(D) Inferior ovary Ray florets have an inferior ovary. Epigynous flowers are found in family Asteraceae, e.g. sunflower, and in an epigynous flower the ovary is inferior.

NEET (UG) 2020 — Code G2 — Answer Key & Solutions

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Free, login-free worked solutions to every question in NEET (UG) 2020 — Code G2. Check your answers below, then predict your rank or practise with full mock tests.

Q1.

Which of the following refer to correct example(s) of organisms which have evolved due to changes in environment brought about by anthropogenic action? (a) Darwin's Finches of Galapagos islands. (b) Herbicide resistant weeds. (c) Drug resistant eukaryotes. (d) Man-created breeds of domesticated animals like dogs.

A. (a) and (c)
B. (b), (c) and (d) ✓
C. only (d)
D. only (a)

Solution: (b), (c) and (d) Herbicide-resistant weeds, drug-resistant eukaryotic cells and man-made breeds of dogs were all selected by human action, so (b), (c) and (d) are correct. (A) Includes (a), but Darwin's finches arose by natural selection (adaptive radiation) on the Galapagos, not by human action. (C) Leaves out (b) and (c); excess use of herbicides and drugs has selected resistant weeds and cells within months or years. (D) Contains only (a), which is evolution by natural selection, not by anthropogenic action.

Q2.

Meiotic division of the secondary oocyte is completed

A. At the time of copulation
B. After zygote formation
C. At the time of fusion of a sperm with an ovum ✓
D. Prior to ovulation

Solution: At the time of fusion of a sperm with an ovum Entry of the sperm into the secondary oocyte induces completion of its second meiotic division, forming a haploid ovum (ootid) and a second polar body.

Q3.

Which of the following is correct about viroids?

A. They have free RNA without protein coat ✓
B. They have DNA with protein coat
C. They have free DNA without protein coat
D. They have RNA with protein coat

Solution: They have free RNA without a protein coat Viroids, found by T.O. Diener to cause potato spindle tuber disease, are free RNA lacking the protein coat that viruses have. (B) and (C) Viroids have no DNA; their genetic material is RNA. (D) RNA enclosed in a protein coat is an RNA virus, not a viroid.

Q4.

The plant parts which consist of two generations - one within the other (a) Pollen grains inside the anther (b) Germinated pollen grain with two male gametes (c) Seed inside the fruit (d) Embryo sac inside the ovule

A. (a), (b) and (c)
B. (c) and (d)
C. (a) and (d) ✓
D. (a) only

Solution: (a) and (d) A haploid pollen grain (male gametophyte) inside the diploid anther and a haploid embryo sac (female gametophyte) inside the diploid ovule are each a gametophyte within a sporophyte. (A) Includes (b) and (c): a germinated pollen grain with its two male gametes is wholly gametophyte, and a seed inside a fruit holds the next sporophyte (the embryo), not a gametophyte. (B) Includes (c) and leaves out (a). (D) Leaves out (d), the embryo sac inside the ovule.

Q5.

Experimental verification of the chromosomal theory of inheritance was done by

A. Sutton
B. Boveri
C. Morgan ✓
D. Mendel

Solution: Morgan Sutton and Boveri proposed the chromosomal theory of inheritance; T.H. Morgan verified it experimentally, working with the fruit fly $Drosophila$ $melanogaster$.

Q6.

Which of the following pairs is of unicellular algae?

A. $Gelidium$ and $Gracilaria$
B. $Anabaena$ and $Volvox$
C. $Chlorella$ and $Spirulina$ ✓
D. $Laminaria$ and $Sargassum$

Solution: $Chlorella$ and $Spirulina$ $Chlorella$ and $Spirulina$ are unicellular algae, rich in proteins and used as food supplements even by space travellers. (A) $Gelidium$ and $Gracilaria$ are multicellular red algae, used to make agar. (B) $Anabaena$ is filamentous and $Volvox$ is colonial. (D) $Laminaria$ and $Sargassum$ are multicellular brown algae.

Q7.

Secondary metabolites such as nicotine, strychnine and caffeine are produced by plants for their

A. Growth response
B. Defence action ✓
C. Effect on reproduction
D. Nutritive value

Solution: Defence action Plants produce nicotine, caffeine, quinine, strychnine and opium as defences against grazers and browsers.

Q8.

By which method was a new breed 'Hisardale' of sheep formed by using Bikaneri ewes and Marino rams?

A. Mutational breeding
B. Cross breeding ✓
C. Inbreeding
D. Out crossing

Solution: Cross breeding In cross-breeding, superior males of one breed are mated with superior females of another; Hisardale was developed in Punjab by crossing Bikaneri ewes with Marino rams.

Q9.

The infectious stage of $Plasmodium$ that enters the human body is

A. Sporozoites ✓
B. Female gametocytes
C. Male gametocytes
D. Trophozoites

Solution: Sporozoites $Plasmodium$ enters the human body as sporozoites, the infectious form, through the bite of an infected female $Anopheles$ mosquito.

Q10.

The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is

A. Root pressure ✓
B. Imbibition
C. Plasmolysis
D. Transpiration

Solution: Root pressure Root pressure is the positive pressure that develops in the xylem of roots; its effect is seen at night and in early morning, when evaporation is low. It forces water out as droplets through special openings of veins near the tips of grass blades; this loss of water in liquid form is guttation.

Q11.

From his experiments, S.L. Miller produced amino acids by mixing the following in a closed flask

A. CH$_3$, H$_2$, NH$_4$ and water vapor at 800$^{\circ}$C
B. CH$_4$, H$_2$, NH$_3$ and water vapor at 600$^{\circ}$C
C. CH$_3$, H$_2$, NH$_3$ and water vapor at 600$^{\circ}$C
D. CH$_4$, H$_2$, NH$_3$ and water vapor at 800$^{\circ}$C ✓

Solution: CH₄, H₂, NH₃ and water vapour at 800 °C In 1953 S.L. Miller passed an electric discharge through a closed flask containing CH₄, H₂, NH₃ and water vapour at 800 °C, and amino acids were formed.

Q12.

In relation to Gross primary productivity and Net primary productivity of an ecosystem, which one of the following statements is correct?

A. Gross primary productivity is always more than net primary productivity ✓
B. Gross primary productivity and Net primary productivity are one and same
C. There is no relationship between Gross primary productivity and Net primary productivity
D. Gross primary productivity is always less than net primary productivity

Solution: Gross primary productivity is always more than net primary productivity Net primary productivity is gross primary productivity minus the loss in respiration: $\text{NPP} = \text{GPP} - R$, so GPP is always more than NPP. (B) The two differ by the energy used in respiration, so they are not the same. (C) NPP is directly related to GPP through $\text{NPP} = \text{GPP} - R$. (D) Respiration loss is never zero, so GPP can never be less than NPP.

Q13.

The sequence that controls the copy number of the linked DNA in the vector, is termed

A. Ori site ✓
B. Palindromic sequence
C. Recognition site
D. Selectable marker

Solution: Ori site The origin of replication (ori) is the sequence where replication starts, and it also controls the copy number of the linked DNA.

Q14.

Cuboidal epithelium with brush border of microvilli is found in

A. Ducts of salivary gland
B. Proximal convoluted tubule of nephron ✓
C. Eustachian tube
D. Lining of intestine

Solution: Proximal convoluted tubule of nephron The proximal convoluted tubule (PCT) is lined by simple cuboidal brush border epithelium, whose microvilli increase the surface area for reabsorption.

Q15.

The body of the ovule is fused within the funicle at

A. Micropyle
B. Nucellus
C. Chalaza
D. Hilum ✓

Solution: Hilum The body of the ovule fuses with the funicle in the region called the hilum, the junction between ovule and funicle.

Q16.

In light reaction, plastoquinone facilitates the transfer of electrons from

A. Cytb$_6$f complex to PS-I
B. PS-I to NADP$^{+}$
C. PS-I to ATP synthase
D. PS-II to Cytb$_6$f complex ✓

Solution: PS-II to Cyt b₆f complex In the Z scheme, electrons from PS II pass from its primary acceptor through plastoquinone to the cytochrome b₆f complex. From the cytochrome b₆f complex, plastocyanin carries them on to PS I.

Q17.

Match the following diseases with the causative organism and select the correct option. Column-I: (a) Typhoid (b) Pneumonia (c) Filariasis (d) Malaria Column-II: (i) $Wuchereria$ (ii) $Plasmodium$ (iii) $Salmonella$ (iv) $Haemophilus$

A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) ✓
B. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
D. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)

Solution: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) (a)-(iii): Typhoid is caused by $Salmonella$ $typhi$. (b)-(iv): Pneumonia is caused by $Streptococcus$ $pneumoniae$ and $Haemophilus$ $influenzae$. (c)-(i): Filariasis (elephantiasis) is caused by the filarial worms $Wuchereria$ $bancrofti$ and $W.$ $malayi$. (d)-(ii): Malaria is caused by different species of $Plasmodium$.

Q18.

Match the following columns and select the correct option. Column-I: (a) $Clostridium$ $butylicum$ (b) $Trichoderma$ $polysporum$ (c) $Monascus$ $purpureus$ (d) $Aspergillus$ $niger$ Column-II: (i) Cyclosporin-A (ii) Butyric Acid (iii) Citric Acid (iv) Blood cholesterol lowering agent

A. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) ✓
B. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
C. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
D. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

Solution: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) (a)-(ii): $Clostridium$ $butylicum$ produces butyric acid. (b)-(i): $Trichoderma$ $polysporum$ produces cyclosporin A, an immunosuppressive agent used in organ-transplant patients. (c)-(iv): $Monascus$ $purpureus$ produces statins, which lower blood cholesterol by competitively inhibiting the enzyme of cholesterol synthesis. (d)-(iii): $Aspergillus$ $niger$ produces citric acid.

Q19.

Which of the following statements are true for the phylum-Chordata? (a) In Urochordata notochord extends from head to tail and it is present throughout their life. (b) In Vertebrata notochord is present during the embryonic period only. (c) Central nervous system is dorsal and hollow. (d) Chordata is divided into 3 subphyla : Hemichordata, Tunicata and Cephalochordata.

A. (c) and (a)
B. (a) and (b)
C. (b) and (c) ✓
D. (d) and (c)

Solution: (b) and (c) In vertebrates the notochord is present only in the embryo and is replaced by the vertebral column (b), and the chordate central nervous system is dorsal and hollow (c). (A) Includes (a), but in Urochordata the notochord is present only in the larval tail; it extends from head to tail throughout life in Cephalochordata. (B) Includes (a), which is wrong for the same reason. (D) Includes (d), but the three subphyla are Urochordata (Tunicata), Cephalochordata and Vertebrata; Hemichordata is now a separate phylum under non-chordates.

Q20.

Goblet cells of alimentary canal are modified from

A. Columnar epithelial cells ✓
B. Chondrocytes
C. Compound epithelial cells
D. Squamous epithelial cells

Solution: Columnar epithelial cells Goblet cells of the alimentary canal are isolated glandular cells (unicellular glands) formed from columnar epithelial cells; they secrete mucus.

Q21.

Which of the following is not an inhibitory substance governing seed dormancy?

A. Abscisic acid
B. Phenolic acid
C. Para-ascorbic acid
D. Gibberellic acid ✓

Solution: Gibberellic acid Gibberellic acid is applied to break seed dormancy; it does not cause it. (A), (B) and (C) Abscisic acid, phenolic acid and para-ascorbic acid are the chemical inhibitors that keep seeds dormant.

Q22.

Name the enzyme that facilitates opening of DNA helix during transcription.

A. DNA helicase
B. DNA polymerase
C. RNA polymerase ✓
D. DNA ligase

Solution: RNA polymerase In transcription, RNA polymerase binds the promoter and itself facilitates opening of the DNA helix, then continues elongation along the template.

Q23.

Match the following Column-I: (a) Inhibitor of catalytic activity (b) Possess peptide bonds (c) Cell wall material in fungi (d) Secondary metabolite Column-II: (i) Ricin (ii) Malonate (iii) Chitin (iv) Collagen Choose the correct option from the following

A. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
C. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
D. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) ✓

Solution: (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) (a)-(ii): Malonate closely resembles succinate and competitively inhibits succinic dehydrogenase. (b)-(iv): Collagen is a protein, so its amino acids are joined by peptide bonds. (c)-(iii): Chitin, a complex polysaccharide, forms the cell wall of fungi and the exoskeleton of arthropods. (d)-(i): Ricin is a toxin, one of the secondary metabolites of plants.

Q24.

Bilaterally symmetrical and acoelomate animals are exemplified by

A. Platyhelminthes ✓
B. Aschelminthes
C. Annelida
D. Ctenophora

Solution: Platyhelminthes Platyhelminthes are bilaterally symmetrical and acoelomate, unlike pseudocoelomate Aschelminthes, coelomate Annelida and radially symmetrical Ctenophora.

