Official Entrance Examination Archive

DUHS 2021 Solved Past Paper

Complete 1:1 authentic annual examination paper (92 MCQs) solved with verified answer keys, step-by-step cognitive error autopsies, and KaTeX mathematical derivations across Biology, Chemistry, Physics, English.

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MCQ #1 of 92 Biology DUHS 2021
[DUHS 2021]

If fertilization occurs, the young embryo is implanted into the:
A
Epimetrium
B
Myometrium
C
Perimetrium
D
Endometrium
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The endometrium is the inner glandular mucosal layer of the uterine wall that undergoes cyclical preparation for blastocyst implantation.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Following fertilization in the fallopian tube, the cleaving blastocyst enters the uterine cavity.


  • Implantation proceeds as the trophoblast penetrates and embeds into the vascularized, receptive endometrial stroma.


Why other options are incorrect:

  • Option A: Epimetrium is an obsolete anatomical term occasionally used synonymously with perimetrium; it is not an implantation surface.
  • Option B: Myometrium is the intermediate muscular tunic composed of smooth muscle bundles responsible for expulsive uterine contractions during parturition.
  • Option C: Perimetrium is the outer serosal coat derived from the peritoneum covering the external uterine surface.
MCQ #2 of 92 Biology DUHS 2021
[DUHS 2021]

Hormones called gonadotropins are released by which gland?
A
Ovaries
B
Hypothalamus
C
Pituitary gland
D
Testes
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Gonadotropins are glycoprotein hormones secreted by the anterior pituitary gland that regulate gonadal endocrine function and gametogenesis.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Gonadotroph cells in the anterior pituitary gland synthesize and secrete follicle-stimulating hormone (FSH) and luteinizing hormone (LH).


  • Because FSH and LH directly stimulate target tissues within the gonads, they are classified as gonadotropins.


Why other options are incorrect:

  • Option A: Ovaries are target organs that produce steroid sex hormones (estrogen and progesterone), not gonadotropins.
  • Option B: The hypothalamus secretes gonadotropin-releasing hormone (GnRH), a peptide neurohormone that acts as a releasing factor rather than a gonadotropin.
  • Option D: Testes are male target gonads that synthesize androgens (principally testosterone) under gonadotropic regulation.
MCQ #3 of 92 Biology DUHS 2021
[DUHS 2021]

Which of the following is found at the ends of long bones, in the nose, larynx, and trachea?
A
Elastic cartilage
B
Fibrous cartilage
C
Perichondrium
D
Hyaline cartilage
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Hyaline cartilage is the most widespread cartilage type, characterized by fine type II collagen fibrils embedded in an amorphous, hydrated glycosaminoglycan matrix.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Hyaline cartilage provides frictionless articulating surfaces as articular cartilage over the epiphyseal ends of long bones.


  • It also forms the cartilaginous skeleton of the external nose, laryngeal thyroid and cricoid plates, and C-shaped rings of the trachea.


Why other options are incorrect:

  • Option A: Elastic cartilage contains dense networks of elastin fibers and is restricted to the auricle (pinna), external acoustic meatus, and epiglottis.
  • Option B: Fibrous cartilage possesses dense parallel bundles of coarse type I collagen fibers, occurring in intervertebral discs and the pubic symphysis.
  • Option C: Perichondrium is the outer fibrous sheath of vascularized dense connective tissue that invests non-articular cartilage, not a cartilage tissue itself.
MCQ #4 of 92 Biology DUHS 2021
[DUHS 2021]

A muscle with characteristics like spontaneous contraction, one nucleus per cell, and no control over contraction is most likely:
A
Smooth muscle
B
Cardiac muscle
C
Skeletal muscle
D
Visceral muscle
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Smooth muscle cells are spindle-shaped, non-striated, uninucleated myocytes under involuntary autonomic regulation capable of sustained myogenic contractions.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Smooth muscle cells possess a single, centrally placed nucleus per cell and lack transverse striations.


  • Single-unit smooth muscle displays intrinsic myogenic rhythmicity (spontaneous depolarization) and operates beyond conscious voluntary control.


  • Although visceral muscle is a contextual synonym, smooth muscle is the canonical histological classification keyed by the examining board.


Why other options are incorrect:

  • Option B: Cardiac muscle fibers are striated, branched, interconnected by intercalated discs, and restricted exclusively to the myocardium.
  • Option C: Skeletal muscle fibers are cylindrical, multi-nucleated, transversely striated, and subject to voluntary somatic motor control.
  • Option D: Visceral muscle is a gross macroscopic anatomical designation rather than the primary histological category.
MCQ #5 of 92 Biology DUHS 2021
[DUHS 2021]

A ball and socket joint is present at the:
A
Shoulder
B
Wrist
C
Elbow
D
Knee
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

A ball and socket joint (spheroidal joint) is a multiaxial synovial joint where a spherical bone head articulates within a cup-shaped cavity to allow movement in all cardinal planes.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The glenohumeral (shoulder) joint is formed by the head of the humerus articulating with the glenoid cavity of the scapula.


  • This morphology permits multiaxial movements: flexion, extension, abduction, adduction, medial rotation, lateral rotation, and circumduction.


Why other options are incorrect:

  • Option B: The radiocarpal (wrist) joint is a biaxial condyloid (ellipsoid) joint permitting motion in two axes without axial rotation.
  • Option C: The humeroulnar (elbow) joint is a uniaxial hinge joint limited to flexion and extension.
  • Option D: The tibiofemoral (knee) joint is a modified bicondylar hinge joint primarily limited to motion in the sagittal plane.
MCQ #6 of 92 Biology DUHS 2021
[DUHS 2021]

At a temperature higher than optimum, enzymatic activity decreases because of:
A
Saturation
B
Inhibition
C
Denaturation
D
Composition
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Excessive thermal energy disrupts non-covalent intramolecular bonds stabilizing the tertiary structure of proteins, causing denaturation of the catalytic active site.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Enzymatic active sites rely on precise three-dimensional conformations stabilized by hydrogen bonds, ionic interactions, and hydrophobic associations.


  • Thermal agitation above the optimum temperature breaks these weak stabilizing interactions, causing irreversible unfolding (denaturation) and loss of substrate binding.


Why other options are incorrect:

  • Option A: Saturation occurs when substrate concentrations are sufficiently elevated such that all catalytic sites are continuously occupied.
  • Option B: Inhibition involves chemical modifiers or ligand analogs that reversibly or irreversibly impair enzyme turnover, not non-specific thermal disruption.
  • Option D: Composition refers to the static covalent primary sequence of amino acids, which remains intact during thermal unfolding.
MCQ #7 of 92 Biology DUHS 2021
[DUHS 2021]

The smaller subunit (\(40\text{S}\)) of a ribosome combines with the larger subunit (\(60\text{S}\)) in the presence of \(\text{Mg}^{2+}\) ions to form an active particle of:
A
\(100\text{S}\)
B
\(208\text{S}\)
C
\(80\text{S}\)
D
\(70\text{S}\)
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Eukaryotic ribosomes sediment at \(80\text{S}\), assembled through the non-additive association of a \(40\text{S}\) small subunit and a \(60\text{S}\) large subunit stabilized by divalent cations.

Formula / Rule / Reaction:

$$\text{Small Subunit }(40\text{S}) + \text{Large Subunit }(60\text{S}) \xrightarrow{\text{Mg}^{2+}} \text{Eukaryotic Ribosome }(80\text{S})$$

Solution:

  • Svedberg units (\(\text{S}\)) quantify sedimentation velocity in an ultracentrifuge, which is a function of both mass and hydrodynamic surface area.


  • Because sedimentation rates are non-additive, the assembly of the eukaryotic \(40\text{S}\) and \(60\text{S}\) subunits in the presence of physiological \(\text{Mg}^{2+}\) produces an \(80\text{S}\) monosome.


Why other options are incorrect:

  • Option A: \(100\text{S}\) corresponds to a dimerized form of bacterial \(70\text{S}\) ribosomes observed under specific stress or high divalent ion concentrations.
  • Option B: \(208\text{S}\) is an arbitrary numerical distractor with no physiological or biophysical basis in ribosomology.
  • Option D: \(70\text{S}\) is the sedimentation coefficient of prokaryotic, mitochondrial, and plastid ribosomes assembled from \(30\text{S}\) and \(50\text{S}\) subunits.
MCQ #8 of 92 Biology DUHS 2021
[DUHS 2021]

The natural bacterial flora is beneficial to humans because it:
A
Interferes with pathogenic colonization
B
Develops resistance against antigens
C
Develops resistance against physical agents
D
Produces antibiotics against pathogens
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Commensal indigenous microflora protects host epithelia through colonization resistance, preventing the adherence, establishment, and proliferation of exogenous pathogens.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Normal microbiota occupies host epithelial surface receptors, outcompeting potential pathogens for cellular binding sites.


  • Commensals also consume accessible luminal nutrients and secrete metabolic byproducts (such as short-chain fatty acids) that lower microenvironmental pH, suppressing pathogenic invasion.


Why other options are incorrect:

  • Option B: Antigens stimulate adaptive host immune defenses; bacteria do not confer physiological resistance against antigenic molecules.
  • Option C: Commensal bacteria provide no protective barrier or resistance mechanisms against physical insults such as mechanical trauma or thermal injury.
  • Option D: Although select species synthesize localized bacteriocins, indigenous microflora does not produce systemic clinical antibiotics.
MCQ #9 of 92 Biology DUHS 2021
[DUHS 2021]

In bacteria, resistance against heat is provided by:
A
Granules
B
Cysts
C
Spores
D
Plasmids
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Bacterial endospores are cryptobiotic, highly dehydrated dormant structures reinforced with calcium dipicolinate and keratin-like protein coats to withstand extreme thermal stress.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Under nutrient starvation, specific genera such as Bacillus and Clostridium initiate sporulation to generate endospores.


  • The endospore core contains high concentrations of dipicolinic acid complexed with calcium ions (calcium dipicolinate) alongside small acid-soluble proteins (SASPs) that protect DNA from heat-induced denaturation.


Why other options are incorrect:

  • Option A: Granules are non-membranous cytoplasmic inclusions used exclusively to store glycogen, polyphosphate, or sulfur.
  • Option B: Bacterial cysts (e.g., in Azotobacter) are dormant cells with thickened coats that resist desiccation, but they remain sensitive to moist heat.
  • Option D: Plasmids are autonomous circular extrachromosomal DNA elements that mediate genetic transfer and antibiotic resistance, not physical heat tolerance.
MCQ #10 of 92 Biology DUHS 2021
[DUHS 2021]

These hormones perform antagonistic functions EXCEPT:
A
Growth hormone and somatostatin
B
Calcitonin and parathormone
C
Adrenaline and noradrenaline
D
Insulin and glucagon
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Antagonistic hormone pairs exert diametrically opposite physiological effects on homeostatic parameters; adrenaline and noradrenaline act cooperatively in sympathetic signaling.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Adrenaline (epinephrine) and noradrenaline (norepinephrine) are catecholamines secreted by the adrenal medulla that act synergistically to execute the acute fight-or-flight response.


  • Because they reinforce rather than counterbalance each other during sympathetic stimulation, they are not antagonistic.


Why other options are incorrect:

  • Option A: Growth hormone promotes somatic development, whereas somatostatin acts as growth hormone-inhibiting hormone (GHIH) to block its release.
  • Option B: Calcitonin lowers circulating blood calcium by stimulating bone deposition, whereas parathormone (PTH) elevates blood calcium via osteolysis.
  • Option D: Insulin decreases blood glucose by stimulating cellular uptake and glycogenesis, whereas glucagon elevates blood glucose through glycogenolysis.
MCQ #11 of 92 Biology DUHS 2021
[DUHS 2021]

Which organelle contains hydrolytic enzymes?
A
Endoplasmic reticulum
B
Ribosomes
C
Lysosomes
D
Cell wall
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Lysosomes are single-membrane eukaryotic organelles packed with acidic hydrolases responsible for intracellular degradation and autophagy.

Formula / Rule / Reaction:

$$\text{Biopolymer} + \text{H}_2\text{O} \xrightarrow{\text{Acid Hydrolases (pH }\approx\text{ 4.5 to 5.0)}} \text{Monomers}$$

Solution:

  • Lysosomes maintain an acidic internal luminal pH (approximately 4.5 to 5.0) via vacuolar \(\text{H}^+\)-ATPase proton pumps.


  • This internal compartment segregates more than forty distinct hydrolytic enzymes (including proteases, nucleases, lipases, and glycosidases) to safely degrade endocytosed material and cellular debris.


Why other options are incorrect:

  • Option A: The endoplasmic reticulum coordinates protein translation, folding, lipid assembly, and calcium storage, but does not sequester broad acid hydrolases.
  • Option B: Ribosomes are non-membranous ribonucleoprotein complexes that catalyze peptide bond formation during translation.
  • Option D: The cell wall is a non-living extracellular exoskeleton of structural polysaccharides outside the plasma membrane.
MCQ #12 of 92 Biology DUHS 2021
[DUHS 2021]

The number and sequence of amino acids in a polypeptide chain is determined by the:
A
Primary structure
B
Secondary structure
C
Tertiary structure
D
Quaternary structure
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The primary structure of a protein is the specific linear sequence and total count of amino acids linked together by covalent peptide bonds.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The linear sequence of amino acids in a polypeptide is genetically dictated by the triplet codon sequence of mRNA transcribed from DNA.


  • This defined sequence constitutes the primary structure, which encodes all thermodynamic information necessary for downstream spatial folding.


