Not yet confirmed against the official key.
Q1.
Match the following genes of the Lac operon with their respective products :
(a) i gene
(b) z gene
(c) a gene
(d) y gene
(i) $\beta$-galactosidase
(ii) Permease
(iii) Repressor
(iv) Transacetylase
Select the correct option.
- A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- B. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
- C. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
- D. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) ✓
Solution: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) In the lac operon the i gene codes for the repressor: (a)-(iii). The z gene codes for β-galactosidase, which hydrolyses lactose: (b)-(i). The a gene codes for transacetylase: (c)-(iv). The y gene codes for permease, which lets β-galactosides into the cell: (d)-(ii).
Q2.
Match the following structures with their respective location in organs
(a) Crypts of Lieberkuhn
(b) Glisson's Capsule
(c) Islets of Langerhans
(d) Brunner's Glands
(i) Pancreas
(ii) Duodenum
(iii) Small intestine
(iv) Liver
Select the correct option from the following
- A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- B. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
- C. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
- D. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) ✓
Solution: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) Crypts of Lieberkuhn lie between the bases of the villi in the small intestine: (a)-(iii). Glisson's capsule is the thin connective tissue sheath covering each lobule of the liver: (b)-(iv). Islets of Langerhans are the endocrine part of the pancreas: (c)-(i). Brunner's glands are found in the submucosa of the duodenum: (d)-(ii).
Q3.
What is the direction of movement of sugars in phloem?
- A. Bi-directional ✓
- B. Non-multidirectional
- C. Upward
- D. Downward
Solution: Bi-directional The source–sink relationship in plants is variable, so sugars in the phloem can move upwards or downwards, i.e. bi-directionally.
Q4.
The ciliated epithelial cells are required to move particles or mucus in a specific direction. In humans, these cells are mainly present in
- A. Bronchioles and Fallopian tubes ✓
- B. Bile duct and Bronchioles
- C. Fallopian tubes and Pancreatic duct
- D. Eustachian tube and Salivary duct
Solution: Bronchioles and Fallopian tubes Ciliated epithelium lines the inner surface of hollow organs such as the bronchioles and Fallopian tubes, where the cilia move mucus or particles in one direction.
Q5.
Which of the following is the most important cause for animals and plants being driven to extinction?
- A. Alien species invasion
- B. Habitat loss and fragmentation ✓
- C. Drought and floods
- D. Economic exploitation
Solution: Habitat loss and fragmentation Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.
Q6.
Which of the following contraceptive methods do involve a role of hormone?
- A. Pills, Emergency contraceptives, Barrier methods.
- B. Lactational amenorrhea, Pills Emergency contraceptives. ✓
- C. Barrier method, Lactational amenorrhea, Pills.
- D. CuT, Pills, Emergency contraceptives.
Solution: Lactational amenorrhea, pills, emergency contraceptives During intense lactation, high prolactin suppresses the gonadotropins, so ovulation does not occur (lactational amenorrhea). Oral pills are progestogens or progestogen–estrogen combinations, and emergency contraceptives give these hormones within 72 hours of coitus. (A) and (C) Barrier methods only stop sperms physically; no hormone is involved. (D) CuT releases copper ions, not hormones.
Q7.
Which of the following pair of organelles does not contain DNA?
- A. Nuclear envelope and Mitochondria
- B. Mitochondria and Lysosomes
- C. Chloroplast and Vacuoles
- D. Lysosomes and Vacuoles ✓
Solution: Lysosomes and vacuoles Lysosomes and vacuoles are membrane-bound sacs that contain no DNA. (A) and (B) Mitochondria are semi-autonomous and carry their own circular DNA. (C) Chloroplasts also carry their own DNA.
Q8.
Placentation in which ovules develop on the inner wall of the ovary or in peripheral part, is
- A. Free central
- B. Basal
- C. Axile
- D. Parietal ✓
Solution: Parietal In parietal placentation the ovules develop on the inner wall of the ovary or on its peripheral part, e.g. mustard and $Argemone$.
Q9.
The Earth Summit held in Rio de Janeiro in 1992 was called
- A. for immediate steps to discontinue use of CFCs that were damaging the ozone layer
- B. to reduce CO$_2$ emissions and global warming
- C. for conservation of biodiversity and sustainable utilization of its benefits ✓
- D. to assess threat posed to native species by invasive weed species
Solution: For conservation of biodiversity and sustainable utilisation of its benefits The Earth Summit (Convention on Biological Diversity, Rio de Janeiro, 1992) called upon all nations to conserve biodiversity and use its benefits sustainably.
Q10.
Purines found both in DNA and RNA are
- A. Cytosine and thymine
- B. Adenine and thymine
- C. Adenine and guanine ✓
- D. Guanine and cytosine
Solution: Adenine and guanine Adenine and guanine are the purines in both DNA and RNA; cytosine and thymine are pyrimidines, and thymine occurs only in DNA.
Q11.
Match the following hormones with the respective disease
(a) Insulin
(b) Thyroxin
(c) Corticoids
(d) Growth Hormone
(i) Addison's disease
(ii) Diabetes insipidus
(iii) Acromegaly
(iv) Goitre
(v) Diabetes mellitus
Select the correct option.
- A. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
- B. (a)-(v), (b)-(i), (c)-(ii), (d)-(iii)
- C. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
- D. (a)-(v), (b)-(iv), (c)-(i), (d)-(iii) ✓
Solution: (a)-(v), (b)-(iv), (c)-(i), (d)-(iii) Insulin deficiency causes diabetes mellitus: (a)-(v). Thyroxine deficiency due to lack of iodine enlarges the thyroid gland, i.e. goitre: (b)-(iv). Underproduction of corticoids by the adrenal cortex causes Addison's disease: (c)-(i). Excess growth hormone in adults causes acromegaly: (d)-(iii). Diabetes insipidus (ii) is due to lack of ADH, so it stays unmatched.
Q12.
The correct sequence of phases of cell cycle is
- A. G$_1$ $\to$ S $\to$ G$_2$ $\to$ M ✓
- B. M $\to$ G$_1$ $\to$ G$_2$ $\to$ S
- C. G$_1$ $\to$ G$_2$ $\to$ S $\to$ M
- D. S $\to$ G$_1$ $\to$ G$_2$ $\to$ M
Solution: G₁ → S → G₂ → M Interphase (G₁, then S, then G₂) comes before the M phase, so the cycle runs G₁ → S → G₂ → M.
Q13.
Which of the following sexually transmitted diseases is not completely curable?
- A. Chlamydiasis
- B. Gonorrhoea
- C. Genital warts
- D. Genital herpes ✓
Solution: Genital herpes Except hepatitis-B, genital herpes and HIV infection, sexually transmitted diseases are completely curable if detected early and treated properly.
Q14.
Polyblend, a fine powder of recycled modified plastic, has proved to be a good material for
- A. Making tubes and pipes
- B. Making plastic sacks
- C. Use as a fertilizer
- D. Construction of roads ✓
Solution: Construction of roads Polyblend, a fine powder of recycled modified plastic, is mixed with the bitumen that is used to lay roads.
Q15.
The shorter and longer arms of a submetacentric chromosome are referred to as
- A. m-arm and n-arm respectively
- B. s-arm and l-arm respectively
- C. p-arm and q-arm respectively ✓
- D. q-arm and p-arm respectively
Solution: p-arm and q-arm respectively In a submetacentric chromosome the centromere lies nearer one end, giving a short arm, called p, and a long arm, called q.
Q16.
Following statements describe the characteristics of the enzyme Restriction Endonuclease. Identify the incorrect statement.
- A. The enzyme recognizes a specific palindromic nucleotide sequence in the DNA.
- B. The enzyme cuts DNA molecule at identified position within the DNA.
- C. The enzyme binds DNA at specific sites and cuts only one of the two strands. ✓
- D. The enzyme cuts the sugar-phosphate backbone at specific sites on each strand.
Solution: The enzyme binds DNA at specific sites and cuts only one of the two strands (the incorrect statement) After binding its recognition sequence, a restriction endonuclease cuts each of the two strands of the double helix, not just one. (A) True: each enzyme recognises a specific palindromic nucleotide sequence. (B) True: it cuts the DNA at a definite position within the DNA. (D) True: it cuts the sugar–phosphate backbone of each strand at specific points.
Q17.
Persistent nucellus in the seed is known as
- A. Tegmen
- B. Chalaza
- C. Perisperm ✓
- D. Hilum
Solution: Perisperm Residual, persistent nucellus in a seed is called perisperm, e.g. black pepper and beet.
Q18.
Identify the cells whose secretion protects the lining of gastro-intestinal tract from various enzymes.
- A. Duodenal Cells
- B. Chief Cells
- C. Goblet Cells ✓
- D. Oxyntic Cells
Solution: Goblet cells Goblet cells secrete mucus, which lubricates the lining and, with bicarbonates, protects the mucosal epithelium from enzymes and the highly concentrated HCl.
Q19.
Which of the following statements is not correct?
- A. Lysosomes are formed by the process of packaging in the endoplasmic reticulum ✓
- B. Lysosomes have numerous hydrolytic enzymes
- C. The hydrolytic enzymes of lysosomes are active under acidic pH
- D. Lysosomes are membrane bound structures
Solution: Lysosomes are formed by the process of packaging in the endoplasmic reticulum (the incorrect statement) Lysosomes are formed by packaging in the Golgi apparatus, not in the endoplasmic reticulum. (B) True: lysosomes are rich in hydrolytic enzymes such as lipases, proteases and carbohydrases. (C) True: these enzymes are optimally active at acidic pH. (D) True: lysosomes are membrane-bound vesicles.
Q20.
Match the following organisms with the products they produce
(a) $Lactobacillus$
(b) $Saccharomyces$ $cerevisiae$
(c) $Aspergillus$ $niger$
(d) $Acetobacter$ $aceti$
(i) Cheese
(ii) Curd
(iii) Citric Acid
(iv) Bread
(v) Acetic Acid
Select the correct option.
- A. (a)-(ii), (b)-(i), (c)-(iii), (d)-(v)
- B. (a)-(ii), (b)-(iv), (c)-(v), (d)-(iii)
- C. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(v) ✓
- D. (a)-(iii), (b)-(iv), (c)-(v), (d)-(i)
Solution: (a)-(ii), (b)-(iv), (c)-(iii), (d)-(v) $Lactobacillus$ (lactic acid bacteria) turns milk into curd: (a)-(ii). $Saccharomyces$ $cerevisiae$ (baker's yeast) is used in bread making: (b)-(iv). $Aspergillus$ $niger$ produces citric acid: (c)-(iii). $Acetobacter$ $aceti$ produces acetic acid: (d)-(v).
Q21.
Which part of the brain is responsible for thermoregulation?
- A. Medulla oblongata
- B. Cerebrum
- C. Hypothalamus ✓
- D. Corpus callosum
Solution: Hypothalamus The hypothalamus contains the centres that control body temperature.
Q22.
In $Antirrhinum$ (Snapdragon), a red flower was crossed with a white flower and in F$_1$ generation pink flowers were obtained. When pink flowers were selfed, the F$_2$ generation showed white, red and pink flowers. Choose the incorrect statement from the following :
- A. Law of Segregation does not apply in this experiment ✓
- B. This experiment does not follow the Principle of Dominance.
- C. Pink colour in F$_1$ is due to incomplete dominance.
- D. Ratio of F$_2$ is $\dfrac{1}{4}$ (Red) : $\dfrac{2}{4}$ (Pink) : $\dfrac{1}{4}$ (White)
Solution: Law of Segregation does not apply in this experiment (the incorrect statement) The law of segregation still applies: the alleles separate during gamete formation, which is why red and white flowers reappear in F₂. (B) True: pink F₁ flowers show that neither allele is dominant, so the principle of dominance is not followed. (C) True: the intermediate pink colour is due to incomplete dominance. (D) True: F₂ is 1 RR (red) : 2 Rr (pink) : 1 rr (white).
Q23.
Which of the following can be used as a biocontrol agent in the treatment of plant disease?
- A. $Lactobacillus$
- B. $Trichoderma$ ✓
- C. $Chlorella$
- D. $Anabaena$
Solution: $Trichoderma$ $Trichoderma$, a free-living fungus common in root ecosystems, is an effective biocontrol agent of several plant pathogens.
Q24.
Select the correct group of biocontrol agents.
- A. $Nostoc$, $Azospirillium$, Nucleopolyhedrovirus
- B. $Bacillus$ $thuringiensis$, Tobacco mosaic virus, Aphids
- C. $Trichoderma$, Baculovirus, $Bacillus$ $thuringiensis$ ✓
- D. $Oscillatoria$, $Rhizobium$, $Trichoderma$
Solution: $Trichoderma$, Baculovirus, $Bacillus$ $thuringiensis$ $Trichoderma$ (a fungus), baculoviruses (e.g. nucleopolyhedrovirus) and $Bacillus$ $thuringiensis$ are all biocontrol agents. (A) $Nostoc$ and $Azospirillum$ are biofertilisers. (B) Tobacco mosaic virus is a plant pathogen and aphids are pests. (D) $Oscillatoria$ and $Rhizobium$ are biofertilisers.
