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NUMS 2026 Solved Past Paper

Complete 1:1 authentic annual examination paper (150 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 150 Biology NUMS 2026
[NUMS 2026]

Identify the type of glycoprotein, responsible for cell fusion process in HIV cycle:
A
gp 40
B
gp 41
C
gp 120
D
gp 121
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The human immunodeficiency virus envelope spikes consist of glycoprotein complexes that mediate host cell recognition and entry. Transmembrane glycoprotein 41 directly facilitates the fusion of the viral envelope with the host plasma membrane.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The envelope spikes of HIV are trimers composed of surface gp120 and transmembrane gp41.


  • Initial attachment occurs when gp120 binds to host CD4 receptors and chemokine co-receptors.


  • This binding produces a conformational change that enables the hydrophobic fusion peptide of gp41 to penetrate the host membrane, facilitating viral fusion and capsid delivery.


Why other options are incorrect:

  • Option A: gp 40 is not a structural glycoprotein component of the HIV envelope.
  • Option C: gp 120 mediates surface attachment and receptor binding, not membrane fusion.
  • Option D: gp 121 does not exist in retroviral envelope nomenclature.
MCQ #2 of 150 Biology NUMS 2026
[NUMS 2026]

Which of the following enzymes converts mRNA to complementary DNA (cDNA)?
A
Integrase
B
Reverse transcriptase
C
Protease
D
DNAase
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Reverse transcriptase is an RNA-dependent DNA polymerase that synthesizes a complementary DNA strand from an RNA template.

Formula / Rule / Reaction:

$$\text{Single-stranded mRNA} \xrightarrow{\text{Reverse transcriptase}} \text{Complementary DNA (cDNA)}$$

Solution:

  • In retroviral replication and biotechnology workflows, reverse transcriptase synthesizes a single-stranded cDNA along an mRNA template.


  • The enzyme then degrades the original RNA strand and synthesizes the complementary DNA strand, producing double-stranded DNA.


Why other options are incorrect:

  • Option A: Integrase inserts retroviral double-stranded DNA into the host cellular chromosome.
  • Option C: Protease cleaves nascent viral precursor polyproteins into individual active proteins.
  • Option D: DNAase hydrolyzes phosphodiester bonds to break down DNA.
MCQ #3 of 150 Biology NUMS 2026
[NUMS 2026]

Edward Jenner used material removed from the lesion on the hand of milkmaid and vaccinated a boy suffering from which of the following disease?
A
Small pox
B
Chicken pox
C
Cow pox
D
Measles
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

In 1796, Edward Jenner developed the first vaccination technique using pus from a cowpox lesion on a milkmaid to confer immunity against smallpox.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Edward Jenner observed that dairymaids who contracted cowpox from cattle lesions were protected from smallpox outbreaks.


  • He collected infectious fluid directly from a cowpox pustule on the hand of milkmaid Sarah Nelmes and inoculated a young boy named James Phipps.


  • The underlying disease causing the lesion on the milkmaid's hand was cowpox.


Why other options are incorrect:

  • Option A: Smallpox was the target disease against which Jenner provided immunity, not the disease present on the milkmaid.
  • Option B: Chickenpox is caused by varicella-zoster virus and was not part of Jenner's inoculation experiments.
  • Option D: Measles is an unrelated morbillivirus infection.
MCQ #4 of 150 Biology NUMS 2026
[NUMS 2026]

Adenovirus are classified as:
A
Helical capsid
B
Polyhedral capsid
C
Enveloped virus
D
Coupler capsid
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Viruses are classified structurally based on capsid symmetry, where adenoviruses are recognized as non-enveloped viruses exhibiting an icosahedral or polyhedral shape.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Adenoviruses possess a naked, non-enveloped protein shell.


  • The capsid has icosahedral symmetry comprising 252 capsomeres, forming a characteristic polyhedral shape.


Why other options are incorrect:

  • Option A: Helical capsids are cylindrical structures characteristic of the tobacco mosaic virus and rabies virus.
  • Option C: Adenoviruses do not possess an outer lipid membrane bilayer envelope.
  • Option D: Coupler capsid is an invalid morphological classification term.
MCQ #5 of 150 Biology NUMS 2026
[NUMS 2026]

If a person is doing strenuous exercise, how many ATPs are produced by cellular respiration?
A
2 ATPs
B
3 ATPs
C
4 ATPs
D
6 ATPs
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

During strenuous exercise, muscle oxygen demand exceeds vascular delivery, forcing cells into anaerobic glycolysis (lactic acid fermentation) which yields a net of 2 ATP per glucose molecule.

Formula / Rule / Reaction:

$$\text{C}_6\text{H}_{12}\text{O}_6 + 2\text{ADP} + 2\text{P}_i \rightarrow 2\text{CH}_3\text{CH(OH)COOH} + 2\text{ATP}$$

Solution:

  • Glycolysis requires an initial investment of 2 ATP and produces 4 ATP via substrate-level phosphorylation.


  • This yields a net gain of 2 ATP per glucose molecule.


  • Under anaerobic conditions, pyruvate is converted to lactate, preventing passage into the Krebs cycle and oxidative phosphorylation.


Why other options are incorrect:

  • Option B: 3 ATP is not a stoichiometric product in either aerobic or anaerobic glucose catabolism.
  • Option C: 4 ATP represents the gross ATP generated by glycolysis before accounting for the 2 ATP invested in the preparatory phase.
  • Option D: 6 ATP represents the theoretical ATP yield from glycolytic NADH under aerobic conditions.
MCQ #6 of 150 Biology NUMS 2026
[NUMS 2026]

The end product of non-cyclic electron pathway is:
A
ATP and NADPH
B
ATP
C
FADH2
D
NAOPH2
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Non-cyclic photophosphorylation (Z-scheme) utilizes both photosystem II and photosystem I to drive the simultaneous synthesis of ATP and NADPH.

Formula / Rule / Reaction:

$$\text{H}_2\text{O} + \text{NADP}^+ + \text{ADP} + \text{P}_i + \text{light} \rightarrow \frac{1}{2}\text{O}_2 + \text{NADPH} + \text{H}^+ + \text{ATP}$$

Solution:

  • Electrons derived from water photolysis travel through the electron transport chain between PS II and PS I to generate a proton gradient that drives ATP synthesis.


  • Electrons excited at PS I are accepted by ferredoxin and transferred via NADP+ reductase to reduce NADP+ into NADPH.


  • Both ATP and NADPH serve as assimilatory reducing equivalents for the Calvin cycle.


Why other options are incorrect:

  • Option B: ATP alone is the product of cyclic photophosphorylation, which lacks photolysis and NADPH production.
  • Option C: FADH2 is a mitochondrial electron carrier produced in the Krebs cycle, not chloroplast light reactions.
  • Option D: NAOPH2 represents an invalid chemical formula.
MCQ #7 of 150 Biology NUMS 2026
[NUMS 2026]

Pyruvic acid is converted into acetaldehyde through which step?
A
Decarboxylation
B
Oxidation
C
Addition of coenzyme A
D
Reduction
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

In alcoholic fermentation, pyruvate undergoes non-oxidative decarboxylation by pyruvate decarboxylase to form acetaldehyde and carbon dioxide.

Formula / Rule / Reaction:

$$\text{CH}_3\text{COCOOH} \xrightarrow{\text{Pyruvate decarboxylase}} \text{CH}_3\text{CHO} + \text{CO}_2$$

Solution:

  • Pyruvate is a 3-carbon alpha-keto acid.


  • The enzyme pyruvate decarboxylase cleaves its terminal carboxyl group, releasing carbon dioxide.


  • The resulting 2-carbon product is acetaldehyde.


Why other options are incorrect:

  • Option B: Acetaldehyde formation involves simple decarboxylation without oxidation.
  • Option C: Addition of coenzyme A converts pyruvate into acetyl-CoA during aerobic cellular respiration.
  • Option D: Reduction occurs in the subsequent reaction when acetaldehyde is reduced to ethanol by alcohol dehydrogenase.
MCQ #8 of 150 Biology NUMS 2026
[NUMS 2026]

Choose the correct sequence of electron carrier in respiratory ETC:
A
Coenzyme Q -> cytochrome a -> cytochrome a3 -> cytochrome b -> cytochrome c
B
Coenzyme Q -> cytochrome b -> cytochrome c -> cytochrome a -> cytochrome a3
C
cytochrome a -> cytochrome a3 -> cytochrome b -> cytochrome c -> coenzyme Q
D
Cytochrome b -> cytochrome c -> coenzyme Q -> cytochrome a -> cytochrome a3
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

In the mitochondrial electron transport chain, electrons flow sequentially down an increasing redox potential gradient from Coenzyme Q to cytochromes b, c, a, and a3.

Formula / Rule / Reaction:

$$\text{CoQ} \rightarrow \text{Cyt } b \rightarrow \text{Cyt } c \rightarrow \text{Cyt } a \rightarrow \text{Cyt } a_3 \rightarrow \text{O}_2$$

Solution:

  • Coenzyme Q collects electrons from complex I and complex II.


  • It transfers electrons to complex III, where cytochrome b resides.


  • From cytochrome b, electrons pass to mobile cytochrome c, and finally to complex IV containing cytochrome a and cytochrome a3.


Why other options are incorrect:

  • Option A: Inverts the positions of cytochromes a and a3 with cytochromes b and c.
  • Option C: Reverses the thermodynamic direction of electron transport.
  • Option D: Places cytochrome b and c upstream of Coenzyme Q.
MCQ #9 of 150 Biology NUMS 2026
[NUMS 2026]

Pumping movement of protons occur from:
A
Outer membrane and inner membrane
B
Matrix of mitochondria and mitochondrial intermembrane space
C
Inner membrane and cisternae
D
Inner membrane and intermembrane space
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

During oxidative phosphorylation, respiratory complexes pump protons across the inner mitochondrial membrane from the mitochondrial matrix into the intermembrane space.

Formula / Rule / Reaction:

$$\text{H}^+_{(\text{matrix})} \xrightarrow{\text{Complexes I, III, IV}} \text{H}^+_{(\text{intermembrane space})}$$

Solution:

  • Energy released by the transfer of electrons along the respiratory chain is coupled to proton translocation.


  • Protons are drawn from the interior mitochondrial matrix and pumped outward across the inner membrane.


  • This accumulates protons within the intermembrane space, generating the proton-motive force.


Why other options are incorrect:

  • Option A: Protons are not pumped between the outer membrane and inner membrane surfaces directly from bulk lipid bilayers.
  • Option C: Cisternae are structures characteristic of the endoplasmic reticulum and Golgi complex.
  • Option D: The source compartment for translocated protons is the mitochondrial matrix, not the inner membrane core.
MCQ #10 of 150 Biology NUMS 2026
[NUMS 2026]

Net product of Krebs cycle is:
A
2CO2, 3ATP, 6NADH, 2FADH
B
4CO2, 3NADH, 6ATP, 3FADH
C
4CO2, 2ATP, 6NADH, 2FADH2
D
CO2, NADH, FADH
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

For each glucose molecule metabolized, two acetyl-CoA molecules enter the Krebs cycle, yielding a combined net total of 4 CO2, 2 ATP, 6 NADH, and 2 FADH2.

Formula / Rule / Reaction:

$$2\text{ Acetyl-CoA} + 6\text{NAD}^+ + 2\text{FAD} + 2\text{ADP} + 2\text{P}_i \rightarrow 4\text{CO}_2 + 6\text{NADH} + 2\text{FADH}_2 + 2\text{ATP} + 2\text{CoA}$$

Solution:

  • A single turn of the citric acid cycle processing one acetyl-CoA produces 2 CO2, 3 NADH, 1 FADH2, and 1 ATP.


  • Since glycolysis breaks down one glucose molecule into two pyruvate units, two turns of the cycle occur per glucose molecule.


  • Multiplying the single-turn output by two yields 4 CO2, 2 ATP, 6 NADH, and 2 FADH2.


Why other options are incorrect:

  • Option A: Lists the stoichiometric output of only one turn for carbon dioxide while misstating ATP numbers.
  • Option B: Inverts the ratios of ATP and NADH.
  • Option D: Fails to state any quantitative stoichiometry.
MCQ #11 of 150 Biology NUMS 2026
[NUMS 2026]

The electron carrier of mitochondria are found in:
A
Outer mitochondrial membrane
B
Stroma
C
Intermembrane space
D
Inner mitochondrial membrane
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The protein complexes and coenzymes of the mitochondrial electron transport chain are integrated directly into the inner mitochondrial membrane.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The inner mitochondrial membrane forms infoldings termed cristae, which contain embedded respiratory complexes (I, II, III, and IV).


  • These cristae house Coenzyme Q, cytochromes, and ATP synthase complexes.


Why other options are incorrect:

  • Option A: The outer mitochondrial membrane contains large pore-forming proteins called porins and lacks respiratory complexes.
  • Option B: The stroma is the fluid matrix found inside chloroplasts, not mitochondria.
  • Option C: The intermembrane space stores the translocated proton gradient, but the electron transport complexes are anchored in the inner membrane.
MCQ #12 of 150 Biology NUMS 2026
[NUMS 2026]

Which one of the following is a monosaccharide?
A
Sucrose
B
Maltose
C
Lactose
D
Fructose
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Monosaccharides are simple monomeric sugars that cannot be hydrolyzed into smaller carbohydrate subunits.

Formula / Rule / Reaction:

$$\text{C}_6\text{H}_{12}\text{O}_6 \quad (\text{Ketohexose})$$

Solution:

  • Fructose is a six-carbon ketohexose monosaccharide.


  • Sucrose, maltose, and lactose are disaccharides composed of two linked monosaccharide units.


Why other options are incorrect:

  • Option A: Sucrose is a disaccharide consisting of glucose and fructose linked by a glycosidic bond.
  • Option B: Maltose is a disaccharide consisting of two glucose units.
  • Option C: Lactose is a disaccharide consisting of glucose and galactose.
MCQ #13 of 150 Biology NUMS 2026
[NUMS 2026]

What is percentage of water in brain cells?
A
20%
B
70%
C
85%
D
30%
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

According to the Punjab Textbook Board, water content varies across tissues based on metabolic activity, with human brain cells containing 85% water.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Tissues with high metabolic rates possess higher aqueous proportions.


  • Brain cells contain 85% water, whereas dense bone cells contain approximately 20% water.


Why other options are incorrect:

  • Option A: 20% represents the water content of bone cells.
  • Option B: 70% represents the average total water composition of bacterial cells and the human body overall.
  • Option D: 30% is not a recognized textbook value for neural tissue.
MCQ #14 of 150 Biology NUMS 2026
[NUMS 2026]

Proportion of carbohydrate in bacterial and mammalian cell is:
A
2%, 4%
B
3%, 4%
C
4%, 2%
D
3%, 2%
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Based on standard chemical composition tables in the Punjab Textbook Board, carbohydrates account for 3% of the total mass in a bacterial cell and 4% in a mammalian cell.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • In typical bacterial cells, carbohydrates constitute 3% of the total cellular mass.


  • In typical mammalian cells, carbohydrates constitute 4% of total cellular mass.


Why other options are incorrect:

  • Option A: 2% is the lipid content of bacterial cells.
  • Option C: Inverts the respective values between bacterial and mammalian cells.
  • Option D: Gives an inaccurate value for mammalian cell carbohydrate content.
MCQ #15 of 150 Biology NUMS 2026
[NUMS 2026]

Water has very high heat capacity due to its:
A
High polarity
B
Hydrogen bonding
C
High heat of vaporization
D
Ionization
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Water exhibits an exceptionally high heat capacity because its extensive intermolecular hydrogen-bonding network absorbs large quantities of thermal energy before kinetic motion increases.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Heat capacity is the thermal energy needed to raise the temperature of a unit mass by one degree Celsius.


  • Thermal energy added to liquid water is consumed in breaking and stretching hydrogen bonds rather than directly increasing molecular translation.


  • This property allows water to function as an effective thermal buffer in living organisms.


Why other options are incorrect:

  • Option A: Polarity determines solvent capability, but thermal buffering is directly mediated by hydrogen bonding.
  • Option C: Heat of vaporization is a separate thermal constant that also arises from hydrogen bonding.
  • Option D: Water undergoes minimal self-ionization, which does not govern bulk heat capacity.
MCQ #16 of 150 Biology NUMS 2026
[NUMS 2026]

Which one of the following biomolecule is involved in the transport of oxygen?
A
Protein
B
Fats
C
Carbohydrate
D
Nucleic acid
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Oxygen is carried in the vertebrate circulatory system bound to hemoglobin, which is a conjugated globular protein.

Formula / Rule / Reaction:

$$\text{Hb} + 4\text{O}_2 \rightleftharpoons \text{Hb}(\text{O}_2)_4$$

Solution:

  • Hemoglobin is a quaternary protein composed of four polypeptide chains (two alpha and two beta globin subunits).


  • Each globin subunit binds a heme prosthetic group containing a ferrous ion (Fe2+) capable of reversibly binding oxygen.


  • Therefore, proteins are the biomolecules responsible for biological oxygen transport.


Why other options are incorrect:

  • Option B: Fats serve as thermal insulators and energy stores.
  • Option C: Carbohydrates function as cellular fuels and structural matrix components.
  • Option D: Nucleic acids store and transmit genetic information.
MCQ #17 of 150 Biology NUMS 2026
[NUMS 2026]

Which is the second most abundant organic molecule on earth?
A
Starch
B
Glycogen
C
Cellulose
D
Chitin
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Cellulose is the most abundant organic biomolecule on Earth, followed by chitin, which forms arthropod exoskeletons and fungal cell walls.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Cellulose is the primary structural carbohydrate of plant cell walls and ranks first in global organic biomass.


  • Chitin, an unbranched polymer of N-acetylglucosamine, is the second most abundant organic compound in the biosphere.


Why other options are incorrect:

  • Option A: Starch is a plant energy storage polymer with total global biomass far below structural polysaccharides.
  • Option B: Glycogen is an animal storage polysaccharide produced in comparatively minor amounts.
  • Option C: Cellulose is the first most abundant organic compound on Earth, not the second.
MCQ #18 of 150 Biology NUMS 2026
[NUMS 2026]

Glycerol reacts with fatty acids to produce:
A
Phospholipids
B
Terpenes
C
Acyl glycerol
D
Steroids
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The condensation reaction between the trihydric alcohol glycerol and fatty acids produces ester bonds, yielding acylglycerols.

Formula / Rule / Reaction:

$$\text{C}_3\text{H}_5(\text{OH})_3 + 3\text{RCOOH} \rightarrow \text{C}_3\text{H}_5(\text{OOCR})_3 + 3\text{H}_2\text{O}$$

Solution:

  • Glycerol contains three hydroxyl groups (-OH).


  • Each hydroxyl group can undergo an esterification reaction with the carboxyl group (-COOH) of a fatty acid.


  • This condensation produces monoacylglycerols, diacylglycerols, or triacylglycerols.


Why other options are incorrect:

  • Option A: Phospholipids contain a phosphate group and an organic head group in addition to glycerol and fatty acids.
  • Option B: Terpenes are synthesized from repeating five-carbon isoprene units.
  • Option D: Steroids are lipids characterized by a four-ring hydrocarbon skeleton.
MCQ #19 of 150 Biology NUMS 2026
[NUMS 2026]

During cell wall formation absence of magnesium results in:
A
Lack of primary cell wall
B
Lack of secondary cell wall
C
Lack of middle lamella
D
No effect on cell wall formation
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The middle lamella is an extracellular cementing layer between adjacent plant cells composed of calcium and magnesium salts of pectin.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • During plant cell cytokinesis, vesicles derived from the Golgi apparatus coalesce to produce the cell plate.


  • Pectic compounds are deposited and cross-linked with divalent magnesium (Mg2+) and calcium (Ca2+) ions to form the middle lamella.


  • The absence of magnesium impairs this cross-linking, causing defective middle lamella formation.


Why other options are incorrect:

  • Option A: The primary cell wall is composed predominantly of cellulose microfibrils in a hemicellulose matrix.
  • Option B: The secondary cell wall consists mainly of cellulose and lignin.
  • Option D: Magnesium deficiency directly disrupts intercellular adherence in plant tissues.
MCQ #20 of 150 Biology NUMS 2026
[NUMS 2026]

Identify the mismatch in the following:
A
Mitochondria - cellular respiration
B
Endoplasmic reticulum - detoxification
C
Mitochondria - deamination
D
Lysosome - autophagy
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Organelles perform specialized metabolic tasks; deamination is an enzymatic catabolic pathway of amino acid breakdown that takes place primarily in the liver, not a defining organelle-level function assigned to mitochondria.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Mitochondria carry out the Krebs cycle and oxidative phosphorylation for cellular respiration.


  • Smooth endoplasmic reticulum contains enzymes responsible for the detoxification of harmful lipid-soluble compounds.


  • Lysosomes digest damaged intracellular organelles through autophagy.


  • Mitochondria are not defined in textbook cytology as the organelle for deamination, making option C the mismatch.