Why other options are incorrect:

  • Option B: Secondary structure refers to localized repeating configurations stabilized by peptide backbone hydrogen bonding, such as alpha-helices and beta-pleated sheets.
  • Option C: Tertiary structure denotes the overall three-dimensional folding of a single polypeptide stabilized by side-chain hydrophobic interactions, hydrogen bonds, salt bridges, and disulfide links.
  • Option D: Quaternary structure describes the multimeric assembly and spatial arrangement of multiple distinct polypeptide subunits.
MCQ #13 of 92 Biology DUHS 2021
[DUHS 2021]

Carbohydrates are organic compounds containing:
A
Carbon, nitrogen, and potassium
B
Carbon, hydrogen, and chlorine
C
Carbon, nitrogen, and oxygen
D
Carbon, hydrogen, and oxygen
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones composed solely of carbon, hydrogen, and oxygen, generally following the stoichiometric ratio of water.

Formula / Rule / Reaction:

$$\text{C}_x(\text{H}_2\text{O})_y \quad \text{or} \quad \text{C}_n\text{H}_{2n}\text{O}_n$$

Solution:

  • Carbohydrates literally designate "hydrates of carbon" because their empirical composition comprises carbon together with hydrogen and oxygen in a 2:1 atomic ratio.


  • Fundamental monosaccharides (such as glucose, \(\text{C}_6\text{H}_{12}\text{O}_6\)) and polysaccharides (such as starch and cellulose) consist strictly of carbon, hydrogen, and oxygen atoms.


Why other options are incorrect:

  • Option A: Nitrogen and potassium are not native constituents of basic carbohydrates; potassium functions as an inorganic electrolyte.
  • Option B: Chlorine is a halogen found as an extracellular anion (chloride), not within carbohydrate structures.
  • Option C: Nitrogen is an elemental hallmark of proteins, amino acids, and nucleic acids, but is absent in standard carbohydrates.
MCQ #14 of 92 Biology DUHS 2021
[DUHS 2021]

Most of the RNA in a cell is present in the:
A
Cell membrane
B
Cytoplasm
C
Nucleolus
D
Lysosomes
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The vast majority of cellular RNA resides within the cytoplasm, primarily distributed as ribosomal RNA (rRNA) in assembled ribosomes and transfer RNA (tRNA).

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Ribosomal RNA accounts for roughly 80% of total cellular RNA, while transfer RNA represents another 10% to 15%.


  • Because functional ribosomes and aminoacyl-tRNAs are localized in the cytosol for translation, up to 90% of a cell's steady-state RNA pool is situated in the cytoplasm.


  • Note on board errata: Past paper answer keys occasionally miskey the nucleolus because it exhibits the highest localized rate of transcription per unit volume; however, standard textbooks confirm that steady-state cellular abundance resides overwhelmingly in the cytoplasm.


Why other options are incorrect:

  • Option A: The cell membrane is a lipid-protein boundary and contains no functional pool of RNA.
  • Option C: The nucleolus is the site of rRNA gene transcription and preribosomal assembly, harboring approximately 5% to 10% of total cellular RNA at any given instant.
  • Option D: Lysosomes contain hydrolytic ribonucleases that degrade endocytosed RNA rather than storing cellular RNA.
MCQ #15 of 92 Biology DUHS 2021
[DUHS 2021]

Turgidity in plant cells is primarily maintained by the:
A
Cell wall pressure
B
Internal pressure
C
Vacuole
D
Osmotic pressure
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The plant central vacuole accumulates inorganic ions and metabolites via tonoplast active transport, creating a hypertonic lumen that drives osmotic water entry to generate turgor pressure.

Formula / Rule / Reaction:

$$\Psi_w = \Psi_s + \Psi_p$$

Solution:

  • Transporters in the tonoplast pump mineral ions into the central vacuolar sap against concentration gradients, significantly depressing the vacuolar solute potential (\(\Psi_s\)).


  • Water flows osmotically into the vacuole down the water potential gradient, expanding the vacuolar volume and driving turgor pressure against the cell wall to support soft tissues.


Why other options are incorrect:

  • Option A: Wall pressure is the opposing mechanical counter-force exerted by the rigid cellulose wall against expanding protoplasm, not the active generating organelle.
  • Option B: Internal pressure is an unspecific mechanical term rather than the specific subcellular organelle maintaining hydrostatic support.
  • Option D: Osmotic pressure is a physical solution property, whereas the vacuole is the biological organelle that establishes and regulates osmotic balance.
MCQ #16 of 92 Biology DUHS 2021
[DUHS 2021]

Chlorophyll pigments are present in which plastid?
A
Chloroplast
B
Chromoplast
C
Elaioplast
D
Amyloplast
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Chlorophyll pigments are embedded within the thylakoid membranes of chloroplasts to absorb light energy during the photosynthetic light reactions.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Chloroplasts are double-membrane plastids containing an internal network of flattened membranous sacs called thylakoids.


  • Hydrophobic phytol tails anchor chlorophyll molecules into thylakoid lipid bilayers, assembling them into antenna complexes and reaction centers.


Why other options are incorrect:

  • Option B: Chromoplasts synthesize and accumulate non-photosynthetic carotenoids (carotenes and xanthophylls) that impart yellow, orange, and red coloration to petals and ripening fruits.
  • Option C: Elaioplasts (sometimes spelled elioplasts) are non-pigmented leucoplasts specialized for lipid and oil storage in seeds.
  • Option D: Amyloplasts are non-pigmented leucoplasts that synthesize and store polymeric starch granules in roots and storage tubers.
MCQ #17 of 92 Biology DUHS 2021
[DUHS 2021]

Xanthophylls are pigments that appear:
A
Yellow
B
Red
C
Orange
D
Green
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Xanthophylls are oxygenated carotenoid derivatives that absorb blue and violet wavelengths of the visible spectrum and transmit yellow light.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Carotenoids are divided into two chemical groups: pure hydrocarbon carotenes and oxygenated xanthophylls.


  • Because xanthophylls (such as lutein and zeaxanthin) absorb light predominantly in the 400 to 500 nm range, they reflect and transmit light perceived as yellow.


Why other options are incorrect:

  • Option B: Red hues in plant organs typically arise from anthocyanin flavonoids stored in vacuoles or specific linear carotenes such as lycopene.
  • Option C: Orange pigmentation is characteristic of pure carotene hydrocarbons, notably beta-carotene.
  • Option D: Green pigmentation is produced by chlorophyll a and chlorophyll b absorbing blue and red wavelengths.
MCQ #18 of 92 Biology DUHS 2021
[DUHS 2021]

Which of the following organelles possesses its own circular DNA?
A
Endoplasmic reticulum
B
Lysosomes
C
Mitochondria
D
Ribosomes
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Mitochondria are semi-autonomous endosymbiotic organelles possessing closed circular double-stranded DNA (mtDNA) and independent prokaryote-like ribosomes.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Derived from an ancestral alpha-proteobacterial endosymbiont, mitochondria retain their own genetic machinery in the matrix.


  • Mitochondrial DNA encodes key polypeptide subunits of the electron transport chain, as well as mitochondrial ribosomal RNAs and tRNAs.


Why other options are incorrect:

  • Option A: The endoplasmic reticulum is a single-membrane endomembrane system lacking intrinsic nucleic acid genomes.
  • Option B: Lysosomes are simple membrane-bound vacuoles containing hydrolytic enzymes and possess no genome.
  • Option D: Ribosomes are non-membranous ribonucleoprotein particles made of rRNA and structural proteins; they do not contain DNA.
MCQ #19 of 92 Biology DUHS 2021
[DUHS 2021]

Which organelle plays a direct role in plant cell wall formation?
A
Ribosomes
B
Cell membrane
C
Plasmids
D
Golgi apparatus
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The Golgi apparatus synthesizes non-cellulosic wall polysaccharides and packages them into vesicles that assemble to establish the phragmoplast and cell plate during cytokinesis.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • During plant cell cytokinesis, Golgi-derived dictyosome vesicles migrate along microtubules to the division plane.


  • These vesicles coalesce to construct the cell plate (middle lamella), discharging matrix components including pectins and hemicellulose via exocytosis.


Why other options are incorrect:

  • Option A: Ribosomes synthesize polypeptides and cytosolic enzymes; they do not synthesize or traffic wall polysaccharides.
  • Option B: While rosette cellulose synthase complexes sit in the plasma membrane, the vesicular synthesis and transport of matrix polysaccharides is directed by the Golgi apparatus.
  • Option C: Plasmids are bacterial extrachromosomal DNA rings that play no physiological role in plant cytokinetic wall assembly.
MCQ #20 of 92 Biology DUHS 2021
[DUHS 2021]

Steroid hormone synthesis occurs predominantly in the:
A
Rough endoplasmic reticulum
B
Smooth endoplasmic reticulum
C
Golgi apparatus
D
Ribosomes
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The smooth endoplasmic reticulum houses the membrane-bound enzymes responsible for lipid metabolism, phospholipid synthesis, and steroidogenesis.

Formula / Rule / Reaction:

$$\text{Cholesterol} \xrightarrow{\text{Smooth ER and Mitochondrial Enzymes}} \text{Steroid Hormones}$$

Solution:

  • The smooth endoplasmic reticulum (SER) is devoid of surface ribosomes and specializes in lipid and carbohydrate biochemical modifications.


  • Endocrine cells of the adrenal cortex and gonads (Leydig and granulosa cells) contain extensive SER networks dedicated to the enzymatic conversion of cholesterol into steroids.


Why other options are incorrect:

  • Option A: The rough endoplasmic reticulum is studded with ribosomes and specializes in the translation, initial core glycosylation, and sorting of secretory and transmembrane proteins.
  • Option C: The Golgi apparatus modifies, sorts, and packages macromolecules received from the ER, but does not synthesize steroid rings.
  • Option D: Ribosomes are exclusively dedicated to peptide bond synthesis during mRNA translation.
MCQ #21 of 92 Biology DUHS 2021
[DUHS 2021]

The master gland of the human endocrine system is the:
A
Pituitary gland
B
Thyroid gland
C
Parathyroid gland
D
Mammary gland
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The pituitary gland (hypophysis) is referred to as the master gland because its anterior trophic hormones regulate the activity of subordinate endocrine glands throughout the body.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The anterior pituitary gland secretes tropic hormones including TSH, ACTH, FSH, and LH under hypothalamic control.


  • These hormones stimulate peripheral endocrine organs (thyroid, adrenal cortex, and gonads) to regulate metabolism, stress responses, and reproduction.


Why other options are incorrect:

  • Option B: The thyroid gland produces thyroid hormones (T3 and T4) and calcitonin; its synthesis is subordinated to pituitary TSH.
  • Option C: Parathyroid glands secrete parathyroid hormone (PTH) in response to ionized calcium levels without pituitary involvement.
  • Option D: Mammary glands are exocrine accessory reproductive structures that produce milk, not endocrine regulator glands.
MCQ #22 of 92 Biology DUHS 2021
[DUHS 2021]

The primary chemical end products of the light-dependent reactions of photosynthesis are:
A
\(\text{ADP}\) and \(\text{NADPH}\)
B
\(\text{ATP}\) and \(\text{NADPH}\)
C
\(\text{CO}_2\) and \(\text{H}_2\text{O}\)
D
\(\text{NADPH}\) and \(\text{AMP}\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The light reactions of photosynthesis capture solar photons within thylakoid membranes to generate chemical assimilatory power in the form of ATP and NADPH.

Formula / Rule / Reaction:

$$2\text{H}_2\text{O} + 2\text{NADP}^+ + 3\text{ADP} + 3\text{P}_i + h\nu \longrightarrow \text{O}_2 + 2\text{NADPH} + 2\text{H}^+ + 3\text{ATP}$$

Solution:

  • Photolysis of water provides electrons to photosystems, while light-driven electron transport reduces \(\text{NADP}^+\) to \(\text{NADPH}\) via ferredoxin-\(\text{NADP}^+\) reductase.


  • Concurrently, chemiosmotic proton translocation across the thylakoid membrane drives ATP synthase to photophosphorylate \(\text{ADP}\) into \(\text{ATP}\). Both products feed into the light-independent Calvin cycle.


Why other options are incorrect:

  • Option A: \(\text{ADP}\) is a low-energy substrate consumed during photophosphorylation rather than a final energy-storing product.
  • Option C: \(\text{CO}_2\) is the inorganic carbon substrate reduced during the Calvin cycle, and \(\text{H}_2\text{O}\) is consumed during photolysis.
  • Option D: \(\text{AMP}\) is not generated by the light-dependent photophosphorylation cascade; the nucleotide produced is ATP.
MCQ #23 of 92 Biology DUHS 2021
[DUHS 2021]

Hydrolysis is defined as the:
A
Combining of molecules due to reaction with hydrogen
B
Combining of molecules due to reaction with water
C
Breaking of chemical bonds in molecules by the addition of water
D
Breaking of chemical bonds in molecules by the addition of hydrogen
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Hydrolysis is a chemical cleavage reaction in which a covalent bond is split through the addition of a water molecule, contributing a proton to one fragment and a hydroxyl group to the other.

Formula / Rule / Reaction:

$$\text{R}_1\text{-}\text{R}_2 + \text{H}_2\text{O} \longrightarrow \text{R}_1\text{-}\text{OH} + \text{R}_2\text{-}\text{H}$$

Solution:

  • The term derives from Greek hydro (water) and lysis (loosening or breaking).


  • During cellular catabolism, hydrolytic enzymes use water molecules to cleave ester, peptide, and glycosidic bonds, breaking biopolymers down into their constituent monomers.