Q25.

Presence of which of the following conditions in urine are indicative of Diabetes Mellitus?

A. Uremia and Renal Calculi
B. Ketonuria and Glycosuria ✓
C. Renal calculi and Hyperglycaemia
D. Uremia and Ketonuria

Solution: Ketonuria and glycosuria Presence of glucose (glycosuria) and ketone bodies (ketonuria) in urine is indicative of diabetes mellitus.

Q26.

Ray florets have

A. Superior ovary
B. Hypogynous ovary
C. Half inferior ovary
D. Inferior ovary ✓

Solution: Inferior ovary Ray florets of sunflower are epigynous: the thalamus grows up around the ovary and fuses with it, so the ovary is inferior.

Q27.

Identify the substances having glycosidic bond and peptide bond, respectively in their structure

A. Glycerol, trypsin
B. Cellulose, lecithin
C. Inulin, insulin ✓
D. Chitin, cholesterol

Solution: Inulin, insulin Inulin is a polymer of fructose whose units are joined by glycosidic bonds. Insulin is a protein whose amino acids are joined by peptide bonds. Glycerol, lecithin (a phospholipid) and cholesterol (a sterol) have neither a glycosidic nor a peptide bond.

Q28.

Which of the following statements is not correct?

A. The proinsulin has an extra peptide called C-peptide.
B. The functional insulin has A and B chains linked together by hydrogen bonds. ✓
C. Genetically engineered insulin is produced in $E.$ $Coli$.
D. In man insulin is synthesised as a proinsulin

Solution: The functional insulin has A and B chains linked together by hydrogen bonds (the incorrect statement) In functional insulin the A and B chains are linked together by disulphide bridges, not hydrogen bonds. (A) Proinsulin does contain an extra stretch, the C-peptide, which is removed when it matures into insulin. (C) Human insulin chains A and B are produced from DNA sequences introduced into plasmids of $E.$ $coli$. (D) In man, insulin is synthesised as a prohormone, proinsulin, which must be processed to become functional.

Q29.

Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (G$_0$). This process occurs at the end of

A. G$_1$ phase ✓
B. S phase
C. G$_2$ phase
D. M phase

Solution: G₁ phase Cells that do not divide further exit G₁ phase and enter an inactive stage, the quiescent stage (G₀), where they stay metabolically active but no longer proliferate.

Q30.

Identify the correct statement with regard to G$_1$ phase (Gap 1) of interphase.

A. Reorganisation of all cell components takes place.
B. Cell is metabolically active, grows but does not replicate its DNA. ✓
C. Nuclear Division takes place.
D. DNA synthesis or replication takes place.

Solution: Cell is metabolically active, grows but does not replicate its DNA In G₁ phase the cell is metabolically active and grows continuously but does not replicate its DNA. (A) Reorganisation of virtually all cell components takes place in the M phase. (C) Nuclear division (karyokinesis) takes place in the M phase. (D) DNA synthesis or replication takes place in the S phase.

Q31.

The QRS complex in a standard ECG represents

A. Depolarisation of auricles
B. Depolarisation of ventricles ✓
C. Repolarisation of ventricles
D. Repolarisation of auricles

Solution: Depolarisation of ventricles The QRS complex represents the depolarisation of the ventricles, which initiates ventricular contraction.

Q32.

If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is $6.6 \times 10^{9}$ bp, then the length of the DNA is approximately

A. 2.5 meters
B. 2.2 meters ✓
C. 2.7 meters
D. 2.0 meters

Solution: 2.2 metres Length of DNA = number of base pairs × distance between consecutive base pairs $$= 6.6 \times 10^{9} \times 0.34 \times 10^{-9}\ \text{m} = 2.244\ \text{m}$$ $\therefore$ the DNA is about 2.2 m long.

Q33.

Which of the following regions of the globe exhibits highest species diversity?

A. Madagascar
B. Himalayas
C. Amazon forests ✓
D. Western Ghats of India

Solution: Amazon forests The largely tropical Amazonian rain forest of South America has the greatest biodiversity on earth.

Q34.

Which of the following is put into Anaerobic sludge digester for further sewage treatment?

A. Floating debris
B. Effluents of primary treatment
C. Activated sludge ✓
D. Primary sludge

Solution: Activated sludge After secondary treatment the bacterial flocs settle in a settling tank as activated sludge. A small part of it is pumped back into the aeration tank as inoculum; the major part is pumped into anaerobic sludge digesters, where anaerobic bacteria digest the bacteria and fungi in it.

Q35.

Dissolution of the synaptonemal complex occurs during

A. Zygotene
B. Diplotene ✓
C. Leptotene
D. Pachytene

Solution: Diplotene Diplotene begins with dissolution of the synaptonemal complex, and the homologous chromosomes of each bivalent start to separate except at the chiasmata.

Q36.

Select the option including all sexually transmitted diseases.

A. Gonorrhoea, Malaria, Genital herpes
B. AIDS, Malaria, Filaria
C. Cancer, AIDS, Syphilis
D. Gonorrhoea, Syphilis, Genital herpes ✓

Solution: Gonorrhoea, Syphilis, Genital herpes Gonorrhoea, syphilis and genital herpes are all sexually transmitted diseases. (A) Malaria is spread by the bite of the female $Anopheles$ mosquito, not by sexual contact. (B) Malaria and filariasis are spread by mosquitoes. (C) Cancer is not an infectious disease.

Q37.

Select the correct statement.

A. Glucagon is associated with hypoglycemia.
B. Insulin acts on pancreatic cells and adipocytes.
C. Insulin is associated with hyperglycemia.
D. Glucocorticoids stimulate gluconeogenesis. ✓

Solution: Glucocorticoids stimulate gluconeogenesis Glucocorticoids stimulate gluconeogenesis, lipolysis and proteolysis. (A) Glucagon raises blood sugar, so it is associated with hyperglycemia. (B) Insulin acts mainly on hepatocytes and adipocytes, not on pancreatic cells. (C) Insulin lowers blood sugar, so it is associated with hypoglycemia.

Q38.

The product(s) of reaction catalyzed by nitrogenase in root nodules of leguminous plants is/are

A. Nitrate alone
B. Ammonia and oxygen
C. Ammonia and hydrogen ✓
D. Ammonia alone

Solution: Ammonia and hydrogen The nitrogenase reaction is N₂ + 8e⁻ + 8H⁺ + 16ATP → 2NH₃ + H₂ + 16ADP + 16Pi. $\therefore$ the products are ammonia and hydrogen.

Q39.

In gel electrophoresis, separated DNA fragments can be visualized with the help of

A. Ethidium bromide in UV radiation ✓
B. Acetocarmine in UV radiation
C. Ethidium bromide in infrared radiation
D. Acetocarmine in bright blue light

Solution: Ethidium bromide in UV radiation Separated DNA fragments are visualised only after staining with ethidium bromide and exposure to UV radiation, which shows them as bright orange bands.

Q40.

In which of the following techniques, the embryos are transferred to assist those females who cannot conceive?

A. GIFT and ZIFT
B. ICSI and ZIFT
C. GIFT and ICSI
D. ZIFT and IUT ✓

Solution: ZIFT and IUT In embryo transfer, the zygote or early embryo (up to 8 blastomeres) goes into the fallopian tube (ZIFT), and embryos with more than 8 blastomeres into the uterus (IUT). (A) GIFT transfers an ovum (a gamete), not an embryo, into the fallopian tube. (B) ICSI injects a sperm directly into an ovum to form an embryo in the laboratory; it does not transfer an embryo. (C) Neither GIFT nor ICSI is an embryo transfer.

Q41.

Select the correct match

A. Phenylketonuria – Autosomal dominant trait
B. Sickle cell anaemia – Autosomal recessive trait, chromosome-11 ✓
C. Thalassemia – X linked
D. Haemophilia – Y linked

Solution: Sickle cell anaemia: autosomal recessive trait, chromosome 11 Sickle-cell anaemia is an autosomal recessive trait caused by a mutation in the β-globin gene on chromosome 11. (A) Phenylketonuria is an autosomal recessive trait, not dominant. (C) Thalassemia is an autosomal recessive disorder, not X-linked. (D) Haemophilia is X-linked recessive, not Y-linked.

Q42.

Which of the following is not an attribute of a population?

A. Natality
B. Mortality
C. Species interaction ✓
D. Sex ratio

Solution: Species interaction Species interaction is an interaction between populations of different species, not an attribute of a single population. (A), (B) and (D) Natality, mortality and sex ratio are attributes that a population has but an individual does not.

Q43.

The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of

A. 1 molecule of 3-C compound ✓
B. 1 molecule of 6-C compound
C. 1 molecule of 4-C compound and 1 molecule of 2-C compound
D. 2 molecules of 3-C compound

Solution: 1 molecule of 3-C compound In photorespiration RuBisCO binds O₂, and RuBP forms one molecule of phosphoglycerate (3-C) and one of phosphoglycolate (2-C) instead of two molecules of PGA. (D) Two molecules of 3-C PGA form only when RuBisCO binds CO₂ (carboxylation). (C) The 2-C phosphoglycolate forms along with a 3-C compound, not a 4-C compound.

Q44.

Match the following concerning essential elements and their functions in plants (a) Iron (b) Zinc (c) Boron (d) Manganese (i) Photolysis of water (ii) Pollen germination (iii) Required for chlorophyll biosynthesis (iv) IAA biosynthesis Select the correct option

A. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
B. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) ✓
C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
D. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Solution: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) (a)-(iii): Iron is essential for the formation of chlorophyll. (b)-(iv): Zinc is needed for the synthesis of auxin (IAA). (c)-(ii): Boron is required for pollen germination. (d)-(i): Manganese is involved in the splitting of water to liberate oxygen during photosynthesis.

Q45.

Which is the important site of formation of glycoproteins and glycolipids in eukaryotic cells?

A. Peroxisomes
B. Golgi bodies ✓
C. Polysomes
D. Endoplasmic reticulum

Solution: Golgi bodies The Golgi apparatus, where proteins and lipids are modified in the cisternae, is the important site of formation of glycoproteins and glycolipids.

Q46.

Select the correct events that occur during inspiration. (a) Contraction of diaphragm (b) Contraction of external inter-costal muscles (c) Pulmonary volume decreases (d) Intra pulmonary pressure increases

A. (c) and (d)
B. (a), (b) and (d)
C. only (d)
D. (a) and (b) ✓

Solution: (a) and (b) Inspiration begins with contraction of the diaphragm (antero-posterior axis) and of the external intercostal muscles (dorso-ventral axis), which enlarges the thoracic chamber. (A) Pulmonary volume increases and intra-pulmonary pressure falls during inspiration; (c) and (d) happen in expiration. (B) Includes (d), but intra-pulmonary pressure falls below atmospheric pressure during inspiration. (C) Contains only (d), which is wrong for the same reason.

Q47.

The roots that originate from the base of the stem are

A. Primary roots
B. Prop roots
C. Lateral roots
D. Fibrous roots ✓

Solution: Fibrous roots In monocots like wheat the short-lived primary root is replaced by many roots that originate from the base of the stem and form the fibrous root system.

Q48.

The ovary is half inferior in :

A. Mustard
B. Sunflower
C. Plum ✓
D. Brinjal

Solution: Plum Plum has perigynous flowers: the other floral parts arise from the rim of the thalamus at about the level of the ovary, which is half inferior. Mustard and brinjal are hypogynous (superior ovary); sunflower is epigynous (inferior ovary).

Q49.

Match the following columns and select the correct option. Column-I: (a) Floating Ribs (b) Acromion (c) Scapula (d) Glenoid cavity Column-II: (i) Located between second and seventh ribs (ii) Head of the Humerus (iii) Clavicle (iv) Do not connect with the sternum

A. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
B. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
C. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) ✓
D. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

Solution: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) (a)-(iv): The 11th and 12th pairs of ribs are not connected ventrally to the sternum, so they are called floating ribs. (b)-(iii): The acromion, the expanded end of the spine of the scapula, articulates with the clavicle. (c)-(i): The scapula is a large triangular flat bone in the dorsal part of the thorax between the second and seventh ribs. (d)-(ii): The glenoid cavity of the scapula articulates with the head of the humerus to form the shoulder joint.

Q50.

If the head of cockroach is removed, it may live for few days because

A. the cockroach does not have nervous system.
B. the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body. ✓
C. the head holds a 1/3$^{\text{rd}}$ of a nervous system while the rest is situated along the dorsal part of its body.
D. the supra-oesophageal ganglia of the cockroach are situated in ventral part of abdomen.

Solution: The head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body The head of the cockroach holds only a bit of the nervous system; the rest lies along the ventral side of the body, so the insect can live for some days after its head is cut off. (A) The cockroach does have a nervous system, spread throughout its body. (C) The rest of the nervous system runs along the ventral side, not the dorsal side. (D) The supra-oesophageal ganglion is the brain, situated in the head.