Why other options are incorrect:

  • Option A: Mitochondria are the primary site of aerobic cellular respiration.
  • Option B: The smooth endoplasmic reticulum performs metabolic detoxification.
  • Option D: Lysosomes mediate normal autophagic digestion.
MCQ #21 of 150 Biology NUMS 2026
[NUMS 2026]

Which organelle will increase in number, in muscles of athlete?
A
Mitochondria
B
Ribosome
C
Golgi apparatus
D
Lysosome
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Athletic endurance training stimulates mitochondrial biogenesis in skeletal muscle fibers to elevate aerobic capacity and sustain ongoing ATP production.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Repetitive muscular contraction increases cellular energy demand and activates signaling cascades for mitochondrial division.


  • Mitochondria proliferate through binary fission inside muscle cells.


  • This physiological adaptation enhances aerobic cellular respiration and delays muscular fatigue.


Why other options are incorrect:

  • Option B: Ribosomal density supports protein translation during hypertrophy, but mitochondrial expansion is the primary physiological adaptation to athletic exercise.
  • Option C: The Golgi apparatus processes secretory proteins, which is not the primary adaptive demand in working myocytes.
  • Option D: Lysosomes degrade macromolecules and do not increase in number to support athletic work.
MCQ #22 of 150 Biology NUMS 2026
[NUMS 2026]

In secretory cells which pathway is followed for the transport of material?
A
Golgi complex -> RER -> SER
B
RER -> golgi complex -> SER
C
RER -> SER -> golgi complex
D
SER -> golgi complex -> RER
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The secretory pathway in eukaryotic cells proceeds directionally from the rough endoplasmic reticulum to the smooth endoplasmic reticulum (transitional ER) and then to the Golgi complex for packaging.

Formula / Rule / Reaction:

$$\text{RER} \rightarrow \text{SER} \rightarrow \text{Golgi complex} \rightarrow \text{Secretory vesicles}$$

Solution:

  • Proteins targeted for secretion are translated by ribosomes attached to the rough endoplasmic reticulum.


  • Nascent proteins enter the lumen and migrate into smooth endoplasmic reticulum or transitional elements.


  • Transport vesicles then carry the molecules to the cis-face of the Golgi complex for final processing and sorting.


Why other options are incorrect:

  • Option A: Inverts the sequence by placing the Golgi complex ahead of the endoplasmic reticulum.
  • Option B: Places the Golgi complex before the smooth endoplasmic reticulum.
  • Option D: Inverts the directional flow by placing the smooth ER before the rough ER.
MCQ #23 of 150 Biology NUMS 2026
[NUMS 2026]

How many number of chromosomes are present in onion?
A
16
B
8
C
14
D
26
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The diploid somatic chromosome number (2n) of the common onion (Allium cepa) is 16.

Formula / Rule / Reaction:

$$2n = 16 \quad (n = 8)$$

Solution:

  • Vegetative somatic cells of Allium cepa possess 8 homologous chromosome pairs.


  • This yields a total somatic diploid count of 16 chromosomes.


Why other options are incorrect:

  • Option B: 8 represents the haploid chromosome number (n) found in onion gametes.
  • Option C: 14 is the diploid chromosome number of the garden pea (Pisum sativum).
  • Option D: 26 is the diploid chromosome number of the common frog.
MCQ #24 of 150 Biology NUMS 2026
[NUMS 2026]

A cell is treated with a toxin that disrupts Golgi apparatus. Which function is affected?
A
Lipid metabolism
B
Protein modification
C
Chromosome separation during cell division
D
ATP production
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The Golgi apparatus is the primary organelle responsible for the post-translational modification, glycosylation, and sorting of proteins.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Proteins synthesized in the rough endoplasmic reticulum are delivered to the Golgi apparatus.


  • Enzymes within the Golgi cisternae add carbohydrate moieties (glycosylation), cleave peptide sequences, and modify side chains.


  • Disruption of this organelle halts cellular protein modification and maturation.


Why other options are incorrect:

  • Option A: Lipid metabolism is handled mainly by the smooth endoplasmic reticulum and peroxisomes.
  • Option C: Chromosome separation is coordinated by the spindle apparatus and centrioles.
  • Option D: ATP synthesis occurs within the mitochondria.
MCQ #25 of 150 Biology NUMS 2026
[NUMS 2026]

Where modification and packaging occur in secretary cells?
A
Golgi apparatus
B
Mitochondria
C
Endoplasmic reticulum
D
Nucleus
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The Golgi apparatus serves as the intracellular sorting, modifying, and packaging hub for molecules destined for secretion.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Materials from the endoplasmic reticulum fuse with the cis-cisternae of the Golgi complex.


  • After chemical modifications are completed, products are sorted and enclosed into membrane-bound secretory vesicles at the trans-Golgi network.


Why other options are incorrect:

  • Option B: Mitochondria generate ATP through oxidative phosphorylation.
  • Option C: The endoplasmic reticulum carries out primary polypeptide synthesis and lipid assembly.
  • Option D: The nucleus contains chromosomal DNA and manages gene transcription.
MCQ #26 of 150 Biology NUMS 2026
[NUMS 2026]

A person continuously faces water imbalance in body due to damage of which part of the brain?
A
Amygdala
B
Thalamus
C
Medulla
D
Hypothalamus
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The hypothalamus contains osmoreceptor centers that regulate thirst and synthesizes antidiuretic hormone to maintain systemic water homeostasis.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Hypothalamic osmoreceptors detect variations in blood osmolarity.


  • The supraoptic and paraventricular nuclei of the hypothalamus produce ADH (vasopressin), which is transported to the posterior pituitary for release.


  • Lesions in the hypothalamus abolish ADH production, leading to severe water loss and fluid imbalance.


Why other options are incorrect:

  • Option A: The amygdala processes emotional memory and fear responses.
  • Option B: The thalamus functions as a relay center for ascending sensory pathways.
  • Option C: The medulla controls autonomic cardiopulmonary rhythms.
MCQ #27 of 150 Biology NUMS 2026
[NUMS 2026]

Which part of brain is activated to release excess CO2 from body produced during strenuous exercise?
A
Pons
B
Medulla
C
Cerebellum
D
Cerebrum
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The respiratory rhythmicity center in the medulla oblongata monitors cerebrospinal fluid acidity and coordinates ventilation to expel carbon dioxide.

Formula / Rule / Reaction:

$$\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$$

Solution:

  • Strenuous exercise elevates metabolic carbon dioxide in the blood.


  • Carbon dioxide diffuses across the blood-brain barrier into cerebrospinal fluid, forming carbonic acid and increasing hydrogen ion concentration.


  • Chemosensitive areas in the medulla oblongata detect this acidosis and stimulate hyperventilation to blow off excess carbon dioxide.


Why other options are incorrect:

  • Option A: The pons contains the pneumotaxic and apneustic centers that modulate respiratory patterns, but the primary chemo-sensing rhythm center resides in the medulla.
  • Option C: The cerebellum coordinates muscular balance and posture.
  • Option D: The cerebrum regulates conscious thought and voluntary movement.
MCQ #28 of 150 Biology NUMS 2026
[NUMS 2026]

The cerebrospinal fluid (CSF) is present between:
A
Dura matter & arachnoid matter
B
Arachnoid matter & Pia matter
C
Dura matter & Pia matter
D
Arachnoid matter & Dura matter
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Cerebrospinal fluid circulates within the subarachnoid space, which is anatomically situated between the middle arachnoid mater and the inner pia mater.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The brain and spinal cord are enclosed by three meningeal layers: the outer dura mater, middle arachnoid mater, and inner pia mater.


  • The fluid-filled anatomical cavity between the arachnoid and pia layers is the subarachnoid space.


  • This space contains the circulating cerebrospinal fluid.


Why other options are incorrect:

  • Option A: The interface between dura mater and arachnoid mater is the subdural space, which contains minimal serous fluid.
  • Option C: The dura mater and pia mater are separated by the intervening arachnoid layer.
  • Option D: Identical to Option A and describes the subdural boundary.
MCQ #29 of 150 Biology NUMS 2026
[NUMS 2026]

What is the role of growth hormone after adolescence?
A
Promotes protein synthesis
B
Inhibits protein synthesis
C
Decreases amino acid uptake
D
Inhibits cell division
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Following epiphyseal plate closure at adolescence, growth hormone maintains tissue mass by stimulating amino acid uptake and protein synthesis.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Growth hormone stimulates cellular uptake of free amino acids across plasma membranes.


  • It promotes translation and transcription to support ongoing protein synthesis for tissue replacement and muscle maintenance.


Why other options are incorrect:

  • Option B: Growth hormone is an anabolic agent; it stimulates rather than inhibits protein synthesis.
  • Option C: Growth hormone accelerates amino acid transport into cells.
  • Option D: Growth hormone supports basal cellular replacement rather than inhibiting division.
MCQ #30 of 150 Biology NUMS 2026
[NUMS 2026]

Which of the following is correct function of calcitonin?
A
Resorption of calcium in bone matrix
B
Inhibits Ca+2 absorption by intestine
C
Increase reabsorption by kidneys
D
Increase blood calcium level
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Calcitonin is a hypocalcemic hormone secreted by the thyroid gland that lowers blood calcium levels by suppressing osteoclastic bone resorption and inhibiting intestinal calcium absorption.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • When systemic blood calcium rises above normal, the thyroid gland secretes calcitonin.


  • Calcitonin acts antagonistically to parathyroid hormone by inhibiting bone breakdown and suppressing intestinal uptake of calcium ions.


  • These coordinated actions return circulating calcium levels back down to baseline.


Why other options are incorrect:

  • Option A: Bone resorption is stimulated by parathyroid hormone, not calcitonin.
  • Option C: Increasing renal calcium reabsorption is mediated by parathyroid hormone.
  • Option D: Calcitonin lowers blood calcium levels.
MCQ #31 of 150 Biology NUMS 2026
[NUMS 2026]

What is secreted when the levels of blood glucose get too low?
A
Calcitonin
B
Glucagon
C
Insulin
D
Corticoid hormones
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Hypoglycemia triggers the alpha cells of the pancreatic islets of Langerhans to release glucagon, which promotes hepatic glycogenolysis.

Formula / Rule / Reaction:

$$\text{Glycogen} \xrightarrow{\text{Glucagon}} \text{Glucose}$$

Solution:

  • When blood glucose falls below physiological set points, pancreatic alpha cells detect the drop.


  • They release glucagon into the portal circulation.


  • Glucagon binds to hepatocytes to stimulate glycogen breakdown and gluconeogenesis, releasing glucose into the blood.


Why other options are incorrect:

  • Option A: Calcitonin regulates systemic calcium balance.
  • Option C: Insulin is secreted by beta cells in response to hyperglycemia.
  • Option D: Corticoids respond to prolonged systemic stress rather than acute blood glucose drops.
MCQ #32 of 150 Biology NUMS 2026
[NUMS 2026]

Pathway followed by nerve impulse during reflex action is:
A
Interneuron -> synapse -> dendrites -> cell body of motor neuron
B
Interneuron -> dendrite -> synapse -> cell body of motor neuron
C
Interneuron -> synapse -> cell body of neuron
D
Interneuron -> dendrites -> cell body of neuron
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

In a spinal reflex arc, the axon terminal of an interneuron releases neurotransmitters across the synapse to the dendrites of the motor neuron, which conduct the signal to its soma.

Formula / Rule / Reaction:

$$\text{Interneuron axon} \rightarrow \text{Synapse} \rightarrow \text{Motor neuron dendrites} \rightarrow \text{Motor neuron soma}$$

Solution:

  • Action potentials reach the presynaptic terminal of the interneuron.


  • Neurotransmitter molecules diffuse across the synaptic cleft to ligand-gated receptors on the postsynaptic dendrites of the motor neuron.


  • The resulting graded potentials travel toward the motor neuron soma to initiate an action potential.


Why other options are incorrect:

  • Option B: Erroneously places the dendrite upstream of the synaptic cleft.
  • Option C: Bypasses the dendrites, which are the main receptive structures of multipolar motor neurons.
  • Option D: Fails to include the synapse.
MCQ #33 of 150 Biology NUMS 2026
[NUMS 2026]

Identify the animals having both mammalian and reptilian features:
A
Mice
B
Spiny ant eater
C
Opossum
D
Bat
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Monotremes (Prototheria), such as the spiny anteater, are primitive egg-laying mammals that display a combination of reptilian and mammalian anatomical traits.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Reptilian characteristics of the spiny anteater include laying shelled cleidoic eggs and possessing a cloaca.


  • Mammalian characteristics include hair covering the body, homeothermy, and milk-secreting mammary glands.


Why other options are incorrect:

  • Option A: Mice are placental eutherians without primitive reptilian reproductive traits.
  • Option C: The opossum is a metatherian marsupial mammal.
  • Option D: Bats are placental flying mammals belonging to the order Chiroptera.
MCQ #34 of 150 Biology NUMS 2026
[NUMS 2026]

An enzyme which requires a cofactor to become active is called:
A
Apoenzyme
B
Holoenzyme
C
Prosthetic group
D
Co-enzyme
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

An apoenzyme is the catalytically inactive protein portion of an enzyme that strictly requires the addition of a specific non-protein cofactor to exhibit catalytic activity.

Formula / Rule / Reaction:

$$\text{Apoenzyme (inactive protein)} + \text{Cofactor} \rightarrow \text{Holoenzyme (active complex)}$$

Solution:

  • According to the Punjab Textbook Board, an enzyme stripped of its cofactor is catalytically inactive and is termed an apoenzyme.


  • When the required cofactor binds to the apoenzyme, it forms the functional holoenzyme.


Why other options are incorrect:

  • Option B: Holoenzyme is the fully assembled, catalytically active enzyme-cofactor complex.
  • Option C: A prosthetic group is a tightly or covalently bound non-protein cofactor.
  • Option D: A co-enzyme is a dialyzable non-protein organic cofactor.
MCQ #35 of 150 Biology NUMS 2026
[NUMS 2026]

Which statement best describes competitive inhibitors?
A
Denature enzyme permanently
B
Used as drugs
C
Prevent enzyme product complex
D
Alter the shape of enzymes
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Competitive inhibitors resemble the substrate structurally and bind directly to the active site, blocking substrate access and preventing enzyme-product complex formation.

Formula / Rule / Reaction:

$$\text{E} + \text{I} \rightleftharpoons \text{EI} \quad (\text{No ES or EP formation})$$

Solution:

  • A competitive inhibitor occupies the catalytic active site of the free enzyme.


  • Because the active site is blocked, the substrate cannot bind to form the enzyme-substrate (ES) complex.


  • Consequently, catalysis cannot proceed and no enzyme-product (EP) complex is formed.


Why other options are incorrect:

  • Option A: Competitive inhibition is reversible and does not denature the enzyme.
  • Option B: Being used as drugs is an application rather than a defining mechanistic description.
  • Option D: Altering enzyme conformation via allosteric binding describes non-competitive inhibition.
MCQ #36 of 150 Biology NUMS 2026
[NUMS 2026]

Due to snake bite a person's heart stops working. Which type of venom will be found in his body?
A
Hemotoxic
B
Hepatotoxic
C
Neurotoxic
D
Cytotoxic
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Hemotoxic venoms target the cardiovascular and circulatory systems, causing hemolysis, extensive hemorrhage, myocardial collapse, and cardiac arrest.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Snake venoms are classified according to their primary physiological target.


  • Hemotoxic venoms attack red blood cells, blood vessels, and myocardial tissue, causing acute circulatory collapse and heart failure.


  • In contrast, neurotoxic venoms target peripheral motor neurons to induce respiratory paralysis.


Why other options are incorrect:

  • Option B: Hepatotoxic substances specifically damage the liver parenchyma.
  • Option C: Neurotoxic venoms disrupt nerve conduction and neuromuscular junctions, leading to respiratory arrest rather than direct primary heart failure.
  • Option D: Cytotoxic venoms cause localized tissue destruction and necrosis at the bite site.
MCQ #37 of 150 Biology NUMS 2026
[NUMS 2026]

Identify the enzyme which shows absolute specificity:
A
Carbonic anhydrase
B
Hexokinase
C
Urease
D
Isomerase
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Absolute specificity describes an enzyme that catalyzes the chemical transformation of only one particular substrate.

Formula / Rule / Reaction:

$$\text{H}_2\text{N-CO-NH}_2 + \text{H}_2\text{O} \xrightarrow{\text{Urease}} 2\text{NH}_3 + \text{CO}_2$$

Solution:

  • Urease exhibits absolute specificity because it acts exclusively on urea to produce ammonia and carbon dioxide.


  • It does not hydrolyze any other chemically related amide or substituted urea compound.


Why other options are incorrect:

  • Option A: Carbonic anhydrase acts on hydration of CO2 and related analogs.
  • Option B: Hexokinase displays group specificity, phosphorylating several different hexose sugars including D-glucose, D-fructose, and D-mannose.
  • Option D: Isomerase is a broad enzyme class encompassing many different isomer-converting catalysts.
MCQ #38 of 150 Biology NUMS 2026
[NUMS 2026]

Some industries are continuously polluting aquatic bodies. This can affect metabolic pathway of plants by changing their pH:
A
More alkaline
B
More acidic
C
Less acidic
D
Less alkaline
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Industrial effluents containing acidic compounds and acid-forming non-metal oxides lower the pH of aquatic bodies, making them more acidic and impairing plant enzyme systems.

Formula / Rule / Reaction:

$$\text{SO}_x / \text{NO}_x + \text{H}_2\text{O} \rightarrow \text{Acidic effluents} \implies \text{pH} < 7$$

Solution:

  • Discharge of untreated chemical and industrial wastes into waterways introduces strong acids and acidic precursors.


  • This reduces the pH of the aquatic environment, shifting it toward more acidic conditions.


  • Acidic conditions denature plant enzymes and disrupt photosynthetic metabolic pathways.


Why other options are incorrect:

  • Option A: Industrial pollution typically lowers environmental pH rather than increasing alkalinity.
  • Option C: Acidic pollution increases acidity rather than making water bodies less acidic.
  • Option D: Less alkaline does not describe the pronounced shift to an acidic environment.
MCQ #39 of 150 Biology NUMS 2026
[NUMS 2026]

All flowering plants are thought to be evolved from a common ancestor on the basis of which homologous structures?
A
Stamen, carpel, sepals, petals
B
Stamen, carpel, thalamus, sepals
C
Stamen, carpel, thalamus, sepals, petals, stalk
D
Stamen, carpel, sepals, thalamus
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Flowering plants share a common evolutionary origin evidenced by the homology of the four concentric floral whorls, which represent modified leaves specialized for reproduction and protection.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Comparative morphological studies demonstrate that sepals, petals, stamens, and carpels are homologous structures derived from ancestral foliar appendages.


  • The uniform presence and developmental arrangement of these four basic whorls across angiosperm families support descent from a common ancestral angiosperm.


Why other options are incorrect:

  • Option B: The thalamus is the receptacle or swollen stem apex, not one of the primary homologous floral appendage whorls.
  • Option C: The stalk (pedicel) and thalamus are modified stem structures rather than the modified homologous leaf whorls.
  • Option D: Omits the corolla (petals) while erroneously substituting the vegetative thalamus.
MCQ #40 of 150 Biology NUMS 2026
[NUMS 2026]

Barriers which effect geographical distribution of life are:
A
Physical and chemical
B
Environmental & Physical
C
Ecological & chemical
D
Physical, Ecological or Environmental
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The geographical distribution of living organisms is restricted by physical barriers, ecological interactions, and environmental or climatic boundaries that limit species dispersal.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Physical barriers include topographical obstacles such as mountain ranges, wide bodies of water, and deserts that physically block migration.


  • Ecological barriers involve biotic limitations, such as competitive exclusion, predation pressure, and absence of mutualistic partners.


  • Environmental barriers comprise climatic constraints including extremes of temperature, humidity, and photoperiod.


Why other options are incorrect:

  • Option A: Omits ecological constraints and presents an incomplete classification.
  • Option B: Lacks ecological competition and habitat-interaction parameters.
  • Option C: Omits physical geography, which is the primary barrier to migration.
MCQ #41 of 150 Biology NUMS 2026
[NUMS 2026]

Why are some genes called pseudoautosomal, even though they are present on sex chromosome?
A
They are present on X chromosomes only
B
They are present on Y chromosomes only
C
They are present on both X and Y chromosomes
D
They are absent from both X and Y chromosomes
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Pseudoautosomal genes reside within homologous regions present on both the X and Y sex chromosomes, enabling them to pair, recombine, and exhibit Mendelian autosomal inheritance patterns.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • The ends of the human sex chromosomes contain homologous sequences designated as pseudoautosomal regions.


  • Because both males and females carry two functional copies of these genes (one on each sex chromosome), their transmission mimics that of standard autosomal traits.


  • This dual presence on both X and Y chromosomes allows regular meiotic crossing over between heteromorphic sex chromosomes.


