layers, of which the inner mucosal lining undergoes cyclical thickening to support blastocyst implantation.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Fertilization takes place in the ampulla of the fallopian tube.
- The resulting blastocyst travels toward the uterine cavity and implants into the glandular endometrium approximately 6 to 8 days post fertilization.
Why other options are incorrect:- Option A: The perimetrium is the outer serosal layer covering the uterus.
- Option B: The myometrium is the thick intermediate muscular layer responsible for contractions during labor.
- Option D: Epimetrium is an obsolete or nonstandard term sometimes confused with perimetrium.
MCQ #2 of 150
Biology
NUMS 2021
[NUMS 2021]
In the human male reproductive system, spermatozoa are produced directly within the:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Spermatogenesis is the process of male gamete formation occurring inside specialized structural units of the testes.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Each testis contains numerous lobules packed with tightly coiled seminiferous tubules.
- Spermatogenic cells line these tubules and undergo meiotic divisions under the support of Sertoli cells to produce immature sperm cells.
Why other options are incorrect:- Option B: The epididymis functions as a storage site where sperm achieve physiological maturation and motility.
- Option C: The vas deferens is an excretory duct transporting mature sperm toward the ejaculatory duct.
- Option D: Seminal vesicles are accessory glands that produce fructose-rich alkaline seminal fluid.
MCQ #3 of 150
Biology
NUMS 2021
[NUMS 2021]
Gonadotropic hormones including FSH and LH are secreted by which endocrine gland?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Gonadotropins are glycoprotein hormones that regulate reproductive gonadal activity and are synthesized by specialized basophil cells in the adenohypophysis.
Formula / Rule / Reaction:$$\text{Hypothalamic GnRH} \rightarrow \text{Anterior Pituitary} \rightarrow \text{FSH and LH}$$
Solution:- Gonadotropin-releasing hormone (GnRH) travels via the hypophyseal portal system from the hypothalamus to the anterior pituitary.
- In response, the anterior pituitary releases Follicle-Stimulating Hormone (FSH) and Luteinizing Hormone (LH) into systemic circulation.
Why other options are incorrect:- Option A: The hypothalamus secretes releasing and inhibiting factors such as GnRH, not the gonadotropins themselves.
- Option C: The adrenal cortex secretes corticosteroid hormones such as aldosterone and cortisol.
- Option D: The testes are target organs that secrete testosterone in response to LH stimulation.
MCQ #4 of 150
Biology
NUMS 2021
[NUMS 2021]
Which type of cartilage is located at the articulating ends of long bones, the nasal septum, larynx, and tracheal rings?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Cartilage is classified into three histological types based on the composition of its extracellular matrix: hyaline, elastic, and fibrocartilage.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Hyaline cartilage possesses a glassy, bluish-white matrix containing fine type II collagen fibrils.
- It provides low-friction articulation surfaces at movable joints and maintains airway patency in the trachea and larynx.
Why other options are incorrect:- Option A: Fibrocartilage contains dense bundles of type I collagen and forms the intervertebral discs and pubic symphysis.
- Option B: Elastic cartilage contains abundant elastic fibers and is found in the pinna of the ear and the epiglottis.
- Option C: Calcified cartilage represents a transitional mineralized state typically seen prior to endochondral ossification.
MCQ #5 of 150
Biology
NUMS 2021
[NUMS 2021]
A muscle characterized by involuntary control, intrinsic autorhythmicity, branching fibers, and typically one centrally placed nucleus per cell is:
C
Multi-unit smooth muscle
D
Single-unit smooth muscle
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Muscle tissue is categorized histologically and functionally into skeletal, cardiac, and smooth muscle variants.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Cardiac muscle cells (cardiomyocytes) are striated, branched, uninucleate, and contract rhythmically without requiring external neural stimulation.
- They are coupled mechanically and electrically via intercalated discs containing gap junctions.
Why other options are incorrect:- Option B: Skeletal muscle fibers are multinucleated, voluntary, and non-branching.
- Option C: Multi-unit smooth muscle fibers are non-striated, lack intrinsic spontaneous rhythmic activity, and depend on neurogenic stimulation.
- Option D: Single-unit smooth muscle lacks striations and organized sarcomere banding.
MCQ #6 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following anatomical locations features a multi-axial ball-and-socket synovial joint?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Ball-and-socket joints permit rotational movement in multiple planes around a central axis.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- The glenohumeral (shoulder) joint is formed by the hemispherical head of the humerus articulating within the glenoid fossa of the scapula.
- This structure provides the widest range of motion of any joint in the human body.
Why other options are incorrect:- Option A: The knee is predominantly a modified hinge synovial joint.
- Option B: The humeroulnar joint of the elbow is a classic uniaxial hinge joint.
- Option D: The radiocarpal joint at the wrist is classified as a biaxial condyloid (ellipsoid) joint.
MCQ #7 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following human physiological processes operates via positive feedback rather than negative feedback?
B
Uterine contractions during labor
C
Regulation of blood glucose concentration
D
Control of blood calcium levels by parathyroid hormone
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Negative feedback counteracts deviations from a set point to preserve homeostasis, whereas positive feedback amplifies an initial stimulus until a discrete physiological terminus is reached.
Formula / Rule / Reaction:$$\text{Cervical stretch} \rightarrow \text{Oxytocin release} \rightarrow \text{Stronger contraction} \rightarrow \text{Greater stretch}$$
Solution:- During parturition, mechanical stretching of the cervix triggers oxytocin secretion from the posterior pituitary.
- Oxytocin intensifies uterine smooth muscle contractions, further stretching the cervix and propagating a self-reinforcing cycle known as the Ferguson reflex.
Why other options are incorrect:- Option A: Core body temperature adjustments use negative feedback loops involving hypothalamic thermoreceptors.
- Option C: Insulin and glucagon maintain blood glucose through reciprocal negative feedback.
- Option D: Parathyroid hormone secretion is inhibited by elevated extracellular calcium via negative feedback.
MCQ #8 of 150
Biology
NUMS 2021
[NUMS 2021]
The gray matter of the spinal cord does NOT contain:
A
Cell bodies of interneurons
B
Cell bodies of somatic motor neurons
C
Unmyelinated axons and dendrites
D
Cell bodies of primary sensory neurons
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The central nervous system segregates neuronal soma and fiber tracts into gray and white matter regions.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Primary sensory neurons are pseudounipolar cells whose soma reside outside the central nervous system within dorsal root ganglia.
- The spinal cord gray matter houses interneuronal cell bodies in the dorsal horn and somatic motor neuron cell bodies in the ventral horn.
Why other options are incorrect:- Option A: Interneuron cell bodies are prominent components of the dorsal and intermediate horns of gray matter.
- Option B: Lower motor neuron cell bodies are situated within the anterior (ventral) horn of the spinal cord gray matter.
- Option C: Synaptic neuropil, including unmyelinated axons, forms the bulk of gray matter.
MCQ #9 of 150
Biology
NUMS 2021
[NUMS 2021]
In enzymatic kinetics, the term 'maximum temperature' designates the point at which:
A
Enzymes begin to denature irreversibly
B
The rate of catalytic activity reaches its optimal peak
C
Enzyme renaturation starts spontaneously
D
Enzyme molecules become completely saturated with substrate
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Thermal energy increases enzyme kinetic activity up to an optimum point, beyond which excessive vibrations disrupt weak non-covalent structural bonds.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- The optimum temperature yields maximal substrate conversion.
- The maximum temperature refers to the upper thermal threshold at which heat-induced thermal unfolding (denaturation) begins, causing the active site geometry to collapse.
Why other options are incorrect:- Option B: The temperature at which an enzyme works with peak efficiency is termed the optimum temperature.
- Option C: Thermal denaturation at high temperatures is generally irreversible; enzymes do not renature upon further heating.
- Option D: Substrate saturation relates to substrate concentration (\(V_{\max}\)), not thermal stability thresholds.
MCQ #10 of 150
Biology
NUMS 2021
[NUMS 2021]
In metabolic pathways, the term 'feedback inhibition' describes a mechanism where:
A
An initial reactant permanently inactivates the first enzyme
B
The accumulated end product allosterically inhibits an early regulatory enzyme
C
The substrate competes reversibly for the active site of the final enzyme
D
A coenzyme is degraded to halt the pathway
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Feedback inhibition (end-product inhibition) is an efficient regulatory cellular mechanism that prevents the wasteful overproduction of metabolic intermediates.
Formula / Rule / Reaction:$$\text{Precursor} \xrightarrow{\text{Enzyme 1}} \text{A} \rightarrow \text{B} \rightarrow \text{End Product} \dashv \text{Enzyme 1}$$
Solution:- When the concentration of the ultimate product exceeds cellular requirements, product molecules bind to an allosteric regulatory site on the initial committing enzyme.
- This induces a conformational change that decreases the enzyme's affinity for its initial substrate, throttling downstream flux.
Why other options are incorrect:- Option A: Reactants stimulate rather than inhibit committing steps, and feedback control is reversible rather than permanent.
- Option C: Feedback inhibition predominantly involves allosteric non-competitive interactions rather than active-site substrate competition.
- Option D: Coenzymes are not routinely degraded to implement rapid metabolic adjustments.
MCQ #11 of 150
Biology
NUMS 2021
[NUMS 2021]
Chemical substances that modulate and control enzymatic reaction velocities are classified as:
A
Only competitive inhibitors
C
Activators and inhibitors
D
Prosthetic groups exclusively
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Enzyme activity in biological systems is subject to dynamic up-regulation and down-regulation via regulatory chemical ligands.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Enzymatic regulators comprise both positive effectors (activators, which enhance catalytic turnover) and negative effectors (inhibitors, which decrease catalytic velocity).
- Together, activators and inhibitors maintain cellular homeostasis.
Why other options are incorrect:- Option A: Regulatory substances include non-competitive inhibitors and activators, not merely competitive inhibitors.
- Option B: Substrates are the transformed reactants, not the regulatory modifying agents.
- Option D: Prosthetic groups are permanently bound non-protein cofactors essential for basal catalysis.
MCQ #12 of 150
Biology
NUMS 2021
[NUMS 2021]
Elevating temperature significantly above the physiological optimum arrests enzymatic function because enzymes:
A
Precipitate out of solution as pure salts
B
Undergo extensive electrolytic ionization
C
Form irreversible covalent complexes with substrate
D
Denature due to disruption of secondary and tertiary structures
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Proteins rely on fragile intramolecular interactions including hydrogen bonds, ionic bridges, and hydrophobic interactions to preserve their functional three-dimensional configuration.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Excessive thermal kinetic energy causes rapid molecular vibration that ruptures stabilizing non-covalent bonds.
- The native tertiary active site collapses into a non-functional, unfolded polypeptide conformation.
Why other options are incorrect:- Option A: Inactivation is driven by structural conformational loss, not salt precipitation.
- Option B: Ionization states change with pH fluctuations; elevated heat causes thermal denaturation.
- Option C: Denatured enzymes lose active-site complementarity and cannot form substrate complexes.
MCQ #13 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following propositions was NOT an original component of Charles Darwin's 1859 theory of natural selection?
A
Heritable variations arise spontaneously through specific gene mutations
B
Organisms produce far more offspring than the environment can support
C
Individuals within a population compete for limited environmental resources
D
Better-adapted individuals exhibit differential survival and reproductive success
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Classical Darwinian evolutionary theory was formulated prior to the development of Mendelian genetics and molecular biology.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Charles Darwin identified overproduction, struggle for existence, and survival of the fittest based on natural phenotypic variation.
- He possessed no working model of genetic inheritance; the role of gene mutations and alleles was integrated later in the Modern Synthesis during the 1930s.
Why other options are incorrect:- Option B: Super-fecundity (overproduction of progeny) was a central premise derived from Thomas Malthus.
- Option C: The struggle for limited resources is a core tenet of Darwin's mechanism.
- Option D: Differential reproductive fitness is the fundamental definition of natural selection.
MCQ #14 of 150
Biology
NUMS 2021
[NUMS 2021]
The vestigial wings of the kiwi, elongated hooves of modern equines, and webbed feet of waterfowl were historically explained using the concept of use and disuse by:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Lamarckism proposed that environmental demands lead to the differential use or disuse of anatomical parts, and that these acquired physiological modifications are transmitted to progeny.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Jean-Baptiste de Lamarck published his evolutionary theory in 1809 in Philosophie Zoologique.
- He used webbed feet of swimming birds and rudimentary wings of flightless birds as canonical examples of structural adaptation through physical use and disuse.
Why other options are incorrect:- Option A: Hugo de Vries introduced the Mutation Theory of evolution in the early 1900s.
- Option C: August Weismann disproved Lamarckian inheritance of acquired traits via his germ-plasm theory.
- Option D: Alfred Russel Wallace co-discovered natural selection alongside Charles Darwin.
MCQ #15 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following statements regarding interferons is NOT correct?
A
They belong to the cytokine family of regulatory proteins
B
They activate natural killer (NK) cells to target abnormal cells
C
They are specialized leukocyte proteins that directly secrete interleukins
D
They induce an antiviral state in neighboring uninfected host cells
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Interferons are signaling glycoproteins produced by virus-infected host cells to inhibit viral replication and modulate innate immune defenses.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Interferons are effector signaling proteins; they do not possess secretory machinery to synthesize or secrete other signaling molecules such as interleukins.
- Interleukins are produced independently by helper T cells, monocytes, and macrophages.
Why other options are incorrect:- Option A: Interferons are biologically classified as cytokine signaling molecules.
- Option B: Type I and Type II interferons stimulate cytotoxic T lymphocytes and natural killer cells.
- Option D: Interferons bind surface receptors on nearby cells, stimulating antiviral proteins that block mRNA translation.
MCQ #16 of 150
Biology
NUMS 2021
[NUMS 2021]
Intrinsic factor, synthesized and released by gastric parietal cells, is essential for the intestinal absorption of:
A
Ascorbic acid (Vitamin C)
B
Dietary iron in the duodenum
D
Cobalamin (Vitamin \(\text{B}_{12}\))
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Vitamin \(\text{B}_{12}\) requires carrier-mediated transport to protect it against luminal degradation and facilitate enterocyte uptake.
Formula / Rule / Reaction:$$\text{Intrinsic Factor} + \text{Vitamin B}_{12} \rightarrow [\text{IF}-\text{B}_{12} \text{ Complex}] \xrightarrow{\text{Cubam receptor}} \text{Ileal Absorption}$$
Solution:- Parietal cells located within the gastric fundus and body secrete intrinsic factor alongside \(\text{HCl}\).
- Intrinsic factor binds cobalamin in the duodenum; the resulting complex resists enzymatic digestion until it binds specific cubilin receptors in the terminal ileum.
Why other options are incorrect:- Option A: Vitamin C is water-soluble and absorbed via sodium-ascorbate cotransporters (SVCT) without intrinsic factor.
- Option B: Inorganic iron absorption is promoted by gastric acid reduction to \(\text{Fe}^{2+}\) and transported by DMT-1.
- Option C: Vitamin D is absorbed via lipid micelle incorporation in the small intestine.
MCQ #17 of 150
Biology
NUMS 2021
[NUMS 2021]
The greatest volume of gastric juice secretion during a meal is triggered primarily during the gastric phase by:
A
The presence of proteinaceous food in the stomach
B
The initial olfactory stimulation from food aromas
C
Visual stimulation upon viewing appetizing food
D
Gustatory stimulation on the tongue
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Gastric secretion is divided into cephalic, gastric, and intestinal phases, with the gastric phase producing approximately 50 to 60 percent of the total volume.
Formula / Rule / Reaction:$$\text{Peptides / Amino acids} \rightarrow \text{G cells} \rightarrow \text{Gastrin release} \rightarrow \text{Massive parietal secretion}$$
Solution:- When ingested proteins enter the stomach, luminal peptides and mechanical distension stimulate antral G cells to release the peptide hormone gastrin.
- Gastrin acts directly on parietal cells and triggers histamine release from ECL cells, inducing maximal gastric acid output.
Why other options are incorrect:- Option B: Olfactory stimulation belongs to the cephalic phase, which accounts for roughly 30 percent of baseline gastric juice secretion.
- Option C: Visual cues initiate vagal reflexes within the cephalic phase only.
- Option D: Taste receptors stimulate parasympathetic fibers during the cephalic phase, yielding smaller volumes than direct gastric luminal presence.
MCQ #18 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following physiological functions is NOT a metabolic role carried out by hepatocytes in the liver?
B
Synthesis of circulating immunoglobulins
D
Deamination of excess amino acids
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The liver performs numerous biochemical conversions and synthesizes the majority of plasma proteins, but does not synthesize antibodies.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Hepatocytes produce plasma albumin, fibrinogen, prothrombin, and acute-phase proteins.
- Immunoglobulins (antibodies) are exclusively produced and secreted by differentiated humoral B-lymphocytes (plasma cells).
Why other options are incorrect:- Option A: Glycogenolysis is the hepatic depolymerization of glycogen reserves into glucose during fasting.
- Option C: Gluconeogenesis is the hepatic synthesis of glucose from non-carbohydrate substrates such as lactate and glycerol.
- Option D: Deamination of amino acids takes place within hepatocytes, yielding ammonia which is converted to urea.
MCQ #19 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following digestive enzymes is NOT produced by the acinar cells of the pancreas?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Digestive enzymes exhibit strict compartmental specialization according to their glandular origin and optimal operating pH.
Formula / Rule / Reaction:$$\text{Pepsinogen} \xrightarrow{\text{Gastric } \text{HCl}} \text{Pepsin (Active at pH 1.5 to 2.0)}$$
Solution:- Pepsin is secreted as the zymogen pepsinogen by chief (peptic) cells located in the gastric glands of the stomach.
- The exocrine pancreas secretes an alkaline mixture containing trypsinogen, chymotrypsinogen, pancreatic lipase, and pancreatic amylase into the duodenum.
Why other options are incorrect:- Option A: Pancreatic lipase is synthesized by pancreatic acinar tissue to hydrolyze dietary triglycerides into monoglycerides and free fatty acids.
- Option B: Trypsinogen is the primary pancreatic endopeptidase zymogen activated by enterokinase in the duodenum.
- Option D: Amylase is produced by both salivary glands and the pancreas to cleave alpha-1,4-glycosidic bonds.
MCQ #20 of 150
Biology
NUMS 2021
[NUMS 2021]
The systemic aorta originates directly from which anatomical chamber of the heart?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The mammalian heart uses two parallel circulatory loops: the pulmonary circuit and the systemic circuit.
Formula / Rule / Reaction:$$\text{Left Ventricle} \xrightarrow{\text{Aortic Semilunar Valve}} \text{Ascending Aorta} \rightarrow \text{Systemic Circulation}$$
Solution:- Oxygenated blood returns from the pulmonary veins to the left atrium, passes through the bicuspid (mitral) valve, and enters the left ventricle.
- Contraction of the left ventricle ejects blood at high pressure through the aortic semilunar valve into the ascending aorta.
Why other options are incorrect:- Option A: The right ventricle pumps deoxygenated blood into the pulmonary trunk toward the lungs.
- Option B: The left atrium receives oxygenated blood from the four pulmonary veins.
- Option C: The right atrium receives systemic venous blood returning via the superior and inferior venae cavae.
MCQ #21 of 150
Biology
NUMS 2021
[NUMS 2021]
The human lymphatic system consists of specialized lymphoid tissues and vessels, but does NOT include the:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The lymphatic system is a specialized vascular network that drains interstitial fluid and houses components of the adaptive immune system.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- The lymphatic system comprises lymphatic capillaries, collecting ducts, lymph nodes, the spleen, tonsils, and the thymus.
- The lungs are primary organs of the respiratory system dedicated to external gas exchange, although they contain resident bronchus-associated lymphoid tissue (BALT).
Why other options are incorrect:- Option B: The spleen is the largest secondary lymphoid organ in the human body, filtering blood-borne antigens.
- Option C: The thymus is a primary lymphoid organ responsible for T-lymphocyte maturation.