Q51.

Identify the incorrect statement.

A. Sapwood is involved in conduction of water and minerals from root to leaf
B. Sapwood is the innermost secondary xylem and is lighter in colour ✓
C. Due to deposition of tannins, resins, oils etc., heart wood is dark in colour
D. Heart wood does not conduct water but gives mechanical support

Solution: Sapwood is the innermost secondary xylem and is lighter in colour (the incorrect statement) Sapwood is the peripheral (outer) region of the secondary xylem; the innermost, darker region is heartwood. (A) Sapwood, which is lighter in colour, does conduct water and minerals from root to leaf. (C) Heartwood is dark brown because tannins, resins, oils, gums and aromatic substances are deposited in it. (D) Heartwood is made of dead, highly lignified elements; it does not conduct water but gives mechanical support.

Q52.

Bt cotton variety that was developed by the introduction of toxin gene of $Bacillus$ $thuringiensis$ (Bt) is resistant to

A. Fungal diseases
B. Plant nematodes
C. Insect predators
D. Insect pests ✓

Solution: Insect pests The Bt toxin genes $cryIAc$ and $cryIIAb$ protect Bt cotton from cotton bollworms, which are insect pests.

Q53.

The number of substrate level phosphorylations in one turn of citric acid cycle is

A. One ✓
B. Two
C. Three
D. Zero

Solution: One In one turn of the citric acid cycle, the only substrate-level phosphorylation is the conversion of succinyl-CoA to succinic acid, which forms one GTP. The GTP then passes its phosphate to ADP in a coupled reaction, forming ATP.

Q54.

Identify the wrong statement with regard to Restriction Enzymes.

A. They cut the strand of DNA at palindromic sites.
B. They are useful in genetic engineering.
C. Sticky ends can be joined by using DNA ligases. ✓
D. Each restriction enzyme functions by inspecting the length of a DNA sequence.

Solution: Sticky ends can be joined by using DNA ligases (true of DNA ligase, not of restriction enzymes) Joining sticky ends is the work of DNA ligase, not of a restriction enzyme, so this statement is the one that does not describe restriction enzymes. (A) Restriction enzymes recognise palindromic nucleotide sequences and cut the DNA there. (B) They are the molecular scissors of genetic engineering, used to make recombinant DNA. (D) Each restriction endonuclease inspects the length of a DNA sequence until it finds its specific recognition sequence.

Q55.

Flippers of Penguins and Dolphins are examples of

A. Convergent evolution ✓
B. Industrial melanism
C. Natural selection
D. Adaptive radiation

Solution: Convergent evolution Flippers of penguins (birds) and dolphins (mammals) are analogous organs, different in origin but similar in function; analogy is a result of convergent evolution.

Q56.

Identify the wrong statement with reference to transport of oxygen

A. Partial pressure of CO$_2$ can interfere with O$_2$ binding with haemoglobin
B. Higher H$^{+}$ conc. in alveoli favours the formation of oxyhaemoglobin ✓
C. Low pCO$_2$ in alveoli favours the formation of oxyhaemoglobin
D. Binding of oxygen with haemoglobin is mainly related to partial pressure of O$_2$

Solution: Higher H⁺ concentration in alveoli favours the formation of oxyhaemoglobin (the incorrect statement) Lower H⁺ concentration favours oxyhaemoglobin formation in the alveoli; higher H⁺ concentration, as in the tissues, favours dissociation of oxygen. (A) pCO₂ does interfere with O₂ binding: high pCO₂ in the tissues favours release of oxygen. (C) Low pCO₂ in the alveoli favours formation of oxyhaemoglobin. (D) Binding of oxygen with haemoglobin is primarily related to the partial pressure of O₂.

Q57.

Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.

A. A person will have only two of the three alleles.
B. When I$^{A}$ and I$^{B}$ are present together, they express same type of sugar. ✓
C. Allele 'i' does not produce any sugar.
D. The gene (I) has three alleles.

Solution: When $I^A$ and $I^B$ are present together, they express the same type of sugar (the incorrect statement) $I^A$ and $I^B$ each produce a slightly different form of the sugar; when present together they are co-dominant and both types of sugar are expressed (blood group AB). (A) Humans are diploid, so each person has only two of the three alleles. (C) Allele $i$ does not produce any sugar. (D) The gene I has three alleles: $I^A$, $I^B$ and $i$.

Q58.

Identify the basic amino acid from the following.

A. Glutamic Acid
B. Lysine ✓
C. Valine
D. Tyrosine

Solution: Lysine Lysine has an extra amino group, which makes it a basic amino acid; glutamic acid is acidic, valine is neutral and tyrosine is aromatic.

Q59.

Name the plant growth regulator which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.

A. Gibberellin ✓
B. Ethylene
C. Abscisic acid
D. Cytokinin

Solution: Gibberellin Spraying sugarcane with gibberellins increases the length of the stem, raising the yield by as much as 20 tonnes per acre.

Q60.

Match the organism with its use in biotechnology. (a) $Bacillus$ $thuringiensis$ (b) $Thermus$ $aquaticus$ (c) $Agrobacterium$ $tumefaciens$ (d) $Salmonella$ $typhimurium$ (i) Cloning vector (ii) Construction of first rDNA molecule (iii) DNA polymerase (iv) Cry proteins Select the correct option from the following:

A. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) ✓
B. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
C. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
D. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

Solution: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) (a)-(iv): $Bacillus$ $thuringiensis$ produces Cry proteins, the insecticidal Bt toxins. (b)-(iii): $Thermus$ $aquaticus$ is the source of the thermostable Taq DNA polymerase used in PCR. (c)-(i): The Ti plasmid of $Agrobacterium$ $tumefaciens$ has been modified into a cloning vector for plants. (d)-(ii): The first recombinant DNA was made by linking an antibiotic-resistance gene with a native plasmid of $Salmonella$ $typhimurium$.

Q61.

Which of the following statements is correct?

A. Adenine pairs with thymine through one H-bond
B. Adenine pairs with thymine through three H-bonds
C. Adenine does not pair with thymine
D. Adenine pairs with thymine through two H-bonds ✓

Solution: Adenine pairs with thymine through two H-bonds Adenine forms two hydrogen bonds with thymine of the opposite strand; guanine forms three with cytosine. (A) and (B) An A-T pair has two hydrogen bonds, not one or three. (C) In DNA adenine always pairs with thymine, a purine with a pyrimidine.

Q62.

Match the following columns and select the correct option. Column-I: (a) Gregarious, polyphagous pest (b) Adult with radial symmetry and larva with bilateral symmetry (c) Book lungs (d) Bioluminescence Column-II: (i) $Asterias$ (ii) Scorpion (iii) $Ctenoplana$ (iv) $Locusta$

A. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓
B. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
C. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
D. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)

Solution: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) (a)-(iv): $Locusta$ (locust) is a gregarious pest. (b)-(i): Echinoderms such as $Asterias$ are radially symmetrical as adults but bilaterally symmetrical as larvae. (c)-(ii): The scorpion, an arachnid, respires through book lungs. (d)-(iii): Bioluminescence, the emission of light by living organisms, is well marked in ctenophores such as $Ctenoplana$.

Q63.

Which of the following would help in prevention of diuresis?

A. Reabsorption of Na$^{+}$ and water from renal tubules due to aldosterone ✓
B. Atrial natriuretic factor causes vasoconstriction
C. Decrease in secretion of renin by JG cells
D. More water reabsorption due to undersecretion of ADH

Solution: Reabsorption of Na⁺ and water from renal tubules due to aldosterone Aldosterone causes reabsorption of Na⁺ and water from the distal parts of the renal tubule, which cuts urine output and so prevents diuresis. (B) Atrial natriuretic factor causes vasodilation, not vasoconstriction, and it promotes loss of Na⁺ and water. (C) Less renin means less angiotensin II and aldosterone, so less reabsorption and more urine. (D) Undersecretion of ADH reduces water reabsorption and causes diuresis, as in diabetes insipidus.

Q64.

Choose the correct pair from the following

A. Polymerases - Break the DNA into fragments
B. Nucleases - Separate the two strands of DNA
C. Exonucleases - Make cuts at specific positions within DNA
D. Ligases - Join the two DNA molecules ✓

Solution: Ligases: join the two DNA molecules DNA ligases join DNA molecules, for example a cut gene with a cut vector. (A) Polymerases synthesise DNA; they do not break it. (B) Nucleases break the phosphodiester bonds of DNA; they do not separate its two strands. (C) Endonucleases make cuts at specific positions within the DNA; exonucleases remove nucleotides from the ends.

Q65.

Identify the correct statement with reference to human digestive system.

A. Serosa is the innermost layer of the alimentary canal
B. Ileum is a highly coiled part ✓
C. Vermiform appendix arises from duodenum
D. Ileum opens into small intestine

Solution: Ileum is a highly coiled part The small intestine has three regions: a C-shaped duodenum, a long coiled jejunum and a highly coiled ileum. (A) Serosa is the outermost layer of the alimentary canal; the mucosa is the innermost. (C) The vermiform appendix arises from the caecum of the large intestine, not the duodenum. (D) The ileum opens into the large intestine.

Q66.

Embryological support for evolution was disapproved by

A. Alfred Wallace
B. Charles Darwin
C. Oparin
D. Karl Ernst von Baer ✓

Solution: Karl Ernst von Baer Ernst Haeckel proposed embryological support for evolution; Karl Ernst von Baer disapproved it, noting that embryos never pass through the adult stages of other animals.

Q67.

Which of the following hormone levels will cause release of ovum (ovulation) from the graffian follicle?

A. High concentration of Progesterone
B. Low concentration of LH
C. Low concentration of FSH
D. High concentration of Estrogen ✓

Solution: High concentration of Estrogen A high level of estrogen in the middle of the cycle acts by positive feedback on the anterior pituitary and brings about a surge of LH. The LH surge, with FSH also at its peak, induces rupture of the Graafian follicle and release of the ovum. (A) Progesterone rises only after ovulation, when the corpus luteum forms. (B) and (C) Ovulation needs high, not low, levels of LH and FSH.

Q68.

The specific palindromic sequence which is recognized by EcoRI is

A. 5$'$ - GGAACC - 3$'$ / 3$'$ - CCTTGG - 5$'$
B. 5$'$ - CTTAAG - 3$'$ / 3$'$ - GAATTC - 5$'$
C. 5$'$ - GGATCC - 3$'$ / 3$'$ - CCTAGG - 5$'$
D. 5$'$ - GAATTC - 3$'$ / 3$'$ - CTTAAG - 5$'$ ✓

Solution: 5′-GAATTC-3′ / 3′-CTTAAG-5′ EcoRI recognises the palindrome 5′-GAATTC-3′, which reads the same on the other strand (3′-CTTAAG-5′) in the 5′→3′ direction, and cuts between G and A. (A) GGAACC is not a palindrome; its complementary strand read 5′→3′ is GGTTCC. (B) This puts CTTAAG on the 5′→3′ strand; it is the EcoRI sequence read 3′→5′, not the site EcoRI recognises. (C) GGATCC is the recognition site of BamHI.

Q69.

The first phase of translation is

A. Recognition of DNA molecule
B. Aminoacylation of tRNA ✓
C. Recognition of an anti-codon
D. Binding of mRNA to ribosome

Solution: Aminoacylation of tRNA Translation begins with activation of amino acids in the presence of ATP and their linkage to their cognate tRNA, called charging or aminoacylation of tRNA.

Q70.

Floridean starch has structure similar to

A. Amylopectin and glycogen ✓
B. Mannitol and algin
C. Laminarin and cellulose
D. Starch and cellulose

Solution: Amylopectin and glycogen Floridean starch, the stored food of red algae, is very similar to amylopectin and glycogen in structure.

Q71.

Strobili or cones are found in

A. $Pteris$
B. $Marchantia$
C. $Equisetum$ ✓
D. $Salvinia$

Solution: $Equisetum$ In some pteridophytes, such as $Selaginella$ and $Equisetum$, the sporophylls form compact structures called strobili or cones.

Q72.

How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?

A. 2
B. 14 ✓
C. 8
D. 4

Solution: 14 Mendel selected 14 true-breeding pea varieties as 7 pairs, each pair similar except for one character with contrasting traits.

Q73.

Snow-blindness in Antarctic region is due to

A. Inflammation of cornea due to high dose of UV-B radiation ✓
B. High reflection of light from snow
C. Damage to retina caused by infra-red rays
D. Freezing of fluids in the eye by low temperature

Solution: Inflammation of cornea due to high dose of UV-B radiation The cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of the cornea, called snow-blindness.

Q74.

The enzyme enterokinase helps in conversion of

A. trypsinogen into trypsin ✓
B. caseinogen into casein
C. pepsinogen into pepsin
D. protein into polypeptides

Solution: Trypsinogen into trypsin Enterokinase, secreted by the intestinal mucosa, activates trypsinogen of the pancreatic juice into trypsin, which in turn activates the other pancreatic enzymes.