Why other options are incorrect:

  • Option A: Genes located solely on the non-homologous region of the X chromosome are X-linked, not pseudoautosomal.
  • Option B: Genes located solely on the Y chromosome are holandric (Y-linked).
  • Option D: Pseudoautosomal genes are physically anchored to the distal segments of both sex chromosomes.
MCQ #42 of 150 Biology NUMS 2026
[NUMS 2026]

As a result of natural disaster, few species continue to survive due to their characteristics. This results in:
A
Increase in favorable alleles
B
Decrease in favorable alleles
C
Increase in unfavorable alleles
D
Decrease in unfavorable alleles
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Natural selection acting via environmental catastrophes selects for phenotypes possessing adaptive survival traits, leading to an increase in the frequency of favorable alleles in surviving generations.

Formula / Rule / Reaction:

$$\Delta q = \frac{s p q^2}{1 - s q^2}$$

Solution:

  • Severe environmental pressures eliminate individuals lacking adaptive survival characteristics.


  • The surviving individuals carry genetic variants that conferred resistance or tolerance to the adverse conditions.


  • Differential reproductive success among the survivors increases the relative representation and frequency of these favorable alleles in the subsequent gene pool.


Why other options are incorrect:

  • Option B: Favorable alleles become more frequent in surviving populations, not less frequent.
  • Option C: Unfavorable alleles lead to selective mortality and consequently diminish in frequency.
  • Option D: While deleterious alleles may decline, natural selection primarily operates to increase the prevalence of favorable alleles among survivors.
MCQ #43 of 150 Biology NUMS 2026
[NUMS 2026]

Which of these correctly order the structure of arteries?
A
Connective tissues -> Smooth muscle -> Elastic tissue -> Endothelium
B
Connective tissues -> Smooth muscle -> Endothelium -> Elastic tissue
C
Elastic tissue -> Smooth muscle -> Connective tissue -> Endothelium
D
Smooth muscle -> Connective tissues -> Elastic tissue -> Endothelium
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Arterial walls consist of three concentric histological tunics: the external connective tissue layer, the intermediate muscular and elastic layer, and the internal endothelial lining.

Formula / Rule / Reaction:

$$\text{Tunica externa} \rightarrow \text{Tunica media} \rightarrow \text{Tunica intima}$$

Solution:

  • Tunica externa forms the outermost sheath and is composed primarily of protective collagenous connective tissue.


  • Tunica media is the middle layer and consists of circular smooth muscle fibers combined with abundant elastic tissue to withstand systolic pressures.


  • Tunica intima forms the innermost surface lined with simple squamous endothelium facing the vascular lumen.


  • From exterior to interior, the histological sequence is connective tissue, smooth muscle, elastic tissue, and endothelium.


Why other options are incorrect:

  • Option B: Places the luminal endothelium outside the elastic tissue lamina.
  • Option C: Incorrectly positions elastic tissue on the outermost external boundary.
  • Option D: Erroneously places smooth muscle at the outer surface rather than connective tissue.
MCQ #44 of 150 Biology NUMS 2026
[NUMS 2026]

Epidermis is outer most layer of the skin, which is composed of:
A
Loosely packed cells
B
Tightly packed cells
C
Scattered cells
D
Ciliated cells
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The epidermis is a keratinized stratified squamous epithelium characterized by tightly packed sheets of epithelial cells joined by intercellular desmosomes to provide a barrier against desiccation and pathogen entry.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Epithelial tissues contain minimal extracellular matrix between adjacent cells.


  • In the epidermis, keratinocytes are tightly packed and anchored together through cell junctions.


  • This dense arrangement creates a physical barrier that resists mechanical abrasion and prevents systemic water loss.


Why other options are incorrect:

  • Option A: Loosely arranged cells with abundant matrix are characteristic of loose connective tissue, not epithelial sheets.
  • Option C: Scattered cells are found in specialized connective tissues and immune niches.
  • Option D: Ciliated epithelial cells line the lumen of the respiratory tract and oviducts, not cutaneous epidermis.
MCQ #45 of 150 Biology NUMS 2026
[NUMS 2026]

Which type of muscle controls the movement of substance through hollow organs?
A
Cardiac muscle
B
Smooth muscles
C
Sphincter muscles
D
Skeleton muscle
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Smooth muscle consists of non-striated, involuntary, spindle-shaped myofilament cells that form the contractile walls of hollow internal viscera.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Hollow visceral organs such as the stomach, intestines, urinary bladder, uterus, and blood vessels require involuntary contractility.


  • Smooth muscle tissue forms concentric and longitudinal layers within these organs.


  • Coordinated, sustained contractions known as peristalsis propel fluids, digested chyme, and excretory products through these lumens.


Why other options are incorrect:

  • Option A: Cardiac muscle tissue is restricted exclusively to the myocardium of the heart.
  • Option C: Sphincter is an anatomical term describing circular ring arrangements of either skeletal or smooth muscle, rather than a distinct primary muscle tissue class.
  • Option D: Skeletal muscle is striated voluntary muscle attached to the skeleton to effect somatic locomotion.
MCQ #46 of 150 Biology NUMS 2026
[NUMS 2026]

Smallest contractile unit of muscle fibers present between two lines is known as:
A
Sarcoplasm
B
Sarcolemma
C
Sarcomere
D
Sarcodina
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The sarcomere is the fundamental structural and functional contractile unit of striated muscle, delimited at each end by a transverse Z-line.

Formula / Rule / Reaction:

$$\text{Contractile Unit} = \text{Segment between two adjacent Z-lines}$$

Solution:

  • Myofibrils in striated skeletal and cardiac muscle are organized into repeating longitudinal subunits.


  • The region of a myofibril situated between two successive Z-lines constitutes one sarcomere.


  • Shortening of sarcomeres via sliding of thin actin filaments over thick myosin filaments produces overall muscle contraction.


Why other options are incorrect:

  • Option A: Sarcoplasm is the cytoplasmic fluid surrounding the myofibrils in a muscle cell.
  • Option B: Sarcolemma is the specialized excitable plasma membrane enclosing a muscle fiber.
  • Option D: Sarcodina is a taxonomic subphylum of amoeboid protozoans.
MCQ #47 of 150 Biology NUMS 2026
[NUMS 2026]

Which ion is responsible for the attachment of tropomyosin with the head of myosin during cross bridge?
A
K+ ions
B
Ca++ ions
C
Na+ ions
D
Mg+
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Excitation-contraction coupling relies on the release of divalent calcium ions from the sarcoplasmic reticulum, which bind troponin to displace tropomyosin and enable cross-bridge cycling.

Formula / Rule / Reaction:

$$\text{Ca}^{2+} + \text{Troponin C} \rightarrow \text{Tropomyosin shift} \rightarrow \text{Actin-Myosin cross-bridge}$$

Solution:

  • Depolarization propagating through transverse tubules triggers the opening of ryanodine channels, releasing calcium ions into the sarcoplasm.


  • Calcium ions bind to the troponin complex.


  • This binding induces a conformational change that shifts inhibitory tropomyosin away from the active myosin-binding sites on actin filaments, initiating cross-bridge attachment.


Why other options are incorrect:

  • Option A: Potassium ions maintain resting membrane potentials across excitable membranes.
  • Option C: Sodium ions drive the depolarizing phase of action potentials along the sarcolemma.
  • Option D: Magnesium serves as an enzymatic cofactor for ATP hydrolysis, but does not mediate steric shifting of troponin-tropomyosin.
MCQ #48 of 150 Biology NUMS 2026
[NUMS 2026]

During muscle contraction which type of events occurs respectively?
A
I band shorten -> H-zone appears
B
I band strengthen -> H-zone disappears
C
I band shorten -> H-zone disappears
D
I band strengthen -> H-zone appears
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

According to the sliding filament hypothesis, muscle shortening occurs as thin actin filaments slide past thick myosin filaments into the central A band, resulting in shortening of the I band and disappearance of the H-zone.

Formula / Rule / Reaction:

$$\text{Contraction} \implies \Delta I < 0, \quad H \rightarrow 0, \quad A = \text{constant}$$

Solution:

  • During myofibril contraction, actin filaments are pulled inward toward the center of the sarcomere by myosin cross-bridge power strokes.


  • The isotropic I bands contain only thin filaments and narrow significantly as the overlap increases.


  • The central H-zone, which represents the bare area of myosin lacking actin overlap at rest, is fully occluded by the converging actin filaments and disappears.


Why other options are incorrect:

  • Option A: The H-zone disappears during contraction rather than appearing.
  • Option B: I band strengthen is non-scientific terminology that fails to describe structural shortening.
  • Option D: Fails to correctly describe the dimensional reduction of either band.
MCQ #49 of 150 Biology NUMS 2026
[NUMS 2026]

Which one of the following organism is responsible for transforming nitrogen gas into ammonium?
A
Cyanobacterium
B
Nitrobacteria
C
Rhizobium
D
Nitrosomonas
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Biological nitrogen fixation is carried out by nitrogen-fixing prokaryotes such as Rhizobium, which utilize the nitrogenase enzyme complex to reduce molecular dinitrogen gas into ammonium.

Formula / Rule / Reaction:

$$\text{N}_2 + 8\text{H}^+ + 8e^- + 16\text{ATP} \xrightarrow{\text{Nitrogenase}} 2\text{NH}_3 + \text{H}_2 + 16\text{ADP} + 16\text{P}_i$$

Solution:

  • Rhizobium species form mutualistic symbiotic root nodules on leguminous plants.


  • Within the bacteroids, nitrogenase reduces inert atmospheric nitrogen gas directly into ammonia, which protonates into ammonium.


  • This organic conversion provides usable assimilable nitrogen to the host plant.


Why other options are incorrect:

  • Option A: While certain cyanobacteria can fix nitrogen, Rhizobium is the designated symbiotic organism emphasized in medical entrance syllabi for direct transformation into plant ammonium.
  • Option B: Nitrobacter is a nitrifying bacterium that oxidizes nitrite into nitrate.
  • Option D: Nitrosomonas is a chemoautotrophic bacterium that oxidizes ammonium into nitrite.
MCQ #50 of 150 Biology NUMS 2026
[NUMS 2026]

Identify one of the following chemo-therapeutic agent used to control bacterial infections:
A
Nitrofurans
B
Hydrogen peroxide
C
Sodium hypochlorite
D
Potassium sorbate
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Chemotherapeutic agents are synthetic or biological chemical compounds administered internally to selectively destroy or inhibit infectious microorganisms without causing host toxicity.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Nitrofurans represent a recognized class of synthetic broad-spectrum chemotherapeutic antibacterial agents.


  • They are metabolized by bacterial nitroreductases into reactive intermediates that induce ribosomal and DNA damage in bacterial pathogens.


  • They are clinically indicated in urinary tract infections and systemic veterinary treatments.


Why other options are incorrect:

  • Option B: Hydrogen peroxide is an oxidizing topical antiseptic and disinfectant, not an internal chemotherapeutic drug.
  • Option C: Sodium hypochlorite is a non-selective chemical disinfectant and bleach unsuitable for internal therapy.
  • Option D: Potassium sorbate is an antifungal and antibacterial chemical food preservative.
MCQ #51 of 150 Biology NUMS 2026
[NUMS 2026]

Why male gonads are present outside the body?
A
They need temperature equal to human body
B
They need 2ºC temperature higher than human body
C
They need 2ºC temperature less than human body
D
They need environmental condition for sperm production
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Spermatogenesis in human males is thermally sensitive and requires an environment 2 to 3 degrees Celsius lower than core abdominal body temperature, which is achieved by housing the testes in the scrotum.

Formula / Rule / Reaction:

$$T_{\text{scrotal}} \approx T_{\text{core}} - (2^\circ\text{C} \text{ to } 3^\circ\text{C}) \approx 34^\circ\text{C} - 35^\circ\text{C}$$

Solution:

  • The internal core body temperature of approximately 37 degrees Celsius causes thermal denaturation of enzymes required for spermatogenic cell division and sperm maturation.


  • The external descent of testes into the scrotal sac permits convective cooling, maintaining a local temperature approximately 2 degrees Celsius below basal core temperature.


  • This cooler environment is mandatory for viable gamete production and storage.


Why other options are incorrect:

  • Option A: Exposure to normal core body temperature impairs meiosis and leads to cryptorchidism-associated infertility.
  • Option B: Elevated temperatures suppress spermatogenesis and induce germinal epithelial degeneration.
  • Option D: Vague non-specific phrasing that lacks physiological temperature values.
MCQ #52 of 150 Biology NUMS 2026
[NUMS 2026]

Which set of events will occur if fusion of gametes does NOT take place?
A
Corpus luteum degenerate, LH increase, progesterone increase
B
Corpus luteum degenerate, LH decline, progesterone increase
C
Corpus luteum degenerate, LH increase, progesterone decline
D
Corpus luteum degenerate, LH decline, progesterone decline
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

In the absence of fertilization, human chorionic gonadotropin is not produced, causing the corpus luteum to regress, which precipitates a sharp drop in both luteinizing hormone and progesterone.

Formula / Rule / Reaction:

$$\text{No fertilization} \rightarrow \text{Corpus albicans} \rightarrow \downarrow [\text{Progesterone}] + \downarrow [\text{LH}]$$

Solution:

  • Without embryonic implantation, there is no trophic stimulation from trophoblastic hCG to rescue the corpus luteum.


  • Luteinizing hormone levels fall during the late secretory phase as negative feedback from prior luteal steroids dampens hypothalamic GnRH pulses.


  • The corpus luteum degenerates into fibrous corpus albicans tissue.


  • Steroidogenesis collapses, causing systemic progesterone concentrations to decline and triggering menstruation.


Why other options are incorrect:

  • Option A: Progesterone levels decline precipitously rather than increasing.
  • Option B: Incorrectly states that progesterone increases following luteal regression.
  • Option C: LH does not increase during luteal regression; it remains suppressed until the next follicular recruitment phase begins.
MCQ #53 of 150 Biology NUMS 2026
[NUMS 2026]

Which of the following is correct order of the transport of bicarbonate from blood to lungs?
A
Capillaries -> Alveoli -> Red blood cells
B
Red blood cells -> Alveoli -> Capillaries
C
Alveoli -> Capillaries -> Red blood cells
D
Red blood cells -> Capillaries -> Alveoli
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

In pulmonary capillaries, bicarbonate stored in venous plasma shifts back into red blood cells, is converted into carbon dioxide gas, diffuses across capillary endothelium, and enters the alveolar air sacs.

Formula / Rule / Reaction:

$$\text{HCO}_3^- \xrightarrow{\text{RBC}} \text{H}_2\text{CO}_3 \xrightarrow{\text{Carbonic anhydrase}} \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Capillary} \rightarrow \text{Alveoli}$$

Solution:

  • At the respiratory surface, plasma bicarbonate re-enters red blood cells via reversed chloride shift.


  • Carbonic anhydrase within erythrocytes catalyzes the dehydration of carbonic acid into molecular carbon dioxide and water.


  • Carbon dioxide gas leaves the erythrocytes, diffuses through the pulmonary capillary wall, and crosses the respiratory membrane into the alveoli for exhalation.


Why other options are incorrect:

  • Option A: Reverses the biological direction of gas diffusion by placing red blood cells downstream of alveoli.
  • Option B: Incorrectly places alveoli before the pulmonary capillary interface.
  • Option C: Describes the pathway of atmospheric gas inhalation rather than excretory bicarbonate clearance.
MCQ #54 of 150 Biology NUMS 2026
[NUMS 2026]

Which set of digestive enzymes play a vital role in breakdown of protein?
A
Pepsin, maltase, erepsin
B
Pepsin, trypsin, amylase
C
Pepsin, trypsin, erepsin
D
Pepsin, amylase, lipase
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Proteolytic enzymes sequentially hydrolyze peptide linkages along protein chains: pepsin acts in the gastric lumen, trypsin acts in the duodenum, and erepsin completes peptide digestion in the small intestine.

Formula / Rule / Reaction:

$$\text{Proteins} \xrightarrow{\text{Pepsin}} \text{Polypeptides} \xrightarrow{\text{Trypsin}} \text{Peptides} \xrightarrow{\text{Erepsin}} \text{Amino Acids}$$

Solution:

  • Pepsin is secreted as inactive pepsinogen by gastric chief cells and cleaves internal peptide bonds under acidic conditions.


  • Trypsin is secreted as trypsinogen by the pancreas and hydrolyzes polypeptide chains into smaller oligopeptides under alkaline conditions.


  • Erepsin is an intestinal peptidase mixture that cleaves small peptides into free amino acids ready for absorption.


Why other options are incorrect:

  • Option A: Maltase is a carbohydrase that hydrolyzes maltose into two glucose units.
  • Option B: Amylase is a glycolytic enzyme that breaks starch down into maltose.
  • Option D: Amylase digests starch and lipase hydrolyzes dietary triglycerides.
MCQ #55 of 150 Biology NUMS 2026
[NUMS 2026]

Gene of one allele which inhibits the effect of other allele at different locus is called as:
A
Hypostatic gene
B
Hylostatic gene
C
Polystatic gene
D
Epistatic gene
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Epistasis is a non-allelic gene interaction in which an allele at one locus masks, suppresses, or inhibits the phenotypic expression of an allele at a completely distinct locus.

Formula / Rule / Reaction:

$$\text{Epistatic Gene (Locus 1)} \xrightarrow{\text{Suppresses}} \text{Hypostatic Gene (Locus 2)}$$

Solution:

  • When an allele masks the phenotype determined by a separate non-allelic gene, the masking locus is termed epistatic.


  • The suppressed or overridden gene located at the alternate locus is designated as hypostatic.


  • Therefore, the gene that carries out the inhibition is the epistatic gene.


Why other options are incorrect:

  • Option A: The hypostatic gene is the one whose phenotypic expression is masked, not the one that causes inhibition.
  • Option B: Hylostatic is a non-biological distractor term.
  • Option C: Polystatic is an invalid genetic term.
MCQ #56 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which of the following is equal to 1 mole of Na+?
A
6.02 x 10^23 atoms
B
6.02 x 10^23 ions
C
6.02 x 10^-23 atoms
D
6.02 x 10^-23 ions
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

One mole of any chemical species corresponds to Avogadro's number of elementary entities, which are discrete ions in the case of ionic charges.

Formula / Rule / Reaction:

$$1\text{ mole of particles} = 6.022 \times 10^{23}\text{ entities}$$

Solution:

  • The chemical species specified is the sodium cation, denoted \(\text{Na}^+\).


  • Because \(\text{Na}^+\) carries a net positive charge due to the loss of a valence electron, its individual units are classified as ions rather than neutral atoms.


  • Therefore, exactly one mole of \(\text{Na}^+\) contains \(6.02 \times 10^{23}\) sodium ions.


Why other options are incorrect:

  • Option A: Neutral metallic sodium comprises atoms, but charged \(\text{Na}^+\) represents ionic species.
  • Option C: Uses a negative exponent of \(10^{-23}\) and misclassifies the ions as atoms.
  • Option D: Inverts Avogadro's number to an infinitesimally small fractional value of \(10^{-23}\).
MCQ #57 of 150 Chemistry NUMS 2026
[NUMS 2026]

The e/m value for positive rays depends upon nature of gas used because:
A
Number of electrons
B
Number of neutrons
C
Number of protons
D
Number of electrons and protons
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Positive rays consist of residual gaseous cations formed by ionization, whose charge-to-mass ratio depends on both the ionic charge (electrons removed) and the total nuclear mass (protons and neutrons).

Formula / Rule / Reaction:

$$\frac{e}{m} = \frac{z \cdot e_0}{M_{\text{ion}}}$$

Solution:

  • Unlike cathode rays which are identical fundamental electrons, positive ray particles originate from gas molecules ionized inside the discharge tube.


  • The net ionic charge \(e\) is governed by the number of electrons stripped from the valence shell relative to the nuclear charge.


  • The mass \(m\) is governed primarily by the number of protons and neutrons making up the nucleus of that specific gas.


  • Consequently, \(e/m\) depends on both the proton content determining atomic identity and mass, and the electronic configuration determining ionization charge.


Why other options are incorrect:

  • Option A: Mass is not determined by electrons, as electron mass is negligible compared to the atomic nucleus.
  • Option B: Neutrons carry no net charge and cannot account for the charge component \(e\).
  • Option C: While protons determine nuclear mass and atomic number, the magnitude of charge \(e\) requires loss of orbital electrons.
MCQ #58 of 150 Chemistry NUMS 2026
[NUMS 2026]

Select the correct increasing energy order of orbitals?
A
4s < 3d < 4p < 5s
B
3d > 4s < 4p < 5s
C
3d > 4s > 5s > 4p
D
4s > 3d > 4p > 5s
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

According to the Aufbau principle and the \((n + l)\) rule, atomic subshells fill in order of increasing total quantum value, with ties resolved by lower principal quantum number \(n\).

Formula / Rule / Reaction:

$$E \propto (n + l)$$

Solution:

  • For \(4s\): \(n = 4, l = 0 \implies n + l = 4\).


  • For \(3d\): \(n = 3, l = 2 \implies n + l = 5\).