Q25.
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by :
- A. Sutton Boveri
- B. T.H. Morgan
- C. Gregor J. Mendel
- D. Alfred Sturtevant ✓
Solution: Alfred Sturtevant Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their positions.
Q26.
Respiratory Quotient (RQ) value of tripalmitin is
- A. 0.09
- B. 0.9
- C. 0.7 ✓
- D. 0.07
Solution: 0.7 Respiratory quotient (RQ) = volume of CO₂ evolved ÷ volume of O₂ consumed. Tripalmitin: 2C₅₁H₉₈O₆ + 145O₂ → 102CO₂ + 98H₂O + energy ∴ $\text{RQ} = \dfrac{102}{145} \approx 0.7$
Q27.
What would be the heart rate of a person if the cardiac output is 5 L, blood volume in the ventricles at the end of diastole is 100 mL and at the end of ventricular systole is 50 mL?
- A. 125 beats per minute
- B. 50 beats per minute
- C. 75 beats per minute
- D. 100 beats per minute ✓
Solution: 100 beats per minute Cardiac output = stroke volume × heart rate. Stroke volume = end-diastolic volume − end-systolic volume = 100 mL − 50 mL = 50 mL ∴ Heart rate $= \dfrac{5000\ \text{mL min}^{-1}}{50\ \text{mL}} = 100$ beats per minute
Q28.
From evolutionary point of view, retention of the female gametophyte with developing young embryo on the parent sporophyte for some time, is first observed in
- A. Gymnosperms
- B. Liverworts
- C. Mosses
- D. Pteridophytes ✓
Solution: Pteridophytes In heterosporous pteridophytes such as $Selaginella$ and $Salvinia$, the female gametophyte is retained on the parent sporophyte for variable periods. The zygote develops into a young embryo within this female gametophyte, a precursor to the seed habit first seen in pteridophytes.
Q29.
Which of the following ecological pyramids is generally inverted?
- A. Pyramid of biomass in a sea ✓
- B. Pyramid of numbers in grassland
- C. Pyramid of energy
- D. Pyramid of biomass in a forest
Solution: Pyramid of biomass in a sea In the sea the biomass of fishes far exceeds that of the phytoplankton, so the pyramid of biomass is generally inverted. (B) The pyramid of numbers in a grassland is upright. (C) The pyramid of energy is always upright; it can never be inverted. (D) The pyramid of biomass in a forest is upright.
Q30.
Colostrum the yellowish fluid, secreted by mother during the initial days of lactation is very essential to impart immunity to the new born infants because it contains
- A. Immunoglobulin A ✓
- B. Natural killer cells
- C. Monocytes
- D. Macrophages
Solution: Immunoglobulin A Colostrum has abundant IgA antibodies, which give the newborn naturally acquired passive immunity.
Q31.
Phloem in gymnosperms lacks :
- A. Both sieve tubes and companion cells ✓
- B. Albuminous cells and sieve cells
- C. Sieve tubes only
- D. Companion cells only
Solution: Both sieve tubes and companion cells Gymnosperm phloem has sieve cells and albuminous cells; it lacks sieve tubes and companion cells.
Q32.
Match the following organisms with their respective characteristics :
(a) $Pila$
(b) $Bombyx$
(c) $Pleurobrachia$
(d) $Taenia$
(i) Flame cells
(ii) Comb plates
(iii) Radula
(iv) Malpighian tubules
Select the correct option from the following :
- A. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
- B. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- C. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) ✓
- D. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
Solution: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) $Pila$ (a mollusc) feeds with a file-like rasping organ, the radula: (a)-(iii). $Bombyx$ (an arthropod) excretes through Malpighian tubules: (b)-(iv). $Pleurobrachia$ (a ctenophore) moves with eight external rows of ciliated comb plates: (c)-(ii). $Taenia$ (a platyhelminth) has flame cells for osmoregulation and excretion: (d)-(i).
Q33.
Use of an artificial kidney during hemodialysis may result in :
(a) Nitrogenous waste build-up in the body
(b) Non-elimination of excess potassium ions
(c) Reduced absorption of calcium ions from gastro-intestinal tract
(d) Reduced RBC production
Which of the following options is the most appropriate?
- A. (a) and (d) are correct
- B. (a) and (b) are correct
- C. (b) and (c) are correct
- D. (c) and (d) are correct ✓
Solution: (c) and (d) are correct Haemodialysis removes urea and excess K⁺ from the blood, so (a) and (b) do not happen. The artificial kidney cannot make erythropoietin, normally secreted by the juxtaglomerular cells, so RBC production falls: (d). It also cannot activate vitamin D (calcitriol) as the kidney does, so less Ca²⁺ is absorbed from the gut: (c).
Q34.
Which of the following statements is correct?
- A. Cornea consists of dense matrix of collagen and is the most sensitive portion the eye. ✓
- B. Cornea is an external, transparent and protective proteinacious covering of the eye-ball.
- C. Cornea consists of dense connective tissue of elastin and can repair itself.
- D. Cornea is convex, transparent layer which is highly vascularised.
Solution: Cornea consists of dense matrix of collagen and is the most sensitive portion of the eye The cornea is a dense matrix of collagen with many nerve endings, which makes it the most sensitive part of the eye. (B) The cornea is only the transparent front part of the outer coat (sclera), not a protein covering of the whole eyeball. (C) Its matrix is collagen, not elastin. (D) The cornea is avascular; it has no blood vessels.
Q35.
Select the incorrect statement.
- A. Human males have one of their sex-chromosome much shorter than the other
- B. Male fruit fly is heterogametic
- C. In male grasshoppers 50% of sperms have no sex-chromosome
- D. In domesticated fowls, sex of progeny depends on the type of sperm rather than egg ✓
Solution: In domesticated fowls, sex of progeny depends on the type of sperm rather than egg (the incorrect statement) In birds the female is heterogametic (ZW) and the male homogametic (ZZ), so the sex of the progeny depends on the egg, not the sperm. (A) True: the Y chromosome of human males is much shorter than the X. (B) True: the male fruit fly is XY, i.e. heterogametic. (C) True: male grasshoppers are XO, so half the sperms carry an X and half carry no sex chromosome.
Q36.
The concept of "$Omnis$ $cellula$-$e$ $cellula$" regarding cell division was first proposed by
- A. Aristotle
- B. Rudolf Virchow ✓
- C. Theodore Schwann
- D. Schleiden
Solution: Rudolf Virchow Rudolf Virchow (1855) explained that new cells are formed from pre-existing cells ($Omnis$ $cellula$-$e$ $cellula$).
Q37.
Which of the statements given below is not true about formation of Annual Rings in trees?
- A. Annual rings are not prominent in trees of temperate region. ✓
- B. Annual ring is a combination of spring wood and autumn wood produced in a year
- C. Differential activity of cambium causes light and dark bands of tissue early and late wood respectively.
- D. Activity of cambium depends upon variation in climate.
Solution: Annual rings are not prominent in trees of temperate region (the incorrect statement) In temperate regions the cambium is very active in spring and less active in winter, so distinct, prominent annual rings are formed. (B) True: one annual ring is the spring (early) wood plus the autumn (late) wood of one year. (C) True: lighter spring wood and darker autumn wood result from the cambium's different activity. (D) True: cambial activity depends on climate, which varies through the year in temperate regions.
Q38.
$Thiobacillus$ is a group of bacteria helpful in carrying out
- A. Denitrification ✓
- B. Nitrogen fixation
- C. Chemoautotrophic fixation
- D. Nitrification
Solution: Denitrification $Thiobacillus$ (with $Pseudomonas$) reduces nitrate in the soil to free nitrogen, i.e. denitrification.
Q39.
Due to increasing air-borne allergens and pollutants, many people in urban areas are suffering from respiratory disorder causing wheezing due to
- A. reduction in the secretion of surfactants by pneumocytes.
- B. benign growth on mucous lining of nasal cavity
- C. inflammation of bronchi and bronchioles ✓
- D. proliferation of fibrous tissues and damage of the alveolar walls
Solution: Inflammation of bronchi and bronchioles Asthma, a difficulty in breathing with wheezing, is due to inflammation of the bronchi and bronchioles, often set off by allergens and pollutants.
Q40.
In some plants, the female gamete develops into embryo without fertilization. This phenomenon is known as
- A. Parthenogenesis ✓
- B. Autogamy
- C. Parthenocarpy
- D. Syngamy
Solution: Parthenogenesis Development of the female gamete into an embryo without fertilisation is parthenogenesis; parthenocarpy is fruit formation without fertilisation.
Q41.
Select the correct option.
- A. There are seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs. ✓
- B. 8$^{\text{th}}$, 9$^{\text{th}}$ and 10$^{\text{th}}$ pairs of ribs articulate directly with the sternum.
- C. 11$^{\text{th}}$ and 12$^{\text{th}}$ pairs of ribs are connected to the sternum with the help of hyaline cartilage.
- D. Each rib is a flat thin bone and all the ribs are connected dorsally to the thoracic vertebrae and ventrally to the sternum.
Solution: Seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs Ribs 1–7 join the sternum through hyaline cartilage (vertebrosternal), ribs 8–10 join the seventh rib (vertebrochondral) and ribs 11–12 are free ventrally (vertebral or floating). (B) The 8th, 9th and 10th pairs join the seventh rib through hyaline cartilage, not the sternum directly. (C) The 11th and 12th pairs are not connected ventrally at all. (D) Only the first seven pairs are joined ventrally to the sternum.
Q42.
How does steroid hormone influence the cellular activities?
- A. Using aquaporin channels as second messenger
- B. Changing the permeability of the cell membrane
- C. Binding to DNA and forming a gene-hormone complex ✓
- D. Activating cyclic AMP located on the cell membrane
Solution: Binding to DNA and forming a gene–hormone complex Steroid hormones pass through the cell membrane and bind intracellular receptors; the hormone–receptor complex then binds the genome and regulates gene expression, with no second messenger such as cAMP.
Q43.
Select the correctly written scientific name of Mango which was first described by Carolus Linnaeus
- A. $Mangifera$ $Indica$
- B. $Mangifera$ $indica$ Car. Linn.
- C. $Mangifera$ $indica$ Linn. ✓
- D. $Mangifera$ $indica$
Solution: $Mangifera$ $indica$ Linn. The genus starts with a capital letter, the specific epithet with a small letter, and the abbreviated name of the author who first described the species (Linn. for Linnaeus) comes at the end. (A) The specific epithet must start with a small letter. (B) Linnaeus is abbreviated as Linn., not Car. Linn. (D) The name is correct but does not show that Linnaeus first described it.
Q44.
What map unit (Centimorgan) is adopted in the construction of genetic maps?
- A. A unit of distance between genes on chromosomes, representing 50% cross over.
- B. A unit of distance between two expressed genes representing 10% cross over.
- C. A unit of distance between two expressed genes representing 100% cross over.
- D. A unit of distance between genes on chromosomes, representing 1% cross over. ✓
Solution: Distance between genes on a chromosome representing 1% crossing over One map unit (centimorgan) is the distance between two genes on a chromosome that gives 1% recombination (crossing over) between them.
Q45.
Cells in G$_0$ phase :
- A. terminate the cell cycle
- B. exit the cell cycle ✓
- C. enter the cell cycle
- D. suspend the cell cycle
Solution: Exit the cell cycle Cells that stop dividing exit G₁ into the quiescent stage G₀; they stay metabolically active but do not proliferate unless called on to do so.
Q46.
Which one of the following statements regarding post-fertilization development in flowering plants is incorrect?
- A. Ovules develop into embryo sac ✓
- B. Ovary develops into fruit
- C. Zygote develops into embryo
- D. Central cell develops into endosperm
Solution: Ovules develop into embryo sac (the incorrect statement) After fertilisation the ovule develops into the seed; the embryo sac forms inside the ovule before fertilisation. (B) True: the ovary develops into the fruit. (C) True: the zygote develops into the embryo. (D) True: the central cell, after triple fusion, becomes the primary endosperm cell, which develops into the endosperm.
Q47.
Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?
- A. Genetic code is specific
- B. Genetic code is not ambiguous
- C. Genetic code is redundant
- D. Genetic code is nearly universal ✓
Solution: Genetic code is nearly universal The genetic code is nearly universal (e.g. UUU codes for phenylalanine from bacteria to humans), so bacteria translate the human insulin gene correctly.
Q48.
Which of the following glucose transporters is insulin-dependent?
- A. GLUT IV ✓
- B. GLUT I
- C. GLUT II
- D. GLUT III
Solution: GLUT IV GLUT IV, in skeletal muscle and adipose tissue, is moved to the cell membrane by insulin, so glucose uptake through it is insulin-dependent; GLUT I, II and III work without insulin.