Q75.

Match the following with respect to meiosis (a) Zygotene (b) Pachytene (c) Diplotene (d) Diakinesis (i) Terminalization (ii) Chiasmata (iii) Crossing over (iv) Synapsis Select the correct option from the following

A. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) ✓
B. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
C. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
D. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

Solution: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) (a)-(iv): In zygotene homologous chromosomes pair up; this is synapsis. (b)-(iii): In pachytene crossing over occurs between non-sister chromatids of homologous chromosomes. (c)-(ii): In diplotene the homologues separate except at the X-shaped crossover sites, the chiasmata. (d)-(i): Diakinesis is marked by terminalisation of the chiasmata.

Q76.

Which of the following statements about inclusion bodies is incorrect?

A. These are involved in ingestion of food particles ✓
B. They lie free in the cytoplasm
C. These represent reserve material in cytoplasm
D. They are not bound by any membrane

Solution: These are involved in ingestion of food particles (the incorrect statement) Inclusion bodies are stored reserve material; they take no part in ingesting food particles. (B) Inclusion bodies lie free in the cytoplasm. (C) They hold reserve material, such as phosphate granules, cyanophycean granules and glycogen granules. (D) They are not bound by any membrane system.

Q77.

Match the following columns and select the correct option. Column-I: (a) Eosinophils (b) Basophils (c) Neutrophils (d) Lymphocytes Column-II: (i) Immune response (ii) Phagocytosis (iii) Release histaminase, destructive enzymes (iv) Release granules containing histamine

A. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
C. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
D. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) ✓

Solution: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) (a)-(iii): Eosinophils resist infections and are associated with allergic reactions; they release histaminase and destructive enzymes. (b)-(iv): Basophils secrete histamine, serotonin and heparin, and are involved in inflammatory reactions. (c)-(ii): Neutrophils are phagocytic cells that destroy foreign organisms entering the body. (d)-(i): B and T lymphocytes are responsible for the immune responses of the body.

Q78.

The transverse section of a plant shows following anatomical features : (a) Large number of scattered vascular bundles surrounded by bundle sheath (b) Large conspicuous parenchymatous ground tissue (c) Vascular bundles conjoint and closed (d) Phloem parenchyma absent Identify the category of plant and its part :

A. Monocotyledonous root
B. Dicotyledonous stem
C. Dicotyledonous root
D. Monocotyledonous stem ✓

Solution: Monocotyledonous stem Many scattered, conjoint and closed vascular bundles, each with a bundle sheath, in a large parenchymatous ground tissue and with no phloem parenchyma, are features of a monocot stem. A dicot stem has open vascular bundles arranged in a ring, and roots have radial, not conjoint, vascular bundles.

Q79.

Match the following columns and select the correct option. Column-I: (a) Pituitary gland (b) Thyroid gland (c) Adrenal gland (d) Pancreas Column-II: (i) Grave's disease (ii) Diabetes mellitus (iii) Diabetes insipidus (iv) Addison's disease

A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) ✓
C. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
D. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

Solution: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) (a)-(iii): Diabetes insipidus results from impaired synthesis or release of ADH from the posterior pituitary. (b)-(i): Graves' disease (exophthalmic goitre) is a form of hyperthyroidism. (c)-(iv): Addison's disease results from underproduction of hormones by the adrenal cortex. (d)-(ii): Diabetes mellitus results from a lack of insulin, secreted by the β-cells of the pancreas.

Q80.

Match the following columns and select the correct option. Column-I: (a) Placenta (b) Zona pellucida (c) Bulbo-urethral glands (d) Leydig cells Column-II: (i) Androgens (ii) Human Chorionic Gonadotropin (hCG) (iii) Layer of the ovum (iv) Lubrication of the Penis

A. (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
B. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
C. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓
D. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

Solution: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (a)-(ii): The placenta secretes human chorionic gonadotropin (hCG), besides hPL, estrogens and progestogens. (b)-(iii): The zona pellucida is a membrane formed by the secondary oocyte around itself, a layer of the ovum. (c)-(iv): Secretions of the bulbourethral glands help in lubrication of the penis. (d)-(i): Leydig (interstitial) cells synthesise and secrete androgens.

Q81.

In water hyacinth and water lily, pollination takes place by :

A. Water currents only
B. Wind and water
C. Insects and water
D. Insects or wind ✓

Solution: Insects or wind In most aquatic plants, such as water hyacinth and water lily, the flowers emerge above the water and are pollinated by insects or wind, as in land plants.

Q82.

According to Robert May, the global species diversity is about

A. 20 million
B. 50 million
C. 7 million ✓
D. 1.5 million

Solution: 7 million Robert May's conservative, scientifically sound estimate places global species diversity at about 7 million; some extreme estimates range from 20 to 50 million.

Q83.

Match the following columns and select the correct option. Column-I: (a) 6-15 pairs of gill slits (b) Heterocercal caudal fin (c) Air Bladder (d) Poison sting Column-II: (i) $Trygon$ (ii) Cyclostomes (iii) Chondrichthyes (iv) Osteichthyes

A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
B. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
C. (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
D. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓

Solution: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (a)-(ii): Cyclostomes have an elongated body bearing 6 to 15 pairs of gill slits for respiration. (b)-(iii): Cartilaginous fishes (Chondrichthyes) have a heterocercal caudal fin. (c)-(iv): Bony fishes (Osteichthyes) have an air bladder that regulates buoyancy. (d)-(i): $Trygon$ (sting ray), a cartilaginous fish, has a poison sting.

Q84.

The process of growth is maximum during

A. Lag phase
B. Senescence
C. Dormancy
D. Log phase ✓

Solution: Log phase Growth is slow at first (lag phase) and then increases rapidly at an exponential rate in the log or exponential phase, the period of maximum growth.

Q85.

Match the following columns and select the correct option. Column-I: (a) Bt cotton (b) Adenosine deaminase deficiency (c) RNAi (d) PCR Column-II: (i) Gene therapy (ii) Cellular defence (iii) Detection of HIV infection (iv) $Bacillus$ $thuringiensis$

A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
C. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
D. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓

Solution: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) (a)-(iv): Bt cotton carries Bt toxin genes from $Bacillus$ $thuringiensis$. (b)-(i): The first clinical gene therapy (1990) was given to a 4-year-old girl with adenosine deaminase (ADA) deficiency. (c)-(ii): RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence. (d)-(iii): PCR is routinely used to detect HIV in suspected AIDS patients.

Q86.

Match the following columns and select the correct option. Column-I: (a) Organ of Corti (b) Cochlea (c) Eustachian tube (d) Stapes Column-II: (i) Connects middle ear and pharynx (ii) Coiled part of the labyrinth (iii) Attached to the oval window (iv) Located on the basilar membrane

A. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B. (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) ✓
C. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
D. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)

Solution: (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) (a)-(iv): The organ of Corti, which contains the auditory hair cells, is located on the basilar membrane. (b)-(ii): The coiled portion of the labyrinth is the cochlea. (c)-(i): The Eustachian tube connects the middle ear cavity with the pharynx. (d)-(iii): The stapes, the last ear ossicle, is attached to the oval window of the cochlea.

Q87.

Which one of the following is the most abundant protein in the animals?

A. Collagen ✓
B. Lectin
C. Insulin
D. Haemoglobin

Solution: Collagen Collagen is the most abundant protein in the animal world; RuBisCO is the most abundant protein in the whole biosphere.

Q88.

Identify the wrong statement with reference to immunity.

A. When ready-made antibodies are directly given, it is called "Passive immunity".
B. Active immunity is quick and gives full response. ✓
C. Foetus receives some antibodies from mother, it is an example for passive immunity.
D. When exposed to antigen (living or dead) antibodies are produced in the host's body. It is called "Active immunity".

Solution: Active immunity is quick and gives full response (the incorrect statement) Active immunity is slow and takes time to give its full effective response; passive immunity, with ready-made antibodies, acts quickly. (A) Giving ready-made antibodies directly is passive immunity. (C) Antibodies passed from mother to foetus through the placenta are an example of passive immunity. (D) Antibodies made in the host's body on exposure to antigens (living or dead microbes or other proteins) give active immunity.

Q89.

Montreal protocol was signed in 1987 for control of

A. Emission of ozone depleting substances ✓
B. Release of Green House gases
C. Disposal of e-wastes
D. Transport of Genetically modified organisms from one country to another

Solution: Emission of ozone depleting substances The Montreal Protocol, signed at Montreal (Canada) in 1987 and effective from 1989, is an international treaty to control the emission of ozone-depleting substances.

Q90.

Match the trophic levels with their correct species examples in grassland ecosystem. (a) Fourth trophic level (b) Second trophic level (c) First trophic level (d) Third trophic level (i) Crow (ii) Vulture (iii) Rabbit (iv) Grass Select the correct option

A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
C. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
D. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓

Solution: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) (c)-(iv): Grass, the producer, is the first trophic level. (b)-(iii): The rabbit, a herbivore (primary consumer), is the second trophic level. (d)-(i): The crow, a primary carnivore, is the third trophic level. (a)-(ii): The vulture, the top carnivore, is the fourth trophic level.

Q91.

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is :

A. 0.25 mm
B. 0.5 mm ✓
C. 1.0 mm
D. 0.01 mm

Solution: 0.5 mm Least count $= \dfrac{\text{pitch}}{\text{number of divisions on the circular scale}}$ $$\therefore \text{pitch} = 0.01\ \text{mm} \times 50 = 0.5\ \text{mm}$$

Q92.

The mean free path for a gas, with molecular diameter d and number density n can be expressed as :

A. $\dfrac{1}{\sqrt{2}\,n\pi d^{2}}$ ✓
B. $\dfrac{1}{\sqrt{2}\,n^{2}\pi d^{2}}$
C. $\dfrac{1}{\sqrt{2}\,n^{2}\pi^{2} d^{2}}$
D. $\dfrac{1}{\sqrt{2}\,n\pi d}$

Solution: $\dfrac{1}{\sqrt{2}\,n\pi d^{2}}$ Mean free path $\lambda = \dfrac{1}{\sqrt{2}\,n\pi d^{2}}$: it falls as the number density $n$ and the collision cross-section $\pi d^2$ increase.

Q93.

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

A. four times
B. one-fourth
C. zero ✓
D. doubled

Solution: Zero Photoelectric emission takes place only if the frequency is at least the threshold frequency $\nu_0$, whatever the intensity. New frequency $\nu' = \dfrac{1.5\,\nu_0}{2} = 0.75\,\nu_0 < \nu_0$ $\therefore$ no photoelectrons are emitted, and the photoelectric current is zero even at double the intensity.

Q94.

In a certain region of space with volume 0.2 m$^{3}$, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is :

A. 0.5 N/C
B. 1 N/C
C. 5 N/C
D. zero ✓

Solution: Zero Electric field is the negative of the potential gradient: $E = -\dfrac{dV}{dr}$ $V = 5$ V everywhere in the region, so $\dfrac{dV}{dr} = 0$ and $E = 0$; the volume of the region does not matter.

Q95.

Which of the following graph represents the variation of resistivity ($\rho$) with temperature (T) for copper?

A. Graph (1)
B. Graph (2) ✓
C. Graph (3)
D. Graph (4)

Solution: Graph (2) For a metal such as copper, resistivity increases with temperature. At temperatures much lower than 0 °C the graph deviates considerably from a straight line, curving upward as the temperature rises.

Q96.

A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L$_1$ when mass M is suspended from its free end. The expression for Young's modulus is :

A. $\dfrac{Mg(L_1 - L)}{AL}$
B. $\dfrac{MgL}{AL_1}$
C. $\dfrac{MgL}{A(L_1 - L)}$ ✓
D. $\dfrac{MgL_1}{AL}$

Solution: $\dfrac{MgL}{A(L_1 - L)}$ Young's modulus $Y = \dfrac{\text{longitudinal stress}}{\text{longitudinal strain}}$ Stress $= \dfrac{Mg}{A}$ and strain $= \dfrac{L_1 - L}{L}$ $$\therefore Y = \dfrac{Mg/A}{(L_1 - L)/L} = \dfrac{MgL}{A(L_1 - L)}$$

Q97.

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be:

A. 524 Hz ✓
B. 536 Hz
C. 537 Hz
D. 523 Hz

Solution: 524 Hz Beat frequency $= |f_A - f_B| = 6$ Hz, so $f_B = 524$ Hz or $536$ Hz. Frequency of a stretched string $f \propto \sqrt{T}$, so decreasing the tension in B lowers $f_B$. The beat frequency rises to 7 Hz, so $f_B$ has moved away from $f_A$: $f_B < f_A$. $$\therefore f_B = 530 - 6 = 524\ \text{Hz}$$

Q98.