  • For \(4p\): \(n = 4, l = 1 \implies n + l = 5\); since \(n=4 > 3\), \(4p\) has higher energy than \(3d\).


  • For \(5s\): \(n = 5, l = 0 \implies n + l = 5\); since \(n=5 > 4\), \(5s\) has higher energy than \(4p\).


  • Arranging in strict ascending order gives: \(4s < 3d < 4p < 5s\).


Why other options are incorrect:

  • Option B: Mixes greater-than and less-than relational signs inconsistently.
  • Option C: Displays an inverted descending sequence with incorrect relative inequalities.
  • Option D: Inverts the inequality signs, depicting a descending rather than increasing sequence.
MCQ #59 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which of the following expressions represent Boyle's law?
A
P1 / V1 = P2 / V2
B
P1 / V1 != P2 / V2
C
P1V1 != P2V2
D
P1V1 = P2V2
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Boyle's law states that at constant temperature, the volume of a fixed mass of an ideal gas is inversely proportional to the applied pressure, yielding a constant pressure-volume product.

Formula / Rule / Reaction:

$$P \propto \frac{1}{V} \implies P V = k \implies P_1 V_1 = P_2 V_2$$

Solution:

  • For an ideal gas held at constant absolute temperature and mole quantity, doubling the pressure halves the volume.


  • The mathematical product of pressure and volume at an initial state must equal the product at a final state.


  • Therefore, the law is represented by the linear relation \(P_1 V_1 = P_2 V_2\).


Why other options are incorrect:

  • Option A: Represents a direct proportionality relation \(P/V = k\), which contradicts inverse gas compression behavior.
  • Option B: States an inequality for an incorrect direct ratio.
  • Option C: Uses a non-equality operator which denies the fundamental constancy of the PV product.
MCQ #60 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which Celsius temperature represents zero kelvin or absolute zero?
A
0ºC
B
+273.16ºC
C
-273.16ºC
D
25ºC
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Absolute zero is the theoretical thermodynamic temperature at which all translational kinetic motion ceases, corresponding to zero kelvin or \(-273.16^\circ\text{C}\) according to textbook standards.

Formula / Rule / Reaction:

$$T (\text{K}) = t (^\circ\text{C}) + 273.16 \implies 0 = t + 273.16 \implies t = -273.16^\circ\text{C}$$

Solution:

  • The absolute temperature scale defines zero kelvin (0 K) as its origin.


  • Converting 0 K to the Celsius scale by subtracting the thermodynamic offset constant 273.16 yields \(-273.16^\circ\text{C}\).


Why other options are incorrect:

  • Option A: \(0^\circ\text{C}\) represents the freezing point of water at standard atmospheric pressure, equal to 273.16 K.
  • Option B: \(+273.16^\circ\text{C}\) represents a positive Celsius temperature equivalent to 546.32 K.
  • Option D: \(25^\circ\text{C}\) corresponds to standard room temperature (298.16 K).
MCQ #61 of 150 Chemistry NUMS 2026
[NUMS 2026]

An egg takes 5 minutes to boil at sea level while 10 minutes at Murree hills because of:
A
Low vapour pressure
B
High vapour pressure
C
High external pressure
D
Low external pressure
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The boiling point of a liquid is directly proportional to external atmospheric pressure; elevated altitudes exhibit reduced atmospheric pressure, lowering the boiling point of water and prolonging cooking times.

Formula / Rule / Reaction:

$$P_{\text{ext}} = P_{\text{vap}} \quad (\text{Condition for boiling})$$

Solution:

  • At high altitudes such as Murree hills, the atmospheric column is shorter, causing lower external atmospheric pressure than at sea level.


  • Water boils when its vapor pressure equals the prevailing external pressure.


  • Because external pressure is reduced, water boils at a lower temperature (below \(100^\circ\text{C}\)).


  • Cooking involves heat transfer that proceeds more slowly at lower boiling temperatures, requiring 10 minutes instead of 5 minutes to boil an egg.


Why other options are incorrect:

  • Option A: The vapor pressure of water depends solely on its temperature, not on altitude.
  • Option B: Liquid vapor pressure is not elevated by high altitude.
  • Option C: High external pressure occurs in deep valleys or pressure cookers, which elevates the boiling point and accelerates cooking.
MCQ #62 of 150 Chemistry NUMS 2026
[NUMS 2026]

Fluorine is a gas and Iodine is a solid at room temperature due to:
A
High electronegativity
B
Low electronegativity
C
High polarizability
D
Low polarizability
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Down the halogen group, increasing atomic size and number of electron shells expand the polarizability of the electron cloud, strengthening intermolecular London dispersion forces.

Formula / Rule / Reaction:

$$\text{Polarizability } (\alpha) \propto \text{Number of electrons} \implies \text{London dispersion forces} \propto \alpha$$

Solution:

  • Fluorine has a compact electron cloud with only 9 electrons per atom, resulting in low polarizability and weak dispersion forces that permit it to exist as a gas.


  • Iodine contains 53 electrons per atom distributed across larger shells far from the nucleus, giving it very high polarizability.


  • This substantial polarizability produces strong temporary dipole attractions that pack \(\text{I}_2\) molecules into a solid crystal lattice at room temperature.


Why other options are incorrect:

  • Option A: Fluorine has the highest electronegativity, but electronegativity does not dictate intermolecular physical aggregation in non-polar halogens.
  • Option B: Low electronegativity is an intrinsic atomic property that does not directly generate cohesive intermolecular forces.
  • Option D: Low polarizability corresponds to weaker dispersion forces, which would favor a gaseous state rather than a solid state.
MCQ #63 of 150 Chemistry NUMS 2026
[NUMS 2026]

Ionic solids are bad conductors of heat and electricity because:
A
They have strong covalent bonds
B
They have strong intermolecular forces
C
Ions can only vibrate about their fixed position
D
Ions are free to move
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Electrical conduction requires mobile charge carriers; in solid ionic lattices, electrostatic attractions bind ions firmly into fixed coordinates, restricting them to vibrational motion.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • In crystalline ionic solids, cations and anions occupy rigid three-dimensional lattice sites held by non-directional electrostatic attractions.


  • There are neither delocalized valence electrons nor mobile ions available to migrate under an applied electric field.


  • The ions are confined to vibrational oscillation about their mean positions, preventing electrical conduction in the solid state.


Why other options are incorrect:

  • Option A: Ionic solids contain ionic bonds rather than covalent bonds.
  • Option B: Ionic solids are macromolecular ionic lattices rather than discrete molecular units held by intermolecular forces.
  • Option D: Free ionic mobility occurs only in the molten state or in aqueous solution, not in the solid state.
MCQ #64 of 150 Chemistry NUMS 2026
[NUMS 2026]

If 10% urea is present in aqueous solution of sodium chloride, what will be the shape of crystals obtained?
A
Octahedral
B
Triclinic
C
Cubic
D
Hexagonal
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Crystal habit describes the external geometry of a crystal; changing growth conditions or introducing impurities such as 10% urea modifies the habit of sodium chloride from cubic to octahedral.

Formula / Rule / Reaction:

$$\text{NaCl}_{\text{(aq)}} \xrightarrow{\text{Pure}} \text{Cubic}; \quad \text{NaCl}_{\text{(aq)}} + 10\%\text{ Urea} \xrightarrow{\text{Crystallization}} \text{Octahedral}$$

Solution:

  • Under normal conditions, sodium chloride crystallizes from pure aqueous solutions as cubic crystals.


  • When 10% urea is added to the mother liquor, urea molecules selectively adsorb onto the (111) faces of the growing crystals.


  • This preferential adsorption suppresses crystal growth along those specific faces, changing the crystal habit into an octahedral geometry.


Why other options are incorrect:

  • Option B: Triclinic is an asymmetric crystal system characterized by unequal axes and non-perpendicular angles.
  • Option C: Cubic shape is obtained only when NaCl crystallizes from pure water in the absence of urea.
  • Option D: Hexagonal geometry belongs to a different crystal family with sixfold rotational symmetry.
MCQ #65 of 150 Chemistry NUMS 2026
[NUMS 2026]

Kc = Kp when delta n is equal to:
A
Zero
B
+1
C
-1
D
-2
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The relation between the gas-phase equilibrium constants \(K_p\) and \(K_c\) is modulated by the term \((RT)^{\Delta n}\), equating the two constants when the change in stoichiometric gaseous moles is zero.

Formula / Rule / Reaction:

$$K_p = K_c (RT)^{\Delta n}$$

Solution:

  • The exponent \(\Delta n\) represents the difference between the moles of gaseous products and gaseous reactants: \(\Delta n = \sum n_{\text{products}} - \sum n_{\text{reactants}}\).


  • When \(\Delta n = 0\), the term becomes \((RT)^0 = 1\).


  • Substituting this into the equation yields \(K_p = K_c(1)\), confirming that \(K_p = K_c\).


Why other options are incorrect:

  • Option B: When \(\Delta n = +1\), \(K_p = K_c(RT)\), so \(K_p > K_c\) at temperatures where \(RT > 1\).
  • Option C: When \(\Delta n = -1\), \(K_p = K_c / (RT)\).
  • Option D: When \(\Delta n = -2\), \(K_p = K_c / (RT)^2\).
MCQ #66 of 150 Chemistry NUMS 2026
[NUMS 2026]

The pH of human blood is maintained at:
A
5.37
B
5.73
C
7.53
D
7.35
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Human blood maintains a slightly alkaline physiological pH regulated within the narrow survival range of 7.35 to 7.45 by the carbonic acid-bicarbonate buffer system.

Formula / Rule / Reaction:

$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = 6.1 + \log(20) \approx 7.40$$

Solution:

  • Extracellular fluid pH is defended by the carbonic acid-bicarbonate buffer system alongside pulmonary and renal compensation.


  • The normal clinical reference range for arterial human blood is 7.35 to 7.45.


  • Among the provided options, 7.35 corresponds to the lower physiological baseline of normal human blood.


Why other options are incorrect:

  • Option A: A pH of 5.37 represents severe acidosis that is incompatible with human cellular life.
  • Option B: A pH of 5.73 represents critical fatal acidosis.
  • Option C: A pH of 7.53 reflects abnormal systemic alkalosis.
MCQ #67 of 150 Chemistry NUMS 2026
[NUMS 2026]

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

The rate of a chemical reaction is expressed in units of concentration per unit time, which matches the units of the rate constant only when the reaction order is zero.

Formula / Rule / Reaction:

$$\text{Units of } k = (\text{mol dm}^{-3})^{1-n} \text{ s}^{-1}$$

Solution:

  • Reaction rate is defined as change in concentration per unit time: \(\text{mol dm}^{-3} \text{ s}^{-1}\).


  • The differential rate equation for an \(n\)-th order reaction is \(\text{Rate} = k[A]^n\).


  • For a zero-order reaction (\(n = 0\)), the rate equation simplifies to \(\text{Rate} = k[A]^0 = k\).


  • Therefore, the unit of \(k\) for a zero-order reaction is identical to the rate of reaction: \(\text{mol dm}^{-3} \text{ s}^{-1}\).


Why other options are incorrect:

  • Option A: For a third-order reaction (\(n=3\)), the unit of \(k\) is \(\text{mol}^{-2} \text{dm}^6 \text{s}^{-1}\).
  • Option B: For a second-order reaction (\(n=2\)), the unit of \(k\) is \(\text{mol}^{-1} \text{dm}^3 \text{s}^{-1}\).
  • Option C: For a first-order reaction (\(n=1\)), the unit of \(k\) is \(\text{s}^{-1}\).
MCQ #68 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which of the following represents Arrhenius constant expression?
A
K = Ae^(Ea/T)
B
K = Ae^(-Ea/RT)
C
K = Ae^(RT/Ea)
D
K = e^(-Ea/RT)
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The Arrhenius equation mathematically quantifies the relationship between the rate constant of a reaction, its activation energy, and thermodynamic temperature.

Formula / Rule / Reaction:

$$k = A e^{-\frac{E_a}{RT}}$$

Solution:

  • In the Arrhenius relationship, \(k\) is the rate constant, \(A\) is the pre-exponential frequency factor, \(E_a\) is activation energy, \(R\) is the universal gas constant, and \(T\) is absolute temperature.


  • The fraction of molecular collisions possessing kinetic energy equal to or greater than \(E_a\) is given by the Boltzmann factor \(e^{-E_a/RT}\).


  • Combining these factors gives the expression \(k = A e^{-E_a/RT}\).


Why other options are incorrect:

  • Option A: Omits the negative sign in the exponent and omits the universal gas constant \(R\).
  • Option C: Inverts the activation energy and thermal energy quotient in the exponent.
  • Option D: Omits the pre-exponential frequency factor \(A\).
MCQ #69 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which expression is according to the Hess's law?
A
\sum \Delta H (\text{cycle}) = 0
B
\sum \Delta H = 2
C
\sum \Delta H (\text{cycle}) = 3
D
\sum \Delta H (\text{Cycle}) = -1
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Hess's law of constant heat summation is an application of the first law of thermodynamics, dictating that the net enthalpy change over a complete closed cyclic path is zero.

Formula / Rule / Reaction:

$$\sum_{\text{cycle}} \Delta H = 0$$

Solution:

  • Enthalpy \(H\) is a thermodynamic state function whose value depends solely on the current state of the system rather than the pathway taken.


  • When a system completes a cycle and returns to its original initial state, the net change in any state function must evaluate to zero.


  • Therefore, summing all stepwise enthalpy changes around a closed Born-Haber or chemical cycle yields \(\sum \Delta H (\text{cycle}) = 0\).


Why other options are incorrect:

  • Option B: Arbitrary non-zero value violating the state function properties of enthalpy.
  • Option C: A closed cycle cannot produce a net enthalpy change of 3.
  • Option D: A closed cycle cannot produce a net enthalpy change of -1.
MCQ #70 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which of the following represents the expression of internal energy?
A
\Delta E = q + t
B
\Delta E = q + w
C
\Delta E = q + v
D
\Delta E = q + p
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

According to the first law of thermodynamics, the change in the internal energy of a closed system equals the net heat added to the system plus the work done on the system.

Formula / Rule / Reaction:

$$\Delta E = q + w$$

Solution:

  • Internal energy \(E\) is the sum of all microscopic kinetic and potential energies within a system.


  • Energy can be transferred across system boundaries either as thermal heat (\(q\)) or mechanical work (\(w\)).


  • Under IUPAC convention adopted in textbook syllabi, the algebraic statement of the first law is \(\Delta E = q + w\).


Why other options are incorrect:

  • Option A: Incorporates time or temperature \(t\) rather than work \(w\).
  • Option C: Replaces thermodynamic work with volume \(v\).
  • Option D: Replaces thermodynamic work with pressure \(p\).
MCQ #71 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which is added in the product side to balance equation?

$$\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^-$$
A
1e-, H+
B
2e-, 3H+
C
5e-
D
3e-
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Balancing redox half-reactions in acidic media requires balancing oxygen using water, hydrogen using protons, and electrical charge using electrons.

Formula / Rule / Reaction:

$$\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^-$$

Solution:

  • Examine the initial atom count: reactants contain \(1\text{ S}\), \(4\text{ O}\), and \(4\text{ H}\); products contain \(1\text{ S}\), \(4\text{ O}\), and \(1\text{ H}\).


  • Balancing hydrogen atoms requires adding \(3\text{H}^+\) to the product side: \(\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+\).


  • The net charge on the reactant side is 0; the net charge on the product side is currently \((-1) + 3(+1) = +2\).


  • To balance charge, add \(2e^-\) to the product side: \(\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^-\).


Why other options are incorrect:

  • Option A: Adding \(1e^-\) and \(1\text{H}^+\) leaves both hydrogen atoms and net electrical charge unbalanced.
  • Option C: Adding 5 electrons fails to supply the required hydrogen counterions.
  • Option D: Fails to balance mass and overcompensates electrical charge.
MCQ #72 of 150 Chemistry NUMS 2026
[NUMS 2026]

The oxidation state of Mn in KMnO4 is:
A
+6
B
+3
C
+7
D
+4
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The oxidation state of an unknown central transition metal is determined by setting the algebraic sum of the oxidation numbers of all atoms in a neutral compound equal to zero.

Formula / Rule / Reaction:

$$\text{ON}(\text{K}) + \text{ON}(\text{Mn}) + 4[\text{ON}(\text{O})] = 0$$

Solution:

  • Potassium is an alkali metal with a fixed oxidation number of \(+1\).


  • Oxygen typically exhibits an oxidation number of \(-2\) in oxoanions.


  • Setting up the equation: \((+1) + x + 4(-2) = 0\).


  • Solving for \(x\): \(1 + x - 8 = 0 \implies x - 7 = 0 \implies x = +7\).


Why other options are incorrect:

  • Option A: \(+6\) is the oxidation state of manganese in potassium manganate, \(\text{K}_2\text{MnO}_4\).
  • Option B: \(+3\) is the oxidation state of manganese in \(\text{Mn}_2\text{O}_3\).
  • Option D: \(+4\) is the oxidation state of manganese in manganese dioxide, \(\text{MnO}_2\).
MCQ #73 of 150 Chemistry NUMS 2026
[NUMS 2026]

When atoms approaches each other for chemical bond formation, it leads to:
A
Decrease in energy
B
Increase in energy
C
Energy remains same
D
Increase in bond length
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Chemical bond formation occurs spontaneously when net attractive electrostatic forces between nuclei and electrons exceed repulsive forces, lowering the potential energy of the system.

Formula / Rule / Reaction:

$$\Delta H_{\text{bond formation}} < 0$$

Solution:

  • As isolated atoms approach from infinite separation, mutual electrostatic attractions between nuclei and electrons lower the potential energy.


  • A stable bond forms at an optimum internuclear distance known as the bond length, where the potential energy curve reaches an absolute minimum.


  • This stabilization releases bond energy, decreasing the overall energy of the system.


Why other options are incorrect:

  • Option B: An increase in potential energy creates repulsive instability, preventing bond formation.
  • Option C: If net energy remained constant, there would be no thermodynamic driving force for chemical bonding.
  • Option D: Bond formation shortens internuclear distances relative to non-bonded van der Waals contact distances.
MCQ #74 of 150 Chemistry NUMS 2026
[NUMS 2026]

The number of bonds in oxygen molecules are:
A
Two sigma bond
B
Two sigma one pi bond
C
One sigma one pi bond
D
Two pi one sigma
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

A diatomic oxygen molecule contains a covalent double bond formed by the sharing of two pairs of valence electrons, comprising one coaxial sigma bond and one lateral pi bond.

Formula / Rule / Reaction:

$$\text{O}=\text{O} \implies 1\sigma + 1\pi$$

Solution:

  • Each oxygen atom has a ground state valence electron configuration of \(2s^2 2p_x^2 2p_y^1 2p_z^1\).


  • Head-on axial overlap between half-filled \(2p_x\) orbitals forms one strong \(\sigma\) bond along the internuclear axis.


  • Parallel sideways overlap between half-filled \(2p_y\) orbitals forms one \(\pi\) bond above and below the axis.


  • Therefore, the double bond in an \(\text{O}_2\) molecule consists of one \(\sigma\) bond and one \(\pi\) bond.


Why other options are incorrect:

  • Option A: Two identical atoms cannot form two coaxial \(\sigma\) bonds between the same nuclear pair.
  • Option B: Two sigma bonds and one pi bond describe an impossible three-bond arrangement.
  • Option D: Two pi bonds and one sigma bond describe the covalent triple bond found in dinitrogen (\(\text{N}_2\)).
MCQ #75 of 150 Chemistry NUMS 2026
[NUMS 2026]

The decreasing order of ionization energy is:
A
O < S < N
B
F > Cl > Br
C
Br < Cl < F
D
C < N < B
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

First ionization energy decreases down a periodic group because increasing principal quantum levels expand atomic radius and augment electron shielding, weakening effective nuclear attraction for valence electrons.

Formula / Rule / Reaction:

$$\text{IE}_1 \propto \frac{Z_{\text{eff}}}{r}$$

Solution:

  • Fluorine, chlorine, and bromine belong to Group 17 (halogens) in Periods 2, 3, and 4 respectively.


  • Down the group, additional electron shells increase atomic radius (\(r_{\text{F}} < r_{\text{Cl}} < r_{\text{Br}}\)) and shielding.


  • Valence electrons become easier to remove, causing first ionization energies to decline: \(\text{F} (1681\text{ kJ/mol}) > \text{Cl} (1251\text{ kJ/mol}) > \text{Br} (1140\text{ kJ/mol})\).


  • Thus, the decreasing order is \(\text{F} > \text{Cl} > \text{Br}\).


Why other options are incorrect:

  • Option A: Uses increasing inequality signs rather than decreasing order, and nitrogen has a higher first ionization energy than oxygen due to half-filled subshell stability.
  • Option C: Displays an ascending (increasing) order rather than a decreasing sequence.
  • Option D: Carbon has a higher ionization energy than boron, making the stated order incorrect.
MCQ #76 of 150 Chemistry NUMS 2026
[NUMS 2026]

The size of Ar is larger than Cl due to:
A
Stearic hindrance
B
Inter-electronic repulsion
C
High ionization energy
D
High shielding effect
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Argon possesses a completely filled valence octet whose electronic density experiences significant inter-electronic repulsion, expanding its electron cloud and resulting in a larger van der Waals radius relative to the covalent radius of chlorine.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Across Period 3, covalent radii decrease from sodium to chlorine as effective nuclear charge rises.