Q49.
Under which of the following conditions will there be no change in the reading frame of following mRNA?
5$'$AACAGCGGUGCUAUU3$'$
- A. Deletion of GGU from 7$^{\text{th}}$, 8$^{\text{th}}$ and 9$^{\text{th}}$ positions ✓
- B. Insertion of G at 5$^{\text{th}}$ position
- C. Deletion of G from 5$^{\text{th}}$ position
- D. Insertion of A and G at 4$^{\text{th}}$ and 5$^{\text{th}}$ positions respectively
Solution: Deletion of GGU from 7th, 8th and 9th positions mRNA is read in triplets, so only an insertion or deletion of three bases (or a multiple of three) keeps the reading frame. Deleting GGU (positions 7–9) gives 5′ AAC AGC GCU AUU 3′: one codon is lost and the rest are read unchanged. (B) and (C) Adding or removing one base shifts every codon after it. (D) Adding two bases also shifts the frame.
Q50.
Select the hormone-releasing Intra-Uterine Devices.
- A. Lippes Loop, Multiload 375
- B. Vaults, LNG-20
- C. Multiload 375, Progestasert
- D. Progestasert, LNG-20 ✓
Solution: Progestasert, LNG-20 Progestasert and LNG-20 are hormone-releasing IUDs; Lippes loop is non-medicated and Multiload 375 releases copper.
Q51.
Variations caused by mutation, as proposed by Hugo de Vries are
- A. small and directionless
- B. random and directional
- C. random and directionless ✓
- D. small and directional
Solution: Random and directionless According to Hugo de Vries, mutations are random and directionless, whereas Darwinian variations are small and directional.
Q52.
Expressed Sequence Tags (ESTs) refers to :
- A. Novel DNA sequences
- B. Genes expressed as RNA ✓
- C. Polypeptide expression
- D. DNA polymorphism
Solution: Genes expressed as RNA Expressed Sequence Tags (ESTs) are the genes that are expressed as RNA, identified in the Human Genome Project.
Q53.
What triggers activation of protoxin to active Bt toxin of $Bacillus$ $thuringiensis$ in boll worm?
- A. Acidic pH of stomach
- B. Body temperature
- C. Moist surface of midgut
- D. Alkaline pH of gut ✓
Solution: Alkaline pH of gut The alkaline pH of the insect gut solubilises the Bt crystals and converts the inactive protoxin into the active toxin.
Q54.
Match the hominids with their correct brain size :
(a) $Homo$ $habilis$
(b) $Homo$ $neanderthalensis$
(c) $Homo$ $erectus$
(d) $Homo$ $sapiens$
(i) 900 cc
(ii) 1350 cc
(iii) 650-800 cc
(iv) 1400 cc
Select the correct option.
- A. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
- B. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
- C. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- D. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) ✓
Solution: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) $Homo$ $habilis$ had a brain of 650–800 cc: (a)-(iii). $Homo$ $neanderthalensis$ had a brain of 1400 cc: (b)-(iv). $Homo$ $erectus$ had a brain of about 900 cc: (c)-(i). $Homo$ $sapiens$ has a brain of about 1350 cc: (d)-(ii).
Q55.
Which of the following pairs of gases is mainly responsible for green house effect?
- A. Carbon dioxide and Methane ✓
- B. Ozone and Ammonia
- C. Oxygen and Nitrogen
- D. Nitrogen and Sulphur dioxide
Solution: Carbon dioxide and methane Carbon dioxide (about 60%) and methane (about 20%) contribute most to global warming, ahead of CFCs (14%) and N₂O (6%).
Q56.
Match Column - I with Column - II
Column - I:
(a) Saprophyte
(b) Parasite
(c) Lichens
(d) Mycorrhiza
Column - II:
(i) Symbiotic association of fungi with plant roots
(ii) Decomposition of dead organic materials
(iii) Living on living plants or animals
(iv) Symbiotic association of algae and fungi
Choose the correct answer from the option given below
- A. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓
- B. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
- C. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- D. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
Solution: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) Saprophytes decompose dead organic matter: (a)-(ii). Parasites live on living plants or animals: (b)-(iii). Lichens are symbiotic associations of algae and fungi: (c)-(iv). Mycorrhizae are symbiotic associations of fungi with plant roots: (d)-(i).
Q57.
Which of the following is true for Golden rice?
- A. It has yellow grains, because of a gene introduced from a primitive variety of rice
- B. It is Vitamin A enriched, with a gene from daffodil ✓
- C. It is pest resistant, with a gene from $Bacillus$ $thuringiensis$
- D. It is drought tolerant, developed using $Agrobacterium$ vector
Solution: It is Vitamin A enriched, with a gene from daffodil Golden rice makes β-carotene (provitamin A) in its grains with the help of a gene from daffodil, so it is vitamin A enriched. (A) Its yellow colour comes from this β-carotene, not from a gene of a primitive rice variety. (C) Pest resistance from a $Bacillus$ $thuringiensis$ gene describes Bt crops. (D) It was not developed for drought tolerance.
Q58.
What is the genetic disorder in which an individual has an overall masculine development gynaecomastia, and is sterile ?
- A. Down's syndrome
- B. Turner's syndrome
- C. Klinefelter's syndrome ✓
- D. Edward syndrome
Solution: Klinefelter's syndrome Klinefelter's syndrome (47, XXY) shows overall masculine development with gynaecomastia, and such individuals are sterile.
Q59.
Extrusion of second polar body from egg nucleus occurs :
- A. simultaneously with first cleavage
- B. after entry of sperm but before fertilization ✓
- C. after fertilization
- D. before entry of sperm into ovum
Solution: After entry of sperm but before fertilisation Entry of the sperm induces the secondary oocyte to complete meiosis II and release the second polar body; only then do the sperm and ovum nuclei fuse (fertilisation).
Q60.
A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?
- A. 0.16(AA); 0.36(Aa); 0.48(aa)
- B. 0.36(AA); 0.48(Aa); 0.16(aa)
- C. 0.16(AA); 0.24(Aa); 0.36(aa)
- D. 0.16(AA); 0.48(Aa); 0.36(aa) ✓
Solution: 0.16 (AA); 0.48 (Aa); 0.36 (aa) Hardy–Weinberg principle: $p^2 + 2pq + q^2 = 1$, with $p = 0.4$ and $q = 1 - 0.4 = 0.6$. AA $= p^2 = (0.4)^2 = 0.16$ Aa $= 2pq = 2(0.4)(0.6) = 0.48$ aa $= q^2 = (0.6)^2 = 0.36$
Q61.
What is the fate of the male gametes discharged in the synergid?
- A. One fuses with the egg and other fuses with central cell nuclei. ✓
- B. One fuses with egg other(s) degenerate(s) in the synergid.
- C. All fuse with the egg.
- D. One fuses with the egg, other(s) fuse(s) with synergid nucleus.
Solution: One fuses with the egg and the other fuses with the central cell nuclei One male gamete fuses with the egg (syngamy) to form the diploid zygote. The other fuses with the two polar nuclei in the central cell (triple fusion) to form the triploid primary endosperm nucleus.
Q62.
In a species, the weight of newborn ranges from 2 to 5 kg. 97% of the newborn with an average weight between 3 to 3.3 kg survive whereas 99% of the infants born with weights from 2 to 2.5 kg or 4.5 to 5 kg die. Which type of selection process is taking place?
- A. Cyclical Selection
- B. Directional Selection
- C. Stabilizing Selection ✓
- D. Disruptive Selection
Solution: Stabilising selection Newborns of average weight survive best while both extremes die, so selection favours the mean value: stabilising selection.
Q63.
Which of the following muscular disorders is inherited?
- A. Botulism
- B. Tetany
- C. Muscular dystrophy ✓
- D. Myasthenia gravis
Solution: Muscular dystrophy Muscular dystrophy, progressive degeneration of skeletal muscle, is mostly due to a genetic disorder; tetany, myasthenia gravis and botulism are not inherited.
Q64.
Which of the following protocols did aim for reducing emission of chlorofluorocarbons into the atmosphere?
- A. Geneva Protocol
- B. Montreal Protocol ✓
- C. Kyoto Protocol
- D. Gothenburg Protocol
Solution: Montreal Protocol The Montreal Protocol, signed at Montreal (Canada) in 1987, controls the emission of ozone-depleting substances such as CFCs.
Q65.
Consider the following statement :
(A) Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
(B) A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme.
Select the correct option.
- A. (A) is false but (B) is true.
- B. Both (A) and (B) are true.
- C. (A) is true but (B) is false.
- D. Both (A) and (B) are false. ✓
Solution: Both (A) and (B) are false (A) False: a prosthetic group is an organic cofactor tightly bound to the apoenzyme; coenzymes are bound only transiently, and metal ions are a separate kind of cofactor. (B) False: the protein part alone is the apoenzyme; the complete, catalytically active enzyme with its bound prosthetic group is the holoenzyme.
Q66.
Consider following features
(a) Organ system level of organisation
(b) Bilateral symmetry
(c) True coelomates with segmentation of body
Select the correct option of animal groups which possess all the above characteristics
- A. Annelida, Mollusca and Chordata
- B. Annelida, Arthropoda and Chordata ✓
- C. Annelida, Arthropoda and Mollusca
- D. Arthropoda, Mollusca and Chordata
Solution: Annelida, Arthropoda and Chordata Annelida, Arthropoda and Chordata are bilaterally symmetrical, coelomate and segmented, with organ-system level of organisation. (A), (C) and (D) all include Mollusca, which are coelomate but unsegmented.
Q67.
$Pinus$ seed cannot germinate and establish without fungal association. This is because :
- A. its seeds contain inhibitors that prevent germination.
- B. its embryo is immature.
- C. it has obligate association with mycorrhizae. ✓
- D. it has very hard seed coat.
Solution: It has obligate association with mycorrhizae $Pinus$ has an obligate association with mycorrhizae, so its seeds cannot germinate and establish without the fungus.
Q68.
Select the correct sequence of organs in the alimentary canal of cockroach starting from mouth
- A. Pharynx $\to$ Oesophagus $\to$ Ileum $\to$ Crop $\to$ Gizzard $\to$ Colon $\to$ Rectum
- B. Pharynx $\to$ Oesophagus $\to$ Crop $\to$ Gizzard $\to$ Ileum $\to$ Colon $\to$ Rectum ✓
- C. Pharynx $\to$ Oesophagus $\to$ Gizzard $\to$ Crop $\to$ Ileum $\to$ Colon $\to$ Rectum
- D. Pharynx $\to$ Oesophagus $\to$ Gizzard $\to$ Ileum $\to$ Crop $\to$ Colon $\to$ Rectum
Solution: Pharynx → Oesophagus → Crop → Gizzard → Ileum → Colon → Rectum In the cockroach the foregut runs pharynx → oesophagus → crop → gizzard, the midgut follows, and the hindgut is ileum → colon → rectum.
Q69.
Which of the following statements regarding mitochondria is incorrect?
- A. Mitochondrial matrix contains single circular DNA molecule and ribosomes.
- B. Outer membrane is permeable to monomers of carbohydrates, fats and proteins.
- C. Enzymes of electron transport are embedded in outer membrane. ✓
- D. Inner membrane is convoluted with infoldings.
Solution: Enzymes of electron transport are embedded in outer membrane (the incorrect statement) The electron transport system is located in the inner mitochondrial membrane, not the outer. (A) True: the matrix has a single circular DNA molecule and ribosomes (70S). (B) True: the outer membrane lets small molecules such as these monomers pass freely. (D) True: the inner membrane forms infoldings called cristae.
Q70.
Drug called 'Heroin' is synthesized by
- A. nitration of morphine
- B. methylation of morphine
- C. acetylation of morphine ✓
- D. glycosylation of morphine
Solution: Acetylation of morphine Heroin (smack) is diacetylmorphine, obtained by acetylation of morphine from the latex of the poppy, $Papaver$ $somniferum$.
Q71.
Conversion of glucose to glucose-6-phosphate, the first irreversible reaction of glycolysis, is catalyzed by
- A. Phosphofructokinase
- B. Aldolase
- C. Hexokinase ✓
- D. Enolase
Solution: Hexokinase Hexokinase phosphorylates glucose to glucose-6-phosphate, the first step of glycolysis.
Q72.
DNA precipitation out of a mixture of biomolecules can be achieved by treatment with
- A. Chilled chloroform
- B. Isopropanol
- C. Chilled ethanol ✓
- D. Methanol at room temperature
Solution: Chilled ethanol Chilled ethanol precipitates the purified DNA, which separates out as fine threads that can be removed by spooling.
Q73.
Which of the following is a commercial blood cholesterol lowering agent?
- A. Lipases
- B. Cyclosporin A
- C. Statin ✓
- D. Streptokinase
Solution: Statin Statins, produced by the yeast $Monascus$ $purpureus$, lower blood cholesterol by competitively inhibiting the enzyme that synthesises cholesterol.
Q74.