A 40 $\mu$F capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly :

A. 2.05 A
B. 2.5 A ✓
C. 25.1 A
D. 1.7 A

Solution: 2.5 A Capacitive reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi \times 50 \times 40 \times 10^{-6}}\ \Omega \approx 79.6\ \Omega$ $$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_C} = \dfrac{200}{79.6}\ \text{A} \approx 2.5\ \text{A}$$

Q99.

A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g = 10 m/s$^{2}$)

A. 340 m
B. 320 m
C. 300 m ✓
D. 360 m

Solution: 300 m Taking downward as positive, $v^2 = u^2 + 2gh$ $$h = \dfrac{v^2 - u^2}{2g} = \dfrac{80^2 - 20^2}{2 \times 10}\ \text{m} = \dfrac{6000}{20}\ \text{m} = 300\ \text{m}$$

Q100.

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2}$ nm, the potential difference is :

A. $10^{2}$ V
B. $10^{3}$ V
C. $10^{4}$ V ✓
D. 10 V

Solution: $10^{4}$ V For an electron accelerated from rest through $V$ volt, $\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{1.227}{\sqrt{V}}\ \text{nm}$ $$\sqrt{V} = \dfrac{1.227}{1.227 \times 10^{-2}} = 10^{2}$$ $$\therefore V = 10^{4}\ \text{V}$$

Q101.

For the logic circuit shown, the truth table is:

A. (A, B) $\to$ Y : (0, 0) $\to$ 0, (0, 1) $\to$ 1, (1, 0) $\to$ 1, (1, 1) $\to$ 1
B. (A, B) $\to$ Y : (0, 0) $\to$ 1, (0, 1) $\to$ 1, (1, 0) $\to$ 1, (1, 1) $\to$ 0
C. (A, B) $\to$ Y : (0, 0) $\to$ 1, (0, 1) $\to$ 0, (1, 0) $\to$ 0, (1, 1) $\to$ 0
D. (A, B) $\to$ Y : (0, 0) $\to$ 0, (0, 1) $\to$ 0, (1, 0) $\to$ 0, (1, 1) $\to$ 1 ✓

Solution: Y = 1 only for A = 1, B = 1 (AND gate) Inputs A and B each pass through a NOT gate, giving $\overline{A}$ and $\overline{B}$, which feed a NOR gate. By De Morgan's theorem, $Y = \overline{\overline{A} + \overline{B}} = \overline{\overline{A \cdot B}} = A \cdot B$ The circuit acts as an AND gate, so Y is 1 only when both A and B are 1.

Q102.

A short electric dipole has a dipole moment of $16 \times 10^{-9}$ C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60$^{\circ}$ with the dipole axis is : $\left(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \text{N m}^{2}/\text{C}^{2}\right)$

A. 200 V ✓
B. 400 V
C. zero
D. 50 V

Solution: 200 V Potential of a short dipole $V = \dfrac{1}{4\pi\epsilon_0}\,\dfrac{p\cos\theta}{r^2}$ $$V = \dfrac{9 \times 10^{9} \times 16 \times 10^{-9} \times \cos 60^{\circ}}{(0.6)^2}\ \text{V} = \dfrac{72}{0.36}\ \text{V} = 200\ \text{V}$$

Q103.

An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m$^{-1}$. The permeability of the material of the rod is : ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)

A. $8.0 \times 10^{-5}$ T m A$^{-1}$
B. $2.4\pi \times 10^{-5}$ T m A$^{-1}$
C. $2.4\pi \times 10^{-7}$ T m A$^{-1}$
D. $2.4\pi \times 10^{-4}$ T m A$^{-1}$ ✓

Solution: $2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$ Relative permeability $\mu_r = 1 + \chi_m = 1 + 599 = 600$ $$\mu = \mu_r\mu_0 = 600 \times 4\pi \times 10^{-7} = 2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$$ The magnetising field ($1200\ \text{A m}^{-1}$) does not enter the calculation of $\mu$.

Q104.

The increase in the width of the depletion region in a p-n junction diode is due to :

A. reverse bias only ✓
B. both forward bias and reverse bias
C. increase in forward current
D. forward bias only

Solution: Reverse bias only Under reverse bias the applied voltage is in the same direction as the barrier potential, so the barrier height increases and the depletion region widens; forward bias narrows it.

Q105.

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is :

A. 5.0 g
B. 10.0 g ✓
C. 20.0 g
D. 2.5 g

Solution: 10.0 g The upward force of surface tension balances the weight of the water raised: $2\pi rT\cos\theta = mg$ $T$ and $\theta$ are the same in both tubes, so $m \propto r$. $$\therefore m_2 = 5.0\ \text{g} \times \dfrac{2r}{r} = 10.0\ \text{g}$$

Q106.

The energy equivalent of 0.5 g of a substance is :

A. $4.5 \times 10^{13}$ J ✓
B. $1.5 \times 10^{13}$ J
C. $0.5 \times 10^{13}$ J
D. $4.5 \times 10^{16}$ J

Solution: $4.5 \times 10^{13}$ J Mass-energy equivalence: $E = mc^2$ $$E = 0.5 \times 10^{-3}\ \text{kg} \times (3 \times 10^{8}\ \text{m s}^{-1})^2 = 4.5 \times 10^{13}\ \text{J}$$

Q107.

The solids which have the negative temperature coefficient of resistance are:

A. insulators only
B. semiconductors only
C. insulators and semiconductors ✓
D. metals

Solution: Insulators and semiconductors In insulators and semiconductors the number of free charge carriers rises rapidly with temperature, so resistivity falls: the temperature coefficient of resistance is negative. In metals the carrier density hardly changes while collisions become more frequent, so resistance rises with temperature (positive coefficient).

Q108.

A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$, then the angle of incidence is nearly equal to :

A. $\dfrac{2A}{\mu}$
B. $\mu A$ ✓
C. $\dfrac{\mu A}{2}$
D. $\dfrac{A}{2\mu}$

Solution: $\mu A$ The ray emerges normally, so the angle of refraction at the second face $r_2 = 0$. For a prism $r_1 + r_2 = A$, so $r_1 = A$. Snell's law at the first face: $\sin i = \mu \sin r_1 = \mu \sin A$ For small angles $\sin\theta \approx \theta$, $\therefore i \approx \mu A$

Q109.

For which one of the following, Bohr model is not valid ?

A. Singly ionised helium atom (He$^{+}$)
B. Deuteron atom
C. Singly ionised neon atom (Ne$^{+}$) ✓
D. Hydrogen atom

Solution: Singly ionised neon atom (Ne⁺) The Bohr model applies only to hydrogen-like species, with a single electron around the nucleus. Ne⁺ still has 9 electrons, so the Bohr model is not valid for it. (A), (B) and (D) He⁺, the deuterium atom and the hydrogen atom each have a single electron.

Q110.

Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is :

A. $1.83 \times 10^{-7}$ rad
B. $7.32 \times 10^{-7}$ rad
C. $6.00 \times 10^{-7}$ rad
D. $3.66 \times 10^{-7}$ rad ✓

Solution: $3.66 \times 10^{-7}$ rad Limit of resolution of a telescope $\Delta\theta = \dfrac{1.22\lambda}{D}$, where $D$ is the diameter of the objective $$\Delta\theta = \dfrac{1.22 \times 600 \times 10^{-9}\ \text{m}}{2\ \text{m}} = 3.66 \times 10^{-7}\ \text{rad}$$

Q111.

A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

A. 32 N ✓
B. 30 N
C. 24 N
D. 48 N

Solution: 32 N Gravitational force $F = \dfrac{GMm}{r^2} \propto \dfrac{1}{r^2}$, where $r$ is the distance from the centre of the earth At $h = \dfrac{R}{2}$, $r = \dfrac{3R}{2}$ $$\therefore F = 72\ \text{N} \times \left(\dfrac{R}{3R/2}\right)^2 = 72 \times \dfrac{4}{9}\ \text{N} = 32\ \text{N}$$

Q112.

A charged particle having drift velocity of $7.5 \times 10^{-4}$ m s$^{-1}$ in an electric field of $3 \times 10^{-10}$ Vm$^{-1}$, has a mobility in m$^{2}$ V$^{-1}$ s$^{-1}$ of :

A. $2.5 \times 10^{6}$ ✓
B. $2.5 \times 10^{-6}$
C. $2.25 \times 10^{-15}$
D. $2.25 \times 10^{15}$

Solution: $2.5 \times 10^{6}$ Mobility $\mu = \dfrac{v_d}{E}$ $$\mu = \dfrac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^{6}\ \text{m}^2\,\text{V}^{-1}\,\text{s}^{-1}$$

Q113.

For transistor action, which of the following statements is correct?

A. Base, emitter and collector regions should have same size.
B. Both emitter junction as well as the collector junction are forward biased.
C. The base region must be very thin and lightly doped. ✓
D. Base, emitter and collector regions should have same doping concentrations.

Solution: The base region must be very thin and lightly doped The base must be very thin and lightly doped so that most carriers injected by the emitter cross it to the collector instead of recombining. (A) The regions differ in size: the base is the thinnest and the collector is larger than the emitter. (B) For transistor action the emitter junction is forward biased and the collector junction reverse biased. (D) The emitter is heavily doped, the collector moderately doped and the base lightly doped.

Q114.

The capacitance of a parallel plate capacitor with air as medium is 6 $\mu$F. With the introduction of a dielectric medium, the capacitance becomes 30 $\mu$F. The permittivity of the medium is : ($\epsilon_0 = 8.85 \times 10^{-12}$ C$^{2}$ N$^{-1}$ m$^{-2}$)

A. $1.77 \times 10^{-12}$ C$^{2}$ N$^{-1}$ m$^{-2}$
B. $0.44 \times 10^{-10}$ C$^{2}$ N$^{-1}$ m$^{-2}$ ✓
C. $5.00$ C$^{2}$ N$^{-1}$ m$^{-2}$
D. $0.44 \times 10^{-13}$ C$^{2}$ N$^{-1}$ m$^{-2}$

Solution: $0.44 \times 10^{-10}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ Dielectric constant $K = \dfrac{C}{C_0} = \dfrac{30\ \mu\text{F}}{6\ \mu\text{F}} = 5$ $$\epsilon = K\epsilon_0 = 5 \times 8.85 \times 10^{-12} \approx 0.44 \times 10^{-10}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$$

Q115.

Taking into account of the significant figures, what is the value of 9.99 m – 0.0099 m?

A. 9.98 m ✓
B. 9.980 m
C. 9.9 m
D. 9.9801 m

Solution: 9.98 m In addition and subtraction, the result keeps only as many decimal places as the term with the fewest: here two, from 9.99 m. $$9.99\ \text{m} - 0.0099\ \text{m} = 9.9801\ \text{m} \approx 9.98\ \text{m}$$

Q116.

Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

A. g/2
B. g/5 ✓
C. g/10
D. g

Solution: g/5 For an Atwood machine, $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$ with $m_1 > m_2$ $$a = \dfrac{(6 - 4)g}{6 + 4} = \dfrac{g}{5}$$

Q117.

A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27$^{\circ}$C. Its density is : (R = 8.3 J mol$^{-1}$ K$^{-1}$)

A. 0.2 kg/m$^{3}$ ✓
B. 0.1 kg/m$^{3}$
C. 0.02 kg/m$^{3}$
D. 0.5 kg/m$^{3}$

Solution: $0.2\ \text{kg m}^{-3}$ From $PV = \dfrac{m}{M}RT$, density $\rho = \dfrac{PM}{RT}$, with $M = 2 \times 10^{-3}\ \text{kg mol}^{-1}$ for H₂ and $T = 300$ K $$\rho = \dfrac{249 \times 10^{3} \times 2 \times 10^{-3}}{8.3 \times 300}\ \text{kg m}^{-3} = 0.2\ \text{kg m}^{-3}$$

Q118.

The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is : (c = speed of electromagnetic waves)

A. 1 : 1 ✓
B. 1 : c
C. 1 : c$^{2}$
D. c : 1

Solution: 1 : 1 In an electromagnetic wave the energy densities of the two fields are equal: $\tfrac{1}{2}\epsilon_0 E^2 = \dfrac{B^2}{2\mu_0}$, since $E = cB$ and $c^2 = \dfrac{1}{\mu_0\epsilon_0}$. $\therefore$ the electric and magnetic fields contribute equally to the intensity, in the ratio 1 : 1.

Q119.

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)

A. $3.14 \times 10^{-4}$ T
B. $6.28 \times 10^{-5}$ T
C. $3.14 \times 10^{-5}$ T
D. $6.28 \times 10^{-4}$ T ✓

Solution: $6.28 \times 10^{-4}$ T Turns per unit length $n = \dfrac{N}{L} = \dfrac{100}{0.5\ \text{m}} = 200\ \text{m}^{-1}$ Field inside a long solenoid $B = \mu_0 nI = 4\pi \times 10^{-7} \times 200 \times 2.5\ \text{T}$ $$\therefore B = 2\pi \times 10^{-4}\ \text{T} = 6.28 \times 10^{-4}\ \text{T}$$

Q120.