  • Argon has a complete octet configuration of \(3s^2 3p^6\) with eight paired electrons in its outermost shell.


  • The mutual electrostatic repulsion among these paired electrons pushes the orbitals outward, causing the measured non-bonded atomic boundary to exceed that of chlorine.


Why other options are incorrect:

  • Option A: Stearic hindrance describes steric clash between bulky functional groups in polyatomic molecules, not isolated monoatomic radii.
  • Option C: High ionization energy is a consequence of stable octet configuration rather than the physical cause of boundary enlargement.
  • Option D: Argon and chlorine possess the same core shielding electrons (10 inner core electrons).
MCQ #77 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which element is lighter than water?
A
Cs
B
Rb
C
Fe
D
K
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The density of an element determines whether it sinks or floats in liquid water; potassium has an unusually low mass-to-volume ratio with a density less than \(1.0\text{ g cm}^{-3}\).

Formula / Rule / Reaction:

$$\rho_{\text{water}} = 1.00\text{ g cm}^{-3}, \quad \rho_{\text{K}} = 0.862\text{ g cm}^{-3}$$

Solution:

  • Water has a standard density of \(1.00\text{ g cm}^{-3}\) at room temperature.


  • Among the alkali metals, lithium (\(0.534\text{ g cm}^{-3}\)), sodium (\(0.968\text{ g cm}^{-3}\)), and potassium (\(0.862\text{ g cm}^{-3}\)) have densities less than water.


  • Rubidium (\(1.532\text{ g cm}^{-3}\)) and cesium (\(1.93\text{ g cm}^{-3}\)) are denser than water and sink.


  • Therefore, potassium floats on water and is lighter than water.


Why other options are incorrect:

  • Option A: Cesium has a density of \(1.93\text{ g cm}^{-3}\), sinking rapidly in water.
  • Option B: Rubidium has a density of \(1.53\text{ g cm}^{-3}\), which is greater than that of water.
  • Option C: Iron has a high density of \(7.87\text{ g cm}^{-3}\).
MCQ #78 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which of the following decreases as we move from top to bottom in Group IA or IIA?
A
Metallic bond strength
B
Electro-positivity
C
Reactivity
D
Ionic radii
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Down main groups IA and IIA, atomic radius increases while valence charge remains constant, decreasing charge density and weakening the electrostatic attraction between metal cations and delocalized electrons.

Formula / Rule / Reaction:

$$\text{Metallic Bond Strength} \propto \frac{\text{Charge on cation}}{\text{Ionic radius}}$$

Solution:

  • Descending Group IA or IIA adds successive electron shells, increasing atomic and ionic dimensions.


  • The larger distance between the positively charged nucleus and the delocalized valence electron sea reduces binding energy.


  • Consequently, lattice binding weakens, causing metallic bond strength and melting points to decrease down the group.


Why other options are incorrect:

  • Option B: Electropositivity increases down the group because valence electrons are held less tightly.
  • Option C: Chemical reactivity toward water and halogens increases down both groups.
  • Option D: Ionic radii increase down the group due to the addition of principal energy levels.
MCQ #79 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which one of the following is the correct order of electronegativity of group II A elements?
A
Be < Mg < Ca < Sr < Ba
B
Be < Ca < Mg < Ba < Sr
C
Ba < Sr < Ca < Mg < Be
D
Be < Sr > Mg < Ca < Ba
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Electronegativity decreases down alkaline earth metals as increasing atomic radius and electron shielding diminish the electrostatic pull of the nucleus on shared bonding pairs.

Formula / Rule / Reaction:

$$\chi \propto \frac{Z_{\text{eff}}}{r}$$

Solution:

  • Pauling electronegativity values for Group IIA elements are: \(\text{Be} = 1.5\), \(\text{Mg} = 1.2\), \(\text{Ca} = 1.0\), \(\text{Sr} = 1.0\), and \(\text{Ba} = 0.9\).


  • Arranging these elements in order of increasing electronegativity gives: \(\text{Ba} < \text{Sr} < \text{Ca} < \text{Mg} < \text{Be}\).


Why other options are incorrect:

  • Option A: Inverts the periodic trend by suggesting electronegativity increases down the group.
  • Option B: Depicts an incorrect random sequence with improper positioning of calcium.
  • Option D: Displays inconsistent inequality directions.
MCQ #80 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which one of the following is heteroatom?
A
Pyrrole
B
Naphthalene
C
Anthracene
D
Cyclohexene
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Heterocyclic compounds contain an atom other than carbon, such as nitrogen, oxygen, or sulfur, within their cyclic ring structure.

Formula / Rule / Reaction:

$$\text{Pyrrole} = \text{C}_4\text{H}_5\text{N} \quad (\text{Five-membered ring containing NH})$$

Solution:

  • A heteroatom is any atom in an organic ring structure that is not carbon.


  • Pyrrole consists of a planar five-membered aromatic ring containing four carbon atoms and one nitrogen heteroatom.


  • Naphthalene, anthracene, and cyclohexene are carbocyclic systems containing only carbon atoms in their ring skeletons.


Why other options are incorrect:

  • Option B: Naphthalene is a fused bicyclic homocyclic hydrocarbon consisting exclusively of carbon ring atoms.
  • Option C: Anthracene is a tricyclic benzenoid hydrocarbon lacking heteroatoms.
  • Option D: Cyclohexene is a carbocyclic cycloalkene ring composed entirely of carbon atoms.
MCQ #81 of 150 Chemistry NUMS 2026
[NUMS 2026]

The function group of acid amide is:
A
\(-\text{C}(=\text{OH})-\text{N}\)
B
\(-\text{C}(=\text{OH})-\text{NH}\)
C
\(-\text{C}(=\text{O})-\text{NH}_2\)
D
\(-\text{C}(=\text{OH})-\text{NH}_2\)
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Acid amides are carboxylic acid derivatives in which the hydroxyl group (\(-\text{OH}\)) of the carboxyl moiety is replaced by an amino group (\(-\text{NH}_2\)).

Formula / Rule / Reaction:

$$\text{R-COOH} \rightarrow \text{R-CONH}_2 \implies -\text{C}(=\text{O})-\text{NH}_2$$

Solution:

  • Carboxylic acid functional groups have the connectivity \(-\text{C}(=\text{O})-\text{OH}\).


  • Nucleophilic substitution replacing \(-\text{OH}\) with \(-\text{NH}_2\) forms the primary carboxamide functional group.


  • This structure consists of a carbonyl group double-bonded to oxygen and single-bonded to an amino group: \(-\text{C}(=\text{O})-\text{NH}_2\).


Why other options are incorrect:

  • Option A: Contains an impossible double-bonded protonated enol-nitrogen system.
  • Option B: Depicts an invalid enolic structure with improper valencies.
  • Option D: Erroneously places a double bond onto an \(-\text{OH}\) group, violating oxygen divalent rules.
MCQ #82 of 150 Chemistry NUMS 2026
[NUMS 2026]

When an alcohol is heated at 170ºC in the presence of sulphuric acid, a water molecule is eliminated resulting in the formation of:
A
Alkyl halide
B
Alkane
C
Alkene
D
Alkyne
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Heating primary alcohols with concentrated sulfuric acid at 170 degrees Celsius induces intramolecular beta-elimination of water, converting the alcohol into an alkene.

Formula / Rule / Reaction:

$$\text{CH}_3-\text{CH}_2\text{OH} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4, \ 170^\circ\text{C}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$$

Solution:

  • Sulfuric acid protonates the hydroxyl group of ethanol to yield an alkyloxonium ion.


  • Loss of water generates an intermediate carbocation.


  • Elimination of a proton from the adjacent beta-carbon at \(170^\circ\text{C}\) establishes a carbon-carbon double bond, yielding an alkene.


Why other options are incorrect:

  • Option A: Alkyl halides require hydrohalic acids (such as \(\text{HCl}\) or \(\text{HBr}\)) or phosphorus halides.
  • Option B: Alkanes require reducing conditions rather than dehydrating conditions.
  • Option D: Alkynes require vicinal dihalide double dehydrohalogenation.
MCQ #83 of 150 Chemistry NUMS 2026
[NUMS 2026]

The IUPAC name of following compound is:

$$\text{CH}_3-\text{CH}=\text{CH}-(\text{CH}_2)_2-\text{CH}_3$$
A
Hex-2-ene
B
Hept-2-ene
C
Hex-4-ene
D
Hept-4-ene
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

IUPAC rules for alkenes require selecting the longest continuous carbon chain containing the double bond and numbering from the end closest to the unsaturation.

Formula / Rule / Reaction:

$$\overset{1}{\text{C}}\text{H}_3-\overset{2}{\text{C}}\text{H}=\overset{3}{\text{C}}\text{H}-\overset{4}{\text{C}}\text{H}_2-\overset{5}{\text{C}}\text{H}_2-\overset{6}{\text{C}}\text{H}_3$$

Solution:

  • Expanding the condensed formula gives a continuous six-carbon unbranched alkylene chain: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\).


  • The six-carbon root is hexene.


  • Numbering from left to right assigns the double bond to locant 2 (carbons 2 and 3), whereas numbering from right to left assigns it to locant 4.


  • Assigning the lowest locant gives the name hex-2-ene.


Why other options are incorrect:

  • Option B: Hept-2-ene specifies a seven-carbon parent chain, whereas this molecule has six carbons.
  • Option C: Hex-4-ene numbers the double bond from the wrong end of the carbon chain.
  • Option D: Hept-4-ene specifies an incorrect seven-carbon chain and incorrect locant.
MCQ #84 of 150 Chemistry NUMS 2026
[NUMS 2026]

Most of the reactions shown by alkenes are:
A
Nucleophilic substitution
B
Nucleophilic addition
C
Electrophilic substitution
D
Addition reactions
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Alkenes possess a nucleophilic carbon-carbon double bond with exposed pi electrons, making addition reactions their characteristic reaction mechanism.

Formula / Rule / Reaction:

$$\text{R-CH}=\text{CH}_2 + \text{X}-\text{Y} \rightarrow \text{R-CH(X)}-\text{CH}_2(\text{Y})$$

Solution:

  • The pi electron cloud in an alkene is loosely held above and below the internuclear sigma framework.


  • These exposed electrons attack electron-deficient species, initiating electrophilic addition reactions.


  • During the reaction, the pi bond is converted into two new strong sigma bonds, producing an addition adduct.


Why other options are incorrect:

  • Option A: Nucleophilic substitution is typical of alkyl halides and alcohols, not electron-rich alkenes.
  • Option B: Nucleophilic addition occurs across polarized carbonyl groups (aldehydes and ketones).
  • Option C: Electrophilic substitution is characteristic of aromatic benzene rings that maintain aromaticity.
MCQ #85 of 150 Chemistry NUMS 2026
[NUMS 2026]

What are the conditions required for polymerization of ethene to polyethene?
A
400ºC and 1000 atm
B
100ºC and 400 atm
C
450ºC and 1000 atm
D
450ºC and 200 atm
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

According to the Punjab Textbook Board, high-pressure free-radical polymerization of ethene requires heating monomeric gas at 400 degrees Celsius under 1000 atmospheres of pressure with trace oxygen.

Formula / Rule / Reaction:

$$n\text{ CH}_2=\text{CH}_2 \xrightarrow{400^\circ\text{C}, \ 1000\text{ atm}, \ 0.1\% \text{ O}_2} -(-\text{CH}_2-\text{CH}_2-)-_n$$

Solution:

  • Low-density polyethylene (LDPE) is manufactured industrially by free-radical addition polymerization.


  • The specified textbook conditions require heating ethene gas to \(400^\circ\text{C}\) under a pressure of \(1000\text{ atm}\) in the presence of trace (\(0.1\%\)) oxygen catalyst.


Why other options are incorrect:

  • Option B: A temperature of \(100^\circ\text{C}\) and \(400\text{ atm}\) is insufficient to initiate non-catalytic free-radical propagation.
  • Option C: \(450^\circ\text{C}\) exceeds standard textbook conditions and promotes thermal cracking.
  • Option D: \(200\text{ atm}\) provides inadequate compression for high-pressure radical propagation.
MCQ #86 of 150 Chemistry NUMS 2026
[NUMS 2026]

The reactivity order of alkyl halide is:
A
R - F > R - Cl > R - Br > R - I
B
R - Cl > R - Br > R - I > R - F
C
R - Br > R - I > R - F > R - Cl
D
R - I > R - Br > R - Cl > R - F
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The chemical reactivity of alkyl halides is governed primarily by carbon-halogen bond dissociation energy, with lower bond strength producing higher reaction rates.

Formula / Rule / Reaction:

$$\text{Reactivity} \propto \frac{1}{\text{Bond Dissociation Energy}} \implies \text{C-I } (238\text{ kJ}) < \text{C-Br } (276\text{ kJ}) < \text{C-Cl } (338\text{ kJ}) < \text{C-F } (484\text{ kJ})$$

Solution:

  • The carbon-iodine bond is the longest and weakest among carbon-halogen bonds due to poor orbital overlap.


  • Iodide also acts as a superior leaving group because of its large ionic radius and low charge density.


  • As bond energy decreases from fluorine to iodine, reactivity in substitution and elimination reactions increases: \(\text{R-I} > \text{R-Br} > \text{R-Cl} > \text{R-F}\).


Why other options are incorrect:

  • Option A: Reflects the trend in carbon-halogen bond polarity rather than actual chemical reactivity.
  • Option B: Erroneously places alkyl chlorides above bromides and iodides.
  • Option C: Out-of-order sequence that places bromides ahead of iodides.
MCQ #87 of 150 Chemistry NUMS 2026
[NUMS 2026]

(CH3)3C-OH is an example of:
A
Primary alcohol
B
Secondary alcohol
C
Tertiary alcohol
D
Iso butyl alcohol
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Alcohols are classified as primary, secondary, or tertiary according to the degree of substitution of the carbon atom directly bonded to the hydroxyl group.

Formula / Rule / Reaction:

$$\text{R}_3\text{C-OH} \implies \text{Tertiary alcohol}$$

Solution:

  • In \((\text{CH}_3)_3\text{C-OH}\) (tert-butanol), the carbon bonded to the \(-\text{OH}\) group is attached to three other methyl carbon atoms.


  • A carbon atom bonded directly to three adjacent alkyl carbons is designated as a tertiary carbon.


  • Therefore, 2-methylpropan-2-ol is a tertiary alcohol.


Why other options are incorrect:

  • Option A: Primary alcohols have the \(-\text{OH}\) group bonded to a carbon attached to at most one other carbon atom (\(\text{R-CH}_2\text{OH}\)).
  • Option B: Secondary alcohols have the \(-\text{OH}\) group bonded to a carbon attached to two other carbon atoms (\(\text{R}_2\text{CHOH}\)).
  • Option D: Isobutyl alcohol is a primary alcohol with structure \((\text{CH}_3)_2\text{CH-CH}_2\text{OH}\).
MCQ #88 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which one of the following is used in the manufacturing of plastics?
A
Methanol
B
Ethanol
C
Propanol
D
Phenol
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Phenol undergoes step-growth condensation polymerization with formaldehyde to produce phenol-formaldehyde resins, commercially known as Bakelite plastics.

Formula / Rule / Reaction:

$$\text{Phenol} + \text{HCHO} \xrightarrow{\text{Acid or Base}} \text{Bakelite (cross-linked plastic)}$$

Solution:

  • Phenol possesses activated ortho and para positions on its benzene ring.


  • Reaction with formaldehyde under heating yields cross-linked phenolic thermosetting polymers known as Bakelite.


  • These materials are widely used in electrical switches, circuit boards, and molded plastics.


Why other options are incorrect:

  • Option A: Methanol is used as an industrial solvent and chemical feedstock, but is not the polymer backbone monomer.
  • Option B: Ethanol serves as a solvent, disinfectant, and biofuel.
  • Option C: Propanol is used primarily as an industrial solvent.
MCQ #89 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which of the following will give formic acid as one of the product of oxidation?
A
Butanone
B
Propanone
C
Pentanone
D
Hexanone
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

According to Popoff's rule, vigorous oxidation of ketones causes carbon-carbon bond cleavage adjacent to the carbonyl group; oxidation of propanone yields ethanoic acid and methanoic (formic) acid.

Formula / Rule / Reaction:

$$\text{CH}_3-\text{CO}-\text{CH}_3 + 3[\text{O}] \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4} \text{CH}_3\text{COOH} + \text{HCOOH}$$

Solution:

  • Propanone (acetone) is a symmetrical three-carbon ketone.


  • Oxidation with acidified potassium dichromate cleaves a carbon-carbon single bond adjacent to the carbonyl group.


  • This cleavage produces a two-carbon carboxylic acid (acetic acid) and a one-carbon carboxylic acid (formic acid).


  • Formic acid may undergo further oxidation to carbon dioxide and water under prolonged heating.


Why other options are incorrect:

  • Option A: Butanone oxidizes to yield two molecules of ethanoic acid (acetic acid).
  • Option C: Pentanone oxidizes to yield mixtures of ethanoic acid and propanoic acid.
  • Option D: Hexanone oxidation cleaves into propanoic or butanoic acids.
MCQ #90 of 150 Chemistry NUMS 2026
[NUMS 2026]

What will be the IUPAC name of the given compound?

$$\begin{matrix} & \text{CH}_3 & \text{O} \\ & | & || \\ \text{CH}_3-\text{CH}_2- & \text{CH}-\text{CH}- & \text{C}-\text{H} \\ & | & \\ & \text{CH}_2-\text{CH}_3 & \end{matrix}$$
A
1 - Ethyl - 2 - methyl pentanal
B
2 - Ethyl - 3 - methyl pentanal
C
3 - Methyl - 4 - ethyl pentanal
D
3 - Methyl pentanal
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

IUPAC naming of aldehydes requires identifying the longest continuous carbon chain containing the formyl carbon (assigned as C1) and listing substituents alphabetically with their lowest locants.

Formula / Rule / Reaction:

$$\overset{5}{\text{C}}\text{H}_3-\overset{4}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\text{H}(\text{CH}_3)-\overset{2}{\text{C}}\text{H}(\text{CH}_2\text{CH}_3)-\overset{1}{\text{C}}\text{HO}$$

Solution:

  • The principal functional group is an aldehyde (\(-\text{CHO}\)), which is assigned locant 1.


  • Tracing the longest continuous carbon chain starting from C1 leads through the ethyl-bearing carbon (C2), the methyl-bearing carbon (C3), and across the ethyl chain (C4-C5), yielding a 5-carbon pentanal backbone.


  • Substituents on this chain are an ethyl group at position 2 and a methyl group at position 3.


  • Alphabetical ordering places ethyl before methyl, giving 2-ethyl-3-methylpentanal.


Why other options are incorrect:

  • Option A: Locant 1 is reserved for the aldehyde carbon itself; an ethyl group cannot be placed at C1 of a pentanal.
  • Option C: Numbers the carbon chain backwards from the alkyl end rather than prioritizing the principal carbonyl carbon.
  • Option D: Omits the ethyl substituent at position 2.
MCQ #91 of 150 Chemistry NUMS 2026
[NUMS 2026]

What is the common name of following compound?

COOHCOOH
A
Benzoic acid
B
Malonic acid
C
Phthalic acid
D
Picric acid
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Benzene-1,2-dicarboxylic acid consists of two adjacent carboxyl groups bonded to an aromatic benzene ring and is known by the common name phthalic acid.

Formula / Rule / Reaction:

$$\text{C}_6\text{H}_4(\text{COOH})_2 \quad (1,2-\text{isomer}) \implies \text{Phthalic acid}$$

Solution:

  • The compound contains a benzene nucleus with two carboxylic acid functional groups located at adjacent ortho positions (1,2-orientation).


  • The systematic IUPAC designation is benzene-1,2-dicarboxylic acid.


  • The universally recognized common name for this aromatic dicarboxylic acid is phthalic acid.


Why other options are incorrect:

  • Option A: Benzoic acid possesses only one carboxyl group bonded to the benzene ring.
  • Option B: Malonic acid is an aliphatic dicarboxylic acid with formula \(\text{CH}_2(\text{COOH})_2\).
  • Option D: Picric acid is 2,4,6-trinitrophenol.
MCQ #92 of 150 Chemistry NUMS 2026
[NUMS 2026]

Acetic anhydride can be prepared when two molecules of carboxylic acids are dehydrated on heating strongly in the presence of:
A
V2O5
B
P2O5
C
H2SO4
D
HgSO4
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Phosphorus pentoxide is an aggressive chemical dehydrating agent that removes a molecule of water from two molecules of acetic acid upon heating to produce acetic anhydride.