Which one of the following equipments is essentially required for growing microbes on a large scale, for industrial production of enzymes?
- A. Bioreactor ✓
- B. BOD incubator
- C. Sludge digester
- D. Industrial oven
Solution: Bioreactor Large volumes of culture (100–1000 litres) for industrial production of enzymes are grown and processed in bioreactors.
Q75.
Which of the following statements is incorrect?
- A. Prions consist of abnormally folded proteins.
- B. Viroids lack a protein coat.
- C. Viruses are obligate parasites.
- D. Infective constituent in viruses is the protein coat. ✓
Solution: Infective constituent in viruses is the protein coat (the incorrect statement) The infective part of a virus is its nucleic acid (DNA or RNA); the protein coat (capsid) only protects it. (A) True: prions consist of abnormally folded proteins. (B) True: viroids are free RNA with no protein coat. (C) True: viruses multiply only inside living cells, so they are obligate parasites.
Q76.
Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following
- A. Tyloses in vessels
- B. Closure of stomata
- C. Flaccidity of bulliform cells ✓
- D. Shrinkage of air spaces in spongy mesophyll
Solution: Flaccidity of bulliform cells Under water stress the bulliform cells of grass leaves become flaccid, so the leaves curl inwards to reduce water loss.
Q77.
Xylem translocates
- A. Water, mineral salts, some organic nitrogen and hormones ✓
- B. Water only
- C. Water and mineral salts only
- D. Water, mineral salts and some organic nitrogen only
Solution: Water, mineral salts, some organic nitrogen and hormones Xylem translocates mainly water, mineral salts, some organic nitrogen and hormones from the roots to the aerial parts.
Q78.
Select the correct sequence for transport of sperm cells in male reproductive system.
- A. Testis $\to$ Epididymis $\to$ Vasa efferentia $\to$ Vas deferens $\to$ Ejaculatory duct $\to$ Inguinal canal $\to$ Urethra $\to$ Urethral meatus
- B. Testis $\to$ Epididymis $\to$ Vasa efferentia $\to$ Rete testis $\to$ Inguinal canal $\to$ Urethra
- C. Seminiferous tubules $\to$ Rete testis $\to$ Vasa efferentia $\to$ Epididymis $\to$ Vas deferens $\to$ Ejaculatory duct $\to$ Urethra $\to$ Urethral meatus ✓
- D. Seminiferous tubules $\to$ Vasa efferentia $\to$ Epididymis $\to$ Inguinal canal $\to$ Urethra
Solution: Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra → Urethral meatus Sperms pass from the seminiferous tubules through the rete testis and vasa efferentia to the epididymis, then through the vas deferens and ejaculatory duct into the urethra, leaving at the urethral meatus.
Q79.
Which of these following methods is the most suitable for disposal of nuclear waste?
- A. Bury the waste within rocks deep below the Earth's surface ✓
- B. Shoot the waste into space
- C. Bury the waste under Antarctic ice-cover
- D. Dump the waste within rocks under deep ocean
Solution: Bury the waste within rocks deep below the Earth's surface Nuclear waste, after pre-treatment, should be stored in suitably shielded containers buried within rocks about 500 m below the earth's surface.
Q80.
Which of the following immune responses is responsible for rejection of kidney graft?
- A. Cell-mediated immune response ✓
- B. Auto-immune response
- C. Humoral immune response
- D. Inflammatory immune response
Solution: Cell-mediated immune response The body distinguishes self from non-self, and graft rejection is due to the cell-mediated immune response (T lymphocytes).
Q81.
What is the site of perception of photoperiod necessary for induction of flowering in plants?
- A. Leaves ✓
- B. Lateral buds
- C. Pulvinus
- D. Shoot apex
Solution: Leaves Leaves perceive the photoperiod; a hormonal substance then moves from the leaves to the shoot apices to induce flowering.
Q82.
Identify the correct pair representing the causative agent of typhoid fever and the confirmatory test for typhoid.
- A. $Salmonella$ $typhi$ / Widal test ✓
- B. $Plasmodium$ $vivax$ / UTI test
- C. $Streptococcus$ $pneumoniae$ / Widal test
- D. $Salmonella$ $typhi$ / Anthrone test
Solution: $Salmonella$ $typhi$ / Widal test Typhoid fever is caused by $Salmonella$ $typhi$ and is confirmed by the Widal test.
Q83.
Concanavalin A is
- A. a pigment
- B. an alkaloid
- C. an essential oil
- D. a lectin ✓
Solution: Lectin Concanavalin A is a lectin, a plant secondary metabolite that agglutinates red blood cells.
Q84.
It takes very long time for pineapple plants to produce flowers. Which combination of hormones can be applied to artificially induce flowering in pineapple plants throughout the year to increase yield?
- A. Cytokinin and Abscisic acid
- B. Auxin and Ethylene ✓
- C. Gibberellin and Cytokinin
- D. Gibberellin and Abscisic acid
Solution: Auxin and Ethylene Auxin induces flowering in pineapple, and ethylene initiates flowering and synchronises fruit-set in pineapple.
Q85.
Select the incorrect statement.
- A. Inbreeding helps in accumulation of superior genes and elimination of undesirable genes
- B. Inbreeding increases homozygosity
- C. Inbreeding is essential to evolve purelines in any animal.
- D. Inbreeding selects harmful recessive genes that reduce fertility and productivity ✓
Solution: Inbreeding selects harmful recessive genes that reduce fertility and productivity (the incorrect statement) Inbreeding exposes harmful recessive genes so that selection can eliminate them; it does not select them. (A) True: with selection at each step, superior genes accumulate and less desirable genes are eliminated. (B) True: mating of closely related individuals increases homozygosity. (C) True: inbreeding is necessary to evolve a pureline in any animal.
Q86.
Which of the following statements is incorrect?
- A. Yeasts have filamentous bodies with long thread-like hyphae. ✓
- B. Morels and truffles are edible delicacies.
- C. $Claviceps$ is a source of many alkaloids and LSD.
- D. Conidia are produced exogenously and ascospores endogenously.
Solution: Yeasts have filamentous bodies with long thread-like hyphae (the incorrect statement) Yeasts are unicellular fungi, with no filamentous body or hyphae. (B) True: morels and truffles are edible ascomycetes considered delicacies. (C) True: $Claviceps$ is a source of many alkaloids and LSD. (D) True: conidia are produced exogenously on conidiophores and ascospores endogenously in asci.
Q87.
Tidal Volume and Expiratory Reserve Volume of an athlete is 500 mL and 1000 mL, respectively. What will be his Expiratory Capacity if the Residual Volume is 1200 mL?
- A. 2700 mL
- B. 1500 mL ✓
- C. 1700 mL
- D. 2200 mL
Solution: 1500 mL Expiratory capacity = tidal volume + expiratory reserve volume = 500 mL + 1000 mL = 1500 mL Residual volume cannot be expired, so it is not part of the expiratory capacity.
Q88.
Which one of the following is not a method of $in$ $situ$ conservation of biodiversity?
- A. Sacred Grove
- B. Biosphere Reserve
- C. Wildlife Sanctuary
- D. Botanical Garden ✓
Solution: Botanical Garden (an ex situ method) A botanical garden conserves plants away from their natural habitat (ex situ); sacred groves, biosphere reserves and wildlife sanctuaries protect them in situ.
Q89.
Which of the following factors is responsible for the formation of concentrated urine?
- A. Hydrostatic pressure during glomerular filtration
- B. Low levels of antidiuretic hormone
- C. Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys. ✓
- D. Secretion of erythropoietin by Juxtaglomerular complex
Solution: Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys The counter-current mechanism of Henle's loop and vasa recta keeps the osmolarity rising towards the inner medulla, from 300 to about 1200 mOsmol L⁻¹. Water leaves the collecting duct into this hyperosmotic interstitium, so urine becomes nearly four times as concentrated as the filtrate.
Q90.
Match the Column-I with Column-II
Column-I:
(a) P - wave
(b) QRS complex
(c) T - wave
(d) Reduction in the size of T-wave
Column-II:
(i) Depolarisation of ventricles
(ii) Repolarisation of ventricles
(iii) Coronary ischemia
(iv) Depolarisation of atria
(v) Repolarisation of atria
Select the correct option.
- A. (a)-(ii), (b)-(iii), (c)-(v), (d)-(iv)
- B. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓
- C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(v)
- D. (a)-(ii), (b)-(i), (c)-(v), (d)-(iii)
Solution: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) The P-wave represents depolarisation of the atria: (a)-(iv). The QRS complex represents depolarisation of the ventricles: (b)-(i). The T-wave represents repolarisation of the ventricles: (c)-(ii). A smaller T-wave indicates insufficient oxygen supply to the heart muscle, i.e. coronary ischaemia: (d)-(iii).
Q91.
When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60$^{\circ}$ with horizontal, it can travel a distance x$_1$ along the plane. But when the inclination is decreased to 30$^{\circ}$ and the same object is shot with the same velocity, it can travel x$_2$ distance. Then x$_1$ : x$_2$ will be:
- A. $1 : 2\sqrt{3}$
- B. $1 : \sqrt{2}$
- C. $\sqrt{2} : 1$
- D. $1 : \sqrt{3}$ ✓
Solution: $x_1 : x_2 = 1 : \sqrt{3}$ On a smooth incline the retardation is $g\sin\theta$, so the distance travelled before stopping is $x = \dfrac{u^2}{2g\sin\theta}$. Same $u$ in both cases, so $x \propto \dfrac{1}{\sin\theta}$. $\dfrac{x_1}{x_2} = \dfrac{\sin 30^\circ}{\sin 60^\circ} = \dfrac{1/2}{\sqrt{3}/2}$ ∴ $x_1 : x_2 = 1 : \sqrt{3}$
Q92.
A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2}$ N/m. The pressure inside the bubble equals at a point Z$_0$ below the free surface of water in a container. Taking g = 10 m/s$^{2}$, density of water = 10$^{3}$ kg/m$^{3}$, the value of Z$_0$ is :
- A. 0.5 cm
- B. 100 cm
- C. 10 cm
- D. 1 cm ✓
Solution: $Z_0 = 1\ \text{cm}$ Excess pressure inside a soap bubble (two surfaces) is $\dfrac{4T}{R}$; excess pressure at depth $Z_0$ in water is $\rho g Z_0$. $\rho g Z_0 = \dfrac{4T}{R} \Rightarrow Z_0 = \dfrac{4T}{R\rho g}$ $Z_0 = \dfrac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^{3} \times 10}\ \text{m} = 10^{-2}\ \text{m}$ ∴ $Z_0 = 1\ \text{cm}$
Q93.
Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is F$_1$. When the space between the two lenses is filled with glycerine (which has the same refractive index ($\mu$ = 1.5) as that of glass) then the equivalent focal length is F$_2$. The ratio F$_1$ : F$_2$ will be :
- A. 3 : 4
- B. 2 : 1
- C. 1 : 2 ✓
- D. 2 : 3
Solution: $F_1 : F_2 = 1 : 2$ Two thin lenses in contact: $\dfrac{1}{F_1} = \dfrac{1}{f} + \dfrac{1}{f} = \dfrac{2}{f}$, so $F_1 = \dfrac{f}{2}$. Glycerine between them forms a concave lens of the same $\mu$ and the same radii, so its focal length is $-f$. $\dfrac{1}{F_2} = \dfrac{1}{f} + \dfrac{1}{f} - \dfrac{1}{f} = \dfrac{1}{f}$, so $F_2 = f$. ∴ $F_1 : F_2 = 1 : 2$
Q94.
$\alpha$-particle consists of :
- A. 2 protons only
- B. 2 protons and 2 neutrons only ✓
- C. 2 electrons, 2 protons and 2 neutrons
- D. 2 electrons and 4 protons only
Solution: 2 protons and 2 neutrons only An α-particle is a helium nucleus (He²⁺): 2 protons and 2 neutrons, with no electrons.
Q95.
Which of the following acts as a circuit protection device?
- A. Fuse ✓
- B. Conductor
- C. Inductor
- D. Switch
Solution: Fuse A fuse wire has a low melting point; excess current heats and melts it, breaking the circuit before damage is done.
Q96.
In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be angle of refraction?
- A. 90$^{\circ}$ ✓
- B. 180$^{\circ}$
- C. 0$^{\circ}$
- D. Equal to angle of incidence
Solution: $90^\circ$ At the critical angle the refracted ray grazes along the interface: $\mu_1\sin i_c = \mu_2\sin 90^\circ$, so the angle of refraction is $90^\circ$.
Q97.
The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path the angle at which he should make his strokes w.r.t. north is given by :
- A. 45$^{\circ}$ west
- B. 30$^{\circ}$ west ✓
- C. 0$^{\circ}$
- D. 60$^{\circ}$ west
Solution: $30^\circ$ west of north For the shortest path the resultant velocity must point due north, so the swimmer heads upstream at $\theta$ west of north. His velocity relative to the water must have a westward component equal to the river speed: $20\sin\theta = 10$ $\sin\theta = \dfrac{10}{20} = \dfrac{1}{2}$ ∴ $\theta = 30^\circ$ west of north
Q98.