- Option D: Lymph nodes are encapsulated secondary lymphoid filtration structures distributed along lymph vessels.
MCQ #22 of 150
Biology
NUMS 2021
[NUMS 2021]
In eukaryotic cells, the small 40S ribosomal subunit joins with the large 60S subunit in the presence of \(\text{Mg}^{2+}\) ions to form a functional translational particle of:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Svedberg (S) units reflect sedimentation velocity during high-speed centrifugation, which depends non-linearly on particle mass, surface area, and shape.
Formula / Rule / Reaction:$$\text{40S Subunit} + \text{60S Subunit} \xrightarrow{\text{Mg}^{2+}} \text{80S Cytoplasmic Ribosome}$$
Solution:- Svedberg coefficients are not additive because complexation alters the total hydrodynamic surface area and frictional coefficient.
- In eukaryotic cytoplasm, the 40S small subunit (containing 18S rRNA) associates with the 60S subunit (containing 28S, 5.8S, and 5S rRNAs) to assemble the mature 80S translation ribosome.
Why other options are incorrect:- Option A: 100S is an arithmetic sum ($40 + 60$), which incorrectly assumes sedimentation values are linearly additive.
- Option C: 70S is the complete ribosome size found in prokaryotes, mitochondria, and chloroplasts (composed of 30S and 50S subunits).
- Option D: 90S corresponds to an early pre-ribosomal intermediate in the nucleolus, not the mature cytoplasmic translation particle.
MCQ #23 of 150
Biology
NUMS 2021
[NUMS 2021]
The resident bacterial flora colonizing human skin and mucous membranes is beneficial because it:
A
Directly produces endogenous broad-spectrum pharmaceutical antibiotics
B
Produces antigens that neutralize human cellular toxins
C
Interferes with pathogenic colonization via niche competition
D
Eliminates the host requirement for adaptive cell-mediated immunity
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The normal human microbiota provides microbial antagonism (competitive exclusion), serving as a crucial component of the innate immune barrier.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Commensal bacteria occupy mucosal surface receptors, consume available local nutrients, and produce bacteriocins as well as acidic waste products.
- This microenvironment directly impedes colonization, proliferation, and tissue invasion by pathogenic microorganisms.
Why other options are incorrect:- Option A: Commensals synthesize bacteriocins for inter-bacterial competition, not clinical pharmaceutical antibiotics.
- Option B: Resident flora do not manufacture host-directed antitoxins.
- Option D: Commensal organisms prime the immune system, but do not replace the host's need for functional T-cell mediated immunity.
MCQ #24 of 150
Biology
NUMS 2021
[NUMS 2021]
In endospore-forming bacteria such as Bacillus and Clostridium, exceptional resistance against extreme heat is conferred by:
A
Chromosomal and extrachromosomal plasmids
B
Intracellular glycogen granules
C
Thick peptidoglycan cell walls alone
D
Dehydrated endospores containing dipicolinic acid and calcium
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Certain Gram-positive bacteria form cryptobiotic resting structures called endospores when subjected to severe nutritional starvation or environmental stress.
Formula / Rule / Reaction:$$\text{Spore heat tolerance} \propto [\text{Calcium-Dipicolinate complex}] + \text{Core Dehydration}$$
Solution:- Bacterial endospores possess an extremely dehydrated protoplasmic core enriched with calcium dipicolinate (dipicolinic acid complexed with \(\text{Ca}^{2+}\)).
- This chemical complex, along with small acid-soluble spore proteins (SASPs) and a multi-layered keratin-like protein coat, stabilizes DNA and cellular enzymes against thermal denaturation up to boiling temperatures.
Why other options are incorrect:- Option A: Plasmids carry accessory genes for antibiotic resistance, not thermal physical resistance.
- Option B: Glycogen granules are simple nutrient storage inclusions with zero protective capacity.
- Option C: Normal vegetative peptidoglycan cell walls denature and lyse readily when boiled.
MCQ #25 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following endocrine hormone pairs does NOT demonstrate antagonistic physiological functions?
A
Adrenaline and Noradrenaline
B
Calcitonin and Parathyroid hormone
D
Growth Hormone and Somatostatin
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Antagonistic hormones exert opposing physiological regulatory effects on a shared parameter, whereas synergistic hormones act in concert to achieve complementary outcomes.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Adrenaline (epinephrine) and noradrenaline (norepinephrine) are co-secreted by the adrenal medulla during sympathetic activation.
- Both act synergistically to mobilize glucose reserves, raise cardiac output, and mediate the fight-or-flight stress response.
Why other options are incorrect:- Option B: Calcitonin lowers blood calcium levels, whereas parathyroid hormone elevates extracellular calcium.
- Option C: Insulin drives cellular glucose uptake to lower glycemia, while glucagon stimulates glycogenolysis to elevate blood glucose.
- Option D: Growth hormone promotes somatic development, while somatostatin acts as growth hormone-inhibiting hormone (GHIH).
MCQ #26 of 150
Biology
NUMS 2021
[NUMS 2021]
Viruses identify and bind to their specific susceptible host cells primarily via:
A
Double-stranded genomic viral DNA
B
Specific surface glycoprotein spikes or capsomeres
C
Reverse transcriptase enzyme complexes
D
Host-derived lipid bilayer envelopes alone
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Viral tropism is determined by stereochemical recognition between viral attachment proteins and complementary surface receptors on the target host cell membrane.
Formula / Rule / Reaction:$$\text{Viral Attachment Protein (Spike / Capsomere)} + \text{Host Membrane Receptor} \rightarrow \text{Adsorption}$$
Solution:- Enveloped viruses display glycoprotein spikes embedded within their envelopes, whereas non-enveloped viruses use surface capsomeres.
- These surface structures bind selectively to corresponding host surface proteins, determining the host range and tissue specificity of the virus.
Why other options are incorrect:- Option A: Viral nucleic acids are sequestered inside the protective protein capsid and cannot interact directly with exterior host receptors.
- Option C: Reverse transcriptase is an internal replication enzyme found in retroviruses, not an external attachment ligand.
- Option D: The lipid envelope is derived non-specifically from host cell membranes during budding and lacks inherent binding specificity without embedded spikes.
MCQ #27 of 150
Biology
NUMS 2021
[NUMS 2021]
Cell-surface markers that facilitate immunological cell-cell recognition and tissue identity are chemically composed of:
A
Insoluble structural lipoproteins exclusively
B
Sphingomyelins and neutral phospholipids
C
Glycoproteins and glycolipids
D
Peripheral channel lipoproteins
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The fluid mosaic model features branched carbohydrate chains covalently linked to proteins or lipids on the outer membrane leaflet, forming the cellular glycocalyx.
Formula / Rule / Reaction:$$\text{Carbohydrate} + \text{Protein / Lipid} \rightarrow \text{Glycoprotein / Glycolipid (Cell Identity Marker)}$$
Solution:- Oligosaccharide chains attached covalently to membrane proteins (glycoproteins) and polar lipids (glycolipids) project into the extracellular space.
- These molecules function as biochemical recognition flags for cell adhesion, blood typing (ABO antigens), and histocompatibility (MHC recognition).
Why other options are incorrect:- Option A: Lipoproteins primarily function as circulating lipid-transport assemblies in plasma, not cell recognition markers.
- Option B: Sphingomyelins and phospholipids form the structural lipid bilayer, but do not provide specific biological recognition codes.
- Option D: Channel proteins mediate transmembrane solute transport rather than antigenic cell identification.
MCQ #28 of 150
Biology
NUMS 2021
[NUMS 2021]
Porin channel proteins are typically present in the outer membranes of which eukaryotic organelles?
A
Endoplasmic reticulum and Golgi apparatus
B
Lysosomes and central vacuoles
C
Ribosomes and peroxisomes
D
Mitochondria and chloroplasts
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Porins are beta-barrel transmembrane proteins that create large, water-filled channels permitting passive diffusion of hydrophilic molecules up to approximately 5 kDa.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Mitochondria and chloroplasts arose via endosymbiosis from ancestral Gram-negative prokaryotes.
- Consequently, their outer envelope membranes contain porin transport proteins that make them freely permeable to small metabolic solutes.
Why other options are incorrect:- Option A: The endomembrane system (ER and Golgi) regulates transport via vesicle fusion and specific permeases rather than open porins.
- Option B: Lysosomal membranes require specialized proton ATPase pumps and transporters to preserve an acidic internal pH.
- Option C: Ribosomes are non-membranous ribonucleoprotein complexes, and peroxisomes use specialized peroxin channels.
MCQ #29 of 150
Biology
NUMS 2021
[NUMS 2021]
The primary physiological function of parietal (oxyntic) cells located in the gastric mucosa is the secretion of:
A
Hydrochloric acid and intrinsic factor
C
Alkaline bicarbonate-rich protective mucus
D
Endocrine regulatory gastrin
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Gastric oxyntic glands contain specialized epithelial cell lineages each responsible for producing specific digestive secretions.
Formula / Rule / Reaction:$$\text{Parietal Cells} \xrightarrow{\text{Stimulation}} \text{HCl (via } \text{H}^+/\text{K}^+\text{-ATPase)} + \text{Intrinsic Factor}$$
Solution:- Parietal cells use active proton pumps (\(\text{H}^+/\text{K}^+\)-ATPase) to secrete hydrochloric acid, which lowers gastric pH to 1.5 to 2.0 to activate pepsinogen and kill pathogens.
- Simultaneously, parietal cells secrete intrinsic factor, which is required for vitamin \(\text{B}_{12}\) absorption in the ileum.
Why other options are incorrect:- Option B: Pepsinogen is secreted by gastric chief (peptic or zymogenic) cells.
- Option C: Protective mucin and bicarbonate are secreted by mucous neck and surface epithelial cells.
- Option D: Gastrin is an endocrine hormone synthesized by neuroendocrine G-cells in the pyloric antrum.
MCQ #30 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following cellular structures is universally present in all prokaryotic and eukaryotic organisms?
C
80S cytoplasmic ribosomes
D
Membrane-bound Golgi apparatus
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:All free-living biological cells require a selectively permeable amphipathic boundary to maintain internal chemical homeostasis distinct from the external environment.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Every living cell, whether prokaryotic (eubacteria, archaea) or eukaryotic (protists, fungi, plants, animals), is enclosed by a phospholipid bilayer membrane.
Why other options are incorrect:- Option A: Cellulose cell walls are restricted to green plants and certain algae; animal cells and mycoplasmas lack cell walls completely.
- Option C: 80S ribosomes are characteristic of eukaryotic cytoplasm; prokaryotes possess smaller 70S ribosomes.
- Option D: The Golgi apparatus is an endomembrane organelle found exclusively in eukaryotic cells.
MCQ #31 of 150
Biology
NUMS 2021
[NUMS 2021]
Which eukaryotic organelle contains an abundance of acid hydrolases active at an internal pH of approximately 4.5 to 5.0?
B
Rough endoplasmic reticulum
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Lysosomes function as the primary catabolic degradation compartments of eukaryotic cells, breaking down macromolecules, engulfed pathogens, and aged organelles.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Lysosomes maintain an acidic interior via vacuolar \(\text{H}^+\)-ATPase proton pumps.
- They contain over 50 different acid hydrolases (including proteases, nucleases, lipases, and sulfatases) that degrade biological polymers at optimal acidic pH.
Why other options are incorrect:- Option A: Peroxisomes contain catalase and oxidative enzymes that generate and degrade hydrogen peroxide at neutral pH.
- Option B: The rough endoplasmic reticulum specializes in protein translation, folding, and post-translational modification.
- Option D: Mitochondria contain enzymes for the citric acid cycle and oxidative phosphorylation within an alkaline matrix (pH ~7.8).
MCQ #32 of 150
Biology
NUMS 2021
[NUMS 2021]
The primary structural level of a protein is fundamentally determined by the:
A
Folding of polypeptide backbones into alpha helices and beta sheets
B
Three-dimensional assembly of multiple distinct polypeptide subunits
C
Overall spatial packing of tertiary globular domains
D
Linear sequence of amino acids linked by covalent peptide bonds
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Proteins exhibit hierarchical structural organization: primary, secondary, tertiary, and quaternary levels.
Formula / Rule / Reaction:$$\text{mRNA transcript} \xrightarrow{\text{Translation}} \text{Amino Acid Sequence (Primary Structure)}$$
Solution:- The primary structure is the specific, genetically encoded linear sequence of amino acids joined together through covalent peptide (amide) linkages.
- This sequence contains all the biochemical information necessary to dictate downstream secondary and tertiary folding.
Why other options are incorrect:- Option A: Localized regular conformations stabilized by backbone hydrogen bonds represent secondary structure.
- Option B: Multi-subunit association via non-covalent or disulfide bonds describes quaternary structure.
- Option C: The complete three-dimensional steric conformation of a single polypeptide chain corresponds to tertiary structure.
MCQ #33 of 150
Biology
NUMS 2021
[NUMS 2021]
Water acts as an outstanding universal biological solvent primarily because of its:
A
High polarity and hydrogen-bonding capability
B
High viscosity and surface tension
C
Neutral physiological pH of exactly 7.0
D
Low specific heat capacity
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The polar geometry and dielectric properties of water allow it to disrupt ionic lattices and form favorable hydration shells around solutes.
Formula / Rule / Reaction:$$\delta^- \text{O} - 2\delta^+ \text{H} \quad (\text{Dipole moment } \mu = 1.85\text{ D})$$
Solution:- Water has a bent molecular geometry with an electronegativity difference between oxygen and hydrogen, producing a permanent electric dipole.
- Its high dielectric constant weakens electrostatic attractions between oppositely charged ions, while its dipole forms hydration shells around polar molecules and ions.
Why other options are incorrect:- Option B: Viscosity and surface tension govern cohesive and capillary behaviors, not solvent dissolution power.
- Option C: A neutral pH reflects balanced autoionization ($[\text{H}^+] = [\text{OH}^-]$) rather than chemical solvation potential.
- Option D: Water possesses an exceptionally high specific heat capacity ($4.184\text{ J/g}^\circ\text{C}$), which stabilizes ambient temperatures.
MCQ #34 of 150
Biology
NUMS 2021
[NUMS 2021]
All carbohydrates are organic compounds that chemically contain which specific combination of elements?
A
Carbon, nitrogen, and oxygen
B
Carbon, hydrogen, and oxygen
C
Carbon, hydrogen, and sulfur
D
Carbon, hydrogen, and nitrogen
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Carbohydrates are polyhydroxy aldehydes, polyhydroxy ketones, or substances that yield these units upon acid hydrolysis.
Formula / Rule / Reaction:$$\text{C}_n(\text{H}_2\text{O})_m \quad \text{or} \quad \text{C}_n\text{H}_{2n}\text{O}_n$$
Solution:- True carbohydrates consist fundamentally of carbon, hydrogen, and oxygen, generally maintaining a $1:2:1$ atomic ratio in simple monosaccharides (e.g., glucose $\text{C}6\text{H}{12}\text{O}_6$).
Why other options are incorrect:- Option A: Nitrogen is an essential component of proteins and nucleic acids, but is absent from basic carbohydrates (except modified derivatives like chitin).
- Option C: Sulfur is found in specific amino acids (cysteine, methionine), not in carbohydrate backbones.
- Option D: Nitrogen is absent from unmodified monosaccharides, oligosaccharides, and polysaccharides.
MCQ #35 of 150
Biology
NUMS 2021
[NUMS 2021]
Within the eukaryotic interphase nucleus, the dense non-membrane-bound structure where ribosomal RNA is primarily transcribed and processed is the:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The nucleolus is a prominent subnuclear compartment organized around nucleolar organizer regions (NORs) of chromosomes containing tandemly repeated rRNA genes.
Formula / Rule / Reaction:$$\text{rRNA genes} \xrightarrow{\text{RNA Pol I}} 45\text{S pre-rRNA} \rightarrow 18\text{S}, 5.8\text{S}, 28\text{S rRNAs}$$
Solution:- The nucleolus is the principal site of ribosomal RNA (rRNA) transcription, enzymatic processing, and assembly with imported ribosomal proteins to build pre-ribosomal subunits.
- It accounts for the majority of continuous RNA transcription within the interphase nucleus.
Why other options are incorrect:- Option A: Nuclear pore complexes are multi-protein octagonal channels that regulate bidirectional nucleocytoplasmic transport.
- Option B: The nuclear envelope is a double membrane continuous with the rough endoplasmic reticulum that demarcates the nucleus.
- Option D: The centrosome is a cytoplasmic microtubule-organizing center containing centrioles.
MCQ #36 of 150
Biology
NUMS 2021
[NUMS 2021]
Turgor pressure in non-woody plant tissues is maintained primarily by the active accumulation of solutes within the:
A
Apoplastic cell wall space
D
Central vacuole bounded by the tonoplast
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The central vacuole functions as a hyperosmotic hydrostatic engine that maintains cellular turgidity and provides mechanical support to herbaceous plants.
Formula / Rule / Reaction:$$\Psi_w = \Psi_s + \Psi_p$$
Solution:- The tonoplast (vacuolar membrane) uses active proton pumps and antiporters to transport inorganic ions (such as \(\text{K}^+\)) into the vacuolar lumen.
- This lowers the vacuolar solute potential ($\Psi_s$), driving water into the vacuole by osmosis and exerting positive turgor pressure ($\Psi_p$) against the rigid cell wall.
Why other options are incorrect:- Option A: The apoplast is the continuum of external cell walls and intercellular spaces through which water diffuses passively.
- Option B: Ribosomes synthesize proteins and play no direct hydrostatic osmoregulatory role.
- Option C: Peroxisomes catabolize fatty acids and toxic peroxides, and do not establish systemic turgidity.
MCQ #37 of 150
Biology
NUMS 2021
[NUMS 2021]
In oxygenic photosynthetic plant cells, green chlorophyll pigments are embedded within the:
A
Thylakoid membranes of chloroplasts
C
Outer chloroplast envelope membrane
D
Inner chloroplast envelope membrane
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Photosynthetic pigments are hydrophobic molecules organized into light-harvesting antenna complexes within specialized internal lipid bilayers.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Chlorophyll molecules possess a hydrophilic porphyrin ring and a lipophilic phytol tail.
- The phytol tail anchors the pigment non-covalently into the thylakoid lipid bilayer, positioning the porphyrin heads to capture incoming photons for photosystems I and II.
Why other options are incorrect:- Option B: The stroma is the hydrophilic fluid matrix where carbon fixation reactions (Calvin cycle) take place.
- Option C: The outer envelope contains non-specific porin channels and lacks photosynthetic reaction centers.
- Option D: The inner membrane regulates metabolite transit and is devoid of pigment-protein complexes.
MCQ #38 of 150
Biology
NUMS 2021
[NUMS 2021]
Carotenoid accessory pigments known as xanthophylls typically impart which characteristic coloration to plant tissues?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Carotenoids are lipid-soluble tetraterpenoid accessory pigments divided into oxygen-free carotenes and oxygenated xanthophylls.
Formula / Rule / Reaction:$$\text{Xanthophylls: } \text{C}_{40}\text{H}_{56}\text{O}_n \quad (\text{Absorb blue-green light, transmit yellow})$$
Solution:- Xanthophylls contain oxygenated functional groups (such as hydroxyl or epoxide groups) on their terminal rings.
- They absorb high-energy blue-violet wavelengths ($400\text{ to }500\text{ nm}$) and transmit yellow wavelengths, providing photoprotection against photo-oxidation.
Why other options are incorrect:- Option A: Blue coloration in plants is typically produced by vacuolar anthocyanin pigments under alkaline pH.
- Option C: Green light is reflected by chlorophyll $a$ and $b$, not by carotenoids.
- Option D: Purple pigmentation is produced by anthocyanin flavonoids, not by lipophilic plastidial carotenoids.
MCQ #39 of 150
Biology
NUMS 2021
[NUMS 2021]
The neural pathway that conducts an involuntary, rapid motor response to an external stimulus without requiring conscious processing in the cerebral cortex is termed a:
A
Sensory ascending tract
C
Corticospinal motor pathway
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:A reflex arc is the anatomical pathway underlying an involuntary reflex action, consisting of a receptor, sensory neuron, integration center, motor neuron, and effector.