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

A. half
B. four times ✓
C. one-fourth
D. double

Solution: Four times Fringe width $\beta = \dfrac{\lambda D}{d}$ $$\beta' = \dfrac{\lambda(2D)}{d/2} = \dfrac{4\lambda D}{d} = 4\beta$$

Q121.

A resistance wire connected in the left gap of a metre bridge balances a 10 $\Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 $\Omega$ of the resistance wire is :

A. $1.0 \times 10^{-1}$ m ✓
B. $1.5 \times 10^{-1}$ m
C. $1.5 \times 10^{-2}$ m
D. $1.0 \times 10^{-2}$ m

Solution: $1.0 \times 10^{-1}$ m Balanced metre bridge: $\dfrac{X}{R} = \dfrac{l_1}{l_2} \Rightarrow \dfrac{X}{10\ \Omega} = \dfrac{3}{2}$, so $X = 15\ \Omega$ Resistance of a uniform wire is proportional to its length, and 1.5 m of this wire has $15\ \Omega$. $$\therefore \text{length of } 1\ \Omega = \dfrac{1.5\ \text{m}}{15} = 0.1\ \text{m} = 1.0 \times 10^{-1}\ \text{m}$$

Q122.

The energy required to break one bond in DNA is $10^{-20}$ J. This value in eV is nearly:

A. 0.6
B. 0.06 ✓
C. 0.006
D. 6

Solution: 0.06 eV $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ $$\therefore 10^{-20}\ \text{J} = \dfrac{10^{-20}}{1.6 \times 10^{-19}}\ \text{eV} = 0.0625\ \text{eV} \approx 0.06\ \text{eV}$$

Q123.

When a uranium isotope $^{235}_{92}$U is bombarded with a neutron, it generates $^{89}_{36}$Kr, three neutrons and :

A. $^{91}_{40}$Zr
B. $^{101}_{36}$Kr
C. $^{103}_{36}$Kr
D. $^{144}_{56}$Ba ✓

Solution: $^{144}_{56}\text{Ba}$ $$^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \rightarrow {}^{89}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n} + {}^{A}_{Z}\text{X}$$ Charge number: $92 + 0 = 36 + 0 + Z \Rightarrow Z = 56$ Mass number: $235 + 1 = 89 + 3 + A \Rightarrow A = 144$ $\therefore$ the other product is $^{144}_{56}\text{Ba}$ (barium).

Q124.

Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is :

A. adiabatic ✓
B. isochoric
C. isobaric
D. isothermal

Solution: Adiabatic The whole system is thermally insulated, so no heat enters or leaves ($\Delta Q = 0$): the process is adiabatic. In this free expansion into vacuum $W = 0$, so $\Delta U = 0$ and the ideal gas ends at its initial temperature, but the sudden expansion is not a slow isothermal process. The volume doubles and the pressure falls, so it is neither isochoric nor isobaric.

Q125.

Light with an average flux of 20 W/cm$^{2}$ falls on a non-reflecting surface at normal incidence having surface area 20 cm$^{2}$. The energy received by the surface during time span of 1 minute is :

A. $12 \times 10^{3}$ J
B. $24 \times 10^{3}$ J ✓
C. $48 \times 10^{3}$ J
D. $10 \times 10^{3}$ J

Solution: $24 \times 10^{3}$ J The surface is non-reflecting, so it absorbs all the incident energy: energy $=$ flux $\times$ area $\times$ time $$= 20\ \text{W cm}^{-2} \times 20\ \text{cm}^2 \times 60\ \text{s} = 24 \times 10^{3}\ \text{J}$$

Q126.

The quantities of heat required to raise the temperature of two solid copper spheres of radii r$_1$ and r$_2$ (r$_1$ = 1.5 r$_2$) through 1 K are in the ratio :

A. $\dfrac{9}{4}$
B. $\dfrac{3}{2}$
C. $\dfrac{5}{3}$
D. $\dfrac{27}{8}$ ✓

Solution: $\dfrac{27}{8}$ Heat required $Q = ms\Delta T$, with mass $m = \tfrac{4}{3}\pi r^3\rho$; $s$, $\rho$ and $\Delta T$ are the same for both spheres. $$\therefore \dfrac{Q_1}{Q_2} = \left(\dfrac{r_1}{r_2}\right)^3 = (1.5)^3 = \dfrac{27}{8}$$

Q127.

The average thermal energy for a mono-atomic gas is : (k$_B$ is Boltzmann constant and T, absolute temperature)

A. $\dfrac{3}{2}k_BT$ ✓
B. $\dfrac{5}{2}k_BT$
C. $\dfrac{7}{2}k_BT$
D. $\dfrac{1}{2}k_BT$

Solution: $\dfrac{3}{2}k_BT$ A monatomic molecule has only 3 translational degrees of freedom, each with average energy $\tfrac{1}{2}k_BT$ (law of equipartition of energy). $$\therefore \text{average thermal energy} = 3 \times \tfrac{1}{2}k_BT = \tfrac{3}{2}k_BT$$

Q128.

A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\dfrac{\pi}{3}$. If instead C is removed from the circuit, the phase difference is again $\dfrac{\pi}{3}$ between current and voltage. The power factor of the circuit is :

A. 0.5
B. 1.0 ✓
C. –1.0
D. zero

Solution: 1.0 With L removed: $\tan\dfrac{\pi}{3} = \dfrac{X_C}{R}$; with C removed: $\tan\dfrac{\pi}{3} = \dfrac{X_L}{R}$ $\therefore X_L = X_C$: the full LCR circuit is at resonance and $Z = R$. $$\text{Power factor} = \cos\phi = \dfrac{R}{Z} = 1$$

Q129.

Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of :

A. 50 cm
B. 67 cm ✓
C. 80 cm
D. 33 cm

Solution: 67 cm Taking the 5 kg particle as the origin, the 10 kg particle is at $x = 1$ m. $$x_{\text{cm}} = \dfrac{5 \times 0 + 10 \times 1}{5 + 10}\ \text{m} = \dfrac{2}{3}\ \text{m} \approx 67\ \text{cm}$$

Q130.

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :

A. $\dfrac{3\pi}{2}$ rad
B. $\dfrac{\pi}{2}$ rad
C. zero
D. $\pi$ rad ✓

Solution: $\pi$ rad Displacement in SHM $x = A\sin\omega t$ Acceleration $a = \dfrac{d^2x}{dt^2} = -\omega^2 A\sin\omega t = \omega^2 A\sin(\omega t + \pi)$ $\therefore$ the phase difference between displacement and acceleration is $\pi$ rad.

Q131.

The Brewsters angle i$_b$ for an interface should be

A. 30$^{\circ}$ < i$_b$ < 45$^{\circ}$
B. 45$^{\circ}$ < i$_b$ < 90$^{\circ}$ ✓
C. i$_b$ = 90$^{\circ}$
D. 0$^{\circ}$ < i$_b$ < 30$^{\circ}$

Solution: $45^{\circ} < i_b < 90^{\circ}$ Brewster's law: $\tan i_b = \mu$ For light going from a rarer to a denser medium (such as air to glass), $1 < \mu < \infty$, so $1 < \tan i_b < \infty$. $$\therefore 45^{\circ} < i_b < 90^{\circ}$$

Q132.

Dimensions of stress are :

A. [ML$^{2}$T$^{-2}$]
B. [ML$^{0}$T$^{-2}$]
C. [ML$^{-1}$T$^{-2}$] ✓
D. [MLT$^{-2}$]

Solution: $[\text{M}\,\text{L}^{-1}\,\text{T}^{-2}]$ $$\text{Stress} = \dfrac{\text{force}}{\text{area}} = \dfrac{[\text{M}\,\text{L}\,\text{T}^{-2}]}{[\text{L}^2]} = [\text{M}\,\text{L}^{-1}\,\text{T}^{-2}]$$

Q133.

The color code of a resistance is given below: the four bands, read in order, are Yellow, Violet, Brown and Gold. The values of resistance and tolerance, respectively, are

A. 47 k$\Omega$, 10%
B. 4.7 k$\Omega$, 5%
C. 470 $\Omega$, 5% ✓
D. 470 k$\Omega$, 5%

Solution: 470 Ω, 5% Colour code: yellow = 4, violet = 7, brown = multiplier $10^{1}$, gold = tolerance ±5% $$R = 47 \times 10^{1}\ \Omega \pm 5\% = 470\ \Omega \pm 5\%$$

Q134.

A spherical conductor of radius 10 cm has a charge of $3.2 \times 10^{-7}$ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? $\left(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \text{Nm}^{2}/\text{C}^{2}\right)$

A. $1.28 \times 10^{5}$ N/C ✓
B. $1.28 \times 10^{6}$ N/C
C. $1.28 \times 10^{7}$ N/C
D. $1.28 \times 10^{4}$ N/C

Solution: $1.28 \times 10^{5}$ N/C Outside a charged spherical conductor the field is as if the whole charge were at the centre: $E = \dfrac{1}{4\pi\epsilon_0}\,\dfrac{Q}{r^2}$ $$E = \dfrac{9 \times 10^{9} \times 3.2 \times 10^{-7}}{(0.15)^2}\ \text{N C}^{-1} = 1.28 \times 10^{5}\ \text{N C}^{-1}$$

Q135.

Find the torque about the origin when a force of $3\hat{j}$ N acts on a particle whose position vector is $2\hat{k}$ m.

A. $6\hat{j}$ Nm
B. $-6\hat{i}$ Nm ✓
C. $6\hat{k}$ Nm
D. $6\hat{i}$ Nm

Solution: $-6\hat{i}$ N m Torque $\vec{\tau} = \vec{r} \times \vec{F} = 2\hat{k} \times 3\hat{j} = 6(\hat{k} \times \hat{j})$ $\hat{k} \times \hat{j} = -\hat{i}$, $\therefore \vec{\tau} = -6\hat{i}\ \text{N m}$

Q136.

The mixture which shows positive deviation from Raoult's law is

A. Benzene + Toluene
B. Acetone + Chloroform
C. Chloroethane + Bromoethane
D. Ethanol + Acetone ✓

Solution: Ethanol + Acetone In pure ethanol the molecules are hydrogen bonded; acetone molecules get in between them and break some of these hydrogen bonds. The weaker interactions let molecules escape more easily, so the vapour pressure is higher than Raoult's law predicts: positive deviation. (A) and (C) Benzene + toluene and chloroethane + bromoethane form nearly ideal solutions. (B) Acetone + chloroform shows negative deviation, as the two form hydrogen bonds with each other.

Q137.

Which of the following is not correct about carbon monoxide ?

A. It reduces oxygen carrying ability of blood.
B. The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin. ✓
C. It is produced due to incomplete combustion.
D. It forms carboxyhaemoglobin

Solution: The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin (the incorrect statement) Carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin, not less. (A) Because CO binds haemoglobin so firmly, it reduces the oxygen-carrying capacity of blood. (C) CO is produced by incomplete combustion of carbon, as in automobile exhaust. (D) CO binds to haemoglobin to form carboxyhaemoglobin.

Q138.

The number of Faradays(F) required to produce 20 g of calcium from molten CaCl$_2$ (Atomic mass of Ca = 40 g mol$^{-1}$) is

A. 2
B. 3
C. 4
D. 1 ✓

Solution: 1 At the cathode Ca²⁺ + 2e⁻ → Ca, so 1 mol of Ca (40 g) needs 2 F. $$\therefore \text{charge for } 20\ \text{g} = \dfrac{20}{40} \times 2\ \text{F} = 1\ \text{F}$$

Q139.

Hydrolysis of sucrose is given by the following reaction. Sucrose $+$ H$_2$O $\rightleftharpoons$ Glucose $+$ Fructose If the equilibrium constant (K$_C$) is $2 \times 10^{13}$ at 300 K, the value of $\Delta_r G^{\ominus}$ at the same temperature will be :

A. 8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(2 $\times$ 10$^{13}$)
B. 8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(3 $\times$ 10$^{13}$)
C. $-$8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(4 $\times$ 10$^{13}$)
D. $-$8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(2 $\times$ 10$^{13}$) ✓

Solution: $-8.314\ \text{J mol}^{-1}\,\text{K}^{-1} \times 300\ \text{K} \times \ln(2 \times 10^{13})$ $\Delta_r G = \Delta_r G^{\circ} + RT\ln Q$, and at equilibrium $\Delta_r G = 0$ with $Q = K$ $$\therefore \Delta_r G^{\circ} = -RT\ln K = -8.314\ \text{J mol}^{-1}\,\text{K}^{-1} \times 300\ \text{K} \times \ln(2 \times 10^{13})$$

Q140.