Formula / Rule / Reaction:

$$2\text{ CH}_3\text{COOH} \xrightarrow{\text{P}_2\text{O}_5, \ \Delta} (\text{CH}_3\text{CO})_2\text{O} + \text{H}_2\text{O}$$

Solution:

  • Two acetic acid molecules undergo bimolecular condensation when heated with phosphorus pentoxide (\(\text{P}_2\text{O}_5\)).


  • Phosphorus pentoxide exhibits high affinity for water, hydrating into metaphosphoric acid (\(\text{HPO}_3\)) and driving the elimination of water.


  • This reaction links the two acyl groups via an oxygen bridge, synthesizing acetic anhydride.


Why other options are incorrect:

  • Option A: \(\text{V}_2\text{O}_5\) is an industrial oxidation catalyst used in the Contact process.
  • Option C: \(\text{H}_2\text{SO}_4\) promotes esterification when alcohol is present, but \(\text{P}_2\text{O}_5\) is the definitive dehydrating agent for anhydride synthesis.
  • Option D: \(\text{HgSO}_4\) catalyzes the hydration of alkynes into aldehydes and ketones.
MCQ #93 of 150 Chemistry NUMS 2026
[NUMS 2026]

Carboxylic acid react with metals to form salt with the evolution of which gas?
A
CO2
B
CO
C
H2
D
CH4
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Carboxylic acids act as Bronsted-Lowry acids, reacting with electropositive active metals via single-displacement redox reactions to produce carboxylate salts and hydrogen gas.

Formula / Rule / Reaction:

$$2\text{ RCOOH} + 2\text{Na} \rightarrow 2\text{ RCOONa} + \text{H}_2\uparrow$$

Solution:

  • The acidic proton on the carboxyl group (\(-\text{COOH}\)) undergoes single displacement by active metals such as sodium, potassium, or magnesium.


  • The metal atom is oxidized to its cation, donating electrons to reduce protons into molecular hydrogen gas.


  • Effervescence of diatomic hydrogen gas (\(\text{H}_2\)) is observed.


Why other options are incorrect:

  • Option A: Carbon dioxide gas is evolved when carboxylic acids react with carbonates or bicarbonates, not elemental metals.
  • Option B: Carbon monoxide is produced during incomplete combustion, not acid-metal displacement.
  • Option D: Methane is not evolved during standard acid-metal displacement reactions.
MCQ #94 of 150 Chemistry NUMS 2026
[NUMS 2026]

Which protein helps in muscles contractions and relaxation in our body?
A
Myosin
B
Albumin
C
Globulin
D
Collagen
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Muscle contraction and relaxation depend on the interaction between the contractile proteins actin and myosin within the sarcomeres of muscle fibers.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • Myosin constitutes the thick filaments of striated and smooth muscle myofibrils.


  • Myosin heads possess ATPase activity and bind actin filaments to form cross-bridges.


  • Power strokes executed by myosin heads shorten the sarcomere, driving muscular contraction and subsequent relaxation.


Why other options are incorrect:

  • Option B: Albumin is a major plasma protein responsible for maintaining intravascular colloidal osmotic pressure.
  • Option C: Globulins are plasma proteins involved in humoral immunity and lipid transport.
  • Option D: Collagen is an extracellular structural fibrous protein providing tensile strength to skin and tendons.
MCQ #95 of 150 Chemistry NUMS 2026
[NUMS 2026]

Various polypeptide chains in haemoglobin remain intact due to:
A
Chemical bonds
B
London dispersion forces
C
Salt bridges
D
Peptide bonds
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The quaternary structure of hemoglobin consists of four separate globin subunits held together by non-covalent electrostatic interactions, specifically salt bridges, alongside hydrogen bonding and hydrophobic contacts.

Formula / Rule / Reaction:

$$\text{R-COO}^- \cdots \text{H}_3\text{N}^+-\text{R}' \quad (\text{Salt bridge / Ionic pair})$$

Solution:

  • Hemoglobin is a tetramer composed of two alpha and two beta polypeptide chains.


  • The quaternary arrangement of these distinct subunits is maintained by non-covalent interactions rather than interchain covalent linkages.


  • Electrostatic salt bridges formed between oppositely charged basic and acidic amino acid side chains hold the chains together in the intact quaternary complex.


Why other options are incorrect:

  • Option A: Non-specific distractor that fails to designate the precise non-covalent interactions maintaining quaternary structure.
  • Option B: London dispersion forces are weak van der Waals forces insufficient on their own to organize the quaternary geometry.
  • Option D: Peptide bonds are strong covalent bonds linking amino acids within individual primary chains, but do not join separate globin chains together.
MCQ #96 of 150 Physics NUMS 2026
[NUMS 2026]

The vertical height and horizontal range of projectiles are equal at angle:
A
30º
B
45º
C
76º
D
90º
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The maximum vertical height of a projectile equals its horizontal range when the projection angle satisfies the trigonometric relation \(\tan\theta = 4\).

Formula / Rule / Reaction:

$$H = \frac{v_0^2 \sin^2\theta}{2g}, \quad R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{2v_0^2 \sin\theta \cos\theta}{g}$$

Solution:

  • Equating height and range: \(H = R\).


  • $$\frac{v_0^2 \sin^2\theta}{2g} = \frac{2v_0^2 \sin\theta \cos\theta}{g}$$


  • Canceling common factors \(v_0^2\), \(g\), and \(\sin\theta\) yields: \(\frac{\sin\theta}{2} = 2\cos\theta\).


  • $$\frac{\sin\theta}{\cos\theta} = 4 \implies \tan\theta = 4$$


  • $$\theta = \arctan(4) \approx 75.96^\circ \approx 76^\circ$$


Why other options are incorrect:

  • Option A: At \(30^\circ\), \(R = 4\sqrt{3}H \approx 6.93H\), so range far exceeds height.
  • Option B: At \(45^\circ\), range reaches its maximum where \(R = 4H\).
  • Option D: At \(90^\circ\), the horizontal range is zero while vertical height is maximized.
MCQ #97 of 150 Physics NUMS 2026
[NUMS 2026]

Range of projectile is maximum at:
A
30º
B
45º
C
60º
D
90º
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Horizontal range is proportional to the sine of twice the launch angle, reaching its absolute mathematical maximum when the argument equals 90 degrees.

Formula / Rule / Reaction:

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$

Solution:

  • The trigonometric factor \(\sin(2\theta)\) reaches its maximum value of 1 when its argument is \(90^\circ\).


  • Setting \(2\theta = 90^\circ\) yields \(\theta = 45^\circ\).


  • Therefore, in the absence of air resistance, a launch angle of \(45^\circ\) maximizes horizontal range.


Why other options are incorrect:

  • Option A: At \(30^\circ\), \(\sin(2\theta) = \sin(60^\circ) = 0.866\), achieving \(86.6\%\) of maximum range.
  • Option C: At \(60^\circ\), \(\sin(2\theta) = \sin(120^\circ) = 0.866\), giving the same sub-maximal range as \(30^\circ\).
  • Option D: At \(90^\circ\), \(\sin(180^\circ) = 0\), producing zero horizontal displacement.
MCQ #98 of 150 Physics NUMS 2026
[NUMS 2026]

A ball is thrown vertically upward, the ball descends towards earth after achieving a certain height because:
A
Earth is exerting more gravitational force than ball
B
The ball is exerting more gravitational force than earth
C
The earth possesses more inertia
D
The ball possesses more inertia
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

By Newton's third law, the gravitational forces between Earth and the ball are strictly equal and opposite; however, the Earth's immense mass gives it vast inertia, making the acceleration of the low-mass ball observable.

Formula / Rule / Reaction:

$$F_{\text{Earth on ball}} = F_{\text{ball on Earth}} = G\frac{M m}{r^2}, \quad a_{\text{ball}} = \frac{F}{m}, \quad a_{\text{Earth}} = \frac{F}{M} \approx 0$$

Solution:

  • Gravitational attraction is a mutual action-reaction pair where both bodies experience identical force magnitudes.


  • Inertia is directly proportional to mass. The mass of Earth is \(5.97 \times 10^{24}\text{ kg}\), giving it vastly greater inertia than the ball.


  • Because of the Earth's huge inertia, its acceleration toward the ball is negligible, while the ball experiences an acceleration of \(9.8\text{ m s}^{-2}\) downward.


Why other options are incorrect:

  • Option A: Stating that Earth exerts more force violates Newton's third law of motion.
  • Option B: The ball cannot exert a greater gravitational force than the Earth.
  • Option D: The ball has very small mass and therefore possesses far less inertia than the Earth.
MCQ #99 of 150 Physics NUMS 2026
[NUMS 2026]

A bomb is exploded into two fragments A and B moving in opposite direction. They conserve:
A
Kinetic Energy
B
Potential Energy
C
Momentum
D
Both kinetic energy and potential energy
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

In an explosion, internal chemical potential energy is converted into mechanical kinetic energy, which does not conserve kinetic energy, while linear momentum is conserved in the absence of net external forces.

Formula / Rule / Reaction:

$$\vec{P}_{\text{initial}} = \vec{P}_{\text{final}} = m_A \vec{v}_A + m_B \vec{v}_B = 0$$

Solution:

  • The explosive event is driven entirely by internal chemical forces with zero net external force acting on the system.


  • According to the law of conservation of linear momentum, the total vector momentum before explosion (zero at rest) must equal the total vector momentum after explosion.


  • Kinetic energy is not conserved because chemical potential energy is converted into kinetic energy of the flying fragments.


Why other options are incorrect:

  • Option A: Kinetic energy increases from zero to a positive non-zero value, so it is not conserved.
  • Option B: Potential energy decreases as chemical bonds are broken during detonation.
  • Option D: Neither mechanical kinetic energy nor chemical potential energy is conserved individually.
MCQ #100 of 150 Physics NUMS 2026
[NUMS 2026]

Work done is negative, when the angle between F and S is:
A
B
60º
C
90º
D
180º
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Mechanical work is the scalar dot product of force and displacement vectors; work is negative when the cosine of the angle between them is negative.

Formula / Rule / Reaction:

$$W = \vec{F} \cdot \vec{S} = F S \cos\theta$$

Solution:

  • For \(\theta = 180^\circ\), the force acts directly opposite to the displacement vector.


  • Evaluating the trigonometric factor: \(\cos(180^\circ) = -1\).


  • Substituting gives \(W = F S (-1) = -FS\), which produces negative work (such as work done by kinetic friction).


Why other options are incorrect:

  • Option A: At \(0^\circ\), \(\cos(0^\circ) = +1\), yielding maximum positive work.
  • Option B: At \(60^\circ\), \(\cos(60^\circ) = +0.5\), yielding positive work.
  • Option C: At \(90^\circ\), \(\cos(90^\circ) = 0\), resulting in zero work done.
MCQ #101 of 150 Physics NUMS 2026
[NUMS 2026]

When the speed of the car is doubled, what will be its kinetic energy?
A
Remains same
B
Doubled
C
Three times
D
Four times
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Kinetic energy varies with the square of translational velocity; doubling the speed increases kinetic energy by a factor of four.

Formula / Rule / Reaction:

$$K = \frac{1}{2} m v^2$$

Solution:

  • Let initial kinetic energy be \(K_1 = \frac{1}{2} m v^2\).


  • When speed is doubled, the new velocity becomes \(v' = 2v\).


  • The new kinetic energy is:


  • $$K_2 = \frac{1}{2} m (2v)^2 = \frac{1}{2} m (4v^2) = 4 \left(\frac{1}{2} m v^2\right) = 4 K_1$$


  • Therefore, the kinetic energy becomes four times its original value.


Why other options are incorrect:

  • Option A: Kinetic energy depends directly on speed and cannot remain unchanged.
  • Option B: Momentum doubles when speed doubles, but kinetic energy scales with speed squared.
  • Option C: Three times corresponds to a velocity increase factor of \(\sqrt{3}\).
MCQ #102 of 150 Physics NUMS 2026
[NUMS 2026]

The work efficiency of a dry cell battery is:
A
30%
B
40%
C
50%
D
90%
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

According to standard energy conversion efficiency tables in the Punjab Textbook Board Physics curriculum, electrochemical dry cells operate with a typical efficiency of 90 percent.

Formula / Rule / Reaction:

$$\eta = \frac{\text{Useful Electrical Energy Output}}{\text{Chemical Energy Input}} \times 100\%$$

Solution:

  • Electrochemical reactions inside a primary dry cell convert chemical potential energy directly into electrical energy without an intermediate thermal thermodynamic cycle.


  • Because they are not constrained by Carnot heat engine limits, electrochemical cells have low internal resistance losses and high efficiency.


  • The textbook value for dry cell battery efficiency is 90%.


Why other options are incorrect:

  • Option A: 30% is typical of internal combustion automobile engines.
  • Option B: 40% is typical of fossil-fuel thermal power stations.
  • Option C: 50% is typical of domestic gas heating systems.
MCQ #103 of 150 Physics NUMS 2026
[NUMS 2026]

When distance r increases than gravitational P.E becomes:
A
Less positive
B
More negative
C
Less negative
D
More positive
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Absolute gravitational potential energy is defined with reference to zero at infinity, having a negative value at finite distances that becomes less negative as radial separation increases.

Formula / Rule / Reaction:

$$U = -\frac{G M m}{r}$$

Solution:

  • Gravitational potential energy in an attractive field is negative everywhere within finite distances.


  • As the radial distance \(r\) increases, the magnitude \(\frac{G M m}{r}\) decreases.


  • Because of the negative sign, a decrease in magnitude means \(U\) shifts closer to zero.


  • Therefore, as \(r\) increases, the gravitational potential energy becomes less negative (more algebraically positive).


Why other options are incorrect:

  • Option A: Gravitational potential energy within a bound system is negative, not positive.
  • Option B: Becoming more negative occurs when \(r\) decreases as an object falls toward Earth.
  • Option D: Potential energy approaches zero from negative values and does not become positive in an attractive field.
MCQ #104 of 150 Physics NUMS 2026
[NUMS 2026]

How many Joules are there in 3KWh?
A
108MJ
B
10.8MJ
C
1.08MJ
D
0.108MJ
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

One kilowatt-hour is an electrical energy unit equal to 3.6 megajoules, which scales proportionally when multiplied by the number of units.

Formula / Rule / Reaction:

$$1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J} = 3.6\text{ MJ}$$

Solution:

  • To find the energy in 3 kWh:


  • $$E = 3\text{ kWh} = 3 \times (3.6 \times 10^6\text{ J}) = 10.8 \times 10^6\text{ J}$$


  • Converting to megajoules (\(1\text{ MJ} = 10^6\text{ J}\)) yields \(10.8\text{ MJ}\).


Why other options are incorrect:

  • Option A: 108 MJ corresponds to 30 kWh, which is an error by a factor of 10.
  • Option C: 1.08 MJ corresponds to 0.3 kWh.
  • Option D: 0.108 MJ corresponds to 0.03 kWh.
MCQ #105 of 150 Physics NUMS 2026
[NUMS 2026]

A car moving on a more steeply banked curves will require:
A
Large speed and large radii
B
Large speed and small radii
C
Small speed and large radii
D
Small speed and small radii
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The required banking angle of a curved track is directly proportional to the square of vehicle speed and inversely proportional to the radius of curvature.

Formula / Rule / Reaction:

$$\tan\theta = \frac{v^2}{r g}$$

Solution:

  • A more steeply banked curve corresponds to a larger banking angle \(\theta\), which increases \(\tan\theta\).


  • For \(\frac{v^2}{r g}\) to have a larger value, the numerator (speed \(v\)) must be large and the denominator (radius \(r\)) must be small.


  • Therefore, negotiating a more steeply banked curve without relying on friction requires higher speed and tighter (smaller) radii.


Why other options are incorrect:

  • Option A: A large radius decreases \(\tan\theta\), requiring a shallower banking angle.
  • Option C: Small speed and large radius result in a small banking angle.
  • Option D: A small speed reduces centripetal demand and does not require steep banking.
MCQ #106 of 150 Physics NUMS 2026
[NUMS 2026]

In a body angular acceleration is produced by:
A
Net force
B
Power
C
Pressure
D
Net torque
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

In rotational dynamics, angular acceleration is produced when an unbalanced net torque acts on a rigid body having rotational inertia.

Formula / Rule / Reaction:

$$\vec{\tau}_{\text{net}} = I \vec{\alpha} \implies \vec{\alpha} = \frac{\vec{\tau}_{\text{net}}}{I}$$

Solution:

  • Newton's second law for rotation states that the net external torque applied to an object is proportional to the resulting angular acceleration.


  • The constant of proportionality is the moment of inertia \(I\).


  • Therefore, a net torque is required to produce an angular acceleration.


Why other options are incorrect:

  • Option A: A net force produces linear acceleration, not angular acceleration unless applied at a distance to generate torque.
  • Option B: Power is the rate of doing work, not the rotational force analog.
  • Option C: Pressure is force per unit area and does not determine rotational acceleration.
MCQ #107 of 150 Physics NUMS 2026
[NUMS 2026]

The analogue of force in circular / angular motion is:
A
Momentum
B
Torque
C
Angular acceleration
D
Angular velocity
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Torque (moment of force) is the rotational analogue of linear force, acting as the dynamic quantity that changes the angular state of motion.

Formula / Rule / Reaction:

$$\text{Linear: } F = m a \quad \longleftrightarrow \quad \text{Rotational: } \tau = I \alpha$$

Solution:

  • Linear force produces linear acceleration and changes linear momentum: \(F = \frac{dp}{dt}\).


  • Torque produces angular acceleration and changes angular momentum: \(\tau = \frac{dL}{dt}\).


  • Comparing the equations shows that torque is the rotational analogue of force.


Why other options are incorrect:

  • Option A: Linear momentum corresponds to angular momentum in rotational motion.
  • Option C: Angular acceleration is the analogue of linear acceleration.
  • Option D: Angular velocity is the analogue of linear velocity.
MCQ #108 of 150 Physics NUMS 2026
[NUMS 2026]

1º is equals to:
A
\pi/90 radians
B
\pi/180 radians
C
\pi radians
D
2\pi radians
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

A full planar circle subtends \(360^\circ\) degrees or \(2\pi\) radians, establishing the conversion factor between degrees and radians.

Formula / Rule / Reaction:

$$180^\circ = \pi\text{ rad} \implies 1^\circ = \frac{\pi}{180}\text{ rad}$$

Solution:

  • A complete revolution contains \(360^\circ = 2\pi\text{ radians}\).


  • Dividing both sides by 180 gives: \(180^\circ = \pi\text{ radians}\).


  • Solving for one degree yields: \(1^\circ = \frac{\pi}{180}\text{ radians} \approx 0.01745\text{ radians}\).


Why other options are incorrect:

  • Option A: \(\pi/90\) radians corresponds to \(2^\circ\).
  • Option C: \(\pi\) radians corresponds to \(180^\circ\).
  • Option D: \(2\pi\) radians corresponds to a full revolution of \(360^\circ\).
MCQ #109 of 150 Physics NUMS 2026
[NUMS 2026]

In uniform circular motion, the angle between centripetal acceleration and linear velocity of a particle is:
A
\pi/2
B
\pi
C
2\pi
D
Zero
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

In uniform circular motion, instantaneous linear velocity is directed tangentially while centripetal acceleration points radially inward, making them perpendicular.

Formula / Rule / Reaction:

$$\vec{a}_c \cdot \vec{v} = 0 \implies \theta = 90^\circ = \frac{\pi}{2}\text{ rad}$$

Solution:

  • Centripetal acceleration points continuously along the radius toward the center of rotation.


  • Linear velocity is directed along the tangent to the circular path at that instant.


  • The tangent to a circle is perpendicular to its radius at every point, so the angle between the two vectors is \(90^\circ\) (\(\pi/2\) radians).


Why other options are incorrect:

  • Option B: An angle of \(\pi\) (\(180^\circ\)) indicates antiparallel vectors, such as velocity and opposing deceleration.
  • Option C: An angle of \(2\pi\) (\(360^\circ\)) represents parallel alignment.
  • Option D: An angle of zero describes parallel vectors, which would change the speed rather than the direction of motion.
MCQ #110 of 150 Physics NUMS 2026
[NUMS 2026]

The distance between two consecutive nodes are:
A
\lambda
B
\lambda/2
C
2\lambda
D
\lambda/4
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

In a standing wave, adjacent nodes are separated by half a wavelength, representing one complete loop between points of zero displacement.

Formula / Rule / Reaction:

$$d_{\text{node-node}} = \frac{\lambda}{2}$$

Solution:

  • A full wavelength (\(\lambda\)) of a standing wave contains two consecutive loops.


  • Each loop is bounded by a node at each end with an antinode located in the center.


  • Therefore, the linear distance between two consecutive nodes is equal to one half-wavelength (\(\lambda/2\)).