Formula / Rule / Reaction:$$\text{Receptor} \rightarrow \text{Sensory Neuron} \rightarrow \text{Spinal Cord Interneuron} \rightarrow \text{Motor Neuron} \rightarrow \text{Effector}$$
Solution:- In a somatic spinal reflex, afferent action potentials enter the dorsal horn of the spinal cord and directly synapse with interneurons or motor neurons.
- Efferent signals exit through the ventral horn to stimulate muscles immediately, avoiding delay from ascending cerebral pathways.
Why other options are incorrect:- Option A: Sensory ascending tracts (such as the spinothalamic tract) carry signals upward to the thalamus and cortex for conscious perception.
- Option C: The corticospinal tract mediates voluntary motor commands descending from the motor cortex.
- Option D: The somatosensory network processes conscious tactile and pain signals in the parietal lobe.
MCQ #40 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following semi-autonomous organelles contains circular double-stranded DNA and 70S ribosomes?
A
Rough endoplasmic reticulum
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:According to the endosymbiotic theory, mitochondria and chloroplasts originated as free-living prokaryotes engulfed by ancestral eukaryotic host cells.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Mitochondria possess their own maternal non-nuclear circular genome (mtDNA) and synthesize some of their own proteins using prokaryotic-like 70S ribosomes.
- They replicate independently within the host cytoplasm via binary fission.
Why other options are incorrect:- Option A: The rough endoplasmic reticulum is an endomembrane organelle studded with host eukaryotic 80S ribosomes and lacks internal DNA.
- Option B: The Golgi apparatus is composed of flattened cisternae that lack genetic material.
- Option D: Lysosomes are single-membrane hydrolytic compartments that contain no autonomous genome or ribosomes.
MCQ #41 of 150
Biology
NUMS 2021
[NUMS 2021]
Which organelle packages and secretes complex non-cellulosic polysaccharides essential for plant cell plate and primary cell wall formation?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:During cytokinesis in plant cells, vesicles derived from dictyosomes (Golgi stacks) coalesce along the equatorial plane to assemble the phragmoplast and cell plate.
Formula / Rule / Reaction:$$\text{Golgi Cisternae} \rightarrow \text{Secretory Vesicles (Pectins and Hemicellulose)} \rightarrow \text{Phragmoplast} \rightarrow \text{Middle Lamella}$$
Solution:- The Golgi complex synthesizes complex matrix polysaccharides such as pectins and hemicelluloses.
- These are packaged into transport vesicles that fuse during telophase to construct the middle lamella and early primary cell wall.
Why other options are incorrect:- Option A: Centrosomes organize microtubule spindles during mitosis and are absent from higher plant cells.
- Option B: Peroxisomes catabolize long-chain fatty acids and manage hydrogen peroxide.
- Option C: The nucleolus synthesizes ribosomal subunits, not wall polysaccharides.
MCQ #42 of 150
Biology
NUMS 2021
[NUMS 2021]
The primary intracellular site responsible for the biosynthesis of steroid hormones, phospholipids, and the detoxification of hydrophobic xenobiotics is the:
A
Smooth endoplasmic reticulum
B
Rough endoplasmic reticulum
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The smooth endoplasmic reticulum (SER) contains integral membrane enzymes that synthesize lipids and metabolize lipophilic chemical compounds.
Formula / Rule / Reaction:$$\text{Cholesterol} \xrightarrow{\text{SER Cytochrome P450}} \text{Steroid Hormones (e.g., Cortisol, Testosterone)}$$
Solution:- The tubular membranes of the SER contain the enzymatic machinery needed to convert cholesterol into steroid derivatives in gonads and adrenal cortex cells.
- In hepatocytes, SER cytochrome P450 monooxygenases hydroxylate lipid-soluble drugs and metabolic toxins to facilitate their renal excretion.
Why other options are incorrect:- Option B: The rough endoplasmic reticulum specializes in translating and processing secretable and membrane-bound proteins.
- Option C: Lysosomes degrade endocytosed macromolecules and damaged organelles.
- Option D: Ribosomes catalyze peptide bond formation during mRNA translation.
MCQ #43 of 150
Biology
NUMS 2021
[NUMS 2021]
Which endocrine structure is historically designated as the 'master gland' because its trophic secretions regulate multiple peripheral endocrine tissues?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The pituitary gland (hypophysis) coordinates the endocrine system by releasing trophic hormones that control target glands such as the thyroid, adrenal cortex, and gonads.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- The anterior pituitary secretes thyroid-stimulating hormone (TSH), adrenocorticotropic hormone (ACTH), and gonadotropins (LH and FSH).
- Although the pituitary itself is regulated by hypothalamic releasing factors, its broad control over peripheral glandular output earned it the classical title of master gland.
Why other options are incorrect:- Option A: The thyroid gland produces metabolic hormones (T3, T4) and calcitonin under upstream TSH direction.
- Option C: The adrenal gland produces corticosteroids and catecholamines, but does not control other distinct endocrine glands.
- Option D: The pineal gland secretes melatonin to govern circadian rhythms.
MCQ #44 of 150
Biology
NUMS 2021
[NUMS 2021]
The high-energy chemical assimilation products synthesized during the light-dependent reactions of photosynthesis and subsequently consumed in the Calvin cycle are:
A
ADP and \(\text{NADP}^+\)
D
\(\text{FADH}_2\) and ATP
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The light reactions of oxygenic photosynthesis convert solar radiant energy into stable chemical energy stored in phosphoanhydride bonds and reducing equivalents.
Formula / Rule / Reaction:$$2\text{H}_2\text{O} + 2\text{NADP}^+ + 3\text{ADP} + 3\text{P}_i \xrightarrow{h\nu} \text{O}_2 + 2\text{NADPH} + 2\text{H}^+ + 3\text{ATP}$$
Solution:- Photolysis of water coupled with non-cyclic electron transport yields reduced NADPH and generates a proton gradient driving ATP synthase.
- Both ATP and NADPH diffuse into the stroma to power carbon reduction in the dark reactions.
Why other options are incorrect:- Option A: ADP and \(\text{NADP}^+\) are the oxidized byproducts returned to the thylakoid membrane from the stroma.
- Option B: Glucose is the ultimate product of the Calvin cycle, not a generated reactant of the light reactions.
- Option D: \(\text{FADH}_2\) is an electron carrier operating in mitochondrial cellular respiration, not chloroplast photosynthetic light reactions.
MCQ #45 of 150
Biology
NUMS 2021
[NUMS 2021]
The biochemical process defined as the cleavage of covalent bonds within a biological macromolecule accompanied by the addition of the elements of water is:
B
Oxidative decarboxylation
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Hydrolysis reactions use water molecules to break covalent linkages that join monomeric units within biological polymers.
Formula / Rule / Reaction:$$\text{R}_1-\text{R}_2 + \text{H}_2\text{O} \xrightarrow{\text{Hydrolase}} \text{R}_1-\text{OH} + \text{R}_2-\text{H}$$
Solution:- During enzymatic digestion of proteins, starches, and triglycerides, a water molecule is consumed per bond broken.
- Hydroxyl (\(-\text{OH}\)) binds one constituent moiety while a hydrogen ion (\(\text{H}^+\)) attaches to the other.
Why other options are incorrect:- Option A: Condensation (dehydration synthesis) joins monomers by eliminating water rather than consuming it.
- Option B: Oxidative decarboxylation removes a carboxyl group as \(\text{CO}_2\) while generating reducing equivalents.
- Option C: Hydrogenation is the chemical addition of elemental hydrogen (\(\text{H}_2\)) across unsaturated double bonds.
MCQ #46 of 150
Biology
NUMS 2021
[NUMS 2021]
Which of the following proteins functions primarily as an extracellular structural component imparting high tensile strength to connective tissues?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Structural proteins form supportive fibers that resist mechanical deformation throughout extracellular matrix networks.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Collagen is an insoluble fibrous glycoprotein characterized by a repeating Gly-X-Y amino acid triplet.
- Three polypeptide alpha chains wind into a right-handed triple helix that provides tensile resistance in bone, cartilage, tendons, and the dermis.
Why other options are incorrect:- Option B: Pepsinogen is an inactive proteolytic zymogen secreted into gastric juice.
- Option C: Casein is a nutrient storage phosphoprotein found in mammalian milk.
- Option D: Insulin is a globular peptide hormone that regulates intermediary carbohydrate metabolism.
MCQ #47 of 150
Biology
NUMS 2021
[NUMS 2021]
In striated skeletal muscle fibers, the release of \(\text{Ca}^{2+}\) ions that couples excitation to myofibrillar contraction is mediated by the:
A
Rough endoplasmic reticulum
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The sarcoplasmic reticulum (SR) is a specialized smooth endoplasmic reticulum in myocytes that stores and releases calcium to regulate the contractile machinery.
Formula / Rule / Reaction:$$\text{Action Potential} \rightarrow \text{T-Tubule DHP Receptor} \rightarrow \text{SR Ryanodine Receptor} \rightarrow \text{Ca}^{2+} \text{ Efflux}$$
Solution:- Depolarization of the sarcolemma travels down transverse tubules, activating voltage-sensing dihydropyridine receptors.
- This opens ryanodine release channels in the terminal cisternae of the sarcoplasmic reticulum, releasing \(\text{Ca}^{2+}\) into the sarcoplasm to bind troponin C.
Why other options are incorrect:- Option A: The rough ER is dedicated to protein translation and lacks high-capacity calcium storage cisternae.
- Option C: The Golgi apparatus modifies and routes secretory vesicles and does not control calcium transients.
- Option D: The lysosomal membrane sequesters acid hydrolases away from the sarcoplasm.
MCQ #48 of 150
Biology
NUMS 2021
[NUMS 2021]
Under normal physiological conditions, the pH of systemic arterial human blood is strictly buffered within the range of:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Human blood is maintained slightly alkaline via the bicarbonate buffer system, pulmonary ventilation, and renal acid-base excretion.
Formula / Rule / Reaction:$$\text{pH} = 6.1 + \log\left(\frac{[\text{HCO}_3^-]}{0.03 \times P_{\text{CO}_2}}\right) \approx 7.40$$
Solution:- Under resting conditions, extracellular fluid pH is kept between 7.35 and 7.45.
- Values below 7.35 represent acidemia, whereas values above 7.45 represent alkalemia, both of which impair cellular enzymatic and cardiovascular function.
Why other options are incorrect:- Option A: A pH of 7.15 to 7.25 reflects severe, life-threatening metabolic or respiratory acidosis.
- Option B: A range of 7.25 to 7.35 represents mild uncompensated systemic acidosis.
- Option D: A pH of 7.45 to 7.55 denotes systemic alkalemia.
MCQ #49 of 150
Biology
NUMS 2021
[NUMS 2021]
In an X-linked recessive condition, if an affected homozygous female marries a phenotypically normal male, what percentage of their daughters will be clinically affected?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:X-linked recessive traits require two mutant alleles for phenotypic expression in biological females (\(X^a X^a\)), whereas a single normal allele (\(X^A\)) masks the mutant phenotype.
Formula / Rule / Reaction:$$\text{Cross: } X^a X^a \times X^A Y \rightarrow \text{Daughters: } X^A X^a \text{ (100\% Carriers, 0\% Affected)}$$
Solution:- The affected mother can only transmit an \(X^a\) allele to her offspring.
- The normal father transmits his single \(X^A\) chromosome to all female progeny.
- Every daughter inherits the genotype \(X^A X^a\); because the dominant \(X^A\) allele compensates, exactly 0 percent of the daughters show clinical symptoms.
Why other options are incorrect:- Option A: A 50 percent frequency would require the father to carry the recessive allele or the mother to be heterozygous.
- Option B: 100 percent of daughters would only be affected if the father also had the disease condition (\(X^a Y\)).
- Option C: A 25 percent ratio is typical of an autosomal recessive intercross, not an X-linked pedigree with a normal father.
MCQ #50 of 150
Biology
NUMS 2021
[NUMS 2021]
How many total molecules of ATP are consumed in the stroma during the Calvin cycle to synthesize one complete molecule of net hexose (glucose)?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Fixation of carbon dioxide into hexose sugar demands chemical energy supplied by ATP and NADPH in the light-independent reactions.
Formula / Rule / Reaction:$$6\text{CO}_2 + 18\text{ATP} + 12\text{NADPH} + 12\text{H}^+ \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 18\text{ADP} + 18\text{P}_i + 12\text{NADP}^+ + 6\text{H}_2\text{O}$$
Solution:- Each single turn of the Calvin cycle fixing one \(\text{CO}_2\) consumes 3 ATP molecules (2 in the reduction phase and 1 in the regeneration phase of RuBP).
- Synthesizing one net mole of glucose requires the fixation of 6 moles of \(\text{CO}_2\), so \(6 \times 3 = 18\text{ ATP}\) are consumed.
Why other options are incorrect:- Option B: 12 ATP accounts only for the phosphorylation of 3-phosphoglycerate to 1,3-bisphosphoglycerate, omitting RuBP regeneration.
- Option C: 9 ATP is the total energetic investment needed to produce a single three-carbon triose phosphate (G3P) intermediate.
- Option D: 36 ATP reflects the gross theoretical ATP yield of complete aerobic cellular respiration, not photosynthetic carbon reduction.
MCQ #51 of 150
Biology
NUMS 2021
[NUMS 2021]
Which structural component projecting outward from the lipid bilayer envelope of viruses facilitates attachment to host cell-surface receptors?
D
Internal nucleoproteins
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Enveloped animal viruses display specialized attachment proteins anchored in their outer lipid membrane that determine tissue tropism.
Formula / Rule / Reaction:$$\text{Viral Glycoprotein Spike (Peplomer)} + \text{Host Cell Membrane Receptor} \rightarrow \text{Adsorption}$$
Solution:- Glycoprotein spikes (peplomers) project from the viral membrane envelope to bind target receptors on host cells.
- Classic examples include hemagglutinin in influenza and the spike glycoprotein in coronaviruses.
Why other options are incorrect:- Option A: Capsomeres are structural protein subunits that form the interior nucleocapsid surrounding the viral genome.
- Option C: Matrix proteins line the inner surface of the viral envelope to provide structural stability.
- Option D: Nucleoproteins are internal basic proteins that package and stabilize the viral nucleic acid genome.
MCQ #52 of 150
Biology
NUMS 2021
[NUMS 2021]
Which aldohexose serves as the universal and primary respiratory fuel across nearly all living organisms?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Monosaccharides are the simplest carbohydrate units, with D-glucose acting as the central substrate for cellular energy production.
Formula / Rule / Reaction:$$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} + 30\text{ to } 32\text{ ATP}$$
Solution:- Glucose enters the universal glycolytic pathway to generate pyruvate, ATP, and NADH.
- Because of its enzymatic compatibility, glucose is the primary circulating sugar in vertebrate blood.
Why other options are incorrect:- Option A: Mannose is an aldohexose epimer predominantly used in glycoprotein synthesis.
- Option B: Galactose is a structural epimer of glucose found in lactose that must be isomerized by liver enzymes.
- Option D: Fructose is a ketohexose found in fruits and seminal fluid that enters glycolysis via downstream phosphorylation intermediates.
MCQ #53 of 150
Biology
NUMS 2021
[NUMS 2021]
Chromosomes that exhibit identical morphology, gene loci, and banding patterns in both biological males and females are designated as:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:In human cytogenetics, the 46 chromosomes are divided into somatic non-sex-determining autosomes and sex-determining allosomes.
Formula / Rule / Reaction:$$\text{Human Karyotype: } 46\text{ Chromosomes} = 22\text{ Pairs of Autosomes (44)} + 1\text{ Pair of Sex Chromosomes (2)}$$
Solution:- Autosomes govern somatic traits and are identical in chromosomal morphology between males and females.
- In humans, chromosome pairs 1 through 22 are homologous autosomes, whereas pair 23 represents the sex chromosomes (XX or XY).
Why other options are incorrect:- Option A: Allosome is the technical synonym for a sex chromosome (X or Y).
- Option B: Sex chromosomes differ in morphology and gene complement between biological sexes (heteromorphic XY in males).
- Option C: Heterosome is another alternate term for allosomes or sex chromosomes.
MCQ #54 of 150
Biology
NUMS 2021
[NUMS 2021]
During the light-dependent stage of photosynthesis, the light-driven catalytic splitting of water molecules to generate protons, electrons, and molecular oxygen is termed:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Water-splitting at the oxygen-evolving complex of photosystem II provides the replacement electrons needed to sustain photosynthetic electron transport.
Formula / Rule / Reaction:$$2\text{H}_2\text{O} \xrightarrow{h\nu, \text{ Mn}_4\text{CaO}_5} 4\text{H}^+ + 4e^- + \text{O}_2$$
Solution:- When reaction center P680 is photo-oxidized to \(\text{P680}^+\), it acquires a strong oxidizing potential.
- The oxygen-evolving complex oxidizes water molecules, releasing protons into the thylakoid lumen and liberating gaseous \(\text{O}_2\).
Why other options are incorrect:- Option B: Photorespiration occurs when RuBisCO binds \(\text{O}_2\) instead of \(\text{CO}_2\), wasting fixed carbon and ATP.
- Option C: Photophosphorylation is the synthesis of ATP from ADP and inorganic phosphate powered by the thylakoid proton gradient.
- Option D: Photoreduction is the transfer of light-energized electrons to terminal acceptors such as \(\text{NADP}^+\).
MCQ #55 of 150
Biology
NUMS 2021
[NUMS 2021]
The light-harvesting reactions, photosynthetic electron transport, and photophosphorylation occur within which chloroplast compartment?
A
Inner envelope membrane
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Chloroplast ultrastructure segregates the light-harvesting apparatus and electron transport chains from the carbon assimilation enzymes.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- The thylakoid network houses photosystems I and II, the cytochrome \(b_6f\) complex, and ATP synthase enzymes.
- In contrast, the dark enzymatic reactions of the Calvin cycle take place in the soluble stroma.
Why other options are incorrect:- Option A: The inner chloroplast envelope regulates solute transport and lacks light-harvesting complexes.
- Option C: The stroma houses soluble Calvin-cycle enzymes, including RuBisCO, but does not perform the primary light reactions.
- Option D: The intermembrane space lies between the inner and outer envelopes and plays no role in photophosphorylation.
MCQ #56 of 150
Biology
NUMS 2021
[NUMS 2021]
Dense regular fibrous connective tissue bands that physically attach skeletal muscles to bones are called:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Dense regular connective tissue contains parallel bundles of type I collagen fibers designed to withstand unidirectional tensile pulling forces.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Tendons transmit mechanical tension from contracting skeletal muscle bellies to the periosteum of bone to produce joint movement.
- Ligaments, by contrast, connect bone directly to bone to stabilize joints.
Why other options are incorrect:- Option A: Ligaments connect bone to bone across articular joints to restrict excessive motion.
- Option B: Articular cartilage provides a smooth, lubricated gliding surface over bone ends.
- Option D: Synovial membranes line the inner joint capsule and secrete lubricating synovial fluid.
MCQ #57 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The outermost living, metabolically active boundary enclosing animal cells is the:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Animal cells lack rigid exterior cell walls, so their selective, living outer barrier is the amphipathic lipid bilayer.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- The plasma membrane (plasmalemma) is an active, living fluid-mosaic structure that regulates selective transport, signal transduction, and ion gradients.
- Unlike plant or fungal cell walls, the plasma membrane is metabolically dynamic.
Why other options are incorrect:- Option A: The glycocalyx is a peripheral carbohydrate coating on the outer leaflet of the membrane, not the boundary itself.
- Option B: Animal cells do not possess a cell wall; where present in plants and fungi, cell walls are non-living extracellular matrices.
- Option C: The nuclear envelope encloses the nucleoplasm within the interior of the cell.