For the reaction, 2Cl(g) $\longrightarrow$ Cl$_2$(g), the correct option is :

A. $\Delta_r$H > 0 and $\Delta_r$S < 0
B. $\Delta_r$H < 0 and $\Delta_r$S > 0
C. $\Delta_r$H < 0 and $\Delta_r$S < 0 ✓
D. $\Delta_r$H > 0 and $\Delta_r$S > 0

Solution: $\Delta_r H < 0$ and $\Delta_r S < 0$ Forming the Cl–Cl bond releases energy, so $\Delta_r H < 0$. Two moles of gaseous atoms give one mole of gaseous molecules, so randomness falls and $\Delta_r S < 0$.

Q141.

Paper chromatography is an example of

A. Partition chromatography ✓
B. Thin layer chromatography
C. Column chromatography
D. Adsorption chromatography

Solution: Partition chromatography Paper chromatography is a type of partition chromatography: water trapped in the chromatography paper is the stationary phase, and the components partition between it and the moving solvent.

Q142.

The rate constant for a first order reaction is $4.606 \times 10^{-3}$ s$^{-1}$. The time required to reduce 2.0 g of the reactant to 0.2 g is :

A. 200 s
B. 500 s ✓
C. 1000 s
D. 100 s

Solution: 500 s First-order rate equation: $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$ $$t = \dfrac{2.303}{4.606 \times 10^{-3}\ \text{s}^{-1}}\log\dfrac{2.0}{0.2} = 500\ \text{s} \times \log 10 = 500\ \text{s}$$

Q143.

Which of the following oxoacid of sulphur has – O – O – linkage?

A. H$_2$SO$_4$, sulphuric acid
B. H$_2$S$_2$O$_8$, peroxodisulphuric acid ✓
C. H$_2$S$_2$O$_7$, pyrosulphuric acid
D. H$_2$SO$_3$, sulphurous acid

Solution: H₂S₂O₈, peroxodisulphuric acid Peroxodisulphuric acid has the structure HO–SO₂–O–O–SO₂–OH, with a peroxide (–O–O–) link between the two sulphur atoms. (C) Pyrosulphuric acid, H₂S₂O₇, has an S–O–S bridge, not an O–O link. (A) and (D) H₂SO₄ and H₂SO₃ each have a single sulphur atom and no O–O bond.

Q144.

Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as

A. Cannizzaro's reaction
B. Cross Cannizzaro's reaction
C. Cross Aldol condensation ✓
D. Aldol condensation

Solution: Cross aldol condensation In dilute alkali, acetophenone (which has α-hydrogens) forms a carbanion that adds to the carbonyl group of benzaldehyde; loss of water gives C₆H₅CH=CHCOC₆H₅ (benzalacetophenone). Aldol condensation between two different carbonyl compounds is called cross aldol condensation. (A) and (B) Cannizzaro reactions need concentrated alkali and an aldehyde without α-hydrogen.

Q145.

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is

A. $\dfrac{\sqrt{2}}{4} \times 288$ pm
B. $\dfrac{4}{\sqrt{3}} \times 288$ pm
C. $\dfrac{4}{\sqrt{2}} \times 288$ pm
D. $\dfrac{\sqrt{3}}{4} \times 288$ pm ✓

Solution: $\dfrac{\sqrt{3}}{4} \times 288$ pm In a bcc cell the atoms touch along the body diagonal: $\sqrt{3}\,a = 4r$ $$\therefore r = \dfrac{\sqrt{3}}{4}\,a = \dfrac{\sqrt{3}}{4} \times 288\ \text{pm}$$

Q146.

Which of the following is a cationic detergent?

A. Sodium stearate
B. Cetyltrimethyl ammonium bromide ✓
C. Sodium dodecylbenzene sulphonate
D. Sodium lauryl sulphate

Solution: Cetyltrimethyl ammonium bromide Cetyltrimethylammonium bromide, CH₃(CH₂)₁₅N⁺(CH₃)₃Br⁻, is a quaternary ammonium salt whose long chain is on the cation, so it is a cationic detergent. (A) Sodium stearate is a soap, the sodium salt of a fatty acid. (C) and (D) Sodium dodecylbenzenesulphonate and sodium lauryl sulphate are anionic detergents: the long chain is on the anion.

Q147.

The calculated spin only magnetic moment of Cr$^{2+}$ ion is

A. 4.90 BM ✓
B. 5.92 BM
C. 2.84 BM
D. 3.87 BM

Solution: 4.90 BM Spin-only magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons Cr is [Ar] 3d⁵ 4s¹, so Cr²⁺ is [Ar] 3d⁴ with $n = 4$. $$\therefore \mu = \sqrt{4 \times 6}\ \text{BM} = \sqrt{24}\ \text{BM} \approx 4.90\ \text{BM}$$

Q148.

HCl was passed through a solution of CaCl$_2$, MgCl$_2$ and NaCl. Which of the following compound(s) crystallise(s)?

A. Only NaCl ✓
B. Only MgCl$_2$
C. NaCl, MgCl$_2$ and CaCl$_2$
D. Both MgCl$_2$ and CaCl$_2$

Solution: Only NaCl HCl gas raises the Cl⁻ concentration (common ion effect), so the least soluble chloride, NaCl, crystallises out. CaCl₂ and MgCl₂ are more soluble than NaCl and remain in solution; this is how pure NaCl is obtained from crude salt.

Q149.

Match the following and identify the correct option. (a) CO(g) + H$_2$(g) (b) Temporary hardness of water (c) B$_2$H$_6$ (d) H$_2$O$_2$ (i) Mg(HCO$_3$)$_2$ + Ca(HCO$_3$)$_2$ (ii) An electron deficient hydride (iii) Synthesis gas (iv) Non-planar structure

A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
C. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
D. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) ✓

Solution: (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) (a)-(iii): A mixture of CO and H₂ is water gas, also called synthesis gas because it is used to make methanol and many hydrocarbons. (b)-(i): Temporary hardness is due to the hydrogencarbonates of magnesium and calcium. (c)-(ii): Diborane, B₂H₆, is an electron-deficient hydride: it has too few electrons for a conventional Lewis structure. (d)-(iv): H₂O₂ has a non-planar, open-book structure.

Q150.

Elimination reaction of 2-Bromo-pentane to form pent-2-ene is (a) $\beta$-Elimination reaction (b) Follows Zaitsev rule (c) Dehydrohalogenation reaction (d) Dehydration reaction

A. (a), (c), (d)
B. (b), (c), (d)
C. (a), (b), (d)
D. (a), (b), (c) ✓

Solution: (a), (b), (c) CH₃CHBrCH₂CH₂CH₃ loses Br and an H from a β-carbon, so this is a β-elimination (a) and a dehydrohalogenation (c). The more substituted alkene, pent-2-ene, is the major product, as Zaitsev's rule predicts (b). (A), (B) and (C) Each includes (d), but HBr, not water, is eliminated, so it is not a dehydration.

Q151.

Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?

A. SCN$^{-}$ < F$^{-}$ < CN$^{-}$ < C$_2$O$_4^{2-}$
B. F$^{-}$ < SCN$^{-}$ < C$_2$O$_4^{2-}$ < CN$^{-}$
C. CN$^{-}$ < C$_2$O$_4^{2-}$ < SCN$^{-}$ < F$^{-}$
D. SCN$^{-}$ < F$^{-}$ < C$_2$O$_4^{2-}$ < CN$^{-}$ ✓

Solution: SCN⁻ < F⁻ < C₂O₄²⁻ < CN⁻ Spectrochemical series: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO $\therefore$ the order of increasing field strength is SCN⁻ < F⁻ < C₂O₄²⁻ < CN⁻.

Q152.

Identify the correct statement from the following :

A. Blister copper has blistered appearance due to evolution of CO$_2$.
B. Vapour phase refining is carried out for Nickel by Van Arkel method.
C. Pig iron can be moulded into a variety of shapes. ✓
D. Wrought iron is impure iron with 4% carbon.

Solution: Pig iron can be moulded into a variety of shapes Iron from the blast furnace, with about 4% carbon and small amounts of S, P, Si and Mn, is pig iron, and it is cast into a variety of shapes. (A) Blister copper gets its blistered look from evolution of SO₂, not CO₂. (B) Nickel is refined in the vapour phase by the Mond process; the van Arkel method is used for Zr and Ti. (D) Wrought iron is the purest form of commercial iron, with very little carbon; it is pig iron that has about 4% carbon.

Q153.

Sucrose on hydrolysis gives

A. $\alpha$-D-Glucose + $\beta$-D-Glucose
B. $\alpha$-D-Glucose + $\beta$-D-Fructose ✓
C. $\alpha$-D-Fructose + $\beta$-D-Fructose
D. $\beta$-D-Glucose + $\alpha$-D-Fructose

Solution: α-D-Glucose + β-D-Fructose In sucrose a glycosidic linkage joins C1 of α-D-glucose to C2 of β-D-fructose, so hydrolysis gives α-D-glucose and β-D-fructose.

Q154.

What is the change in oxidation number of carbon in the following reaction? CH$_4$(g) $+$ 4Cl$_2$(g) $\rightarrow$ CCl$_4$(l) $+$ 4HCl(g)

A. 0 to + 4
B. – 4 to + 4 ✓
C. 0 to – 4
D. + 4 to + 4

Solution: $-4$ to $+4$ In CH₄: $x + 4(+1) = 0 \Rightarrow x = -4$ In CCl₄: $x + 4(-1) = 0 \Rightarrow x = +4$ $\therefore$ the oxidation number of carbon changes from $-4$ to $+4$.

Q155.

The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.

A. Copper
B. Calcium
C. Potassium ✓
D. Iron

Solution: Potassium Potassium ions activate many enzymes, take part in the oxidation of glucose to produce ATP and, with sodium, are responsible for the transmission of nerve signals.

Q156.

Which of the following alkane cannot be made in good yield by Wurtz reaction?

A. 2,3-Dimethylbutane
B. n-Heptane ✓
C. n-Butane
D. n-Hexane

Solution: n-Heptane The Wurtz reaction, 2R–X + 2Na → R–R + 2NaX in dry ether, gives good yields only of symmetrical alkanes with an even number of carbon atoms. n-Heptane (7 C) would need two different alkyl halides (C₃ and C₄), which give a mixture of hexane, heptane and octane. n-Butane, n-hexane and 2,3-dimethylbutane each come from a single halide: ethyl, n-propyl and isopropyl halide.

Q157.

Measuring Zeta potential is useful in determining which property of colloidal solution?

A. Solubility
B. Stability of the colloidal particles ✓
C. Size of the colloidal particles
D. Viscosity

Solution: Stability of the colloidal particles Zeta (electrokinetic) potential is the potential difference between the fixed layer and the diffused layer of opposite charges around a colloidal particle. It measures the charge that keeps the particles repelling one another, so it indicates the stability of the colloid.

Q158.

The freezing point depression constant (K$_f$) of benzene is 5.12 K kg mol$^{-1}$. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) :

A. 0.80 K
B. 0.40 K ✓
C. 0.60 K
D. 0.20 K

Solution: 0.40 K Depression in freezing point $\Delta T_f = K_f m$ $$\Delta T_f = 5.12\ \text{K kg mol}^{-1} \times 0.078\ \text{mol kg}^{-1} = 0.399\ \text{K} \approx 0.40\ \text{K}$$

Q159.

Which of the following amine will give the carbylamine test?

A. C$_6$H$_5$NHCH$_3$
B. C$_6$H$_5$N(CH$_3$)$_2$
C. C$_6$H$_5$NHC$_2$H$_5$
D. C$_6$H$_5$NH$_2$ ✓

Solution: C₆H₅NH₂ Only primary amines give the carbylamine test (a foul-smelling isocyanide on heating with chloroform and ethanolic KOH); aniline, C₆H₅NH₂, is a primary amine. (A) and (C) C₆H₅NHCH₃ and C₆H₅NHC₂H₅ are secondary amines. (B) C₆H₅N(CH₃)₂ is a tertiary amine.

Q160.

Which of the following is a natural polymer?

A. poly (Butadiene-styrene)
B. polybutadiene
C. poly (Butadiene-acrylonitrile)
D. cis-1, 4-polyisoprene ✓

Solution: cis-1,4-polyisoprene Natural rubber is cis-1,4-polyisoprene, a linear polymer of isoprene. Polybutadiene, poly(butadiene-styrene) (Buna-S) and poly(butadiene-acrylonitrile) (Buna-N) are synthetic rubbers.

Q161.

Identify the incorrect statement.

A. The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
B. Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
C. The oxidation states of chromium in CrO$_4^{2-}$ and Cr$_2$O$_7^{2-}$ are not the same. ✓
D. Cr$^{2+}$ (d$^{4}$) is a stronger reducing agent than Fe$^{2+}$ (d$^{6}$) in water.