A parallel plate capacitor of capacitance 20 $\mu$F is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively
- A. Zero, zero
- B. Zero, 60 $\mu$A
- C. 60 $\mu$A, 60 $\mu$A ✓
- D. 60 $\mu$A, zero
Solution: $60\ \mu\text{A}$, $60\ \mu\text{A}$ $q = CV \Rightarrow$ conduction current $i_c = \dfrac{dq}{dt} = C\dfrac{dV}{dt}$ $i_c = 20 \times 10^{-6}\ \text{F} \times 3\ \text{V s}^{-1} = 60 \times 10^{-6}\ \text{A} = 60\ \mu\text{A}$ The displacement current between the plates equals the conduction current in the wires, so $i_d = 60\ \mu\text{A}$.
Q99.
The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:
- A. 3.4 eV, 3.4 eV
- B. –3.4 eV, –3.4 eV
- C. –3.4 eV, –6.8 eV
- D. 3.4 eV, –6.8 eV ✓
Solution: $K = 3.4\ \text{eV}$, $U = -6.8\ \text{eV}$ In Bohr's model, kinetic energy $K = -E$ and potential energy $U = 2E$, where $E$ is the total energy. $K = -(-3.4\ \text{eV}) = 3.4\ \text{eV}$ $U = 2 \times (-3.4\ \text{eV}) = -6.8\ \text{eV}$
Q100.
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where $X = \dfrac{A^{2}B^{1/2}}{C^{1/3}D^{3}}$, will be
- A. 10%
- B. $\left(\dfrac{3}{13}\right)$%
- C. 16% ✓
- D. – 10%
Solution: 16% Percentage errors add, each multiplied by the power of its quantity: $\dfrac{\Delta X}{X} = 2\dfrac{\Delta A}{A} + \dfrac{1}{2}\dfrac{\Delta B}{B} + \dfrac{1}{3}\dfrac{\Delta C}{C} + 3\dfrac{\Delta D}{D}$ $= 2(1\%) + \tfrac{1}{2}(2\%) + \tfrac{1}{3}(3\%) + 3(4\%)$ ∴ Maximum error $= 2\% + 1\% + 1\% + 12\% = 16\%$
Q101.
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre
- A. Decreases as r increases for r < R and for r > R
- B. Increases as r increases for r < R and for r > R
- C. Zero as r increases for r < R, decreases as r increases for r > R ✓
- D. Zero as r increases for r < R, increases as r increases for r > R
Solution: Zero for $r < R$; decreases as $r$ increases for $r > R$ The charge on a hollow metal sphere resides on its outer surface. Inside ($r < R$): Gauss's law with no enclosed charge gives $E = 0$. Outside ($r > R$): $E \cdot 4\pi r^2 = \dfrac{Q}{\epsilon_0} \Rightarrow E = \dfrac{Q}{4\pi\epsilon_0 r^2}$, which decreases as $r$ increases.
Q102.
Two parallel infinite line charges with linear charge densities +$\lambda$ C/m and –$\lambda$ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?
- A. $\dfrac{\lambda}{2\pi\epsilon_0 R}$ N/C
- B. Zero
- C. $\dfrac{2\lambda}{\pi\epsilon_0 R}$ N/C
- D. $\dfrac{\lambda}{\pi\epsilon_0 R}$ N/C ✓
Solution: $\dfrac{\lambda}{\pi\epsilon_0 R}\ \text{N C}^{-1}$ Field of an infinite line charge at distance $r$: $E = \dfrac{\lambda}{2\pi\epsilon_0 r}$ At the midpoint $r = R$ for both lines, and both fields point the same way (away from $+\lambda$, towards $-\lambda$). ∴ $E = 2 \times \dfrac{\lambda}{2\pi\epsilon_0 R} = \dfrac{\lambda}{\pi\epsilon_0 R}\ \text{N C}^{-1}$
Q103.
The unit of thermal conductivity is :
- A. W m$^{-1}$ K$^{-1}$ ✓
- B. J m K$^{-1}$
- C. J m$^{-1}$ K$^{-1}$
- D. W m K$^{-1}$
Solution: $\text{W m}^{-1}\,\text{K}^{-1}$ Rate of heat flow $H = KA\dfrac{\Delta T}{L}$, so $K = \dfrac{HL}{A\,\Delta T}$. ∴ Unit of $K = \dfrac{\text{W} \times \text{m}}{\text{m}^2 \times \text{K}} = \text{W m}^{-1}\,\text{K}^{-1}$
Q104.
The displacement of a particle executing simple harmonic motion is given by
y = A$_0$ + Asin$\omega$t + Bcos$\omega$t
Then the amplitude of its oscillation is given by :
- A. A + B
- B. $A_0 + \sqrt{A^{2} + B^{2}}$
- C. $\sqrt{A^{2} + B^{2}}$ ✓
- D. $\sqrt{A_0^{2} + (A + B)^{2}}$
Solution: $\sqrt{A^2 + B^2}$ $A_0$ only shifts the mean position; the oscillating part is $y - A_0 = A\sin\omega t + B\cos\omega t$. These are two SHMs of the same $\omega$ differing in phase by $90^\circ$. ∴ Amplitude $= \sqrt{A^2 + B^2 + 2AB\cos 90^\circ} = \sqrt{A^2 + B^2}$
Q105.
In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2$^{\circ}$. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? ($\mu_{water}$ = 4/3)
- A. 0.1$^{\circ}$
- B. 0.266$^{\circ}$
- C. 0.15$^{\circ}$ ✓
- D. 0.05$^{\circ}$
Solution: $0.15^\circ$ Angular fringe width $\theta = \dfrac{\lambda}{d}$; in water the wavelength becomes $\dfrac{\lambda}{\mu}$, so $\theta_w = \dfrac{\theta}{\mu}$. $\theta_w = \dfrac{0.2^\circ}{4/3} = 0.2^\circ \times \dfrac{3}{4}$ ∴ $\theta_w = 0.15^\circ$
Q106.
A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth ?
- A. 100 N ✓
- B. 150 N
- C. 200 N
- D. 250 N
Solution: 100 N Acceleration due to gravity at depth $d$: $g' = g\left(1 - \dfrac{d}{R}\right)$ Half way down, $d = \dfrac{R}{2} \Rightarrow g' = \dfrac{g}{2}$ ∴ Weight $= mg' = \dfrac{200\ \text{N}}{2} = 100\ \text{N}$
Q107.
A particle moving with velocity $\vec{V}$ is acted by three forces shown by the vector triangle PQR. The velocity of the particle will :
- A. Change according to the smallest force $\overrightarrow{QR}$
- B. Increase
- C. Decrease
- D. Remain constant ✓
Solution: Remain constant The three forces form the closed triangle PQR taken in the same order, so their vector sum is zero. $\vec{F}_{net} = m\dfrac{d\vec{v}}{dt} = 0$ ∴ $\vec{v}$ is constant, so the velocity remains unchanged.
Q108.
Two particles A and B are moving in uniform circular motion in concentric circles of radii r$_A$ and r$_B$ with speed v$_A$ and v$_B$ respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be :
- A. 1 : 1 ✓
- B. r$_A$ : r$_B$
- C. v$_A$ : v$_B$
- D. r$_B$ : r$_A$
Solution: 1 : 1 Angular speed $\omega = \dfrac{2\pi}{T}$ depends only on the time period. ∴ $T_A = T_B \Rightarrow \omega_A : \omega_B = 1 : 1$
Q109.
A 800 turn coil of effective area 0.05 m$^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{-5}$ T. When the plane of the coil is rotated by 90$^{\circ}$ around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:
- A. 0.02 V ✓
- B. 2 V
- C. 0.2 V
- D. $2 \times 10^{-3}$ V
Solution: 0.02 V Faraday's law: $e = \dfrac{\Delta\phi}{\Delta t}$ (magnitude), with flux $\phi = NBA\cos\theta$. Turning the coil through $90^\circ$ changes the flux from $NBA$ to zero: $\Delta\phi = 800 \times 5 \times 10^{-5} \times 0.05\ \text{Wb} = 2 \times 10^{-3}\ \text{Wb}$ ∴ $e = \dfrac{2 \times 10^{-3}\ \text{Wb}}{0.1\ \text{s}} = 0.02\ \text{V}$
Q110.
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be : (g = 10 m/s$^{2}$)
- A. 10$\pi$ rad/s
- B. $\sqrt{10}$ rad/s
- C. $\dfrac{10}{2\pi}$ rad/s
- D. 10 rad/s ✓
Solution: $10\ \text{rad s}^{-1}$ The normal reaction of the wall gives the centripetal force: $N = mr\omega^2$. Friction must hold up the weight: $\mu N \ge mg \Rightarrow \mu m r\omega^2 \ge mg$ $\omega_{min} = \sqrt{\dfrac{g}{\mu r}} = \sqrt{\dfrac{10}{0.1 \times 1}}\ \text{rad s}^{-1}$ ∴ $\omega_{min} = 10\ \text{rad s}^{-1}$
Q111.
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is :
- A. $\dfrac{1}{2}MgL$
- B. $Mgl$
- C. $MgL$
- D. $\dfrac{1}{2}Mgl$ ✓
Solution: $\dfrac{1}{2}Mgl$ Elastic potential energy of a stretched wire $U = \dfrac{1}{2} \times \text{stretching force} \times \text{extension}$ ∴ $U = \dfrac{1}{2}Mgl$
Q112.
Increase in temperature of a gas filled in a container would lead to :
- A. Decrease in intermolecular distance
- B. Increase in its mass
- C. Increase in its kinetic energy ✓
- D. Decrease in its pressure
Solution: Increase in its kinetic energy Average kinetic energy of a gas molecule $= \dfrac{3}{2}k_BT$, so raising the temperature increases the kinetic energy of the gas.
Q113.
A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance d from the centre of the conductor, is correctly represented by the figure :
- A. Graph (1)
- B. Graph (2)
- C. Graph (3)
- D. Graph (4) ✓
Solution: Graph (4): $B \propto d$ inside, $B \propto \tfrac{1}{d}$ outside Inside the conductor ($d < R$): $B = \dfrac{\mu_0 i d}{2\pi R^2}$, so $B \propto d$, a straight line through the origin. At the surface ($d = R$) B is maximum: $B = \dfrac{\mu_0 i}{2\pi R}$. Outside ($d > R$): $B = \dfrac{\mu_0 i}{2\pi d}$, so $B \propto \dfrac{1}{d}$ (hyperbolic). ∴ The correct plot rises linearly to a peak at $d = R$ and then falls hyperbolically.
Q114.
Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is :
- A. $\dfrac{5}{9}$
- B. $\dfrac{1}{9}$
- C. $\dfrac{8}{9}$ ✓
- D. $\dfrac{4}{9}$
Solution: $\dfrac{8}{9}$ Head-on elastic collision with B at rest: $v_A = \dfrac{m_A - m_B}{m_A + m_B}u = \dfrac{4m - 2m}{6m}u = \dfrac{u}{3}$ A keeps $\left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$ of its kinetic energy. ∴ Fraction lost $= 1 - \dfrac{1}{9} = \dfrac{8}{9}$
Q115.
Which colour of the light has the longest wavelength?
- A. Violet
- B. Red ✓
- C. Blue
- D. Green
Solution: Red In the visible spectrum the wavelength increases from violet to red, so red has the longest wavelength.
Q116.
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is : ($\alpha_{Cu}$ = $1.7 \times 10^{-5}$ K$^{-1}$ and $\alpha_{Al}$ = $2.2 \times 10^{-5}$ K$^{-1}$)
- A. 68 cm ✓
- B. 6.8 cm
- C. 113.9 cm
- D. 88 cm
Solution: 68 cm The difference in lengths stays constant only if both rods expand equally: $L_{Cu}\alpha_{Cu}\Delta T = L_{Al}\alpha_{Al}\Delta T$ $88 \times 1.7 \times 10^{-5} = L_{Al} \times 2.2 \times 10^{-5}$ ∴ $L_{Al} = \dfrac{88 \times 1.7}{2.2}\ \text{cm} = 68\ \text{cm}$
Q117.
For a p-type semiconductor, which of the following statements is true ?
- A. Electrons are the majority carriers and pentavalent atoms are the dopants.
- B. Electrons are the majority carriers and trivalent atoms are the dopants.
- C. Holes are the majority carriers and trivalent atoms are the dopants. ✓
- D. Holes are the majority carriers and pentavalent atoms are the dopants.
Solution: Holes are the majority carriers and trivalent atoms are the dopants A trivalent dopant (e.g. B, Al) has one valence electron too few, creating a hole at each dopant atom, so holes are the majority carriers in a p-type semiconductor. (A) Electrons as majority carriers with pentavalent dopants describes an n-type semiconductor. (B) Trivalent dopants make holes, not electrons, the majority carriers. (D) Pentavalent dopants add free electrons and give an n-type semiconductor.