MCQ #58 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Sensory receptors specialized for the detection of noxious thermal, mechanical, or chemical stimuli that produce the sensation of pain are:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Nociceptors are free nerve endings that respond selectively to potentially damaging stimuli, transmitting nociceptive signals to the central nervous system.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Nociceptors have high activation thresholds triggered by mechanical deformation, extreme temperatures (above \(45^\circ\text{C}\)), or chemical mediators (such as bradykinin, prostaglandins, and \(\text{H}^+\)).
- Their activation generates action potentials carried by A-delta and C fibers to register the sensation of pain.
Why other options are incorrect:- Option B: Proprioceptors (such as muscle spindles and Golgi tendon organs) monitor spatial body position and joint angles.
- Option C: Standard thermoreceptors detect non-noxious, innocuous temperature variations.
- Option D: Baroreceptors monitor stretch and hydrostatic blood pressure changes within arterial walls.
MCQ #59 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The resting electrical membrane potential across the axolemma of a typical mammalian neuron at rest is approximately:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The resting membrane potential reflects the steady-state electrical charge separation maintained across a non-excited neuronal membrane.
Formula / Rule / Reaction:$$V_m = \frac{RT}{F}\ln\left(\frac{P_{\text{K}}[\text{K}^+]_o + P_{\text{Na}}[\text{Na}^+]_o + P_{\text{Cl}}[\text{Cl}^-]_i}{P_{\text{K}}[\text{K}^+]_i + P_{\text{Na}}[\text{Na}^+]_i + P_{\text{Cl}}[\text{Cl}^-]_o}\right) \approx -70\text{ mV}$$
Solution:- At rest, high membrane permeability to \(\text{K}^+\) through leak channels combined with the electrogenic \(\text{Na}^+/\text{K}^+\)-ATPase pump (3 \(\text{Na}^+\) pumped out for every 2 \(\text{K}^+\) pumped in) polarizes the membrane.
- In typical mammalian neurons, the inside of the cell remains negative relative to the outside at approximately \(-70\text{ mV}\).
Why other options are incorrect:- Option A: \(-90\text{ mV}\) is the equilibrium potential for potassium (\(E_{\text{K}}\)) and the resting potential of ventricular cardiac myocytes.
- Option C: \(-50\text{ mV}\) represents the typical threshold potential needed to fire an all-or-none action potential.
- Option D: \(+30\text{ mV}\) represents the peak depolarizing overshoot reached during an action potential.
MCQ #60 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which chemical neurotransmitter operates at all somatic neuromuscular junctions and preganglionic autonomic synapses?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Acetylcholine (ACh) is the primary small-molecule neurotransmitter of the somatic motor division and the peripheral parasympathetic nervous system.
Formula / Rule / Reaction:$$\text{Choline} + \text{Acetyl-CoA} \xrightarrow{\text{Choline Acetyltransferase}} \text{Acetylcholine} + \text{CoA}$$
Solution:- Lower motor neurons release acetylcholine into the synaptic cleft of the neuromuscular junction, where it binds nicotinic receptors on the muscle endplate.
- ACh also serves as the neurotransmitter at all autonomic preganglionic synapses and postganglionic parasympathetic terminals.
Why other options are incorrect:- Option A: Dopamine is a central catecholamine involved in reward pathways and motor coordination in the basal ganglia.
- Option B: Serotonin (5-HT) regulates mood, sleep cycles, and enteric intestinal motility.
- Option D: Norepinephrine is the principal neurotransmitter released by sympathetic postganglionic axons.
MCQ #61 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which of the following real gases behaves most ideally under comparable ambient conditions due to its monoatomic nature and negligible intermolecular forces?
A
Carbon monoxide (\(\text{CO}\))
B
Nitrogen (\(\text{N}_2\))
C
Hydrogen (\(\text{H}_2\))
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Gases approach ideal behavior when intermolecular attractions are minimal and the volume occupied by the gas particles relative to the container volume is negligible.
Formula / Rule / Reaction:$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT \quad (\text{Lowest } a \text{ and } b \text{ yields maximum ideality})$$
Solution:- Helium is a noble gas with a stable duet configuration, a very small atomic radius, and low polarizability.
- Its van der Waals attraction constant (\(a\)) and excluded volume parameter (\(b\)) are exceptionally small, making its behavior closer to the ideal gas law than diatomic or polar gases.
Why other options are incorrect:- Option A: Carbon monoxide is a polar diatomic molecule with dipole-dipole attractions, causing significant non-ideal deviation.
- Option B: Nitrogen is a larger diatomic molecule with higher polarizability and greater London dispersion attractions than helium.
- Option C: Although hydrogen has low mass, helium's closed electronic shell and compact volume minimize intermolecular interactions even further.
MCQ #62 of 150
Chemistry
NUMS 2021
[NUMS 2021]
When two separate ice blocks are firmly pressed against each other and released, they coalesce into a single block primarily because of:
A
Reformation of interfacial hydrogen bonds upon release of pressure
B
London dispersion forces between surface protons
C
Permanent covalent cross-linking across the crystal surface
D
Static electrostatic charge accumulation
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:According to Le Chatelier's principle, applying pressure to ice lowers its melting point, causing a thin film of water to form, which refreezes once pressure is released (regelation).
Formula / Rule / Reaction:$$\text{Ice (Solid)} \xrightarrow{\text{High Pressure}} \text{Water (Liquid)} \xrightarrow{\text{Pressure Released}} \text{Ice (Fused via H-Bonds)}$$
Solution:- Ice has an open tetrahedral structure with a lower density than liquid water; applying mechanical pressure forces the crystal to melt locally at the contacting interface.
- When the pressure is released, the local melting point returns to \(0^\circ\text{C}\), causing the thin interfacial water film to refreeze and form continuous hydrogen-bonded crystal planes.
Why other options are incorrect:- Option B: London dispersion forces are much weaker than directional hydrogen bonds and do not drive ice lattice fusion.
- Option C: Freezing does not form new intramolecular covalent \(\text{O}-\text{H}\) bonds between separate water molecules.
- Option D: Regelation is driven by thermodynamic phase changes, not electrostatic charge accumulation.
MCQ #63 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The rate constant \(k\) possesses the exact same dimensional units as the reaction rate (\(\text{mol dm}^{-3}\text{ s}^{-1}\)) in a:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The dimensional units of the chemical rate constant \(k\) depend directly on the overall order of the reaction \(n\).
Formula / Rule / Reaction:$$\text{Units of } k = (\text{mol dm}^{-3})^{1-n} \cdot \text{s}^{-1}$$
Solution:- For a zero-order reaction, \(n = 0\):
- $$\text{Units of } k = (\text{mol dm}^{-3})^{1-0} \cdot \text{s}^{-1} = \text{mol dm}^{-3}\text{ s}^{-1}$$
- Because reaction rate is defined as change in concentration per unit time (\(\text{mol dm}^{-3}\text{ s}^{-1}\)), the units of \(k\) match the units of rate only when \(n = 0\).
Why other options are incorrect:- Option A: For a first-order reaction (\(n=1\)), the units of \(k\) are \(\text{s}^{-1}\).
- Option C: For a second-order reaction (\(n=2\)), the units of \(k\) are \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).
- Option D: For a third-order reaction (\(n=3\)), the units of \(k\) are \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
MCQ #64 of 150
Chemistry
NUMS 2021
[NUMS 2021]
In both electrolytic and galvanic electrochemical cells, reduction (gain of electrons) invariably takes place at the:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Electrochemical convention defines electrodes based on the chemical half-reaction occurring at their surfaces.
Formula / Rule / Reaction:$$\text{Anode: Oxidation (Loss of } e^-) \quad | \quad \text{Cathode: Reduction (Gain of } e^-)$$
Solution:- By universal convention, the cathode is the electrode where chemical reduction occurs, drawing electrons inward from the circuit.
- The anode is the site of chemical oxidation, releasing electrons outward into the external circuit.
Why other options are incorrect:- Option A: Oxidation occurs at the anode across all electrochemical cells.
- Option B: The salt bridge permits counter-ion migration to preserve electrical neutrality, not redox reactions.
- Option D: The external wire serves as a metallic pathway for physical electron flow.
MCQ #65 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which of the following periodic properties increases progressively down Group 1 (alkali metals) from lithium to cesium?
A
First ionization energy
C
Standard electron affinity
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:For alkali metals, chemical reactivity is governed by how easily the single outer valence electron is lost during oxidation.
Formula / Rule / Reaction:$$\text{M}(s) \rightarrow \text{M}^+(aq) + e^- \quad (\text{Ease of oxidation increases down the group})$$
Solution:- Descending Group 1, additional principal electron shells increase the atomic radius and electron shielding.
- This weakens electrostatic attraction between the nucleus and the valence electron, lowering the ionization energy and making the metal more reactive.
Why other options are incorrect:- Option A: Ionization energy decreases down the group due to increased atomic radius and shielding.
- Option B: Electronegativity decreases down Group 1 as the nucleus is shielded from bonding electrons.
- Option C: Electron affinity becomes less exothermic descending the group due to weaker nuclear pull on incoming electrons.
MCQ #66 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The general decrease in electron affinity when descending a main group in the periodic table is primarily driven by the:
A
Increase in atomic radius and electron shielding
B
Increase in effective nuclear charge
C
Decrease in the principal quantum number
D
Contraction of outer valence orbitals
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Electron affinity measures the enthalpy change when an electron is added to an isolated neutral gaseous atom.
Formula / Rule / Reaction:$$\text{X}(g) + e^- \rightarrow \text{X}^-(g) + \text{Energy} \quad \left(\text{Affinity} \propto \frac{Z_{\text{eff}}}{r}\right)$$
Solution:- Moving down a group, the valence shell is located farther from the nucleus, and inner electron shells provide additional electrostatic shielding.
- The net electrostatic attraction felt by an incoming electron weakens, lowering the energy released upon electron capture.
Why other options are incorrect:- Option B: Increases in effective nuclear charge tend to raise electron affinity, which occurs across a period from left to right.
- Option C: The principal quantum number increases down a group, not decreases.
- Option D: Valence orbitals expand as the principal quantum number increases, rather than contracting.
MCQ #67 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The principle stating that no two electrons within the same atom can possess an identical set of all four quantum numbers is:
A
Hund's rule of maximum multiplicity
B
Pauli's exclusion principle
C
The Aufbau building-up principle
D
Heisenberg's uncertainty principle
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Electrons are fermions and obey Fermi-Dirac statistics, which prevent identical quantum states from existing simultaneously.
Formula / Rule / Reaction:$$\text{State: } (n, l, m_l, m_s) \quad \text{where } m_s = +\frac{1}{2} \text{ or } -\frac{1}{2}$$
Solution:- Wolfgang Pauli formulated the exclusion principle in 1925.
- If two electrons share the same spatial orbital (same \(n\), \(l\), and \(m_l\)), they must have opposite spin quantum numbers (\(m_s = +1/2\) and \(m_s = -1/2\)).
Why other options are incorrect:- Option A: Hund's rule states that degenerate orbitals are occupied singly with parallel spins before pairing occurs.
- Option C: The Aufbau principle states that electrons fill lower-energy atomic orbitals before occupying higher ones.
- Option D: Heisenberg's uncertainty principle sets limits on knowing both the position and momentum of a particle simultaneously.
MCQ #68 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which of the following metal chlorides exhibits the lowest percentage of ionic character (highest covalent character) due to polarization?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Fajans' rules dictate that small, highly charged cations have high charge densities that polarize adjacent electron clouds, introducing significant covalent character into bonds.
Formula / Rule / Reaction:$$\text{Ionic Potential } (\phi) = \frac{\text{Cation Charge}}{\text{Cation Radius}} \quad (\text{Higher } \phi \rightarrow \text{Greater Covalency})$$
Solution:- Comparing cations: \(\text{K}^+\) has \(+1\), \(\text{NaCl}\) has \(+1\), \(\text{Mg}^{2+}\) has \(+2\), and \(\text{Al}^{3+}\) has \(+3\).
- The small \(\text{Al}^{3+}\) ion has the highest charge density, which heavily polarizes the chloride electron clouds and gives anhydrous \(\text{AlCl}_3\) predominant covalent character.
Why other options are incorrect:- Option A: \(\text{KCl}\) contains the large \(\text{K}^+\) cation, giving it very low polarizing power and high ionic character.
- Option B: \(\text{NaCl}\) features a large electronegativity difference (roughly 2.1) and forms a classic ionic crystal.
- Option D: \(\text{MgCl}_2\) is intermediate, but less polarized and more ionic than \(\text{AlCl}_3\).
MCQ #69 of 150
Chemistry
NUMS 2021
[NUMS 2021]
In crystalline solid classification, diamond is characterized as a:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Covalent network solids consist of atoms held together throughout the crystal in a continuous three-dimensional lattice of covalent bonds.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- In diamond, every carbon atom is \(\text{sp}^3\)-hybridized and tetrahedrally bonded to four neighboring carbon atoms by strong sigma bonds (\(1.54\text{ \AA}\)).
- Because there are no discrete molecules, the entire crystal acts as a giant macromolecule, giving diamond high hardness and a very high melting point.
Why other options are incorrect:- Option A: Molecular solids (such as dry ice or iodine) consist of discrete molecules bound by weak intermolecular forces.
- Option B: Metallic solids are held together by delocalized electron seas surrounding metal cations.
- Option C: Ionic solids (such as \(\text{NaCl}\)) are composed of alternating cations and anions held by electrostatic attraction.
MCQ #70 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The ideal gas equation that combines Boyle's, Charles's, and Avogadro's laws for \(n\) moles of an ideal gas is formulated as:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The ideal gas equation represents the equation of state of a hypothetical ideal gas where particle volumes and intermolecular forces are negligible.
Formula / Rule / Reaction:$$V \propto \frac{1}{P} \text{ (Boyle)}, \quad V \propto T \text{ (Charles)}, \quad V \propto n \text{ (Avogadro)} \implies PV = nRT$$
Solution:- Combining these proportionalities yields \(V \propto \frac{nT}{P}\).
- Introducing the universal molar gas constant \(R\) gives the relationship \(PV = nRT\).
Why other options are incorrect:- Option B: \(P = nRTV\) is dimensionally invalid because volume must be in the denominator when isolating pressure.
- Option C: \(PV = \frac{RT}{n}\) places moles in the denominator, contradicting Avogadro's law.
- Option D: \(V = nRTP\) multiplies by pressure rather than dividing by it.
MCQ #71 of 150
Chemistry
NUMS 2021
[NUMS 2021]
How many moles of carbon dioxide (\(\text{CO}_2\)) contain exactly \(16\text{ g}\) of oxygen atoms?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Stoichiometric molar relationships relate the total mass of an element present in a compound to the moles of the chemical formula unit.
Formula / Rule / Reaction:$$n_{\text{O}} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{16\text{ g}}{16\text{ g/mol}} = 1.0\text{ mol of O atoms}$$
Solution:- One mole of \(\text{CO}_2\) contains \(2\) moles of oxygen atoms (\(32\text{ g}\) of oxygen).
- To contain \(16\text{ g}\) of oxygen atoms (\(1.0\text{ mol}\) of \(\text{O}\)), we set up the proportion:
- $$\text{Moles of } \text{CO}_2 = \frac{1.0\text{ mol of O}}{2\text{ mol of O per } \text{CO}_2} = 0.5\text{ mol}$$
Why other options are incorrect:- Option A: \(1.0\text{ mol}\) of \(\text{CO}_2\) contains \(32\text{ g}\) of oxygen atoms.
- Option C: \(2.0\text{ mol}\) of \(\text{CO}_2\) contains \(64\text{ g}\) of oxygen atoms.
- Option D: \(0.25\text{ mol}\) of \(\text{CO}_2\) contains \(8\text{ g}\) of oxygen atoms.
MCQ #72 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The branch of chemistry that deals with the quantitative relationships between reactants and products in a balanced chemical reaction is:
A
Chemical thermodynamics
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Stoichiometry applies the conservation of mass and definite proportions to calculate quantitative relationships in chemical processes.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Stoichiometry uses the molar coefficients of a balanced chemical equation to calculate mole, mass, and volume proportions for reactants and products.
Why other options are incorrect:- Option A: Chemical thermodynamics studies energy conversions, heat exchange, and the spontaneity of reactions.
- Option B: Reaction kinetics investigates reaction rates, mechanisms, and activation energies.
- Option D: Thermochemistry focuses specifically on enthalpy changes and heat transfer during reactions.
MCQ #73 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which gas constitutes the greatest proportion by volume of dry ambient air in Earth's atmosphere?
A
Carbon dioxide (\(\text{CO}_2\))
B
Oxygen (\(\text{O}_2\))
D
Nitrogen (\(\text{N}_2\))
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Earth's atmosphere is a homogeneous mixture of permanent gases maintained by biological and geochemical cycles.
Formula / Rule / Reaction:$$\text{Volumetric Composition: } \text{N}_2 \approx 78.08\%, \quad \text{O}_2 \approx 20.95\%, \quad \text{Ar} \approx 0.93\%, \quad \text{CO}_2 \approx 0.04\%$$
Solution:- Molecular nitrogen (\(\text{N}_2\)) accounts for approximately 78.08 percent of clean, dry air by volume.
- Oxygen is second at roughly 21 percent, followed by argon at 0.93 percent.
Why other options are incorrect:- Option A: Carbon dioxide is a trace atmospheric component at approximately 0.04 percent.
- Option B: Oxygen is the second most abundant gas at approximately 20.95 percent.
- Option C: Argon is the third most abundant gas at approximately 0.93 percent.
MCQ #74 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The rest mass of a proton is approximately how many times heavier than the rest mass of an electron?
C
\(\frac{1}{1836}\) times
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Subatomic nucleons (protons and neutrons) carry the vast majority of atomic mass relative to extranuclear electrons.
Formula / Rule / Reaction:$$\text{Mass ratio} = \frac{m_p}{m_e} = \frac{1.6726 \times 10^{-27}\text{ kg}}{9.1094 \times 10^{-31}\text{ kg}} \approx 1836.15$$
Solution:- A proton has a rest mass of \(1.6726 \times 10^{-27}\text{ kg}\).
- An electron has a rest mass of \(9.1094 \times 10^{-31}\text{ kg}\).
- Dividing the two values shows the proton is approximately 1836 times heavier than the electron.
Why other options are incorrect:- Option B: 184 times underestimates the mass ratio by a factor of ten.
- Option C: \(1/1836\) represents the reciprocal ratio (the mass of an electron relative to a proton).
- Option D: 3672 times is approximately double the actual mass ratio.
MCQ #75 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which quantum number designates the three-dimensional geometric shape of an atomic orbital?
A
Principal quantum number (\(n\))
B
Azimuthal quantum number (\(l\))
C
Magnetic quantum number (\(m_l\))
D
Spin quantum number (\(m_s\))
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The solution to the angular component of the Schrödinger wave equation yields quantum numbers that define orbital geometry and orientation.
Formula / Rule / Reaction:$$l = 0 \rightarrow s \text{ (spherical)}, \quad l = 1 \rightarrow p \text{ (dumbbell)}, \quad l = 2 \rightarrow d \text{ (cloverleaf)}, \quad l = 3 \rightarrow f$$
Solution:- The azimuthal (orbital angular momentum) quantum number \(l\) determines the subshell and geometric shape of the electron cloud.
- Allowed values of \(l\) range from \(0\) to \(n - 1\).
Why other options are incorrect:- Option A: The principal quantum number \(n\) determines the main energy level and radial size of the orbital.
- Option C: The magnetic quantum number \(m_l\) determines the spatial orientation of the orbital in three dimensions.
- Option D: The spin quantum number \(m_s\) indicates the intrinsic angular momentum (spin) of the electron.
MCQ #76 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The maximum number of electrons that can be accommodated within a completely filled \(p\) subshell is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The total electron capacity of an atomic subshell is dictated by its azimuthal quantum number \(l\) and the Pauli exclusion principle.
Formula / Rule / Reaction:$$\text{Capacity} = 2(2l + 1)$$
Solution:- For a \(p\) subshell, the azimuthal quantum number is \(l = 1\).