Solution: The oxidation states of chromium in CrO₄²⁻ and Cr₂O₇²⁻ are not the same (the incorrect statement) Chromium is +6 in both: in CrO₄²⁻, $x + 4(-2) = -2 \Rightarrow x = +6$; in Cr₂O₇²⁻, $2x + 7(-2) = -2 \Rightarrow x = +6$. (A) Transition metals act as catalysts because they can adopt multiple oxidation states and form complexes. (B) Interstitial compounds form when small atoms like H, C or N are trapped inside the crystal lattices of metals. (D) E° is −0.41 V for Cr³⁺/Cr²⁺ but +0.77 V for Fe³⁺/Fe²⁺, so Cr²⁺ is the stronger reducing agent; its d⁴ → d³ change gives a half-filled t₂g level.

Q162.

Which of the following set of molecules will have zero dipole moment?

A. Boron trifluoride, hydrogen fluoride, carbon dioxide, 1,3-dichlorobenzene
B. Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene
C. Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene ✓
D. Ammonia, beryllium difluoride, water, 1,4-dichlorobenzene

Solution: Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene BF₃ (trigonal planar), BeF₂ and CO₂ (linear) and 1,4-dichlorobenzene (two C–Cl dipoles exactly opposite) are symmetrical, so their bond dipoles cancel. (A) HF is polar, and so is 1,3-dichlorobenzene, whose C–Cl dipoles are at 120°. (B) NF₃ (pyramidal), H₂O (bent) and 1,3-dichlorobenzene are polar. (D) NH₃ (pyramidal) and H₂O (bent) are polar.

Q163.

On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be

A. Oxygen gas ✓
B. H$_2$S gas
C. SO$_2$ gas
D. Hydrogen gas

Solution: Oxygen gas In dilute H₂SO₄, oxidation of water (E° = 1.23 V) is preferred at the anode over oxidation of SO₄²⁻ (E° = 1.96 V): 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ At the cathode H⁺ is reduced: 4H⁺(aq) + 4e⁻ → 2H₂(g) $\therefore$ oxygen gas is obtained at the anode.

Q164.

Anisole on cleavage with HI gives

A. Iodobenzene + CH$_3$OH
B. Phenol + C$_2$H$_5$I
C. Iodobenzene + C$_2$H$_5$OH
D. Phenol + CH$_3$I ✓

Solution: Phenol + CH₃I HI protonates the ether oxygen of anisole, C₆H₅–O–CH₃, and I⁻ then attacks the methyl carbon by S$_N$2. The aryl-oxygen bond is more stable, and the sp² ring carbon cannot undergo nucleophilic substitution, so that bond is not broken. $\therefore$ the products are phenol and methyl iodide (CH₃I).

Q165.

The number of protons, neutrons and electrons in $^{175}_{71}$Lu, respectively, are

A. 104, 71 and 71
B. 71, 71 and 104
C. 175, 104 and 71
D. 71, 104 and 71 ✓

Solution: 71, 104 and 71 Protons $=$ electrons $=$ atomic number $= 71$ (neutral atom) Neutrons $=$ mass number $-$ atomic number $= 175 - 71 = 104$

Q166.

Match the following : Oxide: (a) CO (b) BaO (c) Al$_2$O$_3$ (d) Cl$_2$O$_7$ Nature: (i) Basic (ii) Neutral (iii) Acidic (iv) Amphoteric Which of the following is correct option?

A. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) ✓
B. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
C. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
D. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Solution: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) (a)-(ii): CO is a neutral oxide. (b)-(i): BaO, an oxide of an alkaline earth metal, is basic. (c)-(iv): Al₂O₃ is amphoteric: it reacts with both acids and bases. (d)-(iii): Cl₂O₇, the oxide of a non-metal in its highest oxidation state (+7), is acidic.

Q167.

A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following ?

A. + R effect of – CH$_3$ groups
B. – R effect of – CH$_3$ groups
C. Hyperconjugation ✓
D. – I effect of – CH$_3$ groups

Solution: Hyperconjugation Hyperconjugation, the delocalisation of σ electrons of C–H bonds on carbons next to the positive carbon, stabilises a carbocation; more α-hydrogens mean more stabilisation. (CH₃)₃C⁺ has 9 α-hydrogens, while the sec-butyl cation, CH₃CH⁺CH₂CH₃, has 5. (A) and (B) A methyl group has no π bond or lone pair, so it shows no resonance (R) effect. (D) Methyl groups are electron-releasing (+I), not −I.

Q168.

Which one of the followings has maximum number of atoms ?

A. 1 g of Mg(s) [Atomic mass of Mg = 24]
B. 1 g of O$_2$(g) [Atomic mass of O = 16]
C. 1 g of Li(s) [Atomic mass of Li = 7] ✓
D. 1 g of Ag(s) [Atomic mass of Ag = 108]

Solution: 1 g of Li(s) Moles of atoms in 1 g: Mg $= \tfrac{1}{24}$, O₂ $= \tfrac{1}{32} \times 2 = \tfrac{1}{16}$, Li $= \tfrac{1}{7}$, Ag $= \tfrac{1}{108}$ Number of atoms $=$ moles of atoms $\times N_A$, so 1 g of Li, with the largest value ($\tfrac{1}{7}$ mol), has the most atoms.

Q169.

Which of the following is a basic amino acid ?

A. Alanine
B. Tyrosine
C. Lysine ✓
D. Serine

Solution: Lysine Lysine, H₂N(CH₂)₄CH(NH₂)COOH, has two amino groups and one carboxyl group, so it is basic; alanine, tyrosine and serine have one of each and are neutral.

Q170.

The correct option for free expansion of an ideal gas under adiabatic condition is

A. q = 0, $\Delta$T < 0 and w > 0
B. q < 0, $\Delta$T = 0 and w = 0
C. q > 0, $\Delta$T > 0 and w > 0
D. q = 0, $\Delta$T = 0 and w = 0 ✓

Solution: $q = 0$, $\Delta T = 0$ and $w = 0$ Adiabatic means $q = 0$. In free expansion the gas expands against zero external pressure, so $w = -p_{\text{ex}}\Delta V = 0$. First law: $\Delta U = q + w = 0$, and the internal energy of an ideal gas depends only on temperature, so $\Delta T = 0$.

Q171.

Identify the incorrect match. Name: (a) Unnilunium (b) Unniltrium (c) Unnilhexium (d) Unununnium IUPAC Official Name: (i) Mendelevium (ii) Lawrencium (iii) Seaborgium (iv) Darmstadtium

A. (b), (ii)
B. (c), (iii)
C. (d), (iv) ✓
D. (a), (i)

Solution: (d) Unununnium with (iv) Darmstadtium (the incorrect match) Element 111 (un-un-un, unununium) is officially roentgenium (Rg); darmstadtium (Ds) is element 110. (A) Unniltrium (element 103) is lawrencium. (B) Unnilhexium (element 106) is seaborgium. (D) Unnilunium (element 101) is mendelevium.

Q172.

Identify a molecule which does not exist.

A. Li$_2$
B. C$_2$
C. O$_2$
D. He$_2$ ✓

Solution: He₂ He₂ would have the configuration σ1s² σ*1s², with equal numbers of bonding and antibonding electrons. $$\text{Bond order} = \tfrac{1}{2}(N_b - N_a) = \tfrac{1}{2}(2 - 2) = 0$$ $\therefore$ He₂ does not exist; Li₂, C₂ and O₂ have bond orders 1, 2 and 2.

Q173.

Identify the correct statements from the following : (a) CO$_2$(g) is used as refrigerant for ice-cream and frozen food. (b) The structure of C$_{60}$ contains twelve six carbon rings and twenty five carbon rings. (c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline. (d) CO is colorless and odourless gas.

A. (a) and (c) only
B. (b) and (c) only
C. (c) and (d) only ✓
D. (a), (b) and (c) only

Solution: (c) and (d) only ZSM-5, a zeolite, converts alcohols directly into gasoline (c), and CO is a colourless, odourless gas (d). (A) Includes (a), but the refrigerant for ice-cream and frozen food is dry ice, solid CO₂, not CO₂ gas. (B) Includes (b), but C₆₀ has twenty six-membered rings and twelve five-membered rings, not the reverse. (D) Includes (a) and (b), which are both wrong.

Q174.

An alkene on ozonolysis gives methanal as one of the product. Its structure is

A. Structure (1)
B. Structure (2) ✓
C. Structure (3)
D. Structure (4)

Solution: Structure (2), allylcyclohexane Ozonolysis gives methanal (HCHO) only from a terminal =CH₂ group. Allylcyclohexane, a cyclohexane ring carrying a –CH₂–CH=CH₂ side chain, has such a terminal methylene group. On ozonolysis followed by Zn/H₂O it gives cyclohexylacetaldehyde and methanal.

Q175.

Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :

A. Sec. butyl alcohol
B. Tert. butyl alcohol ✓
C. Isobutyl alcohol
D. Isopropyl alcohol

Solution: Tert. butyl alcohol The CH₃ group of the Grignard reagent adds to the carbonyl carbon of acetone, (CH₃)₂C=O, giving the adduct (CH₃)₃C–OMgCl. Hydrolysis gives (CH₃)₃C–OH, tert-butyl alcohol (2-methylpropan-2-ol); a ketone gives a tertiary alcohol.

Q176.

A mixture of N$_2$ and Ar gases in a cylinder contains 7 g of N$_2$ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N$_2$ is : [Use atomic masses (in g mol$^{-1}$) : N = 14, Ar = 40]

A. 12 bar
B. 15 bar ✓
C. 18 bar
D. 9 bar

Solution: 15 bar Moles: N₂ $= \dfrac{7}{28} = 0.25$, Ar $= \dfrac{8}{40} = 0.20$ Dalton's law: partial pressure of N₂ = its mole fraction × total pressure $$= \dfrac{0.25}{0.25 + 0.20} \times 27\ \text{bar} = \dfrac{5}{9} \times 27\ \text{bar} = 15\ \text{bar}$$

Q177.

An increase in the concentration of the reactants of a reaction leads to change in

A. heat of reaction
B. threshold energy
C. collision frequency ✓
D. activation energy

Solution: Collision frequency Collision frequency, the number of collisions per second per unit volume, rises with the concentration of reactants. Heat of reaction, threshold energy and activation energy are properties of the reaction and do not depend on concentration.

Q178.

Find out the solubility of Ni(OH)$_2$ in 0.1 M NaOH. Given that the ionic product of Ni(OH)$_2$ is $2 \times 10^{-15}$

A. $2 \times 10^{-8}$ M
B. $1 \times 10^{-13}$ M
C. $1 \times 10^{8}$ M
D. $2 \times 10^{-13}$ M ✓

Solution: $2 \times 10^{-13}$ M Ni(OH)₂ ⇌ Ni²⁺ + 2OH⁻; in 0.1 M NaOH, [OH⁻] = 0.1 + 2s ≈ 0.1 M (common ion effect). Ionic product [Ni²⁺][OH⁻]² $= s\,(0.1)^2 = 2 \times 10^{-15}$ $$\therefore s = \dfrac{2 \times 10^{-15}}{10^{-2}} = 2 \times 10^{-13}\ \text{M}$$

Q179.

Identify compound X in the following sequence of reactions Toluene $\xrightarrow{\ \text{Cl}_2/h\nu\ }$ X $\xrightarrow[373\ \text{K}]{\ \text{H}_2\text{O}\ }$ Benzaldehyde

A. C$_6$H$_5$CH$_2$Cl
B. C$_6$H$_5$CHCl$_2$ ✓
C. C$_6$H$_5$CCl$_3$
D. C$_6$H$_5$Cl

Solution: C₆H₅CHCl₂ In light, chlorine substitutes the side-chain hydrogens of toluene; with two replaced, X is benzal chloride, C₆H₅CHCl₂. Hydrolysis at 373 K gives the unstable gem-diol C₆H₅CH(OH)₂, which loses water to give benzaldehyde. On hydrolysis C₆H₅CH₂Cl would give benzyl alcohol and C₆H₅CCl₃ benzoic acid.

Q180.

Urea reacts with water to form A which will decompose to form B. B when passed through Cu$^{2+}$ (aq), deep blue colour solution C is formed. What is the formula of C from the following ?

A. [Cu(NH$_3$)$_4$]$^{2+}$ ✓
B. Cu(OH)$_2$
C. CuCO$_3\cdot$Cu(OH)$_2$
D. CuSO$_4$

Solution: [Cu(NH₃)₄]²⁺ Urea is hydrolysed to ammonium carbonate (A): NH₂CONH₂ + 2H₂O → (NH₄)₂CO₃ Ammonium carbonate decomposes to ammonia (B): (NH₄)₂CO₃ → 2NH₃ + CO₂ + H₂O Ammonia with Cu²⁺(aq) forms the deep blue tetraamminecopper(II) ion (C): Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺

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