Why other options are incorrect:

  • Option A: Combining molecules with hydrogen represents a catalytic hydrogenation or reduction reaction, not hydrolysis.
  • Option B: Combining monomeric molecules with the concurrent elimination of water is known as condensation or dehydration synthesis.
  • Option D: Cleaving covalent chemical bonds via molecular hydrogen is termed hydrogenolysis, not hydrolysis.
MCQ #24 of 92 Biology DUHS 2021
[DUHS 2021]

Which of the following is a structural protein?
A
Pepsinogen
B
Casein
C
Enzyme
D
Collagen
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Structural proteins provide mechanical support, physical framework, and tensile strength to biological tissues and cellular structures.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Collagen is an insoluble fibrous protein organized into a triple-helix quaternary conformation.


  • It serves as the principal structural component of extracellular matrices, connective tissues, tendons, ligaments, cartilage, and bone matrix.


Why other options are incorrect:

  • Option A: Pepsinogen is an inactive zymogen precursor of the digestive enzyme pepsin, functioning catalytically rather than structurally.
  • Option B: Casein is a storage phosphoprotein present in mammalian milk that functions as a nutrient source of amino acids and minerals.
  • Option C: Enzyme is a broad functional classification for biological catalysts that accelerate metabolic biochemical reactions.
MCQ #25 of 92 Biology DUHS 2021
[DUHS 2021]

Transmission of impulses is associated with which of the following organelles?
A
Smooth endoplasmic reticulum
B
Rough endoplasmic reticulum
C
Golgi apparatus
D
Ribosomes
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

In excitable muscle cells, the specialized smooth endoplasmic reticulum (sarcoplasmic reticulum) stores, releases, and re-sequesters calcium ions to conduct excitation-contraction impulses.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Standard provincial textbook curricula highlight transmission of impulses among the core functions of the smooth endoplasmic reticulum (SER).


  • In myocytes, the sarcoplasmic reticulum conducts membrane action potentials deep into fibers via terminal cisternae, releasing \(\text{Ca}^{2+}\) to activate troponin and cross-bridge cycling.


Why other options are incorrect:

  • Option B: Rough endoplasmic reticulum is studded with ribosomes and specializes in secretory protein synthesis, folding, and post-translational modification.
  • Option C: The Golgi apparatus functions in glycosylation, sorting, and packaging of macromolecules into secretory vesicles.
  • Option D: Ribosomes are non-membranous ribonucleoprotein complexes responsible for peptide bond formation during translation.
MCQ #26 of 92 Biology DUHS 2021
[DUHS 2021]

The normal pH of human blood is maintained between:
A
7.45 to 7.55
B
7.35 to 7.45
C
7.15 to 7.25
D
7.25 to 7.35
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Human systemic arterial blood pH is tightly buffered by homeostatic mechanisms within a narrow physiological range of 7.35 to 7.45.

Formula / Rule / Reaction:

$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)$$

Solution:

  • The carbonic acid-bicarbonate buffer pair, supported by respiratory regulation of \(\text{CO}_2\) and renal proton excretion, stabilizes extracellular fluid pH.


  • A blood pH below 7.35 defines acidemia (acidosis), while a pH above 7.45 defines alkalemia (alkalosis).


Why other options are incorrect:

  • Option A: A pH range of 7.45 to 7.55 indicates physiological alkalemia, compromising neuromuscular stability and tissue perfusion.
  • Option C: A pH range of 7.15 to 7.25 represents severe uncompensated acidemia, which can induce cardiac depression and central nervous depression.
  • Option D: A pH range of 7.25 to 7.35 reflects mild to moderate acidosis below the normal physiological baseline.
MCQ #27 of 92 Biology DUHS 2021
[DUHS 2021]

In an X-linked recessive disease, an affected woman married to a normal man will have what percentage of affected daughters?
A
0
B
0.25
C
0.5
D
1
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Female offspring inherit one paternal X chromosome and one maternal X chromosome; a phenotypically normal father always transmits a normal dominant X allele to all daughters.

Formula / Rule / Reaction:

$$X^a X^a \text{ (Affected Mother)} \times X^A Y \text{ (Normal Father)} \longrightarrow X^A X^a \text{ (Carrier Daughters)}, \; X^a Y \text{ (Affected Sons)}$$

Solution:

  • An affected female possesses two recessive mutant alleles (\(X^a X^a\)), transmitting \(X^a\) to all progeny.


  • The unaffected father possesses the dominant wild-type allele on his single X chromosome (\(X^A Y\)), transmitting \(X^A\) to 100% of his female offspring.


  • Consequently, 100% of daughters are heterozygous carriers (\(X^A X^a\)) showing normal phenotype, meaning 0% (probability 0) are affected.


Why other options are incorrect:

  • Option B: 0.25 (25%) would only occur in specific autosomal recessive crosses involving two heterozygous carrier parents.
  • Option C: 0.5 (50%) would represent the proportion of affected daughters if the father were affected (\(X^a Y\)) and the mother were a carrier (\(X^A X^a\)).
  • Option D: 1 (100%) represents the proportion of affected sons (\(X^a Y\)), not daughters.
MCQ #28 of 92 Biology DUHS 2021
[DUHS 2021]

The number of ATP used to produce one glucose molecule in the Calvin cycle is:
A
18
B
27
C
9
D
36
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Synthesis of one hexose molecule via the Calvin cycle requires six turns of the cycle to fix six molecules of carbon dioxide, consuming eighteen ATP and twelve NADPH.

Formula / Rule / Reaction:

$$6\text{CO}_2 + 18\text{ATP} + 12\text{NADPH} + 12\text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 18\text{ADP} + 18\text{P}_i + 12\text{NADP}^+$$

Solution:

  • Each single turn of the Calvin cycle fixes one \(\text{CO}_2\) molecule and requires 3 ATP (2 ATP for 3-phosphoglycerate phosphorylation and 1 ATP for RuBP regeneration) along with 2 NADPH.


  • To assemble one net 6-carbon glucose molecule (\(\text{C}_6\text{H}_{12}\text{O}_6\)), six turns are required: \(6 \times 3\text{ ATP} = 18\text{ ATP}\).


Why other options are incorrect:

  • Option B: 27 ATP is an exaggerated count with no biochemical basis in the C3 photosynthetic pathway.
  • Option C: 9 ATP represents the energetic requirement for three turns of the Calvin cycle, which yields only one net triose phosphate (G3P) rather than a complete hexose.
  • Option D: 36 ATP represents the theoretical maximum yield of ATP derived from aerobic respiration of one glucose molecule, not its photosynthetic synthesis.
MCQ #29 of 92 Biology DUHS 2021
[DUHS 2021]

Which hexose is most important from a biological point of view?
A
Mannose
B
Fructose
C
Glucose
D
Lactose
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

D-glucose is the central aldohexose monosaccharide serving as the universal respiratory substrate and primary energetic currency for cellular metabolism.

Formula / Rule / Reaction:

$$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \longrightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} + \text{Energy (ATP)}$$

Solution:

  • Glucose directly enters the universal glycolytic pathway to yield pyruvate, driving substrate-level and oxidative phosphorylation.


  • It serves as the circulating monosaccharide in blood and forms the repeating monomer of storage polymers such as glycogen and starch.


Why other options are incorrect:

  • Option A: Mannose is an epimer of glucose involved primarily in protein N-glycosylation rather than core energetic metabolism.
  • Option B: Fructose is a ketohexose that must be converted into glycolytic intermediates in the liver before complete catabolism.
  • Option D: Lactose is a disaccharide composed of glucose and galactose, not a hexose monosaccharide.
MCQ #30 of 92 Biology DUHS 2021
[DUHS 2021]

Chromosomes that are the same in males and females are called:
A
X chromosomes
B
Y chromosomes
C
Sex chromosomes
D
Autosomes
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Autosomes are non-allosomal chromosomes that occur in morphologically identical homologous pairs in both biological sexes, regulating somatic genetic traits.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Human somatic cells contain 23 pairs of chromosomes (46 total): 22 pairs of autosomes and 1 pair of sex chromosomes (allosomes).


  • Because autosomes are identical in number, banding morphology, and gene loci between males and females, they constitute the shared non-sex chromosome pool.


Why other options are incorrect:

  • Option A: The X chromosome is an allosome present as two copies in typical females (XX) but only one copy in typical males (XY).
  • Option B: The Y chromosome is restricted to biological males, encoding male-determining factors such as the SRY gene.
  • Option C: Sex chromosomes (allosomes) differ structurally and numerically between males (XY) and females (XX).
MCQ #31 of 92 Biology DUHS 2021
[DUHS 2021]

In the light reaction, water is used in:
A
Reduction
B
Oxidation
C
Photolysis
D
Oxygenation
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Photolysis is the light-mediated catalytic splitting of water molecules by the oxygen-evolving complex of photosystem II to release protons, electrons, and molecular oxygen.

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:

  • During non-cyclic photophosphorylation, photon absorption excites the reaction center \(\text{P}_{680}\), generating an electron deficit.


  • Water undergoes photolytic cleavage at the luminal manganese cluster, providing replacement electrons to reduce \(\text{P}_{680}^+\) while liberating \(\text{O}_2\) as a byproduct.


Why other options are incorrect:

  • Option A: Reduction describes the gain of electrons; water donates electrons and is oxidized, whereas \(\text{NADP}^+\) is reduced.
  • Option B: While the chemical cleavage of water is an oxidative process, the specific photochemical pathway in the thylakoid is termed photolysis.
  • Option D: Oxygenation denotes the addition of molecular oxygen to a compound without bond cleavage, which does not define water splitting.
MCQ #32 of 92 Biology DUHS 2021
[DUHS 2021]

Light reaction takes place in which part of the chloroplast?
A
Thylakoid
B
Stroma
C
Lamella
D
Membrane
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The light-dependent reactions occur within the thylakoid membranes and thylakoid lumen where light-harvesting pigment complexes, electron carriers, and ATP synthase reside.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Thylakoid membranes contain Photosystem I, Photosystem II, the cytochrome \(b_6f\) complex, and ATP synthase.


  • Photon capture, electron transport, and the generation of a transmembrane proton motive force occur across this thylakoid membrane system.


Why other options are incorrect:

  • Option B: The stroma is the fluid phase containing soluble enzymes (such as RuBisCO) where the dark reactions (Calvin cycle) take place.
  • Option C: Lamellae (stromal lamellae) are interconnecting thylakoid channels, but "thylakoid" represents the comprehensive functional site.
  • Option D: Membrane is an imprecise descriptor; the outer and inner chloroplast envelope membranes do not harbor photosynthetic complexes.
MCQ #33 of 92 Biology DUHS 2021
[DUHS 2021]

The outermost living boundary of the cell is:
A
Cell wall
B
Cell membrane
C
Cytoplasm
D
Nuclear membrane
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The cell membrane (plasma membrane) is the outermost living, dynamic, semipermeable protoplasmic boundary of all biological cells.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The plasma membrane consists of an amphipathic phospholipid bilayer embedded with functional transport proteins and signaling receptors.


  • In plant, fungal, and bacterial cells, an extracellular cell wall lies exterior to the plasma membrane; however, the cell wall is an inert, non-living secretion.


Why other options are incorrect:

  • Option A: The cell wall is a rigid, non-living mechanical layer composed of cellulose, hemicellulose, and pectin.
  • Option C: Cytoplasm is the internal colloidal substance enclosed within the plasma membrane, not the cellular boundary.
  • Option D: The nuclear membrane is an internal double-membrane envelope that isolates the nucleus within the cytoplasm.
MCQ #34 of 92 Biology DUHS 2021
[DUHS 2021]

The sensation of pain is produced by:
A
Thermoreceptor
B
Photoreceptor
C
Chemoreceptor
D
Nociceptors
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Nociceptors are specialized sensory receptors comprising free nerve endings that respond selectively to noxious tissue-damaging stimuli to elicit pain.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Nociceptors are widely distributed in the skin, articular joints, periosteum, and visceral walls.


  • They detect mechanical injury, extreme thermal changes, and chemical algogens (such as bradykinin, prostaglandins, and \(\text{H}^+\)), transmitting afferent pain impulses via A-delta and C fibers.


Why other options are incorrect:

  • Option A: Thermoreceptors detect non-noxious changes in temperature (warmth and cold).
  • Option B: Photoreceptors (rods and cones) transduce electromagnetic radiation in the retina for visual perception.
  • Option C: Chemoreceptors monitor chemical concentrations such as partial pressures of arterial blood gases (\(\text{pO}_2\), \(\text{pCO}_2\)), pH, and taste/olfactory ligands.
MCQ #35 of 92 Biology DUHS 2021
[DUHS 2021]

The resting membrane potential is:
A
-60 mV
B
-70 mV
C
-50 mV
D
-40 mV
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The resting membrane potential of a typical mammalian neuron is stabilized at approximately \(-70\text{ mV}\), maintained by potassium leak channels and electrogenic sodium-potassium pumps.

Formula / Rule / Reaction:

$$V_m = \frac{RT}{F} \ln\left( \frac{P_{\text{K}}[\text{K}^+]_{\text{out}} + P_{\text{Na}}[\text{Na}^+]_{\text{out}} + P_{\text{Cl}}[\text{Cl}^-]_{\text{in}}}{P_{\text{K}}[\text{K}^+]_{\text{in}} + P_{\text{Na}}[\text{Na}^+]_{\text{in}} + P_{\text{Cl}}[\text{Cl}^-]_{\text{out}}} \right)$$

Solution:

  • Neuronal resting membranes exhibit high permeability to potassium (\(\text{K}^+\)) via nongated leak channels, allowing \(\text{K}^+\) to diffuse out down its chemical gradient.