- The number of degenerate spatial orbitals is \(2l + 1 = 2(1) + 1 = 3\) (corresponding to \(p_x\), \(p_y\), and \(p_z\) where \(m_l = -1, 0, +1\)).
- Because each spatial orbital holds a maximum of two electrons with opposing spins, the maximum capacity is \(3 \times 2 = 6\text{ electrons}\).
Why other options are incorrect:- Option A: 2 electrons is the maximum capacity of an \(s\) subshell (\(l = 0\)).
- Option B: 10 electrons is the maximum capacity of a \(d\) subshell (\(l = 2\)).
- Option D: 14 electrons is the maximum capacity of an \(f\) subshell (\(l = 3\)).
MCQ #77 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which of the following hydrogen halides exhibits the highest bond dissociation energy?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Covalent bond strength in diatomic hydrides correlates inversely with internuclear bond length and directly with electronegativity difference.
Formula / Rule / Reaction:$$\text{Bond Energy: } \text{H}-\text{F} (567\text{ kJ/mol}) > \text{H}-\text{Cl} (431\text{ kJ/mol}) > \text{H}-\text{Br} (366\text{ kJ/mol}) > \text{H}-\text{I} (299\text{ kJ/mol})$$
Solution:- Fluorine has the smallest atomic radius and highest electronegativity of all halogens.
- This produces the shortest internuclear distance (\(0.92\text{ \AA}\)), maximum orbital overlap with hydrogen, and strong electrostatic attraction, giving \(\text{H}-\text{F}\) the highest dissociation energy.
Why other options are incorrect:- Option A: \(\text{HI}\) has the longest bond length and lowest bond dissociation energy, making it the strongest acid in the series.
- Option B: \(\text{HBr}\) has a larger halogen radius and weaker bond overlap than \(\text{HCl}\) and \(\text{HF}\).
- Option C: \(\text{HCl}\) has an intermediate bond energy (\(431\text{ kJ/mol}\)), substantially lower than \(\text{HF}\).
MCQ #78 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The oxidation state of any chemical element in its pure, uncombined free elemental form is:
D
Variable depending on temperature
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Oxidation state represents the hypothetical charge an atom would possess if all bonds to different elements were completely ionic.
Formula / Rule / Reaction:$$\text{Oxidation state of pure element } (\text{e.g., } \text{O}_2, \text{P}_4, \text{S}_8, \text{Na}, \text{Fe}) = 0$$
Solution:- In a free element, homonuclear bonds join identical atoms having zero difference in electronegativity.
- Because bonding electron pairs are shared equally without net transfer, the oxidation number is defined as zero.
Why other options are incorrect:- Option B: +1 is the oxidation state of alkali metals when combined into heteronuclear chemical compounds.
- Option C: -1 is the characteristic oxidation state of halogens in ionic halides.
- Option D: Oxidation state is fixed by electronic bookkeeping rules and does not vary with temperature.
MCQ #79 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which carbon allotrope consists of a single planar layer of \(\text{sp}^2\)-hybridized atoms arranged in a honeycomb lattice, exhibiting immense tensile strength and high electrical conductivity?
D
Fullerene \(\text{C}_{60}\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Two-dimensional nanomaterials can demonstrate physical properties far superior to their bulk crystalline counterparts.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Graphene is an individual, single-atom-thick sheet of \(\text{sp}^2\)-hybridized carbon atoms tightly packed in a two-dimensional hexagonal lattice.
- It is extremely lightweight yet exhibits a tensile strength over 100 times greater than steel, alongside high carrier mobility.
Why other options are incorrect:- Option A: Graphite consists of multiple stacked graphene sheets held together by weak van der Waals forces, making it soft and flaky.
- Option C: Diamond has a three-dimensional tetrahedral network of \(\text{sp}^3\) carbons that behaves as an electrical insulator.
- Option D: Buckminsterfullerene (\(\text{C}_{60}\)) is a zero-dimensional cage-like hollow sphere rather than an extended planar sheet.
MCQ #80 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Liquid water achieves its maximum physical density at a temperature of approximately:
B
\(100.0^\circ\text{C}\)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The density anomaly of water results from a balance between thermal expansion and the collapse of the open, hydrogen-bonded ice lattice.
Formula / Rule / Reaction:$$\rho_{\max} \approx 1.0000\text{ g/cm}^3 \quad \text{at } T = 3.98^\circ\text{C} \text{ (approximately } 4^\circ\text{C})$$
Solution:- When ice melts at \(0^\circ\text{C}\), the open cage-like framework collapses partially, packing molecules closer together and increasing liquid density.
- Above \(3.98^\circ\text{C}\), normal thermal expansion dominates and pushes water molecules farther apart, decreasing density.
Why other options are incorrect:- Option A: At \(0.00^\circ\text{C}\), water retains structured hydrogen-bonded clusters, giving it a lower density (\(0.9998\text{ g/cm}^3\)) than at \(3.98^\circ\text{C}\).
- Option B: At \(100.0^\circ\text{C}\), vigorous thermal kinetic agitation expands the liquid, lowering density to roughly \(0.958\text{ g/cm}^3\).
- Option D: \(2.98^\circ\text{C}\) is below the inflection point where density reaches its peak.
MCQ #81 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Homocyclic (carbocyclic) hydrocarbons are fundamentally subclassified into which two structural divisions?
A
Open-chain and branched-chain
B
Heterocyclic and aromatic
C
Acyclic and heterocyclic
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Homocyclic compounds contain closed rings constructed entirely from carbon atoms without heteroatomic ring substitutions.
Formula / Rule / Reaction:$$\text{Homocyclic Compounds} \rightarrow \begin{cases} \text{Alicyclic (Aliphatic cyclic: e.g., cyclohexane)} \\ \text{Aromatic (Conjugated Huckel systems: e.g., benzene)} \end{cases}$$
Solution:- Alicyclic hydrocarbons resemble open-chain aliphatic compounds in their chemical properties (e.g., cyclobutane, cyclohexane).
- Aromatic hydrocarbons possess delocalized \(\pi\)-electron rings satisfying Huckel's \(4n+2\) rule (e.g., benzene).
Why other options are incorrect:- Option A: Open-chain and branched-chain describe acyclic (aliphatic) non-cyclic compounds.
- Option B: Heterocyclic compounds contain non-carbon atoms (such as O, N, or S) in the ring, so they are not homocyclic.
- Option C: Acyclic refers to non-ring open-chain structures, while heterocyclic rings contain heteroatoms.
MCQ #82 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which of the following hydrocarbon species contains carbon atoms that utilize \(\text{sp}^2\) hybridization?
A
None of these options are correct
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Carbon hybridization is determined by the number of steric domains (sigma bonds plus localized lone pairs) surrounding each carbon atom.
Formula / Rule / Reaction:$$\text{Methane (CH}_4\text{): } \text{sp}^3, \quad \text{Ethane (C}_2\text{H}_6\text{): } \text{sp}^3, \quad \text{Ethyne (C}_2\text{H}_2\text{): } \text{sp}$$
Solution:- In methane (\(\text{CH}_4\)), carbon forms 4 single sigma bonds (\(\text{sp}^3\)).
- In ethane (\(\text{C}_2\text{H}_6\)), both carbons form 4 sigma bonds (\(\text{sp}^3\)).
- In ethyne (alkyne, \(\text{C}_2\text{H}_2\)), each carbon forms 2 sigma bonds and 2 pi bonds (\(\text{sp}\)).
- Because none of the listed hydrocarbons possess \(\text{sp}^2\) hybridization, 'None of these options are correct' is the appropriate choice.
Why other options are incorrect:- Option B: Methane features tetrahedral \(\text{sp}^3\) hybridization with \(109.5^\circ\) bond angles.
- Option C: Ethane features tetrahedral \(\text{sp}^3\) hybridization at both carbon centers.
- Option D: Alkynes like ethyne feature linear \(\text{sp}\) hybridization with \(180^\circ\) bond angles.
MCQ #83 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The industrial solvent and chemical precursor historically known as 'wood spirit' has the molecular formula:
A
\(\text{C}_2\text{H}_5\text{OH}\)
B
\(\text{CH}_3\text{OH}\)
D
\(\text{CH}_3\text{COOH}\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Wood spirit is the historical commercial designation for methanol, derived from its traditional production method.
Formula / Rule / Reaction:$$\text{Destructive distillation of wood} \rightarrow \text{Pyroligneous acid} \rightarrow \text{Methanol (CH}_3\text{OH)}$$
Solution:- Methanol (\(\text{CH}_3\text{OH}\)) was historically prepared by the destructive distillation of wood chips in the absence of air.
- The resulting condensate contained methanol along with acetic acid and acetone.
Why other options are incorrect:- Option A: \(\text{C}_2\text{H}_5\text{OH}\) is ethanol, known commercially as grain alcohol.
- Option C: \(\text{HCHO}\) is formaldehyde (methanal), an aldehyde gas whose aqueous solution is called formalin.
- Option D: \(\text{CH}_3\text{COOH}\) is acetic acid (ethanoic acid), the key acidic component of vinegar.
MCQ #84 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Geometrical (cis-trans) isomerism is most commonly exhibited by which class of hydrocarbons due to restricted bond rotation?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Geometrical isomerism requires restricted rotation about a chemical bond together with two non-identical substituent groups attached to each unsaturated carbon.
Formula / Rule / Reaction:$$\text{Restricted rotation around } \text{C}=\text{C} \text{ double bond} \rightarrow \text{Cis-Trans stereoisomerism}$$
Solution:- The carbon-carbon double bond (\(\text{C}=\text{C}\)) consists of a sigma bond and a pi bond formed by lateral \(p\)-orbital overlap.
- Rotation around this double bond is restricted because it would break the pi bond, allowing distinct cis and trans configurations to exist when each carbon bears two different substituents.
Why other options are incorrect:- Option A: Alkanes have carbon-carbon single bonds (\(\sigma\)) that rotate freely at room temperature, yielding conformational rather than geometric isomers.
- Option B: Alkynes have linear \(\text{sp}\) geometry (\(180^\circ\)), with only one substituent attached per acetylenic carbon.
- Option D: Benzene rings possess symmetric planar delocalization that does not exhibit cis-trans isomerism on the ring itself.
MCQ #85 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The general molecular formula \(\text{C}_n\text{H}_{2n}\text{O}_2\) represents which homologous series of organic compounds?
A
Monohydric aliphatic alcohols
D
Monocarboxylic acids and esters
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A single degree of unsaturation (one double-bond equivalent) combined with two oxygen atoms corresponds to the functional group isomers carboxylic acids and esters.
Formula / Rule / Reaction:$$\text{General Formula: } \text{C}_n\text{H}_{2n}\text{O}_2 \quad (\text{e.g., for } n=2: \text{ CH}_3\text{COOH and HCOOCH}_3)$$
Solution:- A saturated carboxylic acid consists of an alkyl group attached to a carbonyl and a hydroxyl group (\(\text{R}-\text{COOH}\)).
- The carbonyl group introduces one double bond, reducing the hydrogen count by two relative to saturated alkanes, yielding \(\text{C}_n\text{H}_{2n}\text{O}_2\).
Why other options are incorrect:- Option A: Saturated monohydric alcohols have the general formula \(\text{C}_n\text{H}_{2n+2}\text{O}\).
- Option B: Aliphatic ketones have one oxygen atom and one double bond, giving the general formula \(\text{C}_n\text{H}_{2n}\text{O}\).
- Option C: Saturated aliphatic ethers are constitutional isomers of alcohols with the general formula \(\text{C}_n\text{H}_{2n+2}\text{O}\).
MCQ #86 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Which diagnostic test reagent is used to differentiate between aliphatic alcohols and phenols based on nucleophilic substitution?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Aliphatic alcohols undergo nucleophilic substitution of their \(-\text{OH}\) group in acidic conditions, whereas phenols resist \(\text{C}-\text{O}\) bond cleavage due to resonance stabilization.
Formula / Rule / Reaction:$$\text{R}-\text{OH} + \text{HCl} \xrightarrow{\text{Anhydrous ZnCl}_2} \text{R}-\text{Cl} \downarrow \text{ (Turbidity)} + \text{H}_2\text{O}$$
Solution:- Lucas reagent (a solution of anhydrous \(\text{ZnCl}_2\) in concentrated \(\text{HCl}\)) reacts with aliphatic alcohols to produce insoluble alkyl chlorides, visible as cloudy turbidity.
- Phenol does not react with Lucas reagent because the oxygen lone pair is delocalized into the aromatic ring, giving the \(\text{C}-\text{O}\) bond partial double-bond character that resists substitution.
Why other options are incorrect:- Option B: Tollens reagent is an ammoniacal silver nitrate solution used to identify aldehydes by forming a silver mirror.
- Option C: Benedict solution tests for reducing sugars, not alcohols versus phenols.
- Option D: Fehling reagent detects aliphatic aldehydes by precipitating red \(\text{Cu}_2\text{O}\).
MCQ #87 of 150
Chemistry
NUMS 2021
[NUMS 2021]
A 50% concentrated aqueous solution of sodium hydroxide (\(\text{NaOH}\)) is specifically employed in the laboratory to carry out the:
A
Aldol condensation reaction
D
Wolff-Kishner reduction
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Aldehydes lacking alpha-hydrogen atoms undergo base-catalyzed disproportionation (redox disproportionation) when treated with concentrated alkali.
Formula / Rule / Reaction:$$2\text{HCHO} \xrightarrow{50\% \text{ NaOH}} \text{HCOONa} + \text{CH}_3\text{OH}$$
Solution:- The Cannizzaro reaction requires aldehydes without alpha-hydrogens (such as formaldehyde or benzaldehyde) and strong base (typically 50% \(\text{NaOH}\)).
- One molecule of aldehyde is oxidized to a carboxylic acid salt while the second is reduced to a primary alcohol.
Why other options are incorrect:- Option A: Aldol condensation requires aldehydes or ketones with alpha-hydrogens and uses dilute alkali (typically 10% \(\text{NaOH}\)).
- Option C: Clemmensen reduction converts carbonyls to alkanes using zinc amalgam and concentrated \(\text{HCl}\).
- Option D: Wolff-Kishner reduction reduces carbonyls using hydrazine and \(\text{KOH}\) in hot ethylene glycol.
MCQ #88 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Lower aliphatic monocarboxylic acids (such as acetic acid) are completely miscible with water primarily because of:
A
Permanent dipole-induced dipole interactions
B
London dispersion forces
C
Intermolecular hydrogen bonding with solvent molecules
D
Complete electrolytic dissociation into gaseous ions
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Carboxylic acid molecules feature both a polar carbonyl group (\(\text{C}=\text{O}\)) and a polar hydroxyl group (\(-\text{OH}\)), enabling hydrogen bonding with protic solvents.
Formula / Rule / Reaction:$$\text{R}-\text{C}(=\text{O})-\text{O}-\text{H} \cdots :\text{OH}_2 \quad \text{and} \quad \text{R}-\text{C}(=\text{O}\cdots \text{H}-\text{OH})-\text{OH}$$
Solution:- Low molecular mass carboxylic acids form extensive hydrogen bonds with water molecules through both the carbonyl oxygen and hydroxyl hydrogen.
- This favorable solvation energy overcomes solvent cavitation resistance, making lower acids (formic, acetic, propionic) fully miscible in water.
Why other options are incorrect:- Option A: Dipole-induced dipole forces occur between polar and non-polar molecules, which is not the case between acids and water.
- Option B: London dispersion forces are weak non-directional attractions that are secondary to strong hydrogen bonding in aqueous solutions.
- Option D: Carboxylic acids are weak electrolytes that dissociate only slightly in water (typically less than 2 percent).
MCQ #89 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Fischer esterification occurs via an acid-catalyzed condensation reaction between a:
A
Ketone and an aliphatic ether
B
Aldehyde and an aliphatic ketone
C
Carboxylic acid and an aliphatic aldehyde
D
Carboxylic acid and an alcohol
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Direct esterification involves nucleophilic acyl substitution of a carboxylic acid by an alcohol in the presence of an inorganic acid catalyst.
Formula / Rule / Reaction:$$\text{R}-\text{COOH} + \text{R}'-\text{OH} \rightleftharpoons^+} \text{R}-\text{COOR}' + \text{H}_2\text{O}$$
Solution:- Protonation of the carboxylic acid carbonyl oxygen activates the carbonyl carbon toward nucleophilic attack by the alcohol's hydroxyl oxygen.
- Elimination of a water molecule from the tetrahedral intermediate yields the corresponding ester.
Why other options are incorrect:- Option A: Ketones and ethers do not condense to form esters.
- Option B: Aldehydes and ketones undergo crossed aldol additions, forming beta-hydroxy carbonyls.
- Option C: Carboxylic acids and aldehydes do not condense directly to produce esters.
MCQ #90 of 150
Chemistry
NUMS 2021
[NUMS 2021]
In the laboratory, benzene can be synthesized from sodium benzoate via decarboxylation by heating with:
A
Soda lime (\(\text{NaOH} + \text{CaO}\))
B
Anhydrous aluminum chloride (\(\text{AlCl}_3\))
C
Chlorobenzene in dry ether
D
Cumene under catalytic oxidation
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The dry heating of an alkali metal carboxylate salt with a strong alkaline base removes the carboxyl group as a carbonate salt.
Formula / Rule / Reaction:$$\text{C}_6\text{H}_5\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{C}_6\text{H}_6 + \text{Na}_2\text{CO}_3$$
Solution:- Sodium benzoate is mixed with soda lime (a dry blend of \(\text{NaOH}\) and \(\text{CaO}\)) and heated strongly.
- The base removes the carboxylate group to form benzene vapor and solid sodium carbonate, with \(\text{CaO}\) preventing melting and glass erosion.
Why other options are incorrect:- Option B: Anhydrous \(\text{AlCl}_3\) is a Lewis acid catalyst used in Friedel-Crafts alkylation and acylation reactions.
- Option C: Chlorobenzene in dry ether with sodium metal produces biphenyl via the Fittig reaction.
- Option D: Cumene oxidation is an industrial process used to manufacture phenol and acetone.
MCQ #91 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The mathematical relationship that connects the molecular formula of a compound to its empirical formula is:
A
\(\text{Molecular formula} = \frac{\text{Empirical formula}}{n}\)
B
\(\text{Molecular formula} = n \times (\text{Empirical formula})\)
C
\(\text{Molecular formula} = 2n \times (\text{Empirical formula})\)
D
\(\text{Empirical formula} = n \times (\text{Molecular formula})\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The empirical formula represents the simplest whole-number atomic ratio, whereas the molecular formula denotes the actual number of atoms in a molecule.
Formula / Rule / Reaction:$$n = \frac{\text{Molecular Molar Mass}}{\text{Empirical Formula Mass}} \implies \text{Molecular Formula} = [\text{Empirical Formula}]_n$$
Solution:- The scaling integer \(n\) represents the ratio of molecular molar mass to empirical formula mass.
- Multiplying the subscript of each atom in the empirical formula by \(n\) gives the true molecular formula.
Why other options are incorrect:- Option A: Dividing the empirical formula by \(n\) yields fractional atomic values.
- Option C: Including a factor of \(2n\) doubles the true number of constituent atoms.
- Option D: The empirical formula is simpler than the molecular formula, so it cannot equal the molecular formula multiplied by \(n\).
MCQ #92 of 150
Chemistry
NUMS 2021
[NUMS 2021]
Among the elemental halogens, which exhibits the highest chemical reactivity toward standard halogenation reactions?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Halogen reactivity as oxidizing agents and halogenating species decreases down Group 17 as electronegativity and reduction potentials decrease.
Formula / Rule / Reaction:$$\text{Reactivity Trend: } \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \quad (E^\circ_{\text{red}} \text{ for F}_2 = +2.87\text{ V})$$
Solution:- Fluorine (\(\text{F}_2\)) has an exceptionally high standard reduction potential, small atomic size, and a relatively weak \(\text{F}-\text{F}\) bond due to lone-pair repulsion.