Why other options are incorrect:

  • Option A: \(\lambda\) is the distance between three consecutive nodes (two complete loops).
  • Option C: \(2\lambda\) corresponds to four complete loops.
  • Option D: \(\lambda/4\) is the distance between an adjacent node and antinode.
MCQ #111 of 150 Physics NUMS 2026
[NUMS 2026]

If a sound waves moves from air to water, there is NO change in:
A
Velocity
B
Frequency
C
Wavelength
D
Temperature
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

When a wave crosses the boundary between two media, its frequency is determined solely by the vibrating source and remains invariant.

Formula / Rule / Reaction:

$$v = f \lambda \implies f = \text{constant}$$

Solution:

  • The frequency of a wave is determined by the oscillation rate of the source creating the disturbance.


  • When sound transitions from air to water, it enters a medium with higher bulk modulus, causing its speed \(v\) to increase from approximately \(340\text{ m s}^{-1}\) to \(1500\text{ m s}^{-1}\).


  • To satisfy \(v = f \lambda\), the wavelength \(\lambda\) increases proportionally, keeping frequency \(f\) constant.


Why other options are incorrect:

  • Option A: Velocity changes significantly due to differences in elasticity and density between media.
  • Option C: Wavelength changes in direct proportion to velocity.
  • Option D: Temperature is a thermodynamic state variable of each individual medium rather than an intrinsic wave property.
MCQ #112 of 150 Physics NUMS 2026
[NUMS 2026]

The star moving closer to earth will experience:
A
Blue shift
B
Red shift
C
Orange shift
D
Yellow shift
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

According to the Doppler effect for light, relative motion of a source toward an observer shifts observed spectral lines toward higher frequencies and shorter wavelengths (blue shift).

Formula / Rule / Reaction:

$$\Delta \lambda = -\frac{v}{c}\lambda_0 \implies \lambda_{\text{observed}} < \lambda_0$$

Solution:

  • When an astronomical light source moves toward Earth, each successive wave crest is emitted closer to the observer.


  • The wavefronts compress, shortening the observed wavelength.


  • In the visible spectrum, shorter wavelengths correspond to the blue end, producing a blue shift.


Why other options are incorrect:

  • Option B: Red shift occurs when a star is moving away from Earth, shifting spectra toward longer wavelengths.
  • Option C: Orange shift is not a designated astronomical Doppler term.
  • Option D: Yellow shift is not a recognized Doppler term.
MCQ #113 of 150 Physics NUMS 2026
[NUMS 2026]

If the amplitude of a wave is doubled, its intensity is:
A
Doubled
B
Halved
C
Quadrupled
D
One quarter
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The intensity of a mechanical or electromagnetic wave is directly proportional to the square of its wave amplitude.

Formula / Rule / Reaction:

$$I \propto A^2$$

Solution:

  • Wave intensity represents power transmitted per unit cross-sectional area.


  • Let the initial intensity be \(I_1 = k A^2\).


  • When the amplitude is doubled (\(A' = 2A\)), the new intensity is:


  • $$I_2 = k (2A)^2 = k (4A^2) = 4 (k A^2) = 4 I_1$$


  • Therefore, the intensity is quadrupled.


Why other options are incorrect:

  • Option A: Intensity depends on the square of amplitude, so it increases by a factor of four rather than two.
  • Option B: Halving occurs when amplitude decreases by a factor of \(\sqrt{2}\).
  • Option D: Intensity drops to one quarter when the amplitude is halved, not doubled.
MCQ #114 of 150 Physics NUMS 2026
[NUMS 2026]

Which of the following is an example of first law of thermodynamics?
A
Thunderstorm
B
Metabolism
C
Rusting
D
Earthquake
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The first law of thermodynamics is the principle of conservation of energy, which asserts that energy cannot be created or destroyed, only transformed from one form to another within an isolated or biological system.

Formula / Rule / Reaction:

$$\Delta U = Q - W \quad \text{or} \quad \Delta E = q + w$$

Solution:

  • In biological organisms, human metabolism serves as a prime example of the first law of thermodynamics.


  • Chemical potential energy ingested through food is converted into mechanical work by contracting muscles and thermal energy to maintain body temperature.


  • The total internal energy change precisely matches the difference between heat exchanged and work performed.


Why other options are incorrect:

  • Option A: Thunderstorms involve complex macroscopic atmospheric discharges driven by thermodynamic instability and entropy gradients.
  • Option C: Rusting is a spontaneous electrochemical corrosion process typically studied under chemical kinetics and redox equilibria.
  • Option D: Earthquakes represent mechanical strain-energy release along tectonic faults.
MCQ #115 of 150 Physics NUMS 2026
[NUMS 2026]

Thermos flask containing tea as a system is shaken vigorously then there will be:
A
Increase in temperature only
B
Increase in kinetic energy only
C
Increase in temperature & kinetic energy
D
No change in temperature & kinetic energy
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Vigorously shaking an insulated vessel performs mechanical work on the fluid; in an adiabatic system, this work increases internal energy, which elevates molecular kinetic energy and temperature.

Formula / Rule / Reaction:

$$Q = 0 \implies \Delta U = W > 0, \quad \langle K.E. \rangle = \frac{3}{2} k_B T$$

Solution:

  • A thermos flask has insulated walls that prevent heat transfer (\(Q = 0\)).


  • Shaking the liquid performs mechanical work (\(W\)) on the system against internal fluid friction and viscosity.


  • This non-flow work increases the internal energy of the liquid, raising average molecular translational kinetic energy.


  • Because macroscopic absolute temperature is directly proportional to average molecular kinetic energy, both temperature and kinetic energy increase.


Why other options are incorrect:

  • Option A: Omits kinetic energy, which is the direct microscopic manifestation of temperature.
  • Option B: Omits temperature, which rises proportionally with molecular kinetic energy.
  • Option D: Violates the first law of thermodynamics by neglecting the mechanical work done on the fluid.
MCQ #116 of 150 Physics NUMS 2026
[NUMS 2026]

Difference between molar specific heat at constant pressure and molar specific heat at constant volume is equal to:
A
Boltzmann constant
B
Universal gas constant
C
Plank's constant
D
Rydberg constant
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Mayer's relation dictates that the difference between the molar heat capacity of an ideal gas at constant pressure and at constant volume equals the universal molar gas constant.

Formula / Rule / Reaction:

$$C_p - C_v = R$$

Solution:

  • When heat is supplied at constant volume (\(C_v\)), all added energy goes into increasing internal energy because no boundary work is done.


  • When heat is supplied at constant pressure (\(C_p\)), the gas must also expand against external pressure to perform work: \(W = P\Delta V = R\Delta T\).


  • The additional energy required per mole per kelvin is \(R\), yielding \(C_p - C_v = R\).


Why other options are incorrect:

  • Option A: The Boltzmann constant (\(k_B = R/N_A\)) represents gas constant per single molecule, not per mole.
  • Option C: Planck's constant relates photon energy to frequency.
  • Option D: The Rydberg constant relates atomic spectral wavelengths.
MCQ #117 of 150 Physics NUMS 2026
[NUMS 2026]

Coulomb's law of electrostatic induction is valid for:
A
Point charges
B
Stationary charges
C
Stationary & point charges
D
Accelerated point charges
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Coulomb's law mathematically defines the electrostatic force acting between electric charges and is strictly valid only when charges are stationary point charges.

Formula / Rule / Reaction:

$$F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}$$

Solution:

  • The law requires the spatial linear dimensions of the charged bodies to be negligible compared to the separation distance \(r\), satisfying the point charge condition.


  • The charges must be stationary (electrostatic) to eliminate magnetic field generation and electromagnetic radiation.


  • Therefore, Coulomb's formulation is valid strictly for stationary and point charges.


Why other options are incorrect:

  • Option A: Incomplete because moving point charges produce magnetic fields and retarding potentials that alter the interaction force.
  • Option B: Incomplete because extended stationary charges experience charge redistribution via polarization.
  • Option D: Accelerated charges radiate electromagnetic energy, violating static Coulomb equations.
MCQ #118 of 150 Physics NUMS 2026
[NUMS 2026]

If the positive charge is placed in a uniform electric field, the charge will move in the:
A
Opposite direction of electric field
B
Direction of electric field
C
Perpendicular to the direction of electric field
D
Remains at rest
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The electrostatic force experienced by a charge in an electric field is the product of charge and electric field intensity; for a positive charge, force and field vectors are parallel.

Formula / Rule / Reaction:

$$\vec{F} = q\vec{E}$$

Solution:

  • The scalar charge \(q\) is positive (\(q > 0\)).


  • Because \(q\) is positive, the vector \(\vec{F}\) points in the identical direction as \(\vec{E}\).


  • Newton's second law gives \(\vec{a} = \vec{F}/m\), so the positive charge accelerates and moves in the direction of the electric field.


Why other options are incorrect:

  • Option A: A negative charge moves opposite to the electric field because \(q < 0\).
  • Option C: Perpendicular acceleration occurs in magnetic fields via Lorentz force (\(\vec{F} = q\vec{v}\times\vec{B}\)), not static electric fields.
  • Option D: The charge experiences an unbalanced electrostatic force and cannot remain at rest.
MCQ #119 of 150 Physics NUMS 2026
[NUMS 2026]

For each element, value of half-life is:
A
Variable
B
Constant
C
Decrease
D
Increase
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Radioactive decay is a spontaneous nuclear transformation governed by first-order kinetics, making the half-life an intrinsic, invariant constant for each radioactive isotope.

Formula / Rule / Reaction:

$$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} = \text{constant}$$

Solution:

  • The decay constant \(\lambda\) is an immutable nuclear characteristic of a given radioisotope.


  • Half-life does not depend on initial sample mass, temperature, external pressure, or chemical combination.


  • Consequently, the half-life for each specific radioactive element or nuclide is constant.


Why other options are incorrect:

  • Option A: Radioactive half-life does not vary with sample size or environmental conditions.
  • Option C: Half-life remains fixed throughout the lifetime of the radioactive material.
  • Option D: Half-life does not increase as the active nucleus decays.
MCQ #120 of 150 Physics NUMS 2026
[NUMS 2026]

The unit of RC is:
A
Ohm
B
Farad
C
Second
D
Ohm/Farad
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

In capacitive circuits, the product of resistance \(R\) and capacitance \(C\) represents the capacitive time constant, having fundamental dimensions of time measured in seconds.

Formula / Rule / Reaction:

$$\tau = R C \implies [\tau] = [\text{Resistance}] \times [\text{Capacitance}]$$

Solution:

  • Resistance has units of volts per ampere: \(\text{V} / \text{A}\).


  • Capacitance has units of coulombs per volt: \(\text{C} / \text{V}\).


  • Multiplying the units yields:


  • $$[RC] = \left(\frac{\text{V}}{\text{A}}\right) \times \left(\frac{\text{C}}{\text{V}}\right) = \frac{\text{C}}{\text{A}} = \frac{\text{A} \cdot \text{s}}{\text{A}} = \text{second}$$


Why other options are incorrect:

  • Option A: Ohm is the SI unit of electrical resistance and reactance.
  • Option B: Farad is the SI unit of capacitance.
  • Option D: Ohm/Farad represents an incorrect dimensional ratio.
MCQ #121 of 150 Physics NUMS 2026
[NUMS 2026]

The fractional change in resistance per kelvin is known as the temperature coefficient of:
A
Conductance
B
Resistance
C
Inductance
D
Conductivity
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The temperature coefficient of resistance is defined as the fractional change in electrical resistance per unit change in thermodynamic temperature.

Formula / Rule / Reaction:

$$\alpha = \frac{R_t - R_0}{R_0 \Delta T} = \frac{\Delta R}{R_0 \Delta T}$$

Solution:

  • The term \(\Delta R / R_0\) expresses the fractional change in resistance relative to baseline resistance at \(0^\circ\text{C}\).


  • Dividing this fractional ratio by the temperature change in kelvin (\(\Delta T\)) gives the temperature coefficient of resistance (\(\alpha\)).


  • Its SI unit is \(\text{K}^{-1}\).


Why other options are incorrect:

  • Option A: Conductance is the reciprocal of resistance, measured in siemens.
  • Option C: Inductance characterizes magnetic flux linkage per unit current in a coil.
  • Option D: Temperature coefficient of conductivity defines fractional variations in specific electrical conductivity.
MCQ #122 of 150 Physics NUMS 2026
[NUMS 2026]

Which of the following is an example of Ohmic in nature?
A
Thermistor
B
Filament lamp
C
Semiconductor
D
Conductor
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Ohmic devices maintain a strictly linear current-voltage relationship where resistance remains constant independent of applied potential under isothermal conditions.

Formula / Rule / Reaction:

$$V = I R \implies R = \text{constant}$$

Solution:

  • Metallic conductors (such as copper, silver, and aluminum) obey Ohm's law over a wide potential range at constant temperature.


  • Their current-voltage characteristics form a straight line passing through the origin.


  • Therefore, metallic conductors represent ohmic materials.


Why other options are incorrect:

  • Option A: Thermistors are non-ohmic devices whose resistance drops exponentially with rising temperature.
  • Option B: Filament lamps are non-ohmic because resistive heating increases resistance at higher voltages, yielding a curved I-V plot.
  • Option C: Semiconductors display non-linear exponential or threshold conduction characteristics.
MCQ #123 of 150 Physics NUMS 2026
[NUMS 2026]

The material whose resistance decreases with increase in temperature have:
A
Positive temperature coefficient
B
Negative temperature coefficient
C
Thermal coefficient
D
Coefficient of friction
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Materials where thermal excitation liberates intrinsic charge carriers to lower electrical resistance with increasing temperature possess a negative temperature coefficient.

Formula / Rule / Reaction:

$$\alpha = \frac{\Delta R}{R_0 \Delta T} < 0 \quad (\text{when } \Delta R < 0 \text{ for } \Delta T > 0)$$

Solution:

  • In semiconductors and thermistors, elevated temperatures provide energy to break covalent bonds.


  • This releases additional free electrons and holes into the conduction band, increasing electrical conductivity.


  • Because resistance decreases as temperature increases (\(\Delta R < 0\)), the coefficient \(\alpha\) is negative.


Why other options are incorrect:

  • Option A: Metallic conductors have a positive temperature coefficient because thermal lattice vibrations increase electron scattering and resistance.
  • Option C: Thermal coefficient is an incomplete term that lacks sign designation.
  • Option D: Coefficient of friction governs mechanical resistive force between contacting surfaces.
MCQ #124 of 150 Physics NUMS 2026
[NUMS 2026]

The number of magnetic field lines passing through unit area is called:
A
Magnetic field
B
Magnetic flux
C
Magnetic flux density
D
Magnetization
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Magnetic flux density (also known as magnetic induction) is defined as the magnetic flux passing perpendicularly through a unit cross-sectional area.

Formula / Rule / Reaction:

$$B = \frac{\Phi}{A}$$

Solution:

  • Magnetic flux (\(\Phi\)) represents the total number of magnetic field lines passing through an arbitrary surface area.


  • When the total number of lines is normalized per unit area held perpendicular to the field lines, the quantity is magnetic flux density (\(B\)).


  • Its SI unit is the tesla (\(\text{T} = \text{Wb m}^{-2}\)).


Why other options are incorrect:

  • Option A: Magnetic field is a general term describing the region of magnetic influence.
  • Option B: Magnetic flux measures the total lines passing through a given surface without dividing by area.
  • Option D: Magnetization is magnetic dipole moment per unit volume of a material.
MCQ #125 of 150 Physics NUMS 2026
[NUMS 2026]

The magnetic force of a charged particle moving in a uniform magnetic fields depends on:
A
Velocity, area & magnetic field
B
Charge, current & velocity
C
Magnetic field, voltage & charge
D
Velocity, magnetic field & charge
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The magnetic Lorentz force on a moving point charge is directly proportional to the magnitude of the charge, its velocity, and the external magnetic flux density.

Formula / Rule / Reaction:

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

Solution:

  • The scalar magnitude of the magnetic force is determined by four variables:


  • Magnitude of electric charge (\(q\)).


  • Speed of particle motion (\(v\)).


  • Strength of uniform magnetic field (\(B\)).


  • Sine of the orientation angle (\(\theta\)) between velocity and magnetic field vectors.


Why other options are incorrect:

  • Option A: Area is irrelevant to the trajectory force of an isolated point charge.
  • Option B: Current describes macroscopic charge flow over time rather than single-particle kinematics.
  • Option C: Voltage (potential difference) is not an explicit parameter in the Lorentz force equation.
MCQ #126 of 150 Physics NUMS 2026
[NUMS 2026]

The magnetic force acting on a charged particle is maximum when the angle between \(\vec{v}\) and \(\vec{B}\) is:
A
B
45º
C
90º
D
180º
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The magnitude of the magnetic force depends on the sine of the angle between velocity and magnetic field vectors, reaching a maximum when the angle is 90 degrees.

Formula / Rule / Reaction:

$$F = q v B \sin\theta$$

Solution:

  • The trigonometric function \(\sin\theta\) oscillates between \(-1\) and \(+1\).


  • The maximum absolute value occurs when \(\sin\theta = 1\).


  • Solving for angle gives \(\theta = 90^\circ\).


  • Thus, maximum deflecting force occurs when a charged particle enters perpendicular to the magnetic field.


Why other options are incorrect:

  • Option A: At \(0^\circ\), \(\sin(0^\circ) = 0\), resulting in zero magnetic force.
  • Option B: At \(45^\circ\), \(\sin(45^\circ) = 0.707\), delivering \(70.7\%\) of maximum force.
  • Option D: At \(180^\circ\), \(\sin(180^\circ) = 0\), yielding zero magnetic force.
MCQ #127 of 150 Physics NUMS 2026
[NUMS 2026]

Generator effect is practical application of:
A
Static e.m.f
B
Dynamic e.m.f
C
Mutual induction
D
Back emf
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

An electric generator converts mechanical energy into electrical energy by rotating a conductor coil across a magnetic field, inducing a motional or dynamic electromotive force.

Formula / Rule / Reaction:

$$\varepsilon = v B L \sin\theta \quad \text{or} \quad \varepsilon = N B A \omega \sin(\omega t)$$

Solution:

  • Dynamic (motional) emf is induced whenever a conductor moves relative to a stationary magnetic field, sweeping across magnetic flux lines.


  • In electrical AC and DC generators, prime movers rotate armature coils within a magnetic field.


  • This mechanical rotation generates a dynamic induced electromotive force.


Why other options are incorrect:

  • Option A: Static emf is induced without physical motion, as seen across transformer windings via time-varying magnetic fields.
  • Option C: Mutual induction is the operating principle of electrical transformers.
  • Option D: Back emf is the opposing counter-voltage generated within electric motors during operation.
MCQ #128 of 150 Physics NUMS 2026
[NUMS 2026]

If number of turns of coil becomes double, the induced emf will be:
A
Double
B
Reduced to half
C
Remain same
D
Quadruple
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Faraday's law of electromagnetic induction states that the magnitude of induced electromotive force is directly proportional to the total number of turns in the coil.

Formula / Rule / Reaction:

$$\varepsilon = -N \frac{\Delta \Phi}{\Delta t} \implies \varepsilon \propto N$$

Solution:

  • Let initial induced emf be \(\varepsilon_1 = -N \frac{\Delta \Phi}{\Delta t}\).


  • Doubling the number of turns makes \(N' = 2N\).


  • The new induced electromotive force is:


  • $$\varepsilon_2 = - (2N) \frac{\Delta \Phi}{\Delta t} = 2 \left(-N \frac{\Delta \Phi}{\Delta t}\right) = 2 \varepsilon_1$$


  • Therefore, the induced electromotive force doubles.


Why other options are incorrect:

  • Option B: Induced emf is directly proportional to turns; it does not halve.
  • Option C: Emf depends linearly on coil turns and cannot remain constant if turns change.
  • Option D: Quadrupling occurs in self-inductance (\(L \propto N^2\)), not simple linear Faraday emf.
MCQ #129 of 150 Physics NUMS 2026
[NUMS 2026]

Electromagnetic sensor in seismograph converts ground movement into:
A
Magnetic signals
B
Electric signals
C
Audio signals
D
Vibratory signals
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Seismographic electromagnetic transducers operate via Faraday's law of induction, translating relative mechanical ground motion into proportional electrical voltage signals.

Formula / Rule / Reaction:

$$\text{Mechanical ground displacement} \xrightarrow{\text{Electromagnetic induction}} \text{Voltage signal } \varepsilon(t)$$

Solution:

  • A seismometer contains an inertial mass suspended within a frame that moves with seismic ground waves.


  • Relative motion between a coil and magnet fixed to the frame changes magnetic flux linkage.


  • This induces an electrical signal whose amplitude and frequency match the ground vibrations, allowing recording and display.


Why other options are incorrect:

  • Option A: The sensor utilizes existing magnetic fields to generate voltages rather than emitting external magnetic signals.
  • Option C: The direct output is an electrical waveform rather than acoustic sound waves.
  • Option D: Ground movement already is a vibratory signal; the sensor transduces it into electrical signals.
MCQ #130 of 150 Physics NUMS 2026
[NUMS 2026]

A full wave bridge rectifier consists of:
A
No diode
B
Four diodes
C
Two diodes
D
One diode
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

A bridge rectifier utilizes four semiconductor diodes configured in a closed Wheatstone-type bridge network to convert both halves of an AC cycle into pulsating DC.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • A full-wave bridge rectifier circuit routes current through four diodes arranged in a ring.