  • The \(\text{Na}^+/\text{K}^+\)-ATPase actively expels three \(\text{Na}^+\) ions for every two \(\text{K}^+\) ions imported, establishing a stable interior-negative potential of \(-70\text{ mV}\).


Why other options are incorrect:

  • Option A: \(-60\text{ mV}\) is close to the threshold potential for firing action potentials in some specialized cardiac or smooth muscle tissues.
  • Option C: \(-50\text{ mV}\) represents the typical neuronal threshold potential triggering the opening of voltage-gated \(\text{Na}^+\) channels.
  • Option D: \(-40\text{ mV}\) corresponds to the depolarized resting state of non-excitable cells or cardiac pacemaker nodal tissue.
MCQ #36 of 92 Biology DUHS 2021
[DUHS 2021]

Which neurotransmitter is present in the peripheral nervous system?
A
Epinephrine
B
Acetylcholine
C
Dopamine
D
Serotonin
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Acetylcholine is the primary classical neurotransmitter operating throughout the somatic and autonomic branches of the peripheral nervous system.

Formula / Rule / Reaction:

$$\text{Choline} + \text{Acetyl-CoA} \xrightarrow{\text{Choline Acetyltransferase}} \text{Acetylcholine} + \text{CoA}$$

Solution:

  • Acetylcholine is released by all somatic motor neurons at skeletal neuromuscular junctions.


  • It is also the neurotransmitter released by all preganglionic autonomic fibers, postganglionic parasympathetic fibers, and sympathetic fibers innervating sweat glands.


Why other options are incorrect:

  • Option A: Epinephrine functions primarily as a circulating endocrine hormone released into systemic circulation by the adrenal medulla.
  • Option C: Dopamine functions predominantly within central nervous system motor and reward pathways (e.g., substantia nigra, striatum).
  • Option D: Serotonin acts as a major central neurotransmitter regulating mood and sleep, as well as an enteric paracrine signaling agent.
MCQ #37 of 92 Chemistry DUHS 2021
[DUHS 2021]

Two ice blocks stick together due to:
A
London dispersion forces
B
Dipole-dipole forces
C
Hydrogen bonding
D
All of the above options are correct
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

When two ice blocks are pressed together, interfacial pressure melting followed by relief of pressure (regelation) enables water molecules to reform an integrated crystalline lattice via hydrogen bonding.

Formula / Rule / Reaction:

$$\text{O}-\text{H}\cdots\text{O}$$

Solution:

  • Increased pressure at the contact interface lowers the local melting point of ice, producing a micro-film of liquid water.


  • When the external pressure is released, the melting point reverts to \(0^\circ\text{C}\), causing the water film to freeze immediately into a continuous structure knit by hydrogen bonds.


Why other options are incorrect:

  • Option A: London dispersion forces are weak induced dipole-induced dipole interactions present in all molecules, but are insufficient to bond solid macroscopic ice blocks.
  • Option B: Standard dipole-dipole interactions are much weaker than the specific directed hydrogen bonds that govern ice cohesion.
  • Option D: Although dispersion forces exist, hydrogen bonding is the specific intermolecular force responsible for lattice cohesion and regelation.
MCQ #38 of 92 Chemistry DUHS 2021
[DUHS 2021]

The unit of rate constant is the same as for:
A
First order reaction
B
Second order reaction
C
Third order reaction
D
Zero order reaction
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

For a zero-order reaction, the reaction rate is independent of reactant concentrations, making the dimensions of the rate constant identical to those of the reaction rate.

Formula / Rule / Reaction:

$$\text{Rate} = k[\text{A}]^0 = k \implies [k] = \text{mol}\cdot\text{dm}^{-3}\cdot\text{s}^{-1}$$

Solution:

  • The general formula for the unit of a rate constant is \((\text{mol}\cdot\text{dm}^{-3})^{1-n}\cdot\text{s}^{-1}\), where \(n\) is the overall order of the reaction.


  • Substituting \(n = 0\) yields \(\text{mol}\cdot\text{dm}^{-3}\cdot\text{s}^{-1}\), which is the exact unit of the reaction rate (\(d[\text{conc}]/dt\)).


Why other options are incorrect:

  • Option A: For a first-order reaction (\(n = 1\)), the unit of \(k\) is \(\text{s}^{-1}\).
  • Option B: For a second-order reaction (\(n = 2\)), the unit of \(k\) is \(\text{mol}^{-1}\cdot\text{dm}^3\cdot\text{s}^{-1}\).
  • Option C: For a third-order reaction (\(n = 3\)), the unit of \(k\) is \(\text{mol}^{-2}\cdot\text{dm}^6\cdot\text{s}^{-1}\).
MCQ #39 of 92 Chemistry DUHS 2021
[DUHS 2021]

Reduction occurs at:
A
Anode
B
Cathode
C
Salt bridge
D
None of these options are correct
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

By electrochemical convention, the cathode is the electrode where reduction (electron gain) takes place, irrespective of whether the cell is galvanic or electrolytic.

Formula / Rule / Reaction:

$$\text{M}^{n+} + n e^- \longrightarrow \text{M} \quad (\text{Cathodic Reduction})$$

Solution:

  • Cations migrate toward the cathode where they accept electrons, undergoing reduction.


  • Conversely, oxidation (loss of electrons) occurs at the anode in all electrochemical configurations.


Why other options are incorrect:

  • Option A: The anode is the electrode where oxidation (loss of electrons) occurs.
  • Option C: The salt bridge maintains electrical neutrality between half-cells via ion diffusion; it is not an electrode and supports no redox half-reactions.
  • Option D: Option B correctly identifies the electrode.
MCQ #40 of 92 Chemistry DUHS 2021
[DUHS 2021]

Electron affinity decreases down the group due to:
A
Increase in atomic radius
B
Decrease in atomic radius
C
Increase in electronegativity
D
All of the above options are correct
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Descending a group in the periodic table adds electron shells, increasing the atomic radius and shielding effect, which weakens nuclear electrostatic attraction for an incoming electron.

Formula / Rule / Reaction:

$$\text{Electron Affinity} \propto \frac{Z_{\text{eff}}}{r}$$

Solution:

  • As the principal quantum number increases down a group, the distance between the nucleus and the valence shell expands.


  • The shielding effect of intervening core electrons offsets the increase in nuclear charge, resulting in a diminished effective nuclear grip on an added electron and decreasing electron affinity.


Why other options are incorrect:

  • Option B: Atomic radius increases, rather than decreases, as new electronic shells are added down a group.
  • Option C: Electronegativity decreases down a group due to increased bond lengths and shielding.
  • Option D: Only Option A is factually correct.
MCQ #41 of 92 Chemistry DUHS 2021
[DUHS 2021]

Which one is least ionic?
A
\(\text{NaCl}\)
B
\(\text{AlCl}_3\)
C
\(\text{MgCl}_2\)
D
\(\text{KCl}\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

According to Fajan's rules, cations with higher charge density exert greater polarizing power on the electron cloud of an anion, introducing significant covalent character.

Formula / Rule / Reaction:

$$\text{Polarizing Power} \propto \frac{\text{Ionic Charge}}{\text{Ionic Radius}}$$

Solution:

  • The aluminum cation in \(\text{AlCl}_3\) carries a high charge (\(+3\)) and a small ionic radius, giving it the highest charge density among the group.


  • This strong polarizing power distorts the electron cloud of the chloride anions, imparting covalent character and making \(\text{AlCl}_3\) the least ionic (most covalent) chloride listed.


Why other options are incorrect:

  • Option A: \(\text{NaCl}\) contains the univalent \(\text{Na}^+\) ion with lower charge density, exhibiting pronounced ionic character.
  • Option C: \(\text{MgCl}_2\) contains the divalent \(\text{Mg}^{2+}\) ion, which is less polarizing than trivalent \(\text{Al}^{3+}\).
  • Option D: \(\text{KCl}\) contains the large \(\text{K}^+\) ion with very low charge density, making it the most ionic compound in this set.
MCQ #42 of 92 Chemistry DUHS 2021
[DUHS 2021]

Diamond is:
A
Covalent solid
B
Molecular solid
C
Metallic solid
D
Ionic solid
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Diamond is a giant network covalent crystal consisting of carbon atoms bonded to four adjacent carbon atoms via localized \(sp^3\) tetrahedral covalent bonds.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • In diamond, each carbon atom is covalently linked to four other carbons in an infinite three-dimensional tetrahedral lattice.


  • Because the entire crystal is held together by continuous, strong covalent bonds rather than intermolecular forces, diamond is classified as a macromolecular (network) covalent solid.


Why other options are incorrect:

  • Option B: Molecular solids (such as dry ice or iodine) consist of discrete molecules held together by weak van der Waals forces.
  • Option C: Metallic solids are arrays of metal cations enveloped in a sea of delocalized valence electrons.
  • Option D: Ionic solids (such as sodium chloride) consist of alternating positive and negative ions held by electrostatic lattice forces.
MCQ #43 of 92 Chemistry DUHS 2021
[DUHS 2021]

What is the most common gas in the environment?
A
Oxygen
B
Nitrogen
C
Carbon dioxide
D
Helium
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Molecular nitrogen (\(\text{N}_2\)) is the most abundant chemical species in Earth's atmosphere by both volume and mass.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Dry atmospheric air comprises approximately 78.08% nitrogen (\(\text{N}_2\)) by volume.


  • Oxygen makes up roughly 20.95%, argon comprises 0.93%, and carbon dioxide accounts for roughly 0.04%.


Why other options are incorrect:

  • Option A: Oxygen is the second most abundant atmospheric gas at approximately 21% by volume.
  • Option C: Carbon dioxide is a trace atmospheric gas constituting approximately 0.04% (420 ppm) of dry air.
  • Option D: Helium is a noble gas present in trace quantities (around 5.2 ppm).
MCQ #44 of 92 Chemistry DUHS 2021
[DUHS 2021]

Which one of the following has greater bond energy?
A
\(\text{HF}\)
B
\(\text{HCl}\)
C
\(\text{HBr}\)
D
\(\text{HI}\)
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Bond dissociation energy is inversely proportional to internuclear bond length; smaller bonded atoms produce shorter, stronger bonds due to maximal orbital overlap.

Formula / Rule / Reaction:

$$\text{Bond Dissociation Energies: } \text{HF }(568\text{ kJ}\cdot\text{mol}^{-1}) > \text{HCl }(431) > \text{HBr }(366) > \text{HI }(299)$$

Solution:

  • Fluorine has the smallest atomic radius among the halogens, resulting in the shortest internuclear bond length for the \(\text{H}-\text{F}\) covalent bond.


  • The substantial electronegativity difference creates significant ionic resonance stabilization, giving \(\text{HF}\) the highest bond dissociation energy (\(568\text{ kJ}\cdot\text{mol}^{-1}\)).


Why other options are incorrect:

  • Option B: The \(\text{H}-\text{Cl}\) bond has a longer internuclear distance, yielding an intermediate bond energy of \(431\text{ kJ}\cdot\text{mol}^{-1}\).
  • Option C: The \(\text{H}-\text{Br}\) bond has a bond energy of \(366\text{ kJ}\cdot\text{mol}^{-1}\).
  • Option D: The \(\text{H}-\text{I}\) bond is the longest and weakest among the hydrogen halides, with a bond energy of \(299\text{ kJ}\cdot\text{mol}^{-1}\).
MCQ #45 of 92 Chemistry DUHS 2021
[DUHS 2021]

The oxidation number of an element in free state is:
A
1
B
2
C
0
D
-1
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

By definition and convention in redox chemistry, the oxidation state of any atom in its uncombined elemental state is zero.

Formula / Rule / Reaction:

$$\text{Oxidation Number of Uncombined Element} = 0$$

Solution:

  • An element in its free or elemental form (e.g., \(\text{Na}\), \(\text{Fe}\), \(\text{H}_2\), \(\text{O}_2\), \(\text{P}_4\)) has not transferred or unevenly shared electrons with atoms of different electronegativity.


  • Because there is no net redistribution of electron density relative to the neutral atom, its oxidation number is zero.


Why other options are incorrect:

  • Option A: \(+1\) is the characteristic oxidation state of alkali metals in chemical compounds.
  • Option B: \(+2\) is the characteristic oxidation state of alkaline earth metals in chemical compounds.
  • Option D: \(-1\) is the typical oxidation state of halogens in halides.
MCQ #46 of 92 Chemistry DUHS 2021
[DUHS 2021]

Homocyclic hydrocarbons are further divided into:
A
Alicyclic and aromatic
B
Carbocyclic and heterocyclic
C
Acyclic and heterocyclic
D
Heterocyclic and aromatic
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Homocyclic (carbocyclic) compounds contain closed rings composed solely of carbon atoms, which are classified into non-aromatic alicyclic and conjugated aromatic systems.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Homocyclic (or carbocyclic) ring systems contain only ring carbons without heteroatoms.


  • They are subdivided into alicyclic compounds (which resemble aliphatic hydrocarbons in behavior, such as cyclobutane or cyclohexane) and aromatic compounds (which possess delocalized \(\pi\)-electron systems satisfying Hückel's \(4n + 2\) rule, such as benzene).