Q118.
The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the fig.
y - projection of the radius vector of rotating particle P is :
- A. $y(t) = 3\cos\left(\dfrac{\pi t}{2}\right)$, where y in m ✓
- B. $y(t) = -3\cos 2\pi t$, where y in m
- C. $y(t) = 4\sin\left(\dfrac{\pi t}{2}\right)$, where y in m
- D. $y(t) = 3\cos\left(\dfrac{3\pi t}{2}\right)$, where y in m
Solution: $y(t) = 3\cos\left(\dfrac{\pi t}{2}\right)$ m At $t = 0$ the y-displacement is maximum, so $y(t)$ is a cosine function. With $T = 4\ \text{s}$, $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\ \text{rad s}^{-1}$ ∴ $y = a\cos\omega t = 3\cos\left(\dfrac{\pi t}{2}\right)\ \text{m}$
Q119.
A force F = 20 + 10 y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is
- A. 20 J
- B. 30 J
- C. 5 J
- D. 25 J ✓
Solution: 25 J Work done by a variable force $W = \int_{y_1}^{y_2} F\,dy$ $W = \int_0^1 (20 + 10y)\,dy = \left[20y + 5y^2\right]_0^1$ ∴ $W = 20 + 5 = 25\ \text{J}$
Q120.
A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
- A. inclined at an angle of 60$^{\circ}$ from vertical
- B. the mass is at the highest point
- C. the wire is horizontal
- D. the mass is at the lowest point ✓
Solution: The mass is at the lowest point At the lowest point $T - mg = \dfrac{mv^2}{l} \Rightarrow T = mg + \dfrac{mv^2}{l}$ Here the speed is greatest and the tension must also balance the full weight, so the tension is maximum and the wire is most likely to break.
Q121.
Average velocity of a particle executing SHM in one complete vibration is :
- A. Zero ✓
- B. $\dfrac{A\omega}{2}$
- C. $A\omega$
- D. $\dfrac{A\omega^{2}}{2}$
Solution: Zero Average velocity = displacement ÷ time interval. After one complete vibration the particle is back at its starting point, so the displacement, and hence the average velocity, is zero.
Q122.
Pick the wrong answer in the context with rainbow.
- A. Rainbow is a combined effect of dispersion refraction and reflection of sunlight
- B. When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed
- C. The order of colours is reversed in the secondary rainbow
- D. An observer can see a rainbow when his front is towards the sun ✓
Solution: An observer can see a rainbow when his front is towards the sun (the incorrect statement) A rainbow is seen only with the sun behind the observer and the raindrops in front. (A) True: a rainbow results from dispersion and refraction of sunlight in raindrops together with internal reflection. (B) True: two internal reflections in a drop give the secondary rainbow. (C) True: the order of colours in the secondary rainbow is reversed.
Q123.
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (m$_e$ = $9 \times 10^{-31}$ kg)
- A. 12.2 nm
- B. $12.2 \times 10^{-13}$ m
- C. $12.2 \times 10^{-12}$ m ✓
- D. $12.2 \times 10^{-14}$ m
Solution: $12.2 \times 10^{-12}\ \text{m}$ The de Broglie wavelength of an electron accelerated through $V$ volts is $\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{1.227}{\sqrt{V}}\ \text{nm}$ $\lambda = \dfrac{1.227}{\sqrt{10000}}\ \text{nm} = 0.01227\ \text{nm}$ ∴ $\lambda \approx 12.2 \times 10^{-12}\ \text{m}$
Q124.
A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?
- A. 1 J
- B. 3 J ✓
- C. 30 kJ
- D. 2 J
Solution: 3 J Work needed to stop the disc = its total kinetic energy. Rolling disc: $K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{1}{2}mR^2\right)\dfrac{v^2}{R^2} = \dfrac{3}{4}mv^2$ ∴ $K = \dfrac{3}{4} \times 100\ \text{kg} \times (0.2\ \text{m s}^{-1})^2 = 3\ \text{J}$
Q125.
The correct Boolean operation represented by the circuit diagram drawn is :
- A. NOR
- B. AND
- C. OR
- D. NAND ✓
Solution: NAND The LED glows when the voltage across it is high. It glows for inputs (0, 0), (0, 1) and (1, 0), and goes off only when both switches A and B are closed (1, 1). ∴ Y = 1, 1, 1, 0 for AB = 00, 01, 10, 11, which is the truth table of a NAND gate.
Q126.
Ionized hydrogen atoms and $\alpha$-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r$_H$ : r$_\alpha$ will be :
- A. 1 : 4
- B. 2 : 1 ✓
- C. 1 : 2
- D. 4 : 1
Solution: $r_H : r_\alpha = 2 : 1$ Radius of the path $r = \dfrac{mv}{qB} = \dfrac{p}{qB}$; with the same $p$ and $B$, $r \propto \dfrac{1}{q}$. $q_H = e$ and $q_\alpha = 2e$ ∴ $r_H : r_\alpha = 2 : 1$
Q127.
Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :
- A. $\dfrac{4F}{3}$
- B. $F$
- C. $\dfrac{9F}{16}$ ✓
- D. $\dfrac{16F}{9}$
Solution: $\dfrac{9F}{16}$ Initially $F = \dfrac{kQ^2}{r^2}$ (attractive). After transferring $\dfrac{Q}{4}$ from A to B: $q_A = +\dfrac{3Q}{4}$ and $q_B = -Q + \dfrac{Q}{4} = -\dfrac{3Q}{4}$ $F' = \dfrac{k(3Q/4)^2}{r^2} = \dfrac{9}{16} \cdot \dfrac{kQ^2}{r^2}$ ∴ $F' = \dfrac{9F}{16}$
Q128.
In which of the following devices, the eddy current effect is not used?
- A. Electric heater ✓
- B. Induction furnace
- C. Magnetic braking in train
- D. Electromagnet
Solution: Electric heater An electric heater works by Joule heating of a resistance wire, not eddy currents; induction furnaces and magnetic braking in trains rely on eddy currents.
Q129.
At a point A on the earth's surface the angle of dip, $\delta$ = +25$^{\circ}$. At a point B on the earth's surface the angle of dip, $\delta$ = –25$^{\circ}$. We can interpret that:
- A. A and B are both located in the southern hemisphere.
- B. A and B are both located in the northern hemisphere.
- C. A is located in the southern hemisphere and B is located in the northern hemisphere.
- D. A is located in the northern hemisphere and B is located in the southern hemisphere. ✓
Solution: Point A is in the northern hemisphere and point B is in the southern hemisphere Angle of dip is the angle the earth's total magnetic field makes with the horizontal; it is zero at the magnetic equator. Dip is taken as positive in the northern hemisphere (the north end of the needle dips down) and negative in the southern hemisphere. ∴ A ($\delta = +25^\circ$) is in the northern hemisphere and B ($\delta = -25^\circ$) is in the southern hemisphere.
Q130.
Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance.
The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be :
- A. 2 : 1
- B. 4 : 9
- C. 9 : 4 ✓
- D. 1 : 2
Solution: 9 : 4 Power drawn from the source $P = \dfrac{E^2}{R_{eq}}$, so $P \propto \dfrac{1}{R_{eq}}$. (i) Three bulbs in parallel in A, in series with three in parallel in B: $R_{eq} = \dfrac{R}{3} + \dfrac{R}{3} = \dfrac{2R}{3}$ (ii) Two glowing in A and one in B: $R_{eq} = \dfrac{R}{2} + R = \dfrac{3R}{2}$ ∴ $\dfrac{P_1}{P_2} = \dfrac{3R/2}{2R/3} = \dfrac{9}{4}$
Q131.
A small hole of area of cross-section 2 mm$^{2}$ is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s$^{2}$, the rate of flow of water through the open hole would be nearly
- A. $6.4 \times 10^{-6}$ m$^{3}$/s
- B. $12.6 \times 10^{-6}$ m$^{3}$/s ✓
- C. $8.9 \times 10^{-6}$ m$^{3}$/s
- D. $2.23 \times 10^{-6}$ m$^{3}$/s
Solution: $12.6 \times 10^{-6}\ \text{m}^3\,\text{s}^{-1}$ Torricelli's law: speed of efflux $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 2}\ \text{m s}^{-1} = \sqrt{40}\ \text{m s}^{-1} \approx 6.32\ \text{m s}^{-1}$ Rate of flow $Q = av = 2 \times 10^{-6}\ \text{m}^2 \times 6.32\ \text{m s}^{-1}$ ∴ $Q \approx 12.6 \times 10^{-6}\ \text{m}^3\,\text{s}^{-1}$
Q132.
In the circuits shown below, the readings of voltmeters and the ammeters will be
- A. V$_2$ > V$_1$ and i$_1$ > i$_2$
- B. V$_2$ > V$_1$ and i$_1$ = i$_2$
- C. V$_1$ = V$_2$ and i$_1$ > i$_2$
- D. V$_1$ = V$_2$ and i$_1$ = i$_2$ ✓
Solution: $V_1 = V_2$ and $i_1 = i_2$ An ideal voltmeter has infinite resistance and an ideal ammeter has zero resistance. In each circuit $i = \dfrac{10\ \text{V}}{10\ \Omega} = 1\ \text{A}$, so $i_1 = i_2$. Each voltmeter reads $V = iR = 1\ \text{A} \times 10\ \Omega = 10\ \text{V}$, so $V_1 = V_2$.
Q133.
The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:
- A. $\dfrac{3}{2}mgR$
- B. $mgR$
- C. $2mgR$
- D. $\dfrac{1}{2}mgR$ ✓
Solution: $\dfrac{1}{2}mgR$ Gravitational potential energy at distance $r$ from the earth's centre: $U = -\dfrac{GMm}{r}$ $W = U(2R) - U(R) = -\dfrac{GMm}{2R} + \dfrac{GMm}{R} = \dfrac{GMm}{2R}$ ∴ With $GM = gR^2$, $W = \dfrac{1}{2}mgR$
Q134.
In which of the following processes, heat is neither absorbed nor released by a system?
- A. Isochoric
- B. Isothermal
- C. Adiabatic ✓
- D. Isobaric
Solution: Adiabatic In an adiabatic process the system exchanges no heat with its surroundings ($Q = 0$).
Q135.
A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2$\pi$ revolutions is
- A. $2 \times 10^{6}$ N m
- B. $2 \times 10^{-6}$ N m ✓
- C. $2 \times 10^{-3}$ N m
- D. $12 \times 10^{-4}$ N m
Solution: $2 \times 10^{-6}\ \text{N m}$ Work–energy theorem for rotation: the torque must remove all the rotational kinetic energy, $\tau\theta = \dfrac{1}{2}I\omega^2$. $I = \dfrac{1}{2}mr^2 = \dfrac{1}{2} \times 2 \times (0.04)^2 = 1.6 \times 10^{-3}\ \text{kg m}^2$ and $\omega = \dfrac{3 \times 2\pi}{60} = \dfrac{\pi}{10}\ \text{rad s}^{-1}$ $\theta = 2\pi\ \text{rev} = 2\pi \times 2\pi\ \text{rad} = 4\pi^2\ \text{rad}$ ∴ $\tau = \dfrac{I\omega^2}{2\theta} = \dfrac{1.6 \times 10^{-3} \times \pi^2/100}{2 \times 4\pi^2} = 2 \times 10^{-6}\ \text{N m}$
Q136.
Which one is malachite from the following?
- A. CuCO$_3\cdot$Cu(OH)$_2$ ✓
- B. CuFeS$_2$
- C. Cu(OH)$_2$
- D. Fe$_3$O$_4$
Solution: CuCO₃·Cu(OH)₂ Malachite is the green copper ore CuCO₃·Cu(OH)₂; CuFeS₂ is copper pyrites and Fe₃O₄ is magnetite.
Q137.
Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is :
Solution: Mg All enzymes that utilise ATP in phosphate transfer require magnesium (Mg) as the cofactor.
Q138.
For an ideal solution, the correct option is :
- A. $\Delta_{mix}$G = 0 at constant T and P
- B. $\Delta_{mix}$S = 0 at constant T and P
- C. $\Delta_{mix}$V $\neq$ 0 at constant T and P
- D. $\Delta_{mix}$H = 0 at constant T and P ✓
Solution: $\Delta_{mix}H = 0$ at constant T and P An ideal solution obeys Raoult's law at all concentrations and A–B interactions equal A–A and B–B interactions, so $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$. (A) Mixing is spontaneous, so $\Delta_{mix}G < 0$. (B) Randomness increases on mixing, so $\Delta_{mix}S > 0$. (C) For an ideal solution $\Delta_{mix}V = 0$.
Q139.
What is the correct electronic configuration of the central atom in K$_4$[Fe(CN)$_6$] based on crystal field theory?