- These factors lower its activation energy and make it react violently and exothermically, even in the dark.
Why other options are incorrect:- Option A: Iodine is the least reactive halogen in this group due to its large atomic radius and low electronegativity.
- Option B: Bromine is less reactive than fluorine and chlorine, requiring mild heat or light for many additions.
- Option D: Chlorine is a strong oxidizing agent, but its reactivity remains below that of elemental fluorine.
MCQ #93 of 150
Chemistry
NUMS 2021
[NUMS 2021]
One thermochemical calorie (\(\text{cal}\)) is precisely defined in terms of the SI unit of energy as:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A calorie is the amount of thermal energy required to raise the temperature of one gram of pure liquid water by one degree Celsius.
Formula / Rule / Reaction:$$1\text{ cal} = 4.184\text{ Joules (J)}$$
Solution:- By international thermochemical convention, one calorie is defined as exactly \(4.184\text{ J}\).
- This conversion factor equals the specific heat capacity of liquid water at \(15^\circ\text{C}\).
Why other options are incorrect:- Option A: \(4.184\text{ kJ}\) equals one kilocalorie (\(1\text{ kcal} = 1000\text{ cal}\)), commonly used as the dietary Calorie.
- Option B: \(18.0\text{ J}\) is an unrelated numerical value.
- Option C: \(0.4184\text{ J}\) is incorrect by an order of magnitude.
MCQ #94 of 150
Chemistry
NUMS 2021
[NUMS 2021]
The thermochemical law stating that the overall enthalpy change for a chemical reaction is identical whether the reaction occurs in a single step or a series of steps is:
A
Hess's law of constant heat summation
C
Henry's law of gas solubility
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Enthalpy is a thermodynamic state function whose change depends exclusively on the initial and final states of the system, not the reaction path.
Formula / Rule / Reaction:$$\Delta H_{\text{net}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots + \Delta H_n$$
Solution:- Germain Hess formulated this law in 1840 as a direct consequence of the first law of thermodynamics (conservation of energy).
- It allows the enthalpy change of difficult-to-measure reactions to be calculated by summing intermediate steps with known enthalpy values.
Why other options are incorrect:- Option B: Joule's law relates the heat produced in a conductor to the square of the electrical current and resistance.
- Option C: Henry's law states that the solubility of a gas in a liquid is proportional to its partial pressure.
- Option D: The Lavoisier-Laplace law states that the enthalpy change of a reverse reaction is equal in magnitude and opposite in sign to the forward reaction.
MCQ #95 of 150
Physics
NUMS 2021
[NUMS 2021]
Which of the following assertions regarding the rate of a chemical reaction is fundamentally INCORRECT?
A
The rate is affected by changing the temperature of the system
B
The rate is completely independent of the concentrations of reactants
C
The rate depends on the presence and efficiency of a catalyst
D
The rate is related to the overall reaction order
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Chemical kinetics demonstrates that reaction rate is directly proportional to the effective concentrations of reactants raised to their respective reaction orders.
Formula / Rule / Reaction:$$\text{Rate} = k[\text{A}]^x[\text{B}]^y$$
Solution:- According to collision theory and empirical rate laws, raising reactant concentrations increases collision frequency and reaction rate (except in zero-order reactions).
- Therefore, stating that reaction rate is universally independent of concentration is incorrect.
Why other options are incorrect:- Option A: The Arrhenius equation (\(k = A e^{-E_a/RT}\)) confirms that reaction rate increases with rising temperature.
- Option C: Catalysts lower the activation energy, accelerating reaction rates.
- Option D: The mathematical form of the rate law depends directly on reaction orders \(x\) and \(y\).
MCQ #96 of 150
Physics
NUMS 2021
[NUMS 2021]
Which of the following elements belongs to the \(s\)-block of the periodic table?
B
Aluminum (\(\text{Al}\))
C
Potassium (\(\text{K}\))
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Elements are classified into periodic blocks (\(s\), \(p\), \(d\), \(f\)) based on which subshell receives the final valence electron according to the Aufbau principle.
Formula / Rule / Reaction:$$\text{Potassium } (Z = 19): [\text{Ar}]\, 4s^1$$
Solution:- The valence electron of potassium enters the \(4s\) subshell, placing it in Group 1 of the \(s\)-block.
Why other options are incorrect:- Option A: Boron (\(Z = 5\), configuration \([\text{He}]\, 2s^2 2p^1\)) is a \(p\)-block metalloid.
- Option B: Aluminum (\(Z = 13\), configuration \([\text{Ne}]\, 3s^2 3p^1\)) is a \(p\)-block metal.
- Option D: Copper (\(Z = 29\), configuration \([\text{Ar}]\, 3d^{10} 4s^1\)) is a \(d\)-block transition metal.
MCQ #97 of 150
Physics
NUMS 2021
[NUMS 2021]
The bond angle around each carbon atom in a planar ethene molecule (\(\text{H}_2\text{C}=\text{CH}_2\)) is approximately:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:VSEPR theory and hybridization dictate that three steric domains around an \(\text{sp}^2\)-hybridized carbon adopt a trigonal planar geometry.
Formula / Rule / Reaction:$$\text{Hybridization: } \text{sp}^2 \implies \text{Trigonal Planar Geometry with ideal angle } \theta = 120^\circ$$
Solution:- In ethene, each carbon forms two \(\text{C}-\text{H}\) sigma bonds and one \(\text{C}=\text{C}\) double bond (one sigma, one pi).
- The three electron domains orient symmetrically in a single plane, yielding bond angles of approximately \(120^\circ\) (measured \(\text{H}-\text{C}-\text{H}\) is \(117.4^\circ\) and \(\text{H}-\text{C}=\text{C}\) is \(121.3^\circ\)).
Why other options are incorrect:- Option A: \(180^\circ\) is the linear bond angle of \(\text{sp}\)-hybridized alkynes like ethyne.
- Option B: \(109.5^\circ\) is the tetrahedral bond angle of \(\text{sp}^3\)-hybridized alkanes like ethane.
- Option C: \(90^\circ\) bond angles occur in octahedral and square planar geometries, not simple alkenes.
MCQ #98 of 150
Physics
NUMS 2021
[NUMS 2021]
In the rock-salt crystal lattice of sodium chloride (\(\text{NaCl}\)), each \(\text{Na}^+\) cation is octahedrally coordinated by how many nearest \(\text{Cl}^-\) anions?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The coordination number in ionic crystals depends on the cation-to-anion radius ratio (\(r_+ / r_-\)).
Formula / Rule / Reaction:$$\frac{r_{\text{Na}^+}}{r_{\text{Cl}^-}} = \frac{95\text{ pm}}{181\text{ pm}} \approx 0.525 \quad (0.414 \le \text{ratio} < 0.732 \implies \text{Octahedral Coordination Number } = 6)$$
Solution:- Sodium chloride forms a face-centered cubic lattice where each \(\text{Na}^+\) ion is surrounded by 6 equidistant \(\text{Cl}^-\) ions at the vertices of an octahedron.
- By symmetry, each \(\text{Cl}^-\) ion is likewise surrounded by 6 \(\text{Na}^+\) ions, giving a \(6:6\) coordination lattice.
Why other options are incorrect:- Option B: A coordination number of 4 is found in zinc blende (\(\text{ZnS}\)) lattices with smaller radius ratios.
- Option C: A coordination number of 8 occurs in body-centered cubic lattices like cesium chloride (\(\text{CsCl}\)).
- Option D: A coordination number of 12 represents close-packed metallic lattices (FCC and HCP).
MCQ #99 of 150
Physics
NUMS 2021
[NUMS 2021]
Which of the following alkali metal halides possesses the greatest crystal lattice energy?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:According to Born-Lande relation, lattice energy is directly proportional to ionic charges and inversely proportional to the sum of ionic radii.
Formula / Rule / Reaction:$$U_0 \propto \frac{|z_+ z_-|}{r_+ + r_-}$$
Solution:- Comparing sodium halides, the sodium cation (\(\text{Na}^+\)) is common, and all halide anions carry a single negative charge (\(-1\)).
- Fluoride (\(\text{F}^-\)) has the smallest ionic radius in the series, making the internuclear separation \(r_+ + r_-\)) smallest in \(\text{NaF}\) and maximizing its electrostatic lattice energy (\(918\text{ kJ/mol}\)).
Why other options are incorrect:- Option A: Chloride is larger than fluoride, giving \(\text{NaCl}\) a lower lattice energy (\(788\text{ kJ/mol}\)).
- Option C: Bromide is larger than chloride, giving \(\text{NaBr}\) a lower lattice energy (\(751\text{ kJ/mol}\)).
- Option D: Iodide has the largest radius, giving \(\text{NaI}\) the lowest lattice energy (\(700\text{ kJ/mol}\)) in this series.
MCQ #100 of 150
Physics
NUMS 2021
[NUMS 2021]
In the industrial synthesis of ammonia via the Haber-Bosch process, an elevated operating pressure of approximately 200 atm is used primarily to:
A
Prevent catalyst poisoning by sulfur compounds
B
Lower the activation energy barrier
C
Shift the equilibrium forward to maximize ammonia yield
D
Prevent the backward reaction from occurring completely
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Le Chatelier's principle states that an increase in pressure shifts a gaseous equilibrium toward the side with fewer gas molecules.
Formula / Rule / Reaction:$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons} 2\text{NH}_3(g) \quad (\Delta n_g = 2 - 4 = -2)$$
Solution:- Four moles of gaseous reactants yield two moles of gaseous products.
- Increasing operating pressure to 200 atm shifts equilibrium toward the product side, improving the equilibrium yield of \(\text{NH}_3\).
Why other options are incorrect:- Option A: Catalyst poisoning is prevented by gas purification, not high pressure.
- Option B: Activation energy is lowered by the iron catalyst, not by system pressure.
- Option D: High pressure shifts the equilibrium position but does not prevent the reverse reaction in a reversible system.
MCQ #101 of 150
Physics
NUMS 2021
[NUMS 2021]
Introducing a suitable catalyst into a reversible chemical system at equilibrium:
A
Increases the numerical value of the equilibrium constant \(K_c\)
B
Shifts the equilibrium position toward higher product yield
C
Increases the rate of the forward reaction exclusively
D
Accelerates the attainment of equilibrium without shifting its position
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A catalyst provides an alternative reaction pathway with a lower activation energy, accelerating both forward and reverse reaction rates equally.
Formula / Rule / Reaction:$$K_c = \frac{k_f}{k_r} \quad (\text{Both } k_f \text{ and } k_r \text{ increase by identical factors})$$
Solution:- Because the catalyst lowers the energy barrier by the same amount in both directions, forward and backward rates increase by the same factor.
- Consequently, the system reaches equilibrium faster, but the equilibrium concentrations and the value of \(K_c\) remain unchanged.
Why other options are incorrect:- Option A: \(K_c\) depends solely on temperature and is unaffected by catalysts.
- Option B: Catalysts do not alter thermodynamic free energies, so product yield at equilibrium is not changed.
- Option C: Catalysts accelerate both forward and reverse reactions equally.
MCQ #102 of 150
Physics
NUMS 2021
[NUMS 2021]
The minimum kinetic energy that colliding reactant molecules must possess to form the activated transition complex is called:
B
Bond dissociation energy
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:According to collision theory, collisions result in a chemical reaction only if molecules collide with appropriate orientation and sufficient kinetic energy.
Formula / Rule / Reaction:$$k = A e^{-E_a / RT} \implies E_a = \text{Threshold Energy} - \text{Average Reactant Energy}$$
Solution:- The activation energy (\(E_a\)) represents the minimum energy barrier reactants must overcome to rearrange bonds into an unstable transition state.
- Molecules colliding with less energy than \(E_a\) rebound elastically without reacting.
Why other options are incorrect:- Option B: Bond dissociation energy is the energy needed to break one mole of a specific covalent bond homolytically.
- Option C: Enthalpy of reaction is the net difference in enthalpy between products and reactants (\(\Delta H = H_{\text{products}} - H_{\text{reactants}}\)).
- Option D: Lattice energy is the energy released when gaseous ions assemble into a solid crystal lattice.
MCQ #103 of 150
Physics
NUMS 2021
[NUMS 2021]
Which of the following thermodynamic parameters is a path-dependent function rather than a thermodynamic state function?
C
Internal energy (\(U\))
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:State functions depend solely on the present state of a system, whereas path functions depend on the specific path taken between states.
Formula / Rule / Reaction:$$\Delta U = Q - W \quad (U \text{ is a state function; } Q \text{ and } W \text{ are path-dependent})$$
Solution:- Work (\(W = \int P\, dV\)) and heat (\(Q\)) depend on the specific thermodynamic process (isothermal, isobaric, adiabatic).
- In contrast, variables such as internal energy, enthalpy, entropy, pressure, volume, and temperature are independent of the path and qualify as true state functions.
Why other options are incorrect:- Option A: Enthalpy (\(H = U + PV\)) is defined by state functions, making it a state function itself.
- Option C: Internal energy depends only on current temperature and molecular structure, not past history.
- Option D: Entropy is a state function measuring microscopic dispersal of energy.
MCQ #104 of 150
Physics
NUMS 2021
[NUMS 2021]
The time rate of change of linear velocity of a moving object is defined as its:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Acceleration is the vector quantity describing how rapidly an object's velocity changes in magnitude, direction, or both.
Formula / Rule / Reaction:$$\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt}$$
Solution:- By definition, acceleration measures the rate of velocity change with respect to time, measured in \(\text{m/s}^2\).
- Because velocity is a vector, any change in speed or travel direction produces acceleration.
Why other options are incorrect:- Option A: Linear momentum is the product of an object's mass and velocity (\(\vec{p} = m\vec{v}\)).
- Option B: Instantaneous speed is the scalar magnitude of velocity (\(v = |\vec{v}|\)).
- Option D: Impulse is the product of net force and the time duration over which it acts (\(\vec{J} = \vec{F}\Delta t\)).
MCQ #105 of 150
Physics
NUMS 2021
[NUMS 2021]
For a projectile launched from level ground with a fixed initial speed in the absence of air resistance, the horizontal range is maximized at a launch angle of:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The horizontal range of projectile motion depends on the sine of twice the launch angle.
Formula / Rule / Reaction:$$R = \frac{v_0^2 \sin(2\theta)}{g}$$
Solution:- Range \(R\) is maximized when \(\sin(2\theta)\) reaches its maximum possible value of \(1\).
- $$\sin(2\theta) = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ$$
- Therefore, a \(45^\circ\) launch angle balances vertical flight time and horizontal velocity to maximize range.
Why other options are incorrect:- Option A: At \(30^\circ\), \(\sin(2\theta) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866\), yielding less than maximum range.
- Option B: At \(60^\circ\), \(\sin(2\theta) = \sin(120^\circ) = 0.866\), which gives the same sub-maximal range as \(30^\circ\).
- Option C: At \(90^\circ\), \(\sin(180^\circ) = 0\), meaning the projectile travels straight up and lands with zero horizontal range.
MCQ #106 of 150
Physics
NUMS 2021
[NUMS 2021]
Which of the following statements concerning linear momentum is fundamentally INCORRECT?
A
Linear momentum is a scalar physical quantity
B
Linear momentum represents the quantity of motion in a body
C
The SI unit of linear momentum is equivalent to \(\text{N}\cdot\text{s}\)
D
Linear momentum equals the product of mass and velocity
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Linear momentum is a directional vector quantity proportional to mass and velocity.
Formula / Rule / Reaction:$$\vec{p} = m\vec{v}$$
Solution:- Because velocity is a vector and mass is a positive scalar, their product \(\vec{p}\) is inherently a vector pointing in the direction of motion.
- Therefore, calling linear momentum a scalar quantity is false.
Why other options are incorrect:- Option B: Newton historically defined momentum as the total quantity of motion of a body.
- Option C: Dimensional analysis confirms \(\text{kg}\cdot\text{m/s} = (\text{kg}\cdot\text{m/s}^2)\cdot\text{s} = \text{N}\cdot\text{s}\).
- Option D: By standard classical definition, \(\vec{p} = m\vec{v}\).
MCQ #107 of 150
Physics
NUMS 2021
[NUMS 2021]
During a perfectly elastic collision between two isolated bodies, which of the following physical quantities are conserved?
B
Linear momentum, total kinetic energy, and total energy
C
Total kinetic energy only
D
Total energy only, while kinetic energy is always dissipated
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:An elastic collision is an idealized physical interaction where no mechanical kinetic energy is converted into heat, sound, or internal deformation.
Formula / Rule / Reaction:$$\sum \vec{p}_i = \sum \vec{p}_f \quad \text{and} \quad \sum \frac{1}{2}m v_i^2 = \sum \frac{1}{2}m v_f^2$$
Solution:- In all closed, isolated collisions, total linear momentum is conserved due to Newton's third law.
- In a perfectly elastic collision, total kinetic energy is also fully conserved, meaning overall mechanical energy is preserved.
Why other options are incorrect:- Option A: Momentum conservation alone characterizes inelastic collisions where kinetic energy is partially lost.
- Option C: Kinetic energy cannot be conserved without linear momentum also being conserved.
- Option D: This describes an inelastic collision where kinetic energy is dissipated into other energy forms.
MCQ #108 of 150
Physics
NUMS 2021
[NUMS 2021]
The mathematical gradient (slope) of a rectilinear displacement versus time graph represents the object's:
A
Instantaneous acceleration
B
Total distance traveled
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:In kinematics, the derivative of position with respect to time defines instantaneous velocity.
Formula / Rule / Reaction:$$\text{Slope} = \frac{\Delta d}{\Delta t} = \frac{dd}{dt} = v$$
Solution:- Plotting displacement on the vertical axis and time on the horizontal axis makes the slope \(\Delta d / \Delta t\).
- This derivative equals instantaneous velocity, with positive, negative, and zero slopes indicating forward motion, backward motion, and rest respectively.
Why other options are incorrect:- Option A: Acceleration is represented by the slope of a velocity-time graph, not a displacement-time graph.
- Option B: Total distance is the cumulative scalar path length traveled.
- Option D: Mechanical work is obtained by integrating force over displacement (\(W = \int F\, dx\)).
MCQ #109 of 150
Physics
NUMS 2021
[NUMS 2021]
Two solid spheres having masses of \(10\text{ kg}\) and \(50\text{ kg}\) are released simultaneously from rest from the top of an \(80\text{ m}\) high cliff. Neglecting air resistance (\(g = 10\text{ m/s}^2\)), which stone strikes the ground with a greater velocity?
A
The \(10\text{ kg}\) stone
B
The \(50\text{ kg}\) stone
C
The velocity cannot be determined without cross-sectional radii
D
Both stones strike the ground with identical velocities
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Under uniform gravitational acceleration with negligible aerodynamic drag, the acceleration of free fall is independent of an object's mass.
Formula / Rule / Reaction:$$v = \sqrt{2gh}$$
Solution:- Applying conservation of mechanical energy:
- $$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}$$
- Because mass \(m\) cancels out of both sides of the equation, final velocity depends only on gravitational acceleration \(g\) and vertical fall height \(h\).
- $$v = \sqrt{2(10)(80)} = \sqrt{1600} = 40\text{ m/s}$$
- Both stones hit the ground with the same velocity of \(40\text{ m/s}\).
Why other options are incorrect:- Option A: The lighter mass experiences less gravitational force (\(F_g = mg\)), but requires less force to accelerate, keeping acceleration identical.
- Option B: The heavier mass experiences a greater gravitational force, but its greater inertia requires proportionately more force to accelerate.
- Option C: Cross-sectional radius and drag are explicitly neglected in vacuum free-fall conditions.
MCQ #110 of 150
Physics
NUMS 2021
[NUMS 2021]
How much mechanical work is performed by the gravitational force on a stone weighing \(10\text{ N}\) as it falls freely from the top of a \(250\text{ m}\) vertical cliff to its base?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Mechanical work performed by a constant force is defined as the scalar dot product of the force vector and the displacement vector.