Why other options are incorrect:

  • Option B: Carbocyclic is synonymous with homocyclic, and heterocyclic compounds contain at least one heteroatom (such as N, O, or S) in the ring framework.
  • Option C: Acyclic refers to open-chain aliphatic compounds, which do not contain ring structures.
  • Option D: Heterocyclic compounds form an entirely separate primary classification alongside homocyclic compounds.
MCQ #47 of 92 Chemistry DUHS 2021
[DUHS 2021]

Which contains \(sp^2\) hybridization?
A
Methane
B
Ethane
C
Alkyne
D
None of these options are correct
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Orbital hybridization is determined by the steric number of sigma bonds and lone pairs surrounding the central carbon atom; \(sp^2\) hybridization requires three hybrid orbitals oriented at \(120^\circ\).

Formula / Rule / Reaction:

$$\text{Steric Number} = (\text{Number of } \sigma\text{ bonds}) + (\text{Number of lone pairs}) = 3 \implies sp^2$$

Solution:

  • Methane (\(\text{CH}_4\)) possesses four \(\text{C}-\text{H}\) \(\sigma\) bonds, yielding an \(sp^3\) tetrahedral geometry.


  • Ethane (\(\text{C}_2\text{H}_6\)) contains carbon atoms bonded to four other atoms each, forming \(sp^3\) hybridized tetrahedral centers.


  • Alkynes (e.g., ethyne) possess a carbon-carbon triple bond composed of one \(\sigma\) bond and two \(\pi\) bonds, resulting in \(sp\) linear hybridization. Consequently, none of the listed options contain \(sp^2\) hybridization.


Why other options are incorrect:

  • Option A: Methane has four single \(\sigma\) bonds and exhibits \(sp^3\) hybridization.
  • Option B: Both carbon atoms in ethane form four single bonds and are \(sp^3\) hybridized.
  • Option C: Alkynes possess triply bonded carbons with two \(\sigma\) bonds, which is characteristic of \(sp\) hybridization.
MCQ #48 of 92 Chemistry DUHS 2021
[DUHS 2021]

Geometric isomerism is shown by:
A
Alkanes
B
Alkenes
C
Alkynes
D
All of the above options are correct
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Geometric (cis-trans) isomerism arises from restricted rotation about a rigid carbon-carbon double bond when each unsaturated carbon atom bears two different substituent groups.

Formula / Rule / Reaction:

$$\text{Condition: } \text{R}_1\text{R}_2\text{C}=\text{C}\text{R}_3\text{R}_4 \quad (\text{where } \text{R}_1 \neq \text{R}_2 \text{ and } \text{R}_3 \neq \text{R}_4)$$

Solution:

  • The carbon-carbon double bond in alkenes consists of a strong \(\sigma\) bond and an overlapping \(\pi\) bond that prevents free rotation at room temperature without breaking the \(\pi\) overlap.


  • When each double-bonded carbon carries two non-identical ligands, distinct spatial configurations (cis and trans diastereomers) exist.


Why other options are incorrect:

  • Option A: Alkanes possess single carbon-carbon \(\sigma\) bonds capable of free axial rotation, producing conformational rather than geometric isomers.
  • Option C: Alkynes exhibit a linear geometry (\(180^\circ\)) about the triple bond with only one substituent per carbon, precluding cis-trans spatial distinction.
  • Option D: Because alkanes and alkynes cannot exhibit geometric isomerism, this collective option is invalid.
MCQ #49 of 92 Chemistry DUHS 2021
[DUHS 2021]

\(\text{C}_n\text{H}_{2n}\text{O}_2\) is the general molecular formula of:
A
Alcohols
B
Ketones
C
Carboxylic acids
D
Ethers
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Monocarboxylic acids and their functional group isomers, aliphatic esters, contain one carbonyl double bond and two oxygen atoms, adhering to the homologous formula \(\text{C}_n\text{H}_{2n}\text{O}_2\).

Formula / Rule / Reaction:

$$\text{Alkanoic Acid: } \text{R-COOH} \implies \text{C}_n\text{H}_{2n+1}\text{COOH} = \text{C}_{n+1}\text{H}_{2(n+1)}\text{O}_2$$

Solution:

  • A saturated open-chain monocarboxylic acid contains one carbon-oxygen double bond (one degree of unsaturation) and one single carbon-oxygen bond.


  • For example, ethanoic acid (\(\text{CH}_3\text{COOH}\)) has 2 carbons, 4 hydrogens, and 2 oxygens, precisely satisfying \(\text{C}_n\text{H}_{2n}\text{O}_2\) where \(n = 2\).


Why other options are incorrect:

  • Option A: Saturated monohydric aliphatic alcohols conform to the general formula \(\text{C}_n\text{H}_{2n+2}\text{O}\).
  • Option B: Saturated aliphatic ketones possess one oxygen atom and follow the general formula \(\text{C}_n\text{H}_{2n}\text{O}\).
  • Option D: Dialkyl ethers are functional isomers of alcohols and conform to \(\text{C}_n\text{H}_{2n+2}\text{O}\).
MCQ #50 of 92 Chemistry DUHS 2021
[DUHS 2021]

Concentrated 50% \(\text{NaOH}\) is used in:
A
Aldol condensation reaction
B
Cannizzaro reaction
C
Clemmensen reduction reaction
D
Wolff-Kishner reduction reaction
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The Cannizzaro reaction is a base-catalyzed disproportionation of aldehydes lacking alpha-hydrogens, requiring strong concentrated alkaline conditions (50% aqueous or alcoholic \(\text{NaOH}\)).

Formula / Rule / Reaction:

$$2\text{HCHO} + \text{NaOH (50%)} \longrightarrow \text{HCOONa} + \text{CH}_3\text{OH}$$

Solution:

  • Aldehydes with no \(\alpha\)-hydrogen atoms (such as formaldehyde and benzaldehyde) cannot form enolate nucleophiles.


  • When heated with 50% concentrated \(\text{NaOH}\), one molecule undergoes reduction to an alcohol while another molecule undergoes simultaneous oxidation to a carboxylate salt via intermolecular hydride shift.


Why other options are incorrect:

  • Option A: Aldol condensation uses aldehydes containing \(\alpha\)-hydrogens in the presence of dilute base (such as 10% \(\text{NaOH}\)).
  • Option C: Clemmensen reduction utilizes zinc amalgam (\(\text{Zn[Hg]}\)) in concentrated hydrochloric acid (\(\text{HCl}\)), an acidic medium.
  • Option D: Wolff-Kishner reduction employs hydrazine (\(\text{NH}_2\text{NH}_2\)) with \(\text{KOH}\) in high-boiling ethylene glycol at elevated temperatures.
MCQ #51 of 92 Chemistry DUHS 2021
[DUHS 2021]

Carboxylic acids are soluble in water due to:
A
Dipole-dipole forces
B
London dispersion forces
C
Hydrogen bonding
D
Covalent bonding
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Low molecular weight carboxylic acids dissolve in water because the highly polarized hydroxyl and carbonyl groups of the carboxyl moiety form intermolecular hydrogen bonds with water molecules.

Formula / Rule / Reaction:

$$\text{R-C}(=\text{O})\text{---OH}\cdots\text{OH}_2 \quad \text{and} \quad \text{R-C}(=\text{O}\cdots\text{H-OH})\text{OH}$$

Solution:

  • The carboxyl group (\(-\text{COOH}\)) contains both a hydrogen-bond donor (the polarized \(-\text{OH}\) proton) and hydrogen-bond acceptors (carbonyl and hydroxyl oxygens).


  • These sites engage in extensive hydrogen-bonding networks with solvent water molecules, overcoming solute-solute cohesive forces in lower aliphatic acids (up to 4 carbons).


Why other options are incorrect:

  • Option A: Dipole-dipole forces contribute to overall polarity, but the high aqueous solubility of carboxylic acids is governed by hydrogen bonding.
  • Option B: London dispersion forces are weak non-specific attractive forces dominant between non-polar alkyl chains, which oppose water solubility.
  • Option D: Dissolution is a physical solvation phenomenon governed by non-covalent interactions, not the formation of solute-solvent covalent bonds.
MCQ #52 of 92 Chemistry DUHS 2021
[DUHS 2021]

Esterification occurs in the presence of:
A
Aldehydes and ketones
B
Alcohols and ketones
C
Carboxylic acids and alcohols
D
Carboxylic acids and aldehydes
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Fischer esterification is an acid-catalyzed condensation reaction between a carboxylic acid and an alcohol that yields an ester and water.

Formula / Rule / Reaction:

$$\text{R-COOH} + \text{R'--OH} \rightleftharpoons^+} \text{R-COOR'} + \text{H}_2\text{O}$$

Solution:

  • Protonation of the carboxylic acid carbonyl oxygen by a strong mineral acid catalyst (such as concentrated \(\text{H}_2\text{SO}_4\)) activates the carbonyl carbon toward nucleophilic attack.


  • The alcohol attacks the activated carbonyl carbon to form a tetrahedral intermediate, which eliminates water to produce the ester.


Why other options are incorrect:

  • Option A: Combining aldehydes and ketones under basic conditions yields mixed aldol condensation products, not esters.
  • Option B: Alcohols react with ketones under acidic conditions to form hemiketals and ketals, not esters.
  • Option D: Carboxylic acids do not condense with aldehydes to generate ester linkages.
MCQ #53 of 92 Chemistry DUHS 2021
[DUHS 2021]

Benzene can be prepared from sodium benzoate by reaction with:
A
\(\text{AlCl}_3\)
B
\(\text{NaOH}\)
C
Chlorobenzene
D
Cumene
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Decarboxylation of the sodium salt of an aromatic carboxylic acid by heating with soda lime (a dry mixture of \(\text{NaOH}\) and \(\text{CaO}\)) eliminates the carboxylate group to yield benzene.

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:

  • Heating sodium benzoate with soda lime promotes nucleophilic attack by hydroxide on the carbonyl carbon.


  • This causes heterolytic cleavage of the phenyl-carbonyl bond, releasing carbon dioxide as sodium carbonate (\(\text{Na}_2\text{CO}_3\)) and protonating the phenyl carbanion to yield benzene.


Why other options are incorrect:

  • Option A: Anhydrous \(\text{AlCl}_3\) is a Lewis acid catalyst used in electrophilic aromatic substitutions (Friedel-Crafts reactions), not decarboxylation.
  • Option C: Chlorobenzene is an aryl halide that does not induce decarboxylation in benzoate salts.
  • Option D: Cumene (isopropylbenzene) is an alkylbenzene intermediate used industrially for phenol synthesis, not a decarboxylating reagent.
MCQ #54 of 92 Chemistry DUHS 2021
[DUHS 2021]

Empirical formula is related to molecular formula by:
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
\(\text{Molecular formula} = n + (\text{Molecular formula})\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The molecular formula of a compound is a whole-number integer multiple (\(n\)) of its simplest integer-ratio empirical formula.

Formula / Rule / Reaction:

$$\text{Molecular Formula} = n \times (\text{Empirical Formula}), \quad \text{where } n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}$$

Solution:

  • The empirical formula represents the relative simplest ratio of atoms present in a chemical compound.


  • The molecular formula indicates the exact number of atoms of each element in one discrete molecule, obtained by multiplying each empirical subscript by the integer factor \(n\).


Why other options are incorrect:

  • Option A: Dividing the empirical formula by \(n\) would yield fractional subscripts, which is algebraically inverted and chemically meaningless.
  • Option C: Introducing an extra factor of 2 doubles the true stoichiometric molecular mass.
  • Option D: Chemical formulas cannot be summed additively with scalar multipliers; this expression is mathematically unsound.
MCQ #55 of 92 Chemistry DUHS 2021
[DUHS 2021]

Which one is more reactive among halogens?
A
\(\text{F}_2\)
B
\(\text{Cl}_2\)
C
\(\text{Br}_2\)
D
\(\text{I}_2\)
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Elemental fluorine (\(\text{F}_2\)) is the most chemically reactive halogen due to its exceptionally low bond dissociation energy, high electronegativity, and high standard reduction potential.

Formula / Rule / Reaction:

$$\text{Reactivity Order: } \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \quad (E^\circ_{\text{red}} \text{ of } \text{F}_2 = +2.87\text{ V})$$

Solution:

  • The \(\text{F}-\text{F}\) covalent bond is unusually weak (\(158\text{ kJ}\cdot\text{mol}^{-1}\)) due to electrostatic repulsion between the non-bonding lone pairs crowded within small fluorine atoms.


  • Combined with its high electron affinity and high hydration energy of the resulting fluoride ion, fluorine reacts rapidly and exothermically with nearly all elements.


Why other options are incorrect:

  • Option B: Chlorine has a higher bond dissociation energy (\(242\text{ kJ}\cdot\text{mol}^{-1}\)) and a lower reduction potential (\(+1.36\text{ V}\)), making it less reactive than fluorine.
  • Option C: Bromine is a liquid with lower oxidizing ability and electronegativity than chlorine and fluorine.
  • Option D: Iodine has the largest atomic radius, weakest oxidizing capacity, and lowest reactivity among stable halogens.
MCQ #56 of 92 Chemistry DUHS 2021
[DUHS 2021]

One calorie is equal to:
A
\(4.18\text{ kJ}\)
B
\(4.18\text{ J}\)
C
\(18\text{ J}\)
D
\(18\text{ kJ}\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The thermochemical calorie is defined as the amount of thermal energy required to raise the temperature of one gram of pure water by one degree Celsius, equivalent to 4.184 joules.

Formula / Rule / Reaction:

$$1\text{ cal} = 4.184\text{ J} \approx 4.18\text{ J}$$

Solution:

  • The joule (\(\text{J}\)) is the SI unit of energy, while the calorie is a historical metric unit of thermal capacity.


  • By international agreement, \(1\text{ calorie}\) corresponds to \(4.184\text{ joules}\), commonly rounded in provincial textbook problem sets to \(4.18\text{ J}\).