  • During the positive half-cycle, two diagonally opposed diodes conduct in series.


  • During the negative half-cycle, the remaining two diodes conduct, maintaining unipolar current flow across the load.


Why other options are incorrect:

  • Option A: Rectification requires non-linear rectifying diode junctions.
  • Option C: Two diodes are utilized in a center-tapped transformer full-wave rectifier, not a bridge configuration.
  • Option D: A single diode produces half-wave rectification.
MCQ #131 of 150 Physics NUMS 2026
[NUMS 2026]

An electrical component used to convert AC into DC is:
A
Transistor
B
Capacitor
C
Diode
D
Resistor
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

A semiconductor p-n junction diode allows unidirectional current flow under forward bias and blocks current under reverse bias, enabling electrical rectification of AC to DC.

Formula / Rule / Reaction:

$$\text{AC waveform} \xrightarrow{\text{Diode rectification}} \text{Pulsating DC}$$

Solution:

  • Rectification is the process of converting alternating current (AC) into direct current (DC).


  • Because a p-n junction diode conducts current primarily in one forward direction, it blocks alternating reverse half-cycles.


  • This one-way conduction converts alternating waveforms into direct current.


Why other options are incorrect:

  • Option A: Transistors function as electronic amplifiers and switches.
  • Option B: Capacitors store electrostatic charge and act as smoothing filters.
  • Option D: Resistors provide passive opposition to current without rectifying polarities.
MCQ #132 of 150 Physics NUMS 2026
[NUMS 2026]

A photon has:
A
Zero mass, zero momentum
B
Finite mass, finite momentum
C
Zero mass, finite momentum
D
Finite mass, zero momentum
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

In relativistic mechanics, a photon travels at light speed in vacuum with zero rest mass, yet carries a definite, finite momentum proportional to its frequency.

Formula / Rule / Reaction:

$$m_0 = 0, \quad E = p c = h f \implies p = \frac{h}{\lambda}$$

Solution:

  • Special relativity dictates that any entity traveling at speed \(c\) must possess zero invariant rest mass (\(m_0 = 0\)).


  • According to the de Broglie relationship and Compton scattering experiments, photons carry finite momentum given by \(p = h / \lambda\).


  • Therefore, a photon has zero rest mass and finite momentum.


Why other options are incorrect:

  • Option A: Photons exert radiation pressure because they carry finite momentum.
  • Option B: Photons do not possess finite invariant rest mass.
  • Option D: Inverts physical reality; rest mass is zero while momentum is non-zero.
MCQ #133 of 150 Physics NUMS 2026
[NUMS 2026]

Photo-conductive devices like photo cells and solar cells are application of:
A
Photo electric effect
B
Compton's effect
C
Pair production
D
Interference of light
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

Photo-conductive and photovoltaic devices operate by absorbing incident photons whose energy frees bound valence electrons, an application of the photoelectric phenomenon.

Formula / Rule / Reaction:

$$h f = \Phi + K.E_{\text{max}}$$

Solution:

  • Photocells and solar cells rely on light interacting with photosensitive materials.


  • Photons with energy greater than the work function or bandgap promote electrons into conduction states.


  • This internal and external photo-emission underpins photoelectric switches and solar power generation.


Why other options are incorrect:

  • Option B: Compton scattering involves inelastic scattering of high-energy X-ray or gamma photons by atomic electrons.
  • Option C: Pair production is the creation of an electron-positron pair from a high-energy gamma photon near a heavy nucleus.
  • Option D: Interference describes wave superposition that produces fringe patterns without charge liberation.
MCQ #134 of 150 Physics NUMS 2026
[NUMS 2026]

Which of the following particle has smaller De Broglie wave length?
A
Gamma particle
B
Beta particle
C
Proton
D
Alpha particle
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The de Broglie wavelength of a matter wave is inversely proportional to momentum; for particles at comparable speeds, greater mass produces a smaller de Broglie wavelength.

Formula / Rule / Reaction:

$$\lambda = \frac{h}{p} = \frac{h}{m v}$$

Solution:

  • An alpha particle (helium-4 nucleus) has an atomic mass of approximately \(4\text{ u}\) (\(m_\alpha \approx 6.64 \times 10^{-27}\text{ kg}\)).


  • A proton has a mass of approximately \(1\text{ u}\) (\(m_p \approx 1.67 \times 10^{-27}\text{ kg}\)).


  • A beta particle is an electron with mass \(m_e \approx 9.11 \times 10^{-31}\text{ kg}\).


  • Because the alpha particle has the largest mass among the material particles, its momentum \(p\) is greatest, producing the smallest de Broglie wavelength.


Why other options are incorrect:

  • Option A: Gamma radiation consists of massless photons rather than material particles.
  • Option B: Beta particles have small mass, yielding larger matter wavelengths.
  • Option C: Protons are one-fourth the mass of alpha particles and produce longer wavelengths at equal velocities.
MCQ #135 of 150 Physics NUMS 2026
[NUMS 2026]

In the spectrum of hydrogen atom which of the following series has smallest wavelength?
A
Paschen series
B
Balmer series
C
Lyman series
D
Brackett series
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

According to the Rydberg formula, photon wavelength is inversely proportional to transition energy; the Lyman series involves de-excitations to the lowest ground level (n=1), producing the highest energies and smallest wavelengths.

Formula / Rule / Reaction:

$$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad E = \frac{h c}{\lambda}$$

Solution:

  • The Lyman series corresponds to transitions ending at the ground state \(n_1 = 1\).


  • Because \(n_1 = 1\) provides the deepest potential well, transitions into this level release the largest energy quanta.


  • Since wavelength is inversely related to photon energy (\(\lambda = h c / \Delta E\)), the highest energy transitions emit the smallest wavelengths (ultraviolet region).


Why other options are incorrect:

  • Option A: Paschen series transitions end at \(n_1 = 3\) in the infrared region with longer wavelengths.
  • Option B: Balmer series transitions terminate at \(n_1 = 2\) in the visible spectrum.
  • Option D: Brackett series transitions end at \(n_1 = 4\) in the far-infrared region.
MCQ #136 of 150 English NUMS 2026
[NUMS 2026]

The antonym for the word clarity is:
A
Upright
B
Understanding
C
Vagueness
D
Clear
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Clarity denotes the quality of being clear, lucidity, precision, and easy intelligibility; its direct semantic antonym is vagueness, which denotes lack of clearness or precision.

Formula / Rule / Reaction:

$$\text{Clarity } (\text{precision/lucidity}) \iff \text{Antonym: Vagueness } (\text{ambiguity/obscurity})$$

Solution:

  • Clarity refers to expression that is distinct, easily understood, and free from obscurity.


  • Vagueness describes statements or thoughts that are indistinct, imprecise, and poorly defined.


  • Thus, vagueness is the direct antonym.


Why other options are incorrect:

  • Option A: Upright refers to vertical posture or moral integrity.
  • Option B: Understanding is the cognitive capacity to comprehend meaning.
  • Option D: Clear is a direct synonym of clarity, not an antonym.
MCQ #137 of 150 English NUMS 2026
[NUMS 2026]

Yasir Ameen could not pay the rent; accordingly, he was evicted. Meaning of the underlined word is:
A
Expelled from the house
B
Investigated
C
Threatened
D
Rewarded
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The verb to evict means to legally expel or dispossess a tenant from property or leased premises, especially for failure to pay rental arrears.

Formula / Rule / Reaction:

Factual recall / Qualitative concept.

Solution:

  • In real estate and rental agreements, failure to pay rent results in legal dispossession.


  • The word evicted means expelled or removed from a residential property.


Why other options are incorrect:

  • Option B: Investigated means subject to systematic examination or inquiry.
  • Option C: Threatened indicates expressions of intent to inflict harm.
  • Option D: Rewarded signifies receiving recompense or an accolade.
MCQ #138 of 150 English NUMS 2026
[NUMS 2026]

Fill in the blank with the correct verb.

The baby ________ all morning.
A
Is crying
B
Has been crying
C
Cry
D
Have been crying
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The duration adverbial all morning indicates an action that began in the past and has continued uninterrupted into the present, requiring the present perfect continuous tense with third-person singular agreement.

Formula / Rule / Reaction:

$$\text{Subject (singular)} + \text{has been} + \text{verb-ing}$$

Solution:

  • The temporal phrase all morning indicates an ongoing duration spanning from earlier in the morning up to the present.


  • The subject The baby is third-person singular, requiring has been rather than have been.


  • Combining these grammatical requirements yields has been crying.


Why other options are incorrect:

  • Option A: Is crying specifies present continuous action without indicating extended past duration.
  • Option C: Cry is an uninflected base verb that violates third-person singular agreement.
  • Option D: Have been crying uses a plural auxiliary verb with a singular subject.
MCQ #139 of 150 English NUMS 2026
[NUMS 2026]

By 2040, robots ________ many of the jobs that people do today.
A
Will be taking over
B
Will take over
C
Will have taken over
D
Has taken over
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

The preposition by followed by a specified future time marker (By 2040) establishes a deadline prior to which an action will be completed, requiring the future perfect tense.

Formula / Rule / Reaction:

$$\text{By} + [\text{Future Point}] \implies \text{will have} + \text{past participle}$$

Solution:

  • Sentences introduced by the temporal boundary by + [future date] denote perfected future events.


  • The future perfect tense structure is will have + past participle.


  • Applying this rule gives will have taken over.


Why other options are incorrect:

  • Option A: Will be taking over is future continuous, which denotes an ongoing process rather than a completed event.
  • Option B: Will take over is simple future, which lacks the prior completion sense established by by 2040.
  • Option D: Has taken over is present perfect, which cannot modify a future temporal benchmark.
MCQ #140 of 150 English NUMS 2026
[NUMS 2026]

"Each time one of us touches the soil, we feel a sense of personal renewal". The type of sentence is:
A
Compound
B
Complex
C
Compound Complex
D
Simple
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

A complex sentence consists of exactly one independent main clause and at least one dependent subordinate clause introduced by a subordinating conjunction or adverbial clause marker.

Formula / Rule / Reaction:

$$\text{Complex Sentence} = \text{Dependent Subordinate Clause} + \text{Independent Main Clause}$$

Solution:

  • The clause Each time one of us touches the soil is a dependent adverbial clause of time that cannot stand alone as an independent sentence.


  • The clause we feel a sense of personal renewal is an independent main clause containing a subject and a finite predicate.


  • A sentence containing one dependent clause and one independent clause is classified as a complex sentence.


Why other options are incorrect:

  • Option A: Compound sentences require two or more independent clauses joined by coordinating conjunctions (FANBOYS) or semicolons.
  • Option C: Compound-complex sentences require at least two independent clauses and at least one dependent clause.
  • Option D: Simple sentences contain only one independent clause with no dependent clauses.
MCQ #141 of 150 English NUMS 2026
[NUMS 2026]

Identify the correctly punctuated sentence.
A
Milton the great English Poet, was blind
B
Milton, the great English poet, was blind
C
Milton the great English poet was, blind
D
Milton the great, English poet was blind
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

Non-restrictive appositive phrases that rename or provide supplementary information about a proper noun must be enclosed on both sides by commas, and common nouns are not capitalized.

Formula / Rule / Reaction:

$$\text{Proper Noun} + , + [\text{Non-restrictive Appositive}] + , + \text{Predicate}$$

Solution:

  • The phrase the great English poet is a non-defining appositive elaborating on the subject Milton.


  • Punctuation rules require setting off this non-essential phrase with a comma before it and a comma after it.


  • English is capitalized as a proper adjective, whereas poet is a common noun and remains lowercase.


  • Sentence B meets all these syntactical and capitalization criteria.


Why other options are incorrect:

  • Option A: Omits the introductory comma after Milton and capitalizes the common noun Poet.
  • Option C: Omits the opening comma and inserts an erroneous comma between auxiliary was and adjective blind.
  • Option D: Misplaces the comma between adjective great and modifier English.
MCQ #142 of 150 English NUMS 2026
[NUMS 2026]

Identify the correct sentence.
A
A Sindhi woman is going through a bazaar with bear feet.
B
A Sindhi women are going through a bazaar with bare feet.
C
A Sindhi woman is going through a bazaar with bare feet.
D
A Sindhi woman is going through a bazaar with bare foot.
View Answer & Propolis Autopsy
Correct Key: Option C Diagnostic Explanation
Concept:

Subject-verb agreement requires a singular subject (woman) to take a singular auxiliary (is), homophone precision requires bare (uncovered) rather than bear, and plural feet is required for natural human locomotion.

Formula / Rule / Reaction:

Grammatical agreement / Lexical accuracy.

Solution:

  • The singular noun phrase A Sindhi woman agrees with the singular present continuous verb is going.


  • The adjective bare means naked or uncovered, whereas bear refers to an animal or the verb to carry.


  • Locomotion on both feet requires the plural noun feet rather than singular foot.


  • Sentence C satisfies all grammatical, homophonic, and number agreements.


Why other options are incorrect:

  • Option A: Uses the incorrect homophone bear instead of bare.
  • Option B: Combines the singular article A with plural women.
  • Option D: Uses the singular bare foot instead of plural bare feet.
MCQ #143 of 150 English NUMS 2026
[NUMS 2026]

Complete the sentence using correct preposition.

The public are cautioned ________ pick pockets.
A
Against
B
Of
C
For
D
With
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The verb caution functions with the dependent preposition against when warning individuals about potential hazards, criminal acts, or perils.

Formula / Rule / Reaction:

$$\text{To caution} + [\text{person}] + \text{against} + [\text{danger / hazard}]$$

Solution:

  • Idiomatic English syntax pairs the verb caution with against to denote defensive vigilance against a threat.


  • In this sentence, the danger consists of pickpockets.


  • The preposition required is against.


Why other options are incorrect:

  • Option B: Caution of is unidiomatic in this context.
  • Option C: Caution for is grammatically incorrect.
  • Option D: Caution with is unidiomatic when warning against a predatory threat.
MCQ #144 of 150 English NUMS 2026
[NUMS 2026]

Select the correct pair.

Lack of sleep can ________ your health and have a bad ________ on your grades.
A
Affect / affect
B
Effect / effect
C
Effect / affect
D
Affect / effect
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

Affect functions primarily as a verb meaning to influence or produce a change in, whereas effect functions primarily as a noun meaning a result, consequence, or outcome.

Formula / Rule / Reaction:

$$\text{Modal Auxiliary (can)} + \text{Verb (affect)} \dots \text{Adjective (bad)} + \text{Noun (effect)}$$

Solution:

  • The first blank follows the modal auxiliary verb can, requiring the base verb affect.


  • The second blank is preceded by the article a and adjective bad, requiring the noun effect.


  • Therefore, the correct word pair is Affect / effect.


Why other options are incorrect:

  • Option A: Uses the verb affect in the second position where a noun is required.
  • Option B: Uses the noun effect in the first position where an infinitive verb is required.
  • Option C: Reverses both grammatical parts of speech.
MCQ #145 of 150 English NUMS 2026
[NUMS 2026]

Select the correct sentence in style.
A
Having finished his exercise, the books were put away.
B
Having finishing his exercise, the books were put away.
C
Having finishing his exercise, he put his books away.
D
Having finished his exercise, he put away his books.
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

An introductory participial phrase must be followed immediately by the grammatical subject that performs the action described in the participle, avoiding dangling modifiers.

Formula / Rule / Reaction:

$$\text{Perfect Participle } (\text{Having finished } X) + , + \text{Logical Agent } (\text{he}) + \text{Predicate}$$

Solution:

  • The introductory modifier is Having finished his exercise.


  • The agent who performed the action of finishing the exercise is the person (he), not the books.


  • In option D, the subject he directly follows the participial phrase, eliminating any dangling modifier.


Why other options are incorrect:

  • Option A: Contains a dangling modifier because the books cannot perform the exercise.
  • Option B: Contains both an ungrammatical participial construction (Having finishing) and a dangling modifier.
  • Option C: Contains the ungrammatical participle formulation Having finishing.
MCQ #146 of 150 English NUMS 2026
[NUMS 2026]

Select the statement that best describes the given sentence.

"He is anything but a liar."
A
He is a liar.
B
He can be anything but not a liar.
C
He is not a good character and liar as well.
D
He is good character but a liar to some extent.
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The idiomatic expression anything but serves as an emphatic negative meaning definitely not or the complete opposite of the stated noun or quality.

Formula / Rule / Reaction:

$$\text{Subject} + \text{is anything but } X \iff \text{Subject is definitely not } X$$

Solution:

  • The idiom anything but negates the term that follows.


  • Saying He is anything but a liar emphasizes that whatever flaws he may possess, he is definitely not a liar.


  • This corresponds directly to: He can be anything but not a liar.


Why other options are incorrect:

  • Option A: Completely contradicts the meaning of the idiom by stating he is a liar.
  • Option C: Erroneously claims that the individual is a liar.
  • Option D: Contradicts the statement by asserting he is a liar to some extent.
MCQ #147 of 150 English NUMS 2026
[NUMS 2026]

Select the best option to describe the given sentence.

"Unless you apologise I shall punish you."
A
Without apology, you will be forgiven.
B
With apology, you will be punished.
C
There is no need of apology.
D
Without apology you will be punished.
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The conditional conjunction unless introduces an exceptive negative condition equivalent in meaning to if ... not.

Formula / Rule / Reaction:

$$\text{Unless } P, \ Q \iff \text{If not } P, \text{ then } Q \iff \text{Without } P, \ Q$$

Solution:

  • The clause Unless you apologise sets a condition meaning If you do not offer an apology.


  • The consequence of failing to apologize is punishment: I shall punish you.


  • Expressing this relationship concisely yields: Without apology you will be punished.


Why other options are incorrect:

  • Option A: Contradicts the conditional threat by suggesting forgiveness without an apology.
  • Option B: Reverses the condition by asserting punishment occurs with an apology.
  • Option C: An apology is the explicit condition required to avoid punishment.
MCQ #148 of 150 English NUMS 2026
[NUMS 2026]

"From the attitude it is clear that he wants to pay off his old scores." The sentence means that he wants to:
A
Improve his marks.
B
Pay off his debt.
C
Help needy people.
D
Have his revenge.
View Answer & Propolis Autopsy
Correct Key: Option D Diagnostic Explanation
Concept:

The idiom to pay off old scores (or settle old scores) means to retaliate against past grievances, injuries, or insults by exacting revenge.

Formula / Rule / Reaction:

$$\text{To pay off old scores} \iff \text{To exact revenge / retaliate for past wrongs}$$

Solution:

  • The phrase pay off old scores does not relate to academic marks or financial debt.


  • In idiomatic usage, it refers to seeking retribution for previous conflicts or grudges.


  • Therefore, the expression means he wants to have his revenge.


Why other options are incorrect:

  • Option A: Takes scores literally as academic examination marks.
  • Option B: Takes pay off literally as financial loan repayment.
  • Option C: Contradicts the confrontational meaning of the idiom.
MCQ #149 of 150 English NUMS 2026
[NUMS 2026]

A person needs to reflect profoundly on his/her life. The underlined word means:
A
Deeply
B
Superficially
C
Shallow
D
Mirroring
View Answer & Propolis Autopsy
Correct Key: Option A Diagnostic Explanation
Concept:

The adverb profoundly means in a profound manner, characterized by deep intellectual, emotional, or philosophical depth.

Formula / Rule / Reaction:

$$\text{Profoundly} \iff \text{Deeply / Thoroughly / Intensely}$$

Solution:

  • Reflecting profoundly involves introspective examination of meaning and existence.


  • The direct synonym for profoundly in this context is deeply.


Why other options are incorrect:

  • Option B: Superficially means in a shallow or surface-level manner, serving as an antonym.
  • Option C: Shallow is an adjective denoting lack of depth.
  • Option D: Mirroring is a physical optical or behavioral imitation term.
MCQ #150 of 150 English NUMS 2026
[NUMS 2026]

Due to his non-serious attitude, Ali took the advice with a pinch of salt. It means:
A
Took it seriously.
B
Did not take it seriously.
C
Accept it candidly.
D
Implemented it happily.
View Answer & Propolis Autopsy
Correct Key: Option B Diagnostic Explanation
Concept:

The idiom to take something with a pinch of salt (or grain of salt) means to accept information with skepticism, doubt, or without taking it completely seriously.

Formula / Rule / Reaction:

$$\text{Take with a pinch of salt} \iff \text{Regard with skepticism / Not take seriously}$$

Solution:

  • Taking advice with a pinch of salt indicates that the recipient does not treat the advice as fully authoritative or serious.


  • Coupled with Ali's non-serious attitude, the idiom confirms that he did not take the advice seriously.


Why other options are incorrect:

  • Option A: The exact opposite of the skepticism implied by the idiom.
  • Option C: Candid acceptance involves frank and sincere agreement.
  • Option D: Implementing advice happily implies enthusiastic execution rather than skepticism.
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