Formula / Rule / Reaction:$$W = \vec{F} \cdot \vec{d} = F d \cos(\theta)$$
Solution:- The gravitational force acting downward on the stone equals its weight \(w = 10\text{ N}\).
- Displacement is directed downward through \(d = 250\text{ m}\), making the angle \(\theta = 0^\circ\) (\(\cos 0^\circ = 1\)).
- $$W = (10\text{ N})(250\text{ m})(1) = 2500\text{ J}$$
Why other options are incorrect:- Option B: \(250\text{ J}\) corresponds to a displacement of only \(25\text{ m}\).
- Option C: \(25\text{ J}\) corresponds to a displacement of only \(2.5\text{ m}\).
- Option D: \(25000\text{ J}\) results from multiplying by \(g = 10\text{ m/s}^2\) again, forgetting that the weight in Newtons already includes mass times gravity.
MCQ #111 of 150
Physics
NUMS 2021
[NUMS 2021]
If the translational speed of an automobile is reduced to one-half of its original value, by what factor does its kinetic energy decrease relative to the original value?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Translational kinetic energy varies directly with the square of the instantaneous speed.
Formula / Rule / Reaction:$$KE = \frac{1}{2}m v^2$$
Solution:- Let the initial kinetic energy be \(KE_1 = \frac{1}{2}m v^2\).
- When speed is halved, the new speed is \(v' = \frac{v}{2}\).
- $$KE_2 = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{2}m\left(\frac{v^2}{4}\right) = \frac{1}{4}KE_1$$
- Therefore, the new kinetic energy decreases to one-fourth of its initial value.
Why other options are incorrect:- Option A: Linear momentum decreases by a factor of \(1/2\), but kinetic energy depends on velocity squared.
- Option C: \(1/8\) would correspond to a cubic relationship, which is not applicable to classical kinetic energy.
- Option D: \(1/16\) occurs if the speed is reduced to one-fourth of its original value.
MCQ #112 of 150
Physics
NUMS 2021
[NUMS 2021]
If by some internal phenomenon Earth were to contract to one-half of its present radius while keeping its mass strictly constant, the magnitude of the gravitational potential energy of an object resting on its surface would:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The absolute gravitational potential energy of a mass at a planet's surface is inversely proportional to the planet's radial distance from its center of mass.
Formula / Rule / Reaction:$$U = -\frac{G M m}{r} \implies |U| \propto \frac{1}{r}$$
Solution:- Let the initial potential energy magnitude be \(|U_1| = \frac{GMm}{R}\).
- When the radius decreases to \(R' = \frac{R}{2}\) with constant mass \(M\):
- $$|U_2| = \frac{GMm}{R/2} = 2\left(\frac{GMm}{R}\right) = 2|U_1|$$
- Therefore, the magnitude of the surface gravitational potential energy doubles.
Why other options are incorrect:- Option A: The value would halve only if the radius doubled.
- Option B: Changing radial separation directly alters potential energy.
- Option D: Gravitational force (\(F \propto 1/r^2\)) increases four-fold, whereas potential energy varies as \(1/r\) and doubles.
MCQ #113 of 150
Physics
NUMS 2021
[NUMS 2021]
The instantaneous mechanical power delivered by a net force vector \(\vec{F}\) acting on a particle moving with instantaneous velocity vector \(\vec{v}\) is defined by:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Power is the time rate at which mechanical work is performed on a physical system.
Formula / Rule / Reaction:$$P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{s}}{dt} = \vec{F} \cdot \vec{v}$$
Solution:- Mechanical work is given by \(dW = \vec{F} \cdot d\vec{s}\).
- Differentiating with respect to time yields:
- $$P = \vec{F} \cdot \frac{d\vec{s}}{dt} = \vec{F} \cdot \vec{v}$$
- The scalar dot product of force and velocity represents power, measured in Watts (\(\text{J/s}\)).
Why other options are incorrect:- Option A: Kinetic energy is the energy of motion given by \(\frac{1}{2}mv^2\).
- Option B: Impulse is the product of net force and elapsed time (\(\vec{J} = \vec{F}\Delta t\)).
- Option C: Mechanical work represents energy transferred over a spatial displacement (\(W = \int \vec{F}\cdot d\vec{s}\)), not the instantaneous rate.
MCQ #114 of 150
Physics
NUMS 2021
[NUMS 2021]
The angular measure of one degree (\(1^\circ\)) converted into circular radian units is approximately equal to:
A
\(0.0175\text{ radian}\)
B
\(0.1000\text{ radian}\)
C
\(0.1500\text{ radian}\)
D
\(0.2750\text{ radian}\)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Plane angles are converted between sexagesimal degrees and SI radian units using the fundamental geometric identity of a complete circle.
Formula / Rule / Reaction:$$360^\circ = 2\pi\text{ radians} \implies 1^\circ = \frac{\pi}{180}\text{ radians}$$
Solution:- Substituting the value of \(\pi \approx 3.14159\):
- $$1^\circ = \frac{3.14159}{180} \approx 0.017453\text{ radian} \approx 0.0175\text{ radian}$$
Why other options are incorrect:- Option B: \(0.1000\text{ radian}\) equals roughly \(5.73^\circ\).
- Option C: \(0.1500\text{ radian}\) equals roughly \(8.59^\circ\).
- Option D: \(0.2750\text{ radian}\) equals roughly \(15.76^\circ\).
MCQ #115 of 150
Physics
NUMS 2021
[NUMS 2021]
A particle traverses an arc length of \(2\text{ m}\) along the circumference of a circular path of radius \(1\text{ m}\). The angular displacement subtended by the particle at the center is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Angular displacement in radians is the ratio of arc length traversed to the radius of curvature.
Formula / Rule / Reaction:$$\theta = \frac{s}{r}$$
Solution:- Given arc length \(s = 2\text{ m}\) and radius \(r = 1\text{ m}\):
- $$\theta = \frac{2\text{ m}}{1\text{ m}} = 2.0\text{ radians}$$
Why other options are incorrect:- Option A: \(0.5\text{ radian}\) inverts the formula by dividing radius by arc length (\(r/s\)).
- Option C: \(3.0\text{ radians}\) would require an arc length of \(3\text{ m}\).
- Option D: \(0.67\text{ radian}\) corresponds to \(2/3\text{ radian}\), which is dimensionally incorrect for these values.
MCQ #116 of 150
Physics
NUMS 2021
[NUMS 2021]
If both the orbital speed of a body and the radius of its circular trajectory are simultaneously doubled, the required centripetal force will:
D
Increase by eight times
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Centripetal force is directly proportional to the square of tangential speed and inversely proportional to the trajectory radius.
Formula / Rule / Reaction:$$F_c = \frac{m v^2}{r}$$
Solution:- Let the initial force be \(F_1 = \frac{m v^2}{r}\).
- When \(v' = 2v\) and \(r' = 2r\):
- $$F_2 = \frac{m(2v)^2}{2r} = \frac{4 m v^2}{2r} = 2\left(\frac{m v^2}{r}\right) = 2 F_1$$
- Therefore, the centripetal force doubles.
Why other options are incorrect:- Option A: The force would remain unchanged only if radius quadrupled while speed doubled.
- Option B: The force would quadruple if speed doubled while radius remained constant.
- Option D: An eight-fold increase would require speed to double while radius was halved.
MCQ #117 of 150
Physics
NUMS 2021
[NUMS 2021]
The fundamental kinematic relation connecting linear tangential velocity \(v\) and angular velocity \(\omega\) for rotation at radius \(r\) is:
B
\(v = \frac{\omega}{r}\)
C
\(\omega = \frac{v}{r^2}\)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Tangential linear velocity equals the product of the radial distance from the axis of rotation and the instantaneous angular velocity.
Formula / Rule / Reaction:$$\vec{v} = \vec{\omega} \times \vec{r} \implies v = r \omega \quad (\text{for } \vec{\omega} \perp \vec{r})$$
Solution:- Differentiating arc length \(s = r\theta\) with respect to time yields:
- $$\frac{ds}{dt} = r \frac{d\theta}{dt} \implies v = r \omega$$
Why other options are incorrect:- Option A: \(\omega = r v\) is dimensionally inconsistent (\(\text{s}^{-1} \neq \text{m}^2\text{/s}\)).
- Option B: \(v = \omega/r\) incorrectly places radius in the denominator.
- Option C: \(\omega = v/r^2\) is dimensionally incorrect.
MCQ #118 of 150
Physics
NUMS 2021
[NUMS 2021]
In a ripple tank experiment, 100 complete wave crests pass a fixed point in one second. If the wavelength of the water wave is \(1\text{ cm}\), the wave propagation speed is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The speed of any periodic wave equals the product of its temporal frequency and spatial wavelength.
Formula / Rule / Reaction:$$v = f \lambda$$
Solution:- Frequency \(f = 100\text{ waves/s} = 100\text{ Hz}\).
- Wavelength \(\lambda = 1\text{ cm} = 0.01\text{ m}\).
- $$v = (100\text{ s}^{-1})(0.01\text{ m}) = 1.0\text{ m/s}$$
Why other options are incorrect:- Option B: \(2.0\text{ m/s}\) corresponds to double the actual wave speed.
- Option C: \(0.01\text{ m/s}\) results from dividing frequency by wavelength squared incorrectly.
- Option D: \(100.0\text{ m/s}\) fails to convert centimeters to SI meters.
MCQ #119 of 150
Physics
NUMS 2021
[NUMS 2021]
To increase the natural oscillation frequency of an ideal spring-mass system by a factor of four, the attached mass must be:
A
Reduced to one-fourth of its original value
B
Reduced to one-sixteenth of its original value
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The frequency of a simple harmonic oscillator is inversely proportional to the square root of the oscillating mass.
Formula / Rule / Reaction:$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \implies f \propto \frac{1}{\sqrt{m}}$$
Solution:- Let the initial frequency be \(f_1 \propto \frac{1}{\sqrt{m_1}}\).
- For the new frequency to be \(f_2 = 4 f_1\):
- $$\frac{f_2}{f_1} = \sqrt{\frac{m_1}{m_2}} = 4 \implies \frac{m_1}{m_2} = 16 \implies m_2 = \frac{m_1}{16}$$
- Thus, the mass must be reduced to one-sixteenth of its initial value.
Why other options are incorrect:- Option A: Reducing mass to one-fourth only doubles the frequency (\(\sqrt{4} = 2\)).
- Option C: Quadrupling mass halves the frequency.
- Option D: Doubling mass decreases frequency by a factor of \(\sqrt{2}\).
MCQ #120 of 150
Physics
NUMS 2021
[NUMS 2021]
The speed of acoustic sound waves in dry air is measured as \(332\text{ m/s}\) at \(0^\circ\text{C}\). What is its value at a temperature of \(10^\circ\text{C}\)?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Near standard ambient conditions, the speed of sound in air increases approximately linearly with temperature at a rate of \(0.61\text{ m/s}\) per degree Celsius.
Formula / Rule / Reaction:$$v_T = v_0 + 0.61 T$$
Solution:- Given \(v_0 = 332\text{ m/s}\) at \(0^\circ\text{C}\) and temperature \(T = 10^\circ\text{C}\):
- $$v_{10} = 332 + (0.61 \times 10) = 332 + 6.1 = 338.1\text{ m/s}$$
Why other options are incorrect:- Option A: \(332.61\text{ m/s}\) represents a temperature rise of only \(1^\circ\text{C}\).
- Option B: \(334.10\text{ m/s}\) corresponds to a \(3.4^\circ\text{C}\) elevation.
- Option D: \(344.20\text{ m/s}\) corresponds to a temperature of \(20^\circ\text{C}\).
MCQ #121 of 150
Physics
NUMS 2021
[NUMS 2021]
In thermodynamics, Mayer's relation for an ideal gas, \(C_p - C_v = R\), demonstrates that:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:At constant pressure, supplied heat must not only raise internal energy but also perform expansion work against the surroundings.
Formula / Rule / Reaction:$$C_p - C_v = R \implies C_p = C_v + R \quad (\text{where } R > 0)$$
Solution:- Because the universal molar gas constant \(R\) is positive (\(8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\)), \(C_p\) must exceed \(C_v\).
- At constant volume, no boundary work is performed (\(W = P\Delta V = 0\)), so all added heat raises internal energy. At constant pressure, extra heat is required to fuel expansion work.
Why other options are incorrect:- Option A: \(C_p < C_v\) would imply a negative gas constant \(R\), violating the first law of thermodynamics.
- Option B: \(C_p = C_v\) occurs only in incompressible substances where expansion coefficient is zero.
- Option C: Specific heats are positive physical quantities and cannot sum to zero.
MCQ #122 of 150
Physics
NUMS 2021
[NUMS 2021]
A thermodynamic process carried out such that the volume of the working gas remains strictly constant is termed an:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:An isochoric (isovolumetric) process occurs at constant volume, meaning the boundary does not move and mechanical work is zero.
Formula / Rule / Reaction:$$\Delta V = 0 \implies W = \int P\, dV = 0 \implies Q_v = \Delta U$$
Solution:- By definition, an isochoric process maintains \(V = \text{constant}\).
- Because \(\Delta V = 0\), the system performs no \(P\Delta V\) boundary work, and all transferred heat changes internal energy directly.
Why other options are incorrect:- Option B: An isobaric process occurs at constant pressure (\(\Delta P = 0\)).
- Option C: An isothermal process occurs at constant temperature (\(\Delta T = 0\)).
- Option D: An adiabatic process occurs without heat exchange with the environment (\(Q = 0\)).
MCQ #123 of 150
Physics
NUMS 2021
[NUMS 2021]
When two capacitors of capacitances \(2\,\mu\text{F}\) and \(6\,\mu\text{F}\) are connected together in parallel, the equivalent capacitance of the combination is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Capacitors connected in parallel share a common potential difference across their plates, so their equivalent capacitance is the direct algebraic sum of their individual capacitances.
Formula / Rule / Reaction:$$C_{\text{eq}} = C_1 + C_2$$
Solution:- Given \(C_1 = 2\,\mu\text{F}\) and \(C_2 = 6\,\mu\text{F}\):
- $$C_{\text{eq}} = 2\,\mu\text{F} + 6\,\mu\text{F} = 8.0\,\mu\text{F}$$
Why other options are incorrect:- Option A: \(1.5\,\mu\text{F}\) is the equivalent capacitance if connected in series (\(\frac{2 \times 6}{2 + 6} = 1.5\,\mu\text{F}\)).
- Option C: \(4.0\,\mu\text{F}\) represents the simple arithmetic average, not the parallel sum.
- Option D: \(12.0\,\mu\text{F}\) is the product of the two values, which is dimensionally incorrect.
MCQ #124 of 150
Physics
NUMS 2021
[NUMS 2021]
The electrostatic capacitance of an isolated parallel-plate capacitor does NOT depend on which of the following parameters?
A
Surface area of the conducting plates
B
Permittivity of the dielectric medium
C
Thickness of the conducting plates
D
Separation distance between the plates
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Capacitance depends exclusively on the geometry of the plates, their separation distance, and the dielectric permittivity of the intervening medium.
Formula / Rule / Reaction:$$C = \frac{\varepsilon_0 \varepsilon_r A}{d}$$
Solution:- Capacitance is governed by plate area \(A\), plate separation \(d\), and dielectric constant \(\varepsilon_r\).
- Because charges reside entirely on the facing inner surfaces of the conductors, plate thickness has no effect on electrostatic capacitance.
Why other options are incorrect:- Option A: Capacitance is directly proportional to plate surface area \(A\).
- Option B: Capacitance increases linearly with the relative permittivity \(\varepsilon_r\) of the dielectric.
- Option D: Capacitance is inversely proportional to the plate separation distance \(d\).
MCQ #125 of 150
Physics
NUMS 2021
[NUMS 2021]
According to the Maximum Power Transfer Theorem, an electrical source transfers maximum power to an external load resistance \(R\) when:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Maximum power transfer occurs when the load resistance equals the internal resistance of the supplying source.
Formula / Rule / Reaction:$$P_L = I^2 R = \left(\frac{E}{R + r}\right)^2 R \implies \frac{dP_L}{dR} = 0 \implies R = r$$
Solution:- Differentiating power delivered to the load with respect to \(R\) and setting the derivative to zero yields \(R = r\).
- At this matched impedance, exactly 50 percent of total power is delivered to the load while 50 percent is dissipated internally.
Why other options are incorrect:- Option A: When \(R > r\), circuit current drops, reducing total power delivered to the load.
- Option B: When \(R < r\), most of the generated power is dissipated inside the source's internal resistance \(r\).
- Option C: When \(R = 0\) (short circuit), terminal voltage is zero, delivering zero useful power.
MCQ #126 of 150
Physics
NUMS 2021
[NUMS 2021]
A cylindrical metallic wire of initial resistance \(R\) is uniformly drawn and stretched so that its cross-sectional radius is halved. Assuming density and total volume remain constant, its new resistance will be:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Stretching a wire preserves its total volume (\(V = A \cdot L\)), so any decrease in cross-sectional area produces a proportional increase in length.
Formula / Rule / Reaction:$$R = \rho \frac{L}{A} = \rho \frac{V}{A^2} = \rho \frac{V}{\pi^2 r^4} \implies R \propto \frac{1}{r^4}$$
Solution:- When radius is halved (\(r' = r/2\)), the new cross-sectional area becomes:
- $$A' = \pi (r/2)^2 = \frac{A}{4}$$
- Because volume \(V = A L\) is constant, the new length quadruples (\(L' = 4L\)).
- $$R' = \rho \frac{L'}{A'} = \rho \frac{4L}{A/4} = 16\left(\rho \frac{L}{A}\right) = 16 R$$
Why other options are incorrect:- Option B: \(4 R\) accounts only for the length increase or area reduction independently, ignoring conservation of volume.
- Option C: \(8 R\) is an incorrect scaling factor.
- Option D: \(R/4\) occurs if the wire is shortened or its cross-sectional area quadrupled.
MCQ #127 of 150
Physics
NUMS 2021
[NUMS 2021]
When an electrically charged particle enters a uniform magnetic field traveling in a direction parallel to the magnetic field lines, its trajectory will:
A
Deflect in a circular path toward the north pole
B
Continue straight without experiencing any magnetic deflection
C
Spiral along a helical path of increasing pitch
D
Stop immediately due to Lorentz braking
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The magnetic Lorentz force depends on the vector cross product of charge velocity and magnetic flux density.
Formula / Rule / Reaction:$$\vec{F}_B = q(\vec{v} \times \vec{B}) \implies F_B = q v B \sin(\theta)$$
Solution:- Because the velocity vector is parallel to the magnetic field, the angle between them is \(\theta = 0^\circ\).
- $$\sin(0^\circ) = 0 \implies F_B = 0$$
- With zero net magnetic force acting on the particle, it continues along a straight path at constant velocity in accordance with Newton's first law.
Why other options are incorrect:- Option A: Circular deflection requires velocity to be perpendicular (\(\theta = 90^\circ\)) to the magnetic field lines.
- Option C: Helical trajectories require velocity to enter at an oblique angle (\(0^\circ < \theta < 90^\circ\)).
- Option D: Magnetic fields perform zero work on charged particles and cannot decrease kinetic energy to bring them to rest.
MCQ #128 of 150
Physics
NUMS 2021
[NUMS 2021]
The magnetic flux passing through a planar surface is maximized when the angle between the magnetic field vector and the normal area vector is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Magnetic flux is defined as the scalar dot product of the uniform magnetic field vector and the surface area vector.
Formula / Rule / Reaction:$$\Phi_B = \vec{B} \cdot \vec{A} = B A \cos(\theta)$$
Solution:- The vector area \(\vec{A}\) points normal (perpendicular) to the physical plane of the surface.
- Flux is maximized when \(\cos(\theta) = 1\), which occurs when \(\theta = 0^\circ\) (the magnetic field lines run parallel to the surface normal, passing perpendicularly through the plane).
Why other options are incorrect:- Option A: At \(\theta = 90^\circ\), \(\cos(90^\circ) = 0\), meaning the field runs parallel to the surface and zero flux penetrates.