Why other options are incorrect:

  • Option A: \(4.18\text{ kJ}\) represents 1000 calories (one kilocalorie or nutritional Calorie), three orders of magnitude larger than a calorie.
  • Option C: \(18\text{ J}\) is a distractor derived from numeric confusion with the decimal fraction in 4.18.
  • Option D: \(18\text{ kJ}\) has no basis in metric energy conversion.
MCQ #57 of 92 Chemistry DUHS 2021
[DUHS 2021]

The net enthalpy change in a chemical reaction is the same whether it is brought about in two or more different ways in one or several steps. This is known as:
A
Joule's law
B
Law of conservation of energy
C
Hess's law
D
Henry's law
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Hess's Law of Constant Heat Summation states that the total enthalpy change for a chemical reaction is independent of the pathway taken, depending exclusively on initial and final thermodynamic states.

Formula / Rule / Reaction:

$$\Delta H_{\text{net}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots = \sum \Delta H_{\text{steps}}$$

Solution:

  • Enthalpy (\(H\)) is a thermodynamic state function; its change (\(\Delta H\)) depends only on the state of the reactants and products, not on the reaction mechanism.


  • Consequently, the net heat absorbed or evolved across a multi-step reaction equals the algebraic sum of the enthalpy changes of the individual intermediate steps.


Why other options are incorrect:

  • Option A: Joule's law defines the rate at which electrical resistance dissipates energy as heat (\(P = I^2 R\)) or the invariance of ideal gas internal energy with volume.
  • Option B: The law of conservation of energy is the first law of thermodynamics, representing the overarching concept from which Hess's law is derived.
  • Option D: Henry's law states that the mass of a dissolved gas in a liquid is directly proportional to its equilibrium partial pressure.
MCQ #58 of 92 Chemistry DUHS 2021
[DUHS 2021]

Which is not true about the rate of reaction?
A
It is independent of the concentration of reactants.
B
It depends upon the temperature.
C
It does not depend upon the order of reaction.
D
All of the above options are correct.
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

According to chemical kinetics, the instantaneous rate of a chemical reaction generally depends directly on the molar concentrations of the reacting species raised to powers equal to their reaction orders.

Formula / Rule / Reaction:

$$\text{Rate} = k[\text{A}]^m[\text{B}]^n$$

Solution:

  • As reactant concentrations increase, collision frequency between reacting molecules rises, which elevates reaction velocity.


  • Therefore, stating that the reaction rate is independent of reactant concentration is false for all reactions except zero-order systems.


Why other options are incorrect:

  • Option B: This statement is true; reaction rates increase with temperature in accordance with the Arrhenius equation (\(k = A e^{-E_a/RT}\)).
  • Option C: The reaction order is a kinetic parameter, whereas concentration and temperature directly dictate the observable rate of consumption.
  • Option D: Because Option A is a false statement, this collective distractor is incorrect.
MCQ #59 of 92 Chemistry DUHS 2021
[DUHS 2021]

Which one is an s-block element?
A
B
B
K
C
Ca
D
Cumene
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Elements whose differentiating (last) valence electron enters an \(s\)-subshell are classified as \(s\)-block elements, encompassing Groups 1 (alkali metals) and 2 (alkaline earth metals).

Formula / Rule / Reaction:

$$\text{Potassium }(Z = 19): 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1 \quad ([\text{Ar}]\,4s^1)$$

Solution:

  • Potassium (\(\text{K}\), \(Z = 19\)) possesses a single valence electron in its outermost \(4s\) subshell, identifying it as an alkali metal in the \(s\)-block.


  • Note on board errata: Calcium (\(\text{Ca}\), \(Z = 20\), \([\text{Ar}]\,4s^2\)) is also an authentic \(s\)-block element. This item represents a recognized board key defect where Potassium (\(\text{K}\)) was designated as the official key while Boron is \(p\)-block and cumene is an organic compound.


Why other options are incorrect:

  • Option A: Boron (\(\text{B}\), \(Z = 5\)) has the ground-state electron configuration \(1s^2\,2s^2\,2p^1\), placing it in the \(p\)-block.
  • Option C: Calcium is also an \(s\)-block element (Group 2); however, official board grading keyed Option B.
  • Option D: Cumene (isopropylbenzene, \(\text{C}_9\text{H}_{12}\)) is an aromatic hydrocarbon molecule, not a chemical element.
MCQ #60 of 92 Chemistry DUHS 2021
[DUHS 2021]

Bond angle is \(120^\circ\) in:
A
Ethane
B
Ethyne
C
Ethene
D
Cyclopropane
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Molecules with \(sp^2\) hybridized central carbon atoms adopt trigonal planar geometries with ideal bond angles of \(120^\circ\).

Formula / Rule / Reaction:

$$\text{Ethene } (\text{H}_2\text{C}=\text{CH}_2): \quad \text{Trigonal planar geometry, bond angles } \approx 120^\circ$$

Solution:

  • In ethene (\(\text{C}_2\text{H}_4\)), each carbon atom forms three \(\sigma\) bonds (two with hydrogen and one with the adjacent carbon) using three \(sp^2\) hybrid orbitals.


  • To minimize electron repulsion according to VSEPR theory, these three coplanar orbitals are directed toward the corners of an equilateral triangle at \(120^\circ\) angles.


Why other options are incorrect:

  • Option A: Ethane possesses \(sp^3\) hybridized tetrahedral carbon atoms with bond angles of \(109.5^\circ\).
  • Option B: Ethyne contains \(sp\) hybridized linear carbons with bond angles of \(180^\circ\).
  • Option D: Cyclopropane possesses a planar three-membered ring forced into severe ring strain with internal \(\text{C}-\text{C}-\text{C}\) bond angles compressed to \(60^\circ\).
MCQ #61 of 92 Chemistry DUHS 2021
[DUHS 2021]

In a perfectly elastic collision:
A
Only momentum is conserved
B
Only total energy is conserved
C
Only kinetic energy is conserved
D
Momentum, kinetic energy, and total energy are conserved
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

A perfectly elastic collision is an interaction in an isolated system where no mechanical kinetic energy is converted into non-conservative forms such as heat, sound, or permanent deformation.

Formula / Rule / Reaction:

$$\sum \vec{p}_i = \sum \vec{p}_f, \quad \sum \text{KE}_i = \sum \text{KE}_f, \quad \text{and} \quad E_{\text{total}} = \text{constant}$$

Solution:

  • In any isolated system free from external forces, linear momentum is conserved due to Newton's third law.


  • By definition of an elastic collision, kinetic energy is fully conserved across the impact, and total energy is universally conserved in all physical processes.


Why other options are incorrect:

  • Option A: Momentum is conserved, but kinetic energy and total energy are also conserved, rendering this option incomplete.
  • Option B: Total energy is conserved, but momentum and kinetic energy are conserved as well.
  • Option C: Kinetic energy cannot be conserved without simultaneous conservation of momentum.
MCQ #62 of 92 Chemistry DUHS 2021
[DUHS 2021]

The slope of a displacement-time graph is:
A
Velocity
B
Displacement
C
Acceleration
D
Distance
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Velocity is the time rate of change of displacement; graphically, this equals the slope of the tangent to the curve on a displacement versus time plot.

Formula / Rule / Reaction:

$$\text{Slope} = \frac{\Delta d}{\Delta t} = \frac{dd}{dt} = v$$

Solution:

  • The vertical axis of the graph represents displacement (\(d\)), while the horizontal axis represents time (\(t\)).


  • The slope is calculated as the ratio of change in vertical coordinates to horizontal coordinates (\(\Delta d / \Delta t\)), which defines instantaneous velocity.


Why other options are incorrect:

  • Option B: Displacement is the coordinate value on the vertical axis, not the gradient of the curve.
  • Option C: Acceleration is the slope of a velocity-time graph (\(\Delta v / \Delta t\)), which represents the second derivative of displacement.
  • Option D: Distance is a scalar path length, whereas the slope of a scalar distance-time plot yields speed.
MCQ #63 of 92 Physics DUHS 2021
[DUHS 2021]

An object moving along a circular path of radius \(1\text{ m}\) covers a distance of \(2\text{ m}\) along the circumference. What will be the angular displacement covered by it?
A
\(\frac{1}{2}\text{ radian}\)
B
\(2\text{ radians}\)
C
\(3\text{ radians}\)
D
\(\frac{2}{3}\text{ radian}\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Angular displacement (\(\theta\)) is the ratio of arc length (\(s\)) traversed along a circle to the radius (\(r\)) of the circular orbit.

Formula / Rule / Reaction:

$$s = r \theta \implies \theta = \frac{s}{r}$$

Solution:

  • The linear path distance covered along the circumference is \(s = 2\text{ m}\).


  • The radius of the circular path is \(r = 1\text{ m}\).


  • Calculating angular displacement: $$\theta = \frac{2\text{ m}}{1\text{ m}} = 2\text{ radians}$$


Why other options are incorrect:

  • Option A: \(\frac{1}{2}\text{ radian}\) results from inverting the relationship (\(r/s\)).
  • Option C: \(3\text{ radians}\) is a spurious arithmetic distractor.
  • Option D: \(\frac{2}{3}\text{ radian}\) has no physical basis in the given orbital dimensions.
MCQ #64 of 92 Physics DUHS 2021
[DUHS 2021]

Speed of sound is \(332\text{ m/s}\) at \(0^\circ\text{C}\). What will be its value at \(10^\circ\text{C}\)?
A
\(332\text{ m/s}\)
B
\(338.1\text{ m/s}\)
C
\(332.61\text{ m/s}\)
D
\(334.1\text{ m/s}\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The velocity of sound in air increases linearly with temperature near \(0^\circ\text{C}\) by approximately \(0.61\text{ m/s}\) for every degree Celsius rise in temperature.

Formula / Rule / Reaction:

$$v_t = v_0 + 0.61 t$$

Solution:

  • Given base speed \(v_0 = 332\text{ m/s}\) at \(t = 0^\circ\text{C}\) and final temperature \(t = 10^\circ\text{C}\).


  • Calculating the temperature correction: $$\Delta v = 0.61 \times 10 = 6.1\text{ m/s}$$


  • Computing the corrected sound velocity: $$v_{10} = 332 + 6.1 = 338.1\text{ m/s}$$


Why other options are incorrect:

  • Option A: \(332\text{ m/s}\) represents the speed of sound at \(0^\circ\text{C}\) without temperature adjustment.
  • Option C: \(332.61\text{ m/s}\) corresponds to an increase for only \(1^\circ\text{C}\) rather than \(10^\circ\text{C}\).
  • Option D: \(334.1\text{ m/s}\) is an arithmetic error using an incorrect coefficient of \(0.21\text{ m/s}\cdot{}^\circ\text{C}^{-1}\).
MCQ #65 of 92 Physics DUHS 2021
[DUHS 2021]

In molar specific heat, the equation \(C_p - C_v = R\) shows that:
A
\(C_p < C_v\)
B
\(C_p = C_v\)
C
\(C_p > C_v\)
D
\(C_p = \text{Constant}\)
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Molar heat capacity at constant pressure (\(C_p\)) exceeds molar heat capacity at constant volume (\(C_v\)) because heating at constant pressure requires extra energy to perform boundary expansion work.

Formula / Rule / Reaction:

$$C_p - C_v = R \implies C_p = C_v + R$$

Solution:

  • At constant volume, all supplied heat increases the system internal energy (\(dQ_v = dU\)).


  • At constant pressure, gas expansion against surrounding pressure consumes work (\(dW = P\,dV\)), meaning extra heat must be supplied: \(dQ_p = dU + P\,dV\).


  • Because the universal gas constant is positive (\(R > 0\)), \(C_p\) is strictly greater than \(C_v\).


Why other options are incorrect:

  • Option A: \(C_p < C_v\) would require a negative gas constant (\(R < 0\)), which is unphysical.
  • Option B: \(C_p = C_v\) applies only to incompressible condensed phases where thermal expansion work is negligible.
  • Option D: Stating \(C_p = \text{Constant}\) does not express the inequality relationship shown by Mayer's equation.
MCQ #66 of 92 Physics DUHS 2021
[DUHS 2021]

The thermodynamic process during which volume of the system remains constant is:
A
Isothermal
B
Isobaric
C
Isochoric
D
Adiabatic
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

An isochoric (isovolumetric) process is a thermodynamic transformation conducted at constant volume, meaning no boundary displacement work is performed.

Formula / Rule / Reaction:

$$\Delta V = 0 \implies W = \int P \, dV = 0$$

Solution:

  • By definition, an isochoric process occurs within a rigid non-deformable boundary where \(V = \text{constant}\).


  • According to the first law of thermodynamics, since work \(W = P\,\Delta V = 0\), all heat transferred changes internal energy: \(\Delta Q = \Delta U\).


Why other options are incorrect:

  • Option A: An isothermal process is one where the temperature remains constant (\(\Delta T = 0\)).
  • Option B: An isobaric process is one where system pressure remains constant (\(\Delta P = 0\)).
  • Option D: An adiabatic process is one where no heat enters or leaves the system (\(Q = 0\)).
MCQ #67 of 92 Physics DUHS 2021
[DUHS 2021]

If two capacitors of capacitance \(2\;\mu\text{F}\) and \(6\;\mu\text{F}\) are connected in parallel, the equivalent capacitance of the combination will be:
A
\(8\;\mu\text{F}\)
B
\(\frac{3}{2}\;\mu\text{F}\)
C
\(4\;\mu\text{F}\)
D
\(12\;\mu\text{F}\)
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

When capacitors are connected in parallel, the total charge stored is the sum of charges across individual branches, making the equivalent capacitance equal to the algebraic sum of individual capacitances.