- A. $e^{4}t_2^{2}$
- B. $t_{2g}^{4}e_g^{2}$
- C. $t_{2g}^{6}e_g^{0}$ ✓
- D. $e^{3}t_2^{3}$
Solution: $t_{2g}^{6}e_g^{0}$ In K₄[Fe(CN)₆], iron is in the +2 state: Fe²⁺ is 3d⁶. CN⁻ is a strong field ligand, so $\Delta_o > P$ and all six electrons pair up in the lower $t_{2g}$ set. ∴ Configuration $t_{2g}^{6}e_g^{0}$ (low spin, diamagnetic)
Q140.
The number of sigma ($\sigma$) and pi ($\pi$) bonds in pent-2-en-4-yne is
- A. 13$\sigma$ bonds and no $\pi$ bonds
- B. 10$\sigma$ bonds and 3$\pi$ bonds ✓
- C. 8$\sigma$ bonds and 5$\pi$ bonds
- D. 11$\sigma$ bonds and 2$\pi$ bonds
Solution: 10 σ bonds and 3 π bonds Pent-2-en-4-yne is CH₃–CH=CH–C≡CH. σ bonds: 4 C–C + 6 C–H = 10 π bonds: 1 in C=C + 2 in C≡C = 3
Q141.
A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :
- A. C$_4$A$_3$
- B. C$_2$A$_3$
- C. C$_3$A$_2$
- D. C$_3$A$_4$ ✓
Solution: C₃A₄ In hcp, the number of octahedral voids equals the number of anions. Cations fill 75% of them, so C : A $= \dfrac{3}{4} : 1 = 3 : 4$. ∴ Formula C₃A₄
Q142.
The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is :
Solution: 30 N₂ + 3H₂ ⇌ 2NH₃: 3 mol of H₂ give 2 mol of NH₃. ∴ H₂ needed for 20 mol NH₃ $= \dfrac{3}{2} \times 20 = 30$ mol
Q143.
Which of the following is incorrect statement?
- A. SnF$_4$ is ionic in nature
- B. PbF$_4$ is covalent in nature ✓
- C. SiCl$_4$ is easily hydrolysed
- D. GeX$_4$ (X = F, Cl, Br, I) is more stable than GeX$_2$
Solution: PbF₄ is covalent in nature (the incorrect statement) Fluorine is small and highly electronegative, so PbF₄, like SnF₄, is ionic, not covalent. (A) True: SnF₄ is ionic. (C) True: SiCl₄ is easily hydrolysed because Si can accept a lone pair from water into its d orbitals. (D) True: for germanium the +4 state is more stable than +2, so GeX₄ is more stable than GeX₂.
Q144.
Which of the following is an amphoteric hydroxide?
- A. Be(OH)$_2$ ✓
- B. Sr(OH)$_2$
- C. Ca(OH)$_2$
- D. Mg(OH)$_2$
Solution: Be(OH)₂ Be(OH)₂ is amphoteric: it reacts with both acids and bases. Be(OH)₂ + 2HCl → BeCl₂ + 2H₂O Be(OH)₂ + 2NaOH → Na₂[Be(OH)₄] Mg(OH)₂, Ca(OH)₂ and Sr(OH)₂ are basic.
Q145.
The manganate and permanganate ions are tetrahedral, due to :
- A. The $\pi$-bonding involves overlap of d-orbitals of oxygen with d-orbitals of manganese
- B. The $\pi$-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese ✓
- C. There is no $\pi$-bonding
- D. The $\pi$-bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese
Solution: The π-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese In tetrahedral MnO₄²⁻ and MnO₄⁻, the π bonds form by overlap of p orbitals of oxygen with d orbitals of manganese (pπ–dπ bonding).
Q146.
pH of a saturated solution of Ca(OH)$_2$ is 9. The solubility product (K$_{sp}$) of Ca(OH)$_2$ is:
- A. $0.5 \times 10^{-10}$
- B. $0.5 \times 10^{-15}$ ✓
- C. $0.25 \times 10^{-10}$
- D. $0.125 \times 10^{-15}$
Solution: $0.5 \times 10^{-15}$ pH = 9 ⇒ pOH = 5 ⇒ [OH⁻] = 10⁻⁵ M Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻, so [Ca²⁺] = ½[OH⁻] = 0.5 × 10⁻⁵ M ∴ $K_{sp}$ = [Ca²⁺][OH⁻]² $= (0.5 \times 10^{-5})(10^{-5})^2 = 0.5 \times 10^{-15}$
Q147.
The mixture that forms maximum boiling azeotrope is:
- A. Heptane + Octane
- B. Water + Nitric acid ✓
- C. Ethanol + Water
- D. Acetone + Carbon disulphide
Solution: Water + Nitric acid Solutions showing a large negative deviation from Raoult's law form a maximum boiling azeotrope. Nitric acid and water show negative deviation; their azeotrope is about 68% HNO₃ by mass and boils at 393.5 K. Ethanol + water and acetone + carbon disulphide show positive deviation (minimum boiling), and heptane + octane is nearly ideal.
Q148.
Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code:
Column-I:
(a) XeF$_4$
(b) XeF$_6$
(c) XeOF$_4$
(d) XeO$_3$
Column-II:
(i) Pyramidal
(ii) Square planar
(iii) Distorted octahedral
(iv) Square pyramidal
- A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- B. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
- C. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓
- D. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
Solution: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) XeF₄ has 4 bond pairs and 2 lone pairs, so it is square planar: (a)-(ii). XeF₆ has 6 bond pairs and 1 lone pair, so it is distorted octahedral: (b)-(iii). XeOF₄ has 5 bond pairs and 1 lone pair, so it is square pyramidal: (c)-(iv). XeO₃ has 3 bond pairs and 1 lone pair, so it is pyramidal: (d)-(i).
Q149.
Which of the following reactions are disproportionation reaction?
(a) 2Cu$^{+}$ $\longrightarrow$ Cu$^{2+}$ + Cu$^{0}$
(b) 3MnO$_4^{2-}$ + 4H$^{+}$ $\longrightarrow$ 2MnO$_4^{-}$ + MnO$_2$ + 2H$_2$O
(c) 2KMnO$_4$ $\xrightarrow{\Delta}$ K$_2$MnO$_4$ + MnO$_2$ + O$_2$
(d) 2MnO$_4^{-}$ + 3Mn$^{2+}$ + 2H$_2$O $\longrightarrow$ 5MnO$_2$ + 4H$^{+}$
Select the correct option from the following
- A. (a) and (d) only
- B. (a) and (b) only ✓
- C. (a), (b) and (c)
- D. (a), (c) and (d)
Solution: (a) and (b) only In disproportionation the same element in one species is both oxidised and reduced. (a) Cu⁺ (+1) gives Cu²⁺ (+2) and Cu (0): disproportionation. (b) Mn in MnO₄²⁻ (+6) gives MnO₄⁻ (+7) and MnO₂ (+4): disproportionation. (c) Mn is only reduced (+7 to +6 and +4) while oxygen is oxidised (−2 to 0), so it is not disproportionation. (d) Mn (+7) and Mn (+2) both give MnO₂ (+4): comproportionation, the reverse process.
Q150.
Conjugate base for Br$\ddot{\text{o}}$nsted acids H$_2$O and HF are :
- A. H$_3$O$^{+}$ and H$_2$F$^{+}$, respectively
- B. OH$^{-}$ and H$_2$F$^{+}$, respectively
- C. H$_3$O$^{+}$ and F$^{-}$, respectively
- D. OH$^{-}$ and F$^{-}$, respectively ✓
Solution: OH⁻ and F⁻, respectively A conjugate base is what remains when a Brönsted acid loses H⁺. H₂O loses H⁺ to give OH⁻, and HF loses H⁺ to give F⁻. H₃O⁺ and H₂F⁺ are conjugate acids, formed by gaining H⁺.
Q151.
Among the following, the reaction that proceeds through an electrophilic substitution, is:
- A. C$_6$H$_5$CH$_2$OH + HCl $\xrightarrow{\text{heat}}$ C$_6$H$_5$CH$_2$Cl + H$_2$O
- B. C$_6$H$_5$N$_2^{+}$Cl$^{-}$ $\xrightarrow{\text{Cu}_2\text{Cl}_2}$ C$_6$H$_5$Cl + N$_2$
- C. C$_6$H$_6$ + Cl$_2$ $\xrightarrow{\text{AlCl}_3}$ C$_6$H$_5$Cl + HCl ✓
- D. C$_6$H$_6$ + Cl$_2$ $\xrightarrow{\text{UV light}}$ C$_6$H$_6$Cl$_6$ (benzene hexachloride)
Solution: C₆H₆ + Cl₂ → C₆H₅Cl + HCl, with anhydrous AlCl₃ Anhydrous AlCl₃ generates the electrophile Cl⁺ (Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻), which substitutes an H on the benzene ring. (A) Cl⁻ replaces the OH of benzyl alcohol on the side-chain carbon: nucleophilic substitution. (B) In the Sandmeyer reaction Cl⁻ replaces the –N₂⁺ group with the help of Cu(I); no electrophile attacks the ring. (D) In UV light, Cl₂ adds to benzene by a free radical mechanism to give benzene hexachloride: addition, not substitution.
Q152.
An alkene "A" on reaction with O$_3$ and Zn–H$_2$O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:
- A. CH$_3$–CHCl–CH(CH$_3$)$_2$
- B. Cl–CH$_2$–CH$_2$–CH(CH$_3$)$_2$
- C. CH$_3$–CH$_2$–CH(CH$_2$Cl)–CH$_3$
- D. CH$_3$–CH$_2$–CCl(CH$_3$)–CH$_3$ ✓
Solution: CH₃–CH₂–CCl(CH₃)–CH₃ (2-chloro-2-methylbutane) Ozonolysis giving propanone and ethanal in a 1 : 1 ratio means A is (CH₃)₂C=CH–CH₃, 2-methylbut-2-ene. HCl adds by Markovnikov's rule: H⁺ adds to the =CH– carbon, giving the more stable tertiary carbocation. Cl⁻ then adds to that tertiary carbon, so B is CH₃–CH₂–CCl(CH₃)–CH₃, 2-chloro-2-methylbutane.
Q153.
A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor (Z) is :
- A. Z < 1 and repulsive forces are dominant
- B. Z > 1 and attractive forces are dominant
- C. Z > 1 and repulsive forces are dominant
- D. Z < 1 and attractive forces are dominant ✓
Solution: $Z < 1$ and attractive forces are dominant Compressibility factor $Z = \dfrac{V_{real}}{V_{ideal}} = 0.8$ here, since the molar volume is 20% smaller than ideal. $Z < 1$ means attractive forces dominate, so the gas is more compressible than an ideal gas.
Q154.
Among the following, the one that is not a green house gas is
- A. Sulphur dioxide ✓
- B. Nitrous oxide
- C. Methane
- D. Ozone
Solution: Sulphur dioxide (not a greenhouse gas) SO₂ causes acid rain but is not a greenhouse gas; methane, nitrous oxide and ozone are greenhouse gases.
Q155.
In which case change in entropy is negative?
- A. 2H(g) $\to$ H$_2$(g) ✓
- B. Evaporation of water
- C. Expansion of a gas at constant temperature
- D. Sublimation of solid to gas
Solution: 2H(g) → H₂(g) Two gas atoms combine into one molecule, so the number of gaseous particles and the randomness fall: $\Delta S < 0$. Evaporation, isothermal expansion of a gas and sublimation all increase randomness ($\Delta S > 0$).
Q156.
Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (Given that 1 L bar = 100 J)
- A. 30 J
- B. –30 J ✓
- C. 5 kJ
- D. 25 J
Solution: $-30\ \text{J}$ Work in expansion against a constant external pressure: $w = -p_{ex}\Delta V$ $w = -2\ \text{bar} \times (0.25 - 0.1)\ \text{L} = -0.3\ \text{L bar}$ ∴ $w = -0.3 \times 100\ \text{J} = -30\ \text{J}$ (the gas does 30 J of work)
Q157.
Which of the following species is not stable?
- A. [SiCl$_6$]$^{2-}$ ✓
- B. [SiF$_6$]$^{2-}$
- C. [GeCl$_6$]$^{2-}$
- D. [Sn(OH)$_6$]$^{2-}$
Solution: [SiCl₆]²⁻ (not stable) Six large Cl⁻ ions cannot be accommodated around the small Si⁴⁺, and the lone pair–Si⁴⁺ interaction is weak, so [SiCl₆]²⁻ is not formed. (B) [SiF₆]²⁻ exists, since six small F⁻ ions fit around Si. (C) and (D) Ge and Sn are larger, so [GeCl₆]²⁻ and [Sn(OH)₆]²⁻ are stable.
Q158.