- Option B: At \(\theta = 45^\circ\), \(\cos(45^\circ) \approx 0.707\), yielding only 70.7 percent of maximum flux.
- Option D: At \(\theta = 180^\circ\), \(\cos(180^\circ) = -1\), yielding maximum negative flux rather than maximum positive flux.
MCQ #129 of 150
Physics
NUMS 2021
[NUMS 2021]
The standard domestic alternating current (AC) electrical grid supply throughout Pakistan operates at a nominal frequency of:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:National electrical grids distribute alternating current synchronized at standardized frequencies.
Formula / Rule / Reaction:$$f = \frac{N_s \cdot P}{120} = 50\text{ Hz} \quad (\text{Standard Pakistani Domestic Supply: } 230\text{ V, } 50\text{ Hz})$$
Solution:- Pakistan follows the British and European electrical distribution standard, operating at a single-phase nominal voltage of \(230\text{ V}\) and a line frequency of \(50\text{ Hz}\).
Why other options are incorrect:- Option A: \(70\text{ Hz}\) is not used in standard utility grids.
- Option B: \(60\text{ Hz}\) is the standard line frequency in North America and parts of Japan.
- Option C: \(100\text{ Hz}\) represents the ripple frequency of a full-wave rectified \(50\text{ Hz}\) signal, not the grid line frequency.
MCQ #130 of 150
Physics
NUMS 2021
[NUMS 2021]
An electrical device that steps up or steps down alternating electromotive force (EMF) through mutual electromagnetic induction is a:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Transformers transfer electrical energy between circuits via mutual magnetic flux linkage without changing signal frequency.
Formula / Rule / Reaction:$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$$
Solution:- A transformer consists of two insulated coils wound around a shared ferromagnetic core.
- Alternating current in the primary winding creates a time-varying magnetic flux that induces a stepped-up or stepped-down alternating EMF in the secondary winding according to Faraday's law.
Why other options are incorrect:- Option B: A commutator is a rotary mechanical switch in DC motors and dynamos that reverses current direction.
- Option C: A DC generator converts mechanical rotation into unidirectional direct current.
- Option D: A potentiometer measures unknown EMFs without drawing current from the circuit.
MCQ #131 of 150
Physics
NUMS 2021
[NUMS 2021]
The physical law establishing that an induced electric current always flows in such a direction that its magnetic effect opposes the change that produces it is:
C
Faraday's Law of Electrolysis
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Lenz's law provides the physical basis for the negative sign in Faraday's law of electromagnetic induction, enforcing conservation of energy.
Formula / Rule / Reaction:$$\mathcal{E} = -N \frac{d\Phi_B}{dt}$$
Solution:- Heinrich Lenz formulated this law in 1834.
- The negative sign shows that the induced magnetic field counteracts the change in primary magnetic flux, ensuring that mechanical work must be expended to generate electrical energy.
Why other options are incorrect:- Option A: Ampere's circuital law relates the integrated magnetic field around a closed loop to the electric current passing through it.
- Option C: Faraday's laws of electrolysis govern mass deposition at electrodes during electroplating.
- Option D: Joule's law relates heat dissipation in a conductor to the square of current and resistance (\(H = I^2 R t\)).
MCQ #132 of 150
Physics
NUMS 2021
[NUMS 2021]
An electronic circuit assembly that converts alternating current (AC) into unidirectional direct current (DC) is called a:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Rectification uses semiconductor diodes to permit current flow in one direction while blocking reverse current.
Formula / Rule / Reaction:$$\text{AC Input (Bidirectional)} \xrightarrow{\text{PN Junction Diode Circuit}} \text{Pulsating DC Output (Unidirectional)}$$
Solution:- A rectifier employs one or more \(p\)-\(n\) junction diodes configured in half-wave or full-wave bridge arrangements.
- Because diodes conduct only under forward bias, the alternating waveform is converted into unidirectional current.
Why other options are incorrect:- Option A: An operational amplifier increases the voltage, current, or power of an input signal.
- Option B: A filter circuit smooths voltage ripples but does not perform the primary AC-to-DC conversion.
- Option D: A modulator encodes information onto a high-frequency carrier wave.
MCQ #133 of 150
English
NUMS 2021
[NUMS 2021]
For a projectile launched at an angle \(\theta\) over horizontal terrain, its velocity vector components at the highest point of its trajectory satisfy:
A
\(v_x = 0, \quad v_y = 0\)
B
\(v_x = 0, \quad v_y = \text{Constant}\)
C
\(v_x = \text{Variable}, \quad v_y = 0\)
D
\(v_x = \text{Constant}, \quad v_y = 0\)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:In two-dimensional projectile motion neglecting air drag, the horizontal velocity component remains constant while the vertical velocity component changes under uniform gravity.
Formula / Rule / Reaction:$$v_x(t) = v_0 \cos(\theta) = \text{Constant}, \quad v_y(t) = v_0 \sin(\theta) - gt$$
Solution:- At peak height (\(H_{\max}\)), the projectile temporarily stops rising before descending, so \(v_y = 0\).
- Because no horizontal forces act on the projectile, its horizontal velocity remains non-zero and constant: \(v_x = v_0 \cos(\theta)\).
Why other options are incorrect:- Option A: \(v_x = 0\) occurs only in purely vertical launches (\(\theta = 90^\circ\)).
- Option B: \(v_x\) cannot be zero while \(v_y\) remains constant at the apex.
- Option C: \(v_x\) does not vary because horizontal acceleration is zero.
MCQ #134 of 150
English
NUMS 2021
[NUMS 2021]
A radioactive sample containing \(32\text{ g}\) of a phosphorus radioisotope decays until only \(2\text{ g}\) of the original parent nuclide remains after \(60\text{ days}\). The half-life of this isotope is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Radioactive decay follows first-order kinetics, where the remaining undecayed mass halves after each elapsed half-life period.
Formula / Rule / Reaction:$$N(t) = N_0 \left(\frac{1}{2}\right)^n \implies n = \frac{t}{T_{1/2}}$$
Solution:- Calculate the fraction of undecayed mass remaining:
- $$\frac{N(t)}{N_0} = \frac{2\text{ g}}{32\text{ g}} = \frac{1}{16} = \left(\frac{1}{2}\right)^4$$
- Because \(n = 4\) half-lives elapsed over \(60\text{ days}\):
- $$T_{1/2} = \frac{t}{n} = \frac{60\text{ days}}{4} = 15\text{ days}$$
Why other options are incorrect:- Option B: \(10\text{ days}\) would correspond to \(6\) half-lives, leaving only \(0.5\text{ g}\) remaining.
- Option C: \(6\text{ days}\) would correspond to \(10\) half-lives.
- Option D: \(5\text{ days}\) would correspond to \(12\) half-lives.
MCQ #135 of 150
English
NUMS 2021
[NUMS 2021]
The SI unit of gravitational potential \(V_g\) is expressed as:
A
\(\text{J}\cdot\text{s}\)
C
\(\text{J}\cdot\text{kg}\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Gravitational potential at a point in a field is defined as the work done per unit mass in bringing a test mass from infinity to that point.
Formula / Rule / Reaction:$$V_g = \frac{W}{m}$$
Solution:- Work \(W\) is measured in Joules (\(\text{J}\)) and mass \(m\) is measured in kilograms (\(\text{kg}\)).
- Therefore, the derived SI unit of gravitational potential is \(\text{J/kg}\) (or \(\text{m}^2\text{/s}^2\)).
Why other options are incorrect:- Option A: \(\text{J}\cdot\text{s}\) is the SI unit of action and Planck's constant \(h\).
- Option C: \(\text{J}\cdot\text{kg}\) is dimensionally incorrect.
- Option D: \(\text{N/m}\) is the SI unit of surface tension and spring constant.
MCQ #136 of 150
English
NUMS 2021
[NUMS 2021]
Complete the sentence using the correct grammatical tense: When we arrived at the cinema, the feature film _________.
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:When two past actions occur sequentially, the past perfect tense is used for the earlier event, while the simple past is used for the subsequent event.
Formula / Rule / Reaction:$$\text{Past Perfect (had + past participle)} \rightarrow \text{Earlier past event before another past action}$$
Solution:- The arrival at the cinema occurred in the past ('when we arrived').
- The screening began prior to this arrival, requiring the past perfect form: 'had already started'.
Why other options are incorrect:- Option A: 'Already started' lacks the auxiliary 'had' required to form the past perfect tense.
- Option B: 'Would already start' denotes a past conditional or habitual action, not an accomplished prior event.
- Option D: 'Starts already' uses the present tense, conflicting with the past frame 'when we arrived'.
MCQ #137 of 150
English
NUMS 2021
[NUMS 2021]
Select the correct form to complete the sentence: The hot soup _________ delicious.
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Sensory linking verbs describing states rather than dynamic actions use simple present forms and must agree in number with their subject.
Formula / Rule / Reaction:$$\text{Singular Subject (Soup)} + \text{Singular Stative Verb (Tastes)}$$
Solution:- 'Soup' is an uncountable singular noun.
- 'Taste' functions here as a copular (linking) verb connecting the subject to the predicate adjective 'delicious', taking the third-person singular inflection 'tastes'.
Why other options are incorrect:- Option A: 'Taste' is plural or base form and fails subject-verb agreement with singular 'soup'.
- Option B: 'Is tasting' implies an active, dynamic tasting action rather than a sensory property.
- Option C: 'Have tasted' takes a plural auxiliary 'have' and shifts meaning to an active past experience.
MCQ #138 of 150
English
NUMS 2021
[NUMS 2021]
Fill in the blank with the appropriate verb form: She _________ unconscious since 4 o'clock.
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The preposition 'since' specifying a starting point in past time requires a present perfect verb construction to denote a state continuing into the present.
Formula / Rule / Reaction:$$\text{Subject} + \text{has/have been} + \text{Adjective} + \text{since} + \text{Specific Point in Time}$$
Solution:- Because the state began at 4 o'clock and persists into the current moment, the present perfect 'has been' is grammatically correct.
Why other options are incorrect:- Option B: 'Was' is simple past, which indicates a completed state disconnected from the present.
- Option C: 'Is' is simple present, which cannot be paired with a 'since' duration phrase.
- Option D: 'Had' lacks the past participle predicate 'been'.
MCQ #139 of 150
English
NUMS 2021
[NUMS 2021]
Complete the sentence expressing an unreal past wish: I wish I _________ the correct answer to this dilemma.
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Wishes regarding present hypothetical or counterfactual states require the unreal past (subjunctive past tense) in the subordinate clause.
Formula / Rule / Reaction:$$\text{Subject} + \text{wish} + (\text{that}) + \text{Subject} + \text{Simple Past Verb (Subjunctive)}$$
Solution:- The speaker does not know the answer at present.
- To express a hypothetical desire contrary to present reality, the verb moves one step back into the past tense: 'knew'.
Why other options are incorrect:- Option A: 'Know' uses present indicative, which cannot follow counterfactual 'wish'.
- Option C: 'Will know' expresses future prediction rather than present counterfactual desire.
- Option D: 'Have known' uses present perfect, which does not convey the subjunctive mood.
MCQ #140 of 150
English
NUMS 2021
[NUMS 2021]
Choose the most accurate synonym for the word: ANALOGUE
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:An analogue is a person, structure, or thing that exhibits parallel attributes or functions comparable to another.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Derived from the Greek analogos (proportionate), 'analogue' denotes an entity that is similar, parallel, or comparable in function or form to something else.
Why other options are incorrect:- Option A: 'Opposite' is an antonym denoting contrary properties.
- Option B: 'Behind' is a spatial or temporal preposition.
- Option D: 'Mechanical' refers to machines or physical tools.
MCQ #141 of 150
English
NUMS 2021
[NUMS 2021]
Choose the most accurate synonym for the word: PULSATING
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Pulsating describes rhythmic, periodic expansion, contraction, or beating movements.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- 'Pulsating' describes rhythmic beating or throbbing variations in intensity, such as an arterial pulse or heart beat.
Why other options are incorrect:- Option A: 'Fluid' denotes continuous liquid flow rather than rhythmic pulsation.
- Option B: 'Static' is an antonym meaning motionless or stationary.
- Option C: 'Pressurized' refers to continuous application of force per unit area.
MCQ #142 of 150
English
NUMS 2021
[NUMS 2021]
Choose the word that correctly completes the sentence: The administrative head of an academic institution is called a:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Homophones are words that share identical pronunciations but differ in spelling, etymology, and definition.
Formula / Rule / Reaction:$$\text{Principal (Noun)} = \text{Head of an institution} \quad | \quad \text{Principle (Noun)} = \text{Fundamental rule or moral truth}$$
Solution:- 'Principal' functions as a noun designating the chief administrator or headmaster of an educational institution.
- 'Principle' denotes a foundational doctrine, standard, or ethical guideline.
Why other options are incorrect:- Option B: 'Principle' refers to a fundamental law, rule, or moral belief.
- Option C: 'Premier' denotes a prime minister or head of government.
- Option D: 'Prior' refers to an officer in a monastic religious order or something preceding in time.
MCQ #143 of 150
English
NUMS 2021
[NUMS 2021]
The common theatrical idiom 'break a leg' is colloquially used to:
A
Threaten an adversary with physical harm
B
Wish good luck to someone before a public performance
C
Acknowledge irreversible defeat
D
Express disciplinary reprimand
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:An idiom is an established figurative expression whose intended meaning cannot be deduced literally from its constituent words.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- In theatrical tradition, directly wishing a performer 'good luck' was thought to bring misfortune due to superstition.
- Consequently, performers adopted the ironic phrase 'break a leg' as an idiom meaning 'good luck'.
Why other options are incorrect:- Option A: The phrase is not used as a literal or figurative physical threat.
- Option C: Acknowledging defeat is described by idioms such as 'throw in the towel'.
- Option D: Reprimanding someone is captured by phrases like 'read the riot act'.
MCQ #144 of 150
English
NUMS 2021
[NUMS 2021]
The idiomatic expression 'bite the bullet' means to:
A
Act rashly without considering danger
B
Retaliate aggressively against an opponent
C
Accept an unpleasant and inevitable situation with fortitude
D
Refuse to yield in the face of logical persuasion
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Idiomatic figurative expressions convey cultural meanings distinct from their literal wording.
Formula / Rule / Reaction:Factual recall / Qualitative concept.Solution:- Originating from military field surgery where wounded soldiers bit on lead bullets to endure pain without anesthesia, 'bite the bullet' means to face a grim or difficult situation with resilience.
Why other options are incorrect:- Option A: Acting rashly without caution is described by idioms like 'leap before you look'.
- Option B: Aggressive retaliation is conveyed by 'an eye for an eye'.
- Option D: Stubborn refusal to change is captured by 'digging one's heels in'.
MCQ #145 of 150
English
NUMS 2021
[NUMS 2021]
Identify the correctly spelled word referring to a person who organizes and operates a business venture:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:English words borrowed directly from French preserve their native morphological spelling patterns.
Formula / Rule / Reaction:$$\text{Root: French } \textit{entreprendre} \rightarrow \text{Entrepreneur}$$
Solution:- The correct orthographic spelling is 'Entrepreneur' (beginning with 'Entre-' and ending with '-neur').
Why other options are incorrect:- Option A: 'Enterpreneur' misspells the initial prefix as 'enter-'.
- Option B: 'Entripreneur' substitutes an incorrect internal vowel 'i'.
- Option C: 'Entreprineur' corrupts the penultimate vowel into 'i'.
MCQ #146 of 150
English
NUMS 2021
[NUMS 2021]
Complete the sentence using the proper conditional form: Unless we _________ immediately, we cannot arrive on time.
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The subordinating conjunction 'unless' means 'if not' and intrinsically carries a negative meaning, so the dependent clause takes an affirmative verb.
Formula / Rule / Reaction:$$\text{Unless} + \text{Present Simple Affirmative Verb} \dots \text{Main Clause (Modal + Bare Infinitive)}$$
Solution:- Because 'unless' already means 'if we do not', adding a negative auxiliary ('do not start') creates an ungrammatical double negative.
- In a first conditional construction, the present simple affirmative verb 'start' is required.
Why other options are incorrect:- Option B: 'Will start' places a future modal inside a conditional subordinate time clause, which violates standard grammar rules.
- Option C: 'Do not start' produces an unintended double negative ('if we do not not start').
- Option D: 'Are starting' uses present continuous, which does not fit habitual conditional statements.
MCQ #147 of 150
English
NUMS 2021
[NUMS 2021]
Select the grammatically correct comparative structure: Professor Daud is more dedicated than _________ in the college.
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:When comparing an individual to other members of a group to which that individual belongs, the word 'other' must be included to avoid an illogical self-comparison.
Formula / Rule / Reaction:$$\text{Comparative Degree} + \text{than} + \text{all other} + \text{Plural Noun}$$
Solution:- Because Professor Daud is himself a member of the faculty, comparing him to 'all teachers' illogically includes himself.
- Adding 'other' ('all other teachers') excludes him from the comparison set, maintaining logical consistency.
Why other options are incorrect:- Option A: 'All teachers' illogically compares Professor Daud to himself.
- Option C: 'Any teachers' uses an incorrect plural; idiomatically it should be 'any other teacher' (singular).
- Option D: 'Other all teachers' violates standard English determiner word order.
MCQ #148 of 150
English
NUMS 2021
[NUMS 2021]
Choose the correct dependent preposition: Responsible drivers must always abide _________ established traffic regulations.
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Certain English verbs require fixed dependent prepositions to form idiomatic phrasal verbs with specific meanings.
Formula / Rule / Reaction:$$\text{Abide} + \text{by} = \text{To comply with, conform to, or obey a rule}$$
Solution:- The verb 'abide' pairs specifically with the preposition 'by' to mean obeying rules, laws, or decisions.
Why other options are incorrect:- Option A: 'Abide with' is an archaic construction meaning to stay or dwell alongside someone.
- Option B: 'Abide to' is an ungrammatical combination in standard English.
- Option D: 'Abide at' is an obsolete expression referring to physical residence.
MCQ #149 of 150
English
NUMS 2021
[NUMS 2021]
Choose the correct preposition to complete the sentence: A true soldier prefers an honorable death _________ dishonor.
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The verb 'prefer' takes the preposition 'to' rather than 'than' when comparing two nouns or gerunds.
Formula / Rule / Reaction:$$\text{Prefer} + [\text{Noun A}] + \text{to} + [\text{Noun B}]$$
Solution:- In formal English, preferences between two entities are constructed using 'prefer X to Y'.
- Although 'prefer X over Y' appears in colloquial speech, 'prefer X to Y' is the standard formal construction tested in entrance examinations.
Why other options are incorrect:- Option A: 'Prefer than' is ungrammatical; 'than' is used with comparative adjectives (e.g., 'better than'), not with the verb 'prefer'.
- Option B: 'Over' is an informal variant avoided in formal grammar examinations.
- Option C: 'Against' is incorrect following 'prefer'.
MCQ #150 of 150
English
NUMS 2021
[NUMS 2021]
Select the verb form that satisfies the rule of proximity: Neither the students nor their teacher _________ present at the ceremony yesterday.
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:When two subjects are joined by correlative conjunctions such as 'neither... nor' or 'either... or', the verb agrees in number and person with the closer subject.
Formula / Rule / Reaction:$$\text{Neither} + [\text{Subject 1 (Plural)}] + \text{nor} + [\text{Subject 2 (Singular)}] + \text{Singular Verb}$$
Solution:- The compound subject consists of plural 'the students' and singular 'their teacher'.
- According to the rule of proximity, the verb agrees with 'their teacher' (singular).
- Because the temporal adverb 'yesterday' specifies past time, the singular past verb 'was' is required.
Why other options are incorrect:- Option B: 'Were' agrees with the further plural subject 'students', violating the rule of proximity.
- Option C: 'Are' is a present plural verb, conflicting with the singular subject and the past adverb 'yesterday'.
- Option D: 'Have been' is present perfect plural, violating both the proximity rule and the simple past timeframe.