Formula / Rule / Reaction:

$$C_{\text{eq}} = C_1 + C_2 + C_3 + \cdots + C_n$$

Solution:

  • Given branch capacitances: \(C_1 = 2\;\mu\text{F}\) and \(C_2 = 6\;\mu\text{F}\).


  • In a parallel circuit, both capacitors experience the same potential difference \(V\).


  • Calculating equivalent capacitance: $$C_{\text{eq}} = 2\;\mu\text{F} + 6\;\mu\text{F} = 8\;\mu\text{F}$$


Why other options are incorrect:

  • Option B: \(\frac{3}{2}\;\mu\text{F}\) (\(1.5\;\mu\text{F}\)) is the equivalent capacitance if the two capacitors were connected in series: \(C_{\text{eq}} = \frac{2 \times 6}{2 + 6} = 1.5\;\mu\text{F}\).
  • Option C: \(4\;\mu\text{F}\) is the arithmetic difference between the two values, which has no circuit basis.
  • Option D: \(12\;\mu\text{F}\) is the product of the two values.
MCQ #68 of 92 Physics DUHS 2021
[DUHS 2021]

Power transfer will be maximum when:
A
\(R > r\)
B
\(R < r\)
C
\(R = r\)
D
\(R = \frac{1}{r}\)
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

According to the Maximum Power Transfer Theorem, an electric power source transfers maximum power to an external load when the load resistance equals the internal resistance of the 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 load power \(P_L\) with respect to load resistance \(R\) and equating to zero establishes the condition for maximum power dissipation.


  • When \(R = r\), internal power dissipation inside the source equals power delivered to the load, achieving maximum transfer efficiency of 50%.


Why other options are incorrect:

  • Option A: When \(R > r\), circuit current drops, reducing total power dissipated across the load.
  • Option B: When \(R < r\), internal resistance dissipates the majority of the source energy as heat within the source.
  • Option D: \(R = \frac{1}{r}\) is dimensionally inconsistent because resistance cannot equal conductance.
MCQ #69 of 92 Physics DUHS 2021
[DUHS 2021]

A wire is stretched so that its radius of cross-section becomes half. Then resistance of the wire will be:
A
\(4R\)
B
\(\frac{R}{4}\)
C
\(8R\)
D
\(16R\)
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

When a wire is stretched, its total volume remains constant; halving the cross-sectional radius reduces the area by a factor of 4 and quadruples the length, compounding resistance sixteen-fold.

Formula / Rule / Reaction:

$$R = \rho \frac{L}{A} \quad \text{and} \quad \text{Volume } V = A \times L = \pi r^2 L = \text{constant} \implies R \propto \frac{1}{r^4}$$

Solution:

  • When stretched to half radius, the new cross-sectional area becomes: $$A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{A}{4}$$


  • Because volume is conserved (\(V = A \cdot L = A' \cdot L'\)), the length increases fourfold: $$L' = 4L$$


  • Calculating new resistance: $$R' = \rho \frac{L'}{A'} = \rho \frac{4L}{A/4} = 16 \left(\rho \frac{L}{A}\right) = 16R$$


Why other options are incorrect:

  • Option A: \(4R\) accounts only for the length increase or area reduction independently, ignoring volume conservation.
  • Option B: \(\frac{R}{4}\) represents a resistance decrease, which contradicts the elongation and thinning of the wire.
  • Option C: \(8R\) is an arithmetic error confusing cubic with fourth-power geometric scaling.
MCQ #70 of 92 Physics DUHS 2021
[DUHS 2021]

When a charged particle enters a magnetic field parallel to the field lines, it:
A
Deflects towards North
B
Deflects towards South
C
Moves straight
D
Moves in a circular path
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The magnetic Lorentz force on a moving charge is proportional to the vector cross product of its velocity and the magnetic flux density; when velocity is parallel to the magnetic field, the net force is zero.

Formula / Rule / Reaction:

$$\vec{F} = q(\vec{v} \times \vec{B}) \implies F = q v B \sin\theta = q v B \sin(0^\circ) = 0$$

Solution:

  • Because the charged particle travels parallel to the magnetic flux lines, the angle of entry is \(\theta = 0^\circ\).


  • Substituting \(\sin(0^\circ) = 0\) yields zero magnetic Lorentz force, meaning no transverse acceleration acts on the particle.


  • By Newton's first law, the particle continues along its original linear trajectory at constant velocity without deflection.


Why other options are incorrect:

  • Option A: Deflection northward requires an unbalanced perpendicular Lorentz force vector.
  • Option B: Deflection southward similarly requires a non-zero transverse magnetic deflecting force.
  • Option D: Circular motion requires a continuous centripetal force, which occurs only when entry is perpendicular (\(\theta = 90^\circ\)).
MCQ #71 of 92 Physics DUHS 2021
[DUHS 2021]

Magnetic flux is maximum when the angle between the magnetic field and the vector area is:
A
\(0^\circ\)
B
\(45^\circ\)
C
\(90^\circ\)
D
\(180^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Magnetic flux measures the total number of magnetic field lines penetrating a surface, quantified as the scalar dot product of the magnetic field vector and the area vector.

Formula / Rule / Reaction:

$$\Phi_B = \vec{B} \cdot \vec{A} = B A \cos\theta$$

Solution:

  • The vector area \(\vec{A}\) is defined normal (perpendicular) to the planar surface.


  • When the magnetic field \(\vec{B}\) is aligned parallel to the normal vector area \(\vec{A}\), the angle is \(\theta = 0^\circ\).


  • Because \(\cos(0^\circ) = 1\) (its maximum algebraic value), magnetic flux reaches its absolute peak: \(\Phi_{B,\text{max}} = B A\).


Why other options are incorrect:

  • Option B: At \(\theta = 45^\circ\), \(\cos(45^\circ) \approx 0.707\), yielding roughly 71% of maximum flux.
  • Option C: At \(\theta = 90^\circ\), \(\cos(90^\circ) = 0\), producing zero magnetic flux because field lines graze the surface plane without crossing.
  • Option D: At \(\theta = 180^\circ\), \(\cos(180^\circ) = -1\), which represents maximum flux in the reverse direction rather than the standard positive maximum.
MCQ #72 of 92 Physics DUHS 2021
[DUHS 2021]

The domestic electricity supply in Pakistan has a frequency of:
A
\(70\text{ Hz}\)
B
\(50\text{ Hz}\)
C
\(100\text{ Hz}\)
D
\(30\text{ Hz}\)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

National utility distribution grids deliver single-phase alternating current at standardized parameters; Pakistan operates under the 230 V, 50 Hz utility standard.

Formula / Rule / Reaction:

$$f = 50\text{ Hz} \implies T = \frac{1}{f} = \frac{1}{50} = 0.02\text{ s} = 20\text{ ms}$$

Solution:

  • The national power grid follows International Electrotechnical Commission (IEC) technical standards.


  • Domestic wall outlets provide alternating voltage completing 50 full sinusoidal cycles per second, corresponding to a frequency of \(50\text{ Hz}\).


Why other options are incorrect:

  • Option A: 70 Hz is an arbitrary value not used in standard commercial power transmission grids.
  • Option C: 100 Hz is the ripple frequency of full-wave rectified 50 Hz AC, not the line transmission frequency.
  • Option D: 30 Hz is not utilized in utility distribution grids.
MCQ #73 of 92 Physics DUHS 2021
[DUHS 2021]

The device that increases or decreases the electromotive force (emf) is called a:
A
AC generator
B
Motor
C
Transformer
D
DC generator
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

A transformer is a static electromagnetic machine that transforms alternating voltages and currents to higher or lower levels through mutual induction across magnetically coupled windings.

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 copper coils (primary and secondary) wound around a common laminated soft-iron core.


  • Alternating current in the primary winding creates a time-varying magnetic flux that links with the secondary winding, inducing a stepped-up or stepped-down alternating emf determined by the turns ratio.


Why other options are incorrect:

  • Option A: An AC generator converts mechanical energy into alternating electrical energy through electromagnetic induction.
  • Option B: An electric motor converts electrical energy into mechanical rotational work.
  • Option D: A DC generator produces unidirectional electrical current using a split-ring commutator.
MCQ #74 of 92 Physics DUHS 2021
[DUHS 2021]

The induced current flows in such a direction as to oppose the cause that produces it. This statement is:
A
Ampere's Law
B
Faraday's Law
C
Lenz's Law
D
Joule's Law
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Lenz's Law states that the polarity of an induced electromotive force produces an induced current whose magnetic field directly opposes the initial change in magnetic flux that generated it.

Formula / Rule / Reaction:

$$\mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t}$$

Solution:

  • Lenz's Law is a physical embodiment of the law of conservation of energy applied to electromagnetic induction.


  • The minus sign in Faraday's mathematical equation represents Lenz's Law, ensuring that work must be done against the opposing induced magnetic field to generate electricity.


Why other options are incorrect:

  • Option A: Ampere's Law relates magnetic field circulation along a closed path to the net enclosed electric current (\(\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}\)).
  • Option B: Faraday's Law quantifies the magnitude of induced emf as proportional to the time rate of change of magnetic flux.
  • Option D: Joule's Law quantifies the rate of thermal energy dissipation by an electric current through a resistive conductor (\(P = I^2 R\)).
MCQ #75 of 92 Physics DUHS 2021
[DUHS 2021]

The device used for conversion of AC into DC is a:
A
Diode
B
Amplifier
C
Rectifier
D
Transistor
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

A rectifier is an electrical circuit or apparatus composed of semiconductor diodes designed specifically to convert bidirectional alternating current into unidirectional direct current.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • While individual p-n junction diodes provide the unidirectional conduction mechanism, the operational apparatus or complete circuit performing AC-to-DC conversion is designated as a rectifier.


  • Rectifiers are categorized into half-wave configurations (using a single diode) and full-wave configurations (center-tapped or bridge rectifiers using multiple diodes).


Why other options are incorrect:

  • Option A: A diode is the discrete two-terminal electronic component; the assembled functional circuit that converts AC to DC is called a rectifier.
  • Option B: An amplifier increases the amplitude, voltage, or power of an incoming electrical signal without rectifying its waveform.
  • Option D: A transistor is a three-terminal semiconductor device used for switching and signal amplification.
MCQ #76 of 92 Physics DUHS 2021
[DUHS 2021]

The process of conversion of AC into DC is called:
A
Rectification
B
Amplification
C
Rectifier
D
Conversion
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Rectification is the physical electronic process of converting a periodically reversing alternating current (AC) waveform into a unidirectional direct current (DC) waveform.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The process utilizing unidirectional conducting elements to suppress or invert alternate half-cycles of alternating voltage is termed rectification.


  • The circuit or hardware unit performing this task is termed the rectifier, whereas the process itself is rectification.


Why other options are incorrect:

  • Option B: Amplification is the process of magnifying signal power or amplitude.
  • Option C: Rectifier is the noun designating the physical device, not the process.
  • Option D: Conversion is a broad generic category that does not identify the specific electronic process.
MCQ #77 of 92 Physics DUHS 2021
[DUHS 2021]

A \(32\text{ g}\) phosphorus sample decays and \(2\text{ g}\) remains undecayed after \(60\text{ days}\). The half-life of phosphorus is:
A
\(5\text{ days}\)
B
\(6\text{ days}\)
C
\(10\text{ days}\)
D
\(15\text{ days}\)
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Radioactive decay obeys first-order kinetics; the fraction of undecayed parent nuclei remaining after \(n\) half-lives is given by \((1/2)^n\).

Formula / Rule / Reaction:

$$N(t) = N_0 \left(\frac{1}{2}\right)^n \quad \text{and} \quad n = \frac{t}{T_{1/2}}$$

Solution:

  • Calculate the ratio of surviving mass to initial mass: $$\frac{N(t)}{N_0} = \frac{2\text{ g}}{32\text{ g}} = \frac{1}{16} = \left(\frac{1}{2}\right)^4$$


  • This reveals that exactly \(n = 4\) half-lives have elapsed across the 60-day period.


  • Determine the half-life: $$T_{1/2} = \frac{t}{n} = \frac{60\text{ days}}{4} = 15\text{ days}$$


Why other options are incorrect:

  • Option A: \(5\text{ days}\) would mean 12 half-lives elapsed in 60 days, leaving only \(32/4096\text{ g}\).
  • Option B: \(6\text{ days}\) corresponds to 10 elapsed half-lives, leaving \(0.031\text{ g}\).
  • Option C: \(10\text{ days}\) corresponds to 6 elapsed half-lives, leaving \(0.5\text{ g}\).
MCQ #78 of 92 Physics DUHS 2021
[DUHS 2021]

The SI unit of gravitational potential is:
A
\(\text{J}/\text{kg}\)
B
\(\text{J}/(\text{m}\cdot\text{kg})\)
C
\(\text{J}/(\text{m}\cdot\text{s})\)
D
\(\text{J}/\text{s}\)
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Gravitational potential at a point in space is defined as the work done per unit mass in bringing a test body from infinity to that point within the gravitational field.

Formula / Rule / Reaction:

$$V_g = \frac{W}{m} = -\frac{G M}{r} \implies [V_g] = \frac{\text{J}}{\text{kg}} = \text{m}^2\cdot\text{s}^{-2}$$

Solution:

  • Work (energy) has the SI derived unit of joules (\(\text{J}\)).


  • Mass has the SI fundamental base unit of kilograms (\(\text{kg}\)).


  • Dividing work by mass yields the SI unit of gravitational potential as joules per kilogram (\(\text{J}/\text{kg}\)).


Why other options are incorrect:

  • Option B: Dividing by meters introduces an extraneous length dimension, which represents gravitational field strength per unit mass rather than potential.
  • Option C: Incorporating seconds in the denominator is dimensionally invalid for potential.
  • Option D: Joules per second (\(\text{J}/\text{s}\)) defines the watt (\(\text{W}\)), the SI unit of power.
MCQ #79 of 92 Physics DUHS 2021
[DUHS 2021]

In nuclear radiography for medical diagnosis, which rays are used?
A
Alpha particles
B
UV rays
C
Gamma rays
D
X-rays
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Nuclear medicine diagnostic techniques (such as scintigraphy and SPECT) use internally administered radioisotopes that emit penetrating gamma rays detected externally by scintillation cameras.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • In nuclear radiography, radioactive tracers (such as Technetium-99m) are introduced into the physiological system.


  • These tracers emit high-energy gamma photons that readily penetrate surrounding soft tissue and bone, allowing external gamma cameras to image organ function and pathology.


Why other options are incorrect:

  • Option A: Alpha particles possess high ionizing power but minimal tissue penetration (stopped by a few micrometers of tissue), causing localized radiation necrosis without escaping the body for imaging.
  • Option B: Ultraviolet radiation is non-ionizing optical radiation unable to penetrate beneath the stratum corneum of the epidermis.
  • Option D: X-rays are generated by high-voltage vacuum tubes for external transmission radiography (CT and plain radiographs), whereas nuclear radiography relies on internally administered radionuclides emitting gamma rays.
MCQ #80 of 92 Physics DUHS 2021
[DUHS 2021]

The SI unit of linear momentum is:
A
\(\text{N}\cdot\text{m}\)
B
\(\text{N}/\text{s}\)
C
\(\text{N}\cdot\text{s}\)
D
\(\text{N}\cdot\text{s}^2\)
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Linear momentum is defined as the product of mass and velocity; by Newton's second law, momentum change equals mechanical impulse, giving it units of newton-seconds.

Formula / Rule / Reaction:

$$\vec{p} = m\vec{v} \implies \vec{J} = \Delta\vec{p} = \vec{F}\Delta t \implies [p] = \text{kg}\cdot\text{m}\cdot\text{s}^{-1} = \text{N}\cdot\text{s}$$

Solution:

  • In SI base units, momentum is expressed as kilogram meters per second (\(\text{kg}\cdot\text{m}\cdot\text{s}^{-1}\)).


  • Recalling that one newton is \(1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}\), multiplying newtons by seconds yields: $$\text{N}\cdot\text{s} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{s} = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}$$


Why other options are incorrect:

  • Option A: Newton-meter (\(\text{N}\cdot\text{m}\)) is the unit of torque or energy (joule).
  • Option B: Newton per second (\(\text{N}/\text{s}\)) represents the time rate of change of force (yank).
  • Option D: Newton-second squared (\(\text{N}\cdot\text{s}^2\)) corresponds to mass times distance, which is dimensionally incorrect.
MCQ #81 of 92 Physics DUHS 2021
[DUHS 2021]

Fill in the blanks with the most appropriate option:

When we went to cinema, the film __________.
A
Already started
B
Had already started
C
Would already start
D
Started already
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

When two past actions are related in time, the earlier completed action requires the past perfect tense (had + past participle), while the later action is stated in the simple past.

Formula / Rule / Reaction:

$$\text{Action 1 (Earlier in past)}: \text{had} + \text{past participle} \quad \text{vs.} \quad \text{Action 2 (Later in past)}: \text{simple past (went)}$$

Solution:

  • The arrival at the cinema is a past event denoted by the simple past verb "went".


  • Because the screening of the movie began prior to this arrival, the past perfect construction "had already started" is grammatically required.


Why other options are incorrect:

  • Option A: "Already started" lacks the auxiliary "had", failing to signal the earlier completed past action.
  • Option C: "Would already start" expresses a conditional or habitual aspect, which is incorrect in this temporal sequence.
  • Option D: "Started already" uses simple past tense and misplaced adverbial positioning, failing to convey temporal precedence.
MCQ #82 of 92 Physics DUHS 2021
[DUHS 2021]

The soup _______ good.
A
Taste
B
Tastes
C
Is tasting
D
Will tasting
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Verbs of sensory perception (taste, smell, sound, look) function as stative linking verbs taking simple present tense and an adjective complement when describing inherent qualities.

Formula / Rule / Reaction:

$$\text{Singular Subject (3rd Person)} + \text{Stative Verb (-s/-es)} + \text{Predicate Adjective}$$

Solution:

  • The subject "The soup" is an uncountable singular noun requiring a third-person singular verb ending in "-s".


  • Because "taste" denotes a static quality of the soup rather than a deliberate physical action, stative simple present "tastes" is grammatically required.


Why other options are incorrect:

  • Option A: "Taste" is the plural or base form, violating subject-verb agreement with the singular noun "soup".
  • Option C: "Is tasting" uses the continuous aspect, which is used when an active agent samples food, not for an intrinsic quality.
  • Option D: "Will tasting" is an ungrammatical modal auxiliary construction.
MCQ #83 of 92 Physics DUHS 2021
[DUHS 2021]

She _______ unconscious since 4 o’clock.
A
Is
B
Was
C
Has been
D
Would
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The present perfect tense is required to express a state that began at a specified point in the past and continues unbroken into the present, especially when governed by the preposition "since".

Formula / Rule / Reaction:

$$\text{Subject} + \text{has/have been} + \text{Complement} + \text{since} + \text{Point in Time}$$

Solution:

  • The temporal marker "since 4 o'clock" establishes a starting point in the past that continues up to the present moment.


  • The singular subject "She" requires the third-person singular present perfect form "has been".


Why other options are incorrect:

  • Option A: "Is" is simple present tense and cannot be paired with a past temporal starting boundary marked by "since".
  • Option B: "Was" is simple past tense, implying that the period of unconsciousness has ended and no longer connects to the present.
  • Option D: "Would" is a modal auxiliary verb requiring a base verb complement, leaving the clause grammatically incomplete.
MCQ #84 of 92 Physics DUHS 2021
[DUHS 2021]

We will stay here _______ Monday.
A
Until
B
Till
C
Still
D
Since
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Prepositions of time indicate terminal limits of an action; "till" and "until" denote continuity up to a specified end point.

Formula / Rule / Reaction:

$$\text{Action} + \text{till / until} + \text{Terminal Boundary Time}$$

Solution:

  • The sentence denotes an action in the future ("will stay") lasting up to a terminal day ("Monday").


  • While both "until" and "till" convey this meaning, "Till" represents the officially keyed choice in the provincial examination paper.


Why other options are incorrect:

  • Option A: Although "until" is formal and grammatically synonymous, the examining board rubric keyed Option B.
  • Option C: "Still" is an adverb expressing continuity or static condition, not a temporal preposition.
  • Option D: "Since" marks the beginning of an event from a past reference point, not an endpoint in the future.
MCQ #85 of 92 English DUHS 2021
[DUHS 2021]

The word curiosity means:
A
Cruel
B
Keen
C
Angry
D
Killing
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Curiosity denotes an eager desire to learn, investigate, or acquire knowledge, matching the semantic definition of keenness or inquisitiveness.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Curiosity stems from Latin curiosus (inquisitive, careful).


  • Among the provided distractors, "keen" denotes an eager, intense interest or sharpness of attention, representing the closest semantic match.


Why other options are incorrect:

  • Option A: "Cruel" means willfully causing pain or suffering to others.
  • Option C: "Angry" describes feeling or showing strong resentment or wrath.
  • Option D: "Killing" denotes the act of causing death.
MCQ #86 of 92 English DUHS 2021
[DUHS 2021]

DRAWBACK, this word is a combination of:
A
Verb + Adjective
B
Noun + Adjective
C
Verb + Noun
D
Verb + Adverb
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Compound nouns are frequently constructed by combining a base verb stem with a spatial or directional adverb particle.

Formula / Rule / Reaction:

$$\text{Draw (Verb)} + \text{Back (Adverb Particle)} \longrightarrow \text{Drawback (Compound Noun)}$$

Solution:

  • The constituent morpheme "draw" functions as a verb meaning to pull or retract.


  • The morpheme "back" functions as an adverb particle denoting reverse direction.


  • Their combination produces the compound noun "drawback", denoting a disadvantage, defect, or hindrance.


Why other options are incorrect:

  • Option A: "Back" does not function as an adjective in this verbal-particle compounding construction.
  • Option B: "Draw" is an action verb root rather than an initial noun.
  • Option C: "Back" operates as an adverbial particle rather than a noun in this compound.
MCQ #87 of 92 English DUHS 2021
[DUHS 2021]

I wish I _______ the answer.
A
Will know
B
Knew
C
Have known
D
Know
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Clauses following the verb "wish" expressing an unreal, counterfactual desire about a present state require the past subjunctive (simple past tense).

Formula / Rule / Reaction:

$$\text{Subject} + \text{wish} + (\text{that}) + \text{Subject} + \text{Past Subjunctive (Simple Past)}$$

Solution:

  • The sentence expresses a counterfactual longing about the present (the speaker currently does not know the answer).


  • Under English grammatical rules governing hypothetical wish clauses, the verb must shift back into the simple past tense ("knew").


Why other options are incorrect:

  • Option A: "Will know" expresses simple future indicative certainty, which is incorrect in an unreal wish clause.
  • Option C: "Have known" is present perfect indicative, which cannot represent an unreal present condition.
  • Option D: "Know" is present indicative, failing to reflect the subjunctive mood.
MCQ #88 of 92 English DUHS 2021
[DUHS 2021]

What does the word "Analogue" mean?
A
Opposite
B
Comparable
C
Behind
D
Clock
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

An analogue is a person, structure, or thing that is analogous, parallel, or comparable to another in key functions or characteristics.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The term "analogue" derives from Greek analogos (proportionate, corresponding).


  • When two entities share equivalent structural or functional attributes, they are considered comparable analogues.


Why other options are incorrect:

  • Option A: "Opposite" denotes complete contradiction or inverse relation, the antonym of analogue.
  • Option C: "Behind" indicates spatial or chronological lag.
  • Option D: "Clock" is an instrument that may display time using an analog dial, but the word "analogue" itself means comparable.
MCQ #89 of 92 English DUHS 2021
[DUHS 2021]

What does the word "Pulsating" mean?
A
Throbbing
B
Pressure
C
Fluid
D
Heart beat
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Pulsating is a participial adjective describing rhythmic expansion and contraction, vibration, or throbbing.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Pulsating originates from the Latin verb pulsare, meaning to beat or strike repeatedly.


  • "Throbbing" describes the same repetitive, rhythmic pulsation, making it the accurate lexical synonym.


Why other options are incorrect:

  • Option B: "Pressure" is a physical force per unit area, not an oscillatory rhythmic movement.
  • Option C: "Fluid" denotes a state of matter (liquid or gas) that flows, not an oscillation.
  • Option D: "Heart beat" is a compound noun naming the cardiac cycle rather than a participial adjective describing rhythmic movement.
MCQ #90 of 92 English DUHS 2021
[DUHS 2021]

The person in charge of a school is called:
A
Principal
B
Principle
C
Head
D
Staff
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Homophones share identical pronunciation but differ in spelling and definition; "principal" denotes a chief executive or school administrator, whereas "principle" refers to a rule or moral doctrine.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • "Principal" functions as a noun designating the chief administrative head of an educational institution.


  • In contrast, "principle" is a noun signifying a fundamental truth, law, or guiding doctrine.


Why other options are incorrect:

  • Option B: "Principle" is a fundamental law, belief, or ethical standard, never a person.
  • Option C: "Head" is an informal descriptor, whereas "principal" is the formal institutional title tested.
  • Option D: "Staff" refers collectively to the employees of an organization rather than the single administrative head.
MCQ #91 of 92 English DUHS 2021
[DUHS 2021]

Break a leg is commonly used to:
A
Curse someone
B
Physically harm someone
C
Wish good luck to someone
D
Teach someone a lesson
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Idiomatic expressions carry non-literal figurative meanings; "break a leg" is a traditional theatrical idiom used to wish performers success and good fortune.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Rooted in theatrical superstition that directly wishing someone good luck invites misfortune, performers say "break a leg" instead.


  • In contemporary English, the phrase is widely used beyond the theater to wish someone good luck before a major endeavor or presentation.


Why other options are incorrect:

  • Option A: Despite its literal wording, the idiom is never used as an offensive curse or malediction.
  • Option B: The idiom does not denote actual physical bone injury or bodily assault.
  • Option D: Teaching someone a lesson means punishing someone for an error, unrelated to this idiom.
MCQ #92 of 92 English DUHS 2021
[DUHS 2021]

Bite the bullet is commonly used to:
A
Accept defeat
B
Accept consequences
C
Accept death
D
Accept bad faith
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

"Bite the bullet" is an idiom meaning to face a painful, grim, or unavoidable situation with courage, endurance, and stoic fortitude.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The expression originates from historical military surgery, where wounded soldiers bit on lead bullets to endure painful procedures before anesthesia was available.


  • In modern figurative usage, it signifies accepting difficult circumstances or facing unavoidable consequences stoically.


Why other options are incorrect:

  • Option A: "Accept defeat" corresponds to idioms like "throw in the towel" or "wave the white flag".
  • Option C: "Accept death" signifies passive resignation to mortality, whereas biting the bullet denotes enduring hardship to move forward.
  • Option D: "Bad faith" refers to intentional deception or duplicity, which has no connection to this idiom.
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