For a cell involving one electron E$^{\circ}_{cell}$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : $\left[\text{Given that } \dfrac{2.303\,RT}{F} = 0.059\ \text{V at } T = 298\ \text{K}\right]$
- A. $1.0 \times 10^{30}$
- B. $1.0 \times 10^{2}$
- C. $1.0 \times 10^{5}$
- D. $1.0 \times 10^{10}$ ✓
Solution: $1.0 \times 10^{10}$ At equilibrium $E_{cell} = 0$ and $Q = K$, so $E^\circ_{cell} = \dfrac{0.059}{n}\log K$. $\log K = \dfrac{nE^\circ_{cell}}{0.059} = \dfrac{1 \times 0.59}{0.059} = 10$ ∴ $K = 1.0 \times 10^{10}$
Q159.
The method used to remove temporary hardness of water is :
- A. Synthetic resins method
- B. Calgon's method
- C. Clark's method ✓
- D. Ion-exchange method
Solution: Clark's method In Clark's method a calculated amount of lime is added to hard water, which removes temporary hardness. Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃↓ + 2H₂O Calgon's method, ion exchange and synthetic resins are used to remove permanent hardness.
Q160.
Which will make basic buffer?
- A. 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
- B. 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH$_3$COOH
- C. 100 mL of 0.1 M CH$_3$COOH + 100 mL of 0.1 M NaOH
- D. 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH$_4$OH ✓
Solution: 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH₄OH 10 mmol HCl converts 10 of the 20 mmol NH₄OH into NH₄Cl, leaving a weak base with its salt of a strong acid: a basic buffer. (A) 10 mmol HCl and 10 mmol NaOH neutralise completely to NaCl. (B) 2.5 mmol CH₃COOH with 5 mmol NaOH leaves excess strong base, NaOH. (C) 10 mmol of each give only CH₃COONa, a salt solution, not a buffer.
Q161.
The most suitable reagent for the following conversion, is :
CH$_3$–C$\equiv$C–CH$_3$ $\longrightarrow$ cis-2-butene
- A. Hg$^{2+}$/H$^{+}$, H$_2$O
- B. Na/liquid NH$_3$
- C. H$_2$, Pd/C, quinoline ✓
- D. Zn/HCl
Solution: H₂, Pd/C, quinoline Lindlar's catalyst (palladised charcoal partially deactivated with quinoline) adds H₂ to the alkyne from one side (syn addition). ∴ But-2-yne gives cis-but-2-ene. (A) Hg²⁺/H⁺ with water hydrates the alkyne to a ketone (butan-2-one). (B) Na in liquid NH₃ gives trans-but-2-ene.
Q162.
The compound that is most difficult to protonate is :
- A. Ph–O–H (phenol) ✓
- B. H–O–H (water)
- C. CH$_3$–O–H (methanol)
- D. CH$_3$–O–CH$_3$ (dimethyl ether)
Solution: Ph–O–H (phenol) In phenol the lone pair on oxygen is delocalised into the benzene ring by resonance, leaving the oxygen with a partial positive charge. So a proton attaches least readily to phenol's oxygen; in water, methanol and dimethyl ether the lone pairs on oxygen stay available.
Q163.
Which is the correct thermal stability order for H$_2$E (E = O, S, Se, Te and Po)?
- A. H$_2$Se < H$_2$Te < H$_2$Po < H$_2$O < H$_2$S
- B. H$_2$S < H$_2$O < H$_2$Se < H$_2$Te < H$_2$Po
- C. H$_2$O < H$_2$S < H$_2$Se < H$_2$Te < H$_2$Po
- D. H$_2$Po < H$_2$Te < H$_2$Se < H$_2$S < H$_2$O ✓
Solution: H₂Po < H₂Te < H₂Se < H₂S < H₂O Down group 16 the size of E increases, so the H–E bond dissociation enthalpy decreases and the hydrides become less stable to heat. ∴ H₂Po < H₂Te < H₂Se < H₂S < H₂O
Q164.
The correct structure of tribromooctaoxide is
- A. Structure (1)
- B. Structure (2) ✓
- C. Structure (3)
- D. Structure (4)
Solution: Structure (2): the chain O₃Br–BrO₂–BrO₃ Br₃O₈ is a chain of three bromine atoms, Br–Br–Br. Each terminal Br carries three doubly bonded O atoms and the central Br carries two (3 + 2 + 3 = 8 O). No oxygen atom carries a formal negative charge.
Q165.
The major product of the following reaction is:
Benzene-1,2-dicarboxylic acid (phthalic acid) $+$ NH$_3$ $\xrightarrow{\text{strong heating}}$
- A. A benzene ring bearing two –NH$_2$ groups at the 1,2-positions
- B. A benzene ring bearing –COOH and –CONH$_2$ at the 1,2-positions
- C. The cyclic imide in which the two ring carbonyls are joined through a single –NH– group (phthalimide) ✓
- D. A benzene ring bearing –COOH and –NH$_2$ at the 1,2-positions
Solution: Phthalimide (the cyclic imide) Phthalic acid and NH₃ first form ammonium phthalate. On heating it loses 2H₂O to give phthalamide (benzene-1,2-dicarboxamide). On strong heating phthalamide loses NH₃ and the ring closes to give the cyclic imide, phthalimide.
Q166.
Match the following :
(a) Pure nitrogen
(b) Haber process
(c) Contact process
(d) Deacon's process
(i) Chlorine
(ii) Sulphuric acid
(iii) Ammonia
(iv) Sodium azide or Barium azide
Which of the following is the correct option?
- A. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) ✓
- B. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
- C. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
- D. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Solution: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) Very pure nitrogen is obtained by thermal decomposition of sodium azide or barium azide: (a)-(iv). The Haber process makes ammonia from N₂ and H₂: (b)-(iii). The Contact process makes sulphuric acid: (c)-(ii). Deacon's process makes chlorine by oxidising HCl gas with air over CuCl₂: (d)-(i).
Q167.
For the chemical reaction
N$_2$(g) + 3H$_2$(g) $\rightleftharpoons$ 2NH$_3$(g)
The correct option is:
- A. $3\dfrac{d[\text{H}_2]}{dt} = 2\dfrac{d[\text{NH}_3]}{dt}$
- B. $-\dfrac{1}{3}\dfrac{d[\text{H}_2]}{dt} = -\dfrac{1}{2}\dfrac{d[\text{NH}_3]}{dt}$
- C. $-\dfrac{d[\text{N}_2]}{dt} = 2\dfrac{d[\text{NH}_3]}{dt}$
- D. $-\dfrac{d[\text{N}_2]}{dt} = \dfrac{1}{2}\dfrac{d[\text{NH}_3]}{dt}$ ✓
Solution: $-\dfrac{d[\text{N}_2]}{dt} = \dfrac{1}{2}\dfrac{d[\text{NH}_3]}{dt}$ Rate $= -\dfrac{d[\text{N}_2]}{dt} = -\dfrac{1}{3}\dfrac{d[\text{H}_2]}{dt} = +\dfrac{1}{2}\dfrac{d[\text{NH}_3]}{dt}$: each rate is divided by its coefficient, with a minus sign for reactants. (A) The correct relation is $-2\dfrac{d[\text{H}_2]}{dt} = 3\dfrac{d[\text{NH}_3]}{dt}$. (B) NH₃ is a product, so its term must be positive. (C) The factor on the NH₃ term is $\tfrac{1}{2}$, not 2.
Q168.
The structure of intermediate A in the following reaction, is
Cumene $\xrightarrow{\ \text{O}_2\ }$ A $\xrightarrow[\text{H}_2\text{O}]{\ \text{H}^{+}\ }$ Phenol $+$ CH$_3$COCH$_3$
- A. C$_6$H$_5$–CH(CH$_3$)–CH$_2$–O–O–H
- B. C$_6$H$_5$–O–CH(CH$_3$)$_2$
- C. C$_6$H$_5$–C(CH$_3$)$_2$–O–O–H ✓
- D. C$_6$H$_5$–O–O–CH(CH$_3$)$_2$
Solution: C₆H₅–C(CH₃)₂–O–O–H (cumene hydroperoxide) Cumene is oxidised by air to cumene hydroperoxide, C₆H₅–C(CH₃)₂–O–O–H, which is A. Cumene hydroperoxide with dilute acid gives phenol and acetone.
Q169.
The non-essential amino acid among the following is:
- A. Lysine
- B. Valine
- C. Leucine
- D. Alanine ✓
Solution: Alanine Alanine can be made in the body, so it is non-essential; lysine, valine and leucine are essential amino acids that must come from the diet.
Q170.
Which of the following diatomic molecular species has only $\pi$ bonds according to Molecular Orbital Theory?
- A. Be$_2$
- B. O$_2$
- C. N$_2$
- D. C$_2$ ✓
Solution: C₂ C₂ (12 electrons): $\sigma 1s^2\,\sigma^*1s^2\,\sigma 2s^2\,\sigma^*2s^2\,(\pi 2p_x^2 = \pi 2p_y^2)$ The σ and σ* electrons cancel, leaving 4 bonding electrons in π orbitals: bond order 2, made of two π bonds only. (A) Be₂ has bond order 0, so there is no bond at all. (B) O₂ has a filled $\sigma 2p_z$ orbital, so its double bond is one σ and one π. (C) N₂ has a filled $\sigma 2p_z$ orbital, so its triple bond is one σ and two π.
Q171.
The correct order of the basic strength of methyl substituted amines in aqueous solution is :
- A. CH$_3$NH$_2$ > (CH$_3$)$_2$NH > (CH$_3$)$_3$N
- B. (CH$_3$)$_2$NH > CH$_3$NH$_2$ > (CH$_3$)$_3$N ✓
- C. (CH$_3$)$_3$N > CH$_3$NH$_2$ > (CH$_3$)$_2$NH
- D. (CH$_3$)$_3$N > (CH$_3$)$_2$NH > CH$_3$NH$_2$
Solution: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N In water, the basic strength of amines depends on the +I effect of alkyl groups, solvation of the ammonium ion by H-bonding and steric hindrance together. For methyl amines the balance gives (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N. The tertiary amine comes last because its ion has only one N–H for H-bonding and is crowded.
Q172.
Which mixture of the solutions will lead to the formation of negatively charged colloidal [AgI]I$^{-}$ sol ?
- A. 50 mL of 0.1 M AgNO$_3$ + 50 mL of 0.1 M KI
- B. 50 mL of 1 M AgNO$_3$ + 50 mL of 1.5 M KI
- C. 50 mL of 1 M AgNO$_3$ + 50 mL of 2 M KI ✓
- D. 50 mL of 2 M AgNO$_3$ + 50 mL of 1.5 M KI
Solution: 50 mL of 1 M AgNO₃ + 50 mL of 2 M KI A sol takes the charge of the ion it adsorbs, so [AgI]I⁻ forms when KI is in excess and I⁻ is adsorbed on AgI. 50 mmol AgNO₃ with 100 mmol KI reacts 1 : 1 and leaves the largest excess of I⁻ (50 mmol). (A) 5 mmol of each react completely, leaving no excess ion to charge the sol. (B) 50 mmol AgNO₃ with 75 mmol KI leaves a smaller excess of I⁻ (25 mmol). (D) 100 mmol AgNO₃ with 75 mmol KI leaves Ag⁺ in excess, giving the positive sol [AgI]Ag⁺.
Q173.
Identify the incorrect statement related to PCl$_5$ from the following:
- A. PCl$_5$ molecule is non-reactive ✓
- B. Three equatorial P–Cl bonds make an angle of 120$^{\circ}$ with each other
- C. Two axial P–Cl bonds make an angle of 180$^{\circ}$ with each other
- D. Axial P–Cl bonds are longer than equatorial P–Cl bonds
Solution: PCl₅ molecule is non-reactive (the incorrect statement) The axial P–Cl bonds (240 pm) are longer and weaker than the equatorial ones (202 pm), which makes PCl₅ reactive. (B) True: the three equatorial bonds lie in one plane at 120° to each other. (C) True: the two axial bonds are at 180° to each other. (D) True: axial bonds suffer more repulsion from the equatorial bond pairs, so they are longer.
Q174.
Among the following, the narrow spectrum antibiotic is :
- A. Chloramphenicol
- B. Penicillin G ✓
- C. Ampicillin
- D. Amoxycillin
Solution: Penicillin G Penicillin G acts mainly on Gram-positive bacteria, so it is narrow spectrum; chloramphenicol, ampicillin and amoxycillin are broad spectrum.
Q175.
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:
- A. t = 2.303/k
- B. t = 0.693/k
- C. t = 6.909/k
- D. t = 4.606/k ✓
Solution: $t = \dfrac{4.606}{k}$ First order: $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ For 99% completion $[A] = \dfrac{[A]_0}{100}$, so $\log\dfrac{[A]_0}{[A]} = \log 100 = 2$. ∴ $t = \dfrac{2.303 \times 2}{k} = \dfrac{4.606}{k}$
Q176.
For the second period elements the correct increasing order of first ionisation enthalpy is:
- A. Li < Be < B < C < O < N < F < Ne
- B. Li < Be < B < C < N < O < F < Ne
- C. Li < B < Be < C < O < N < F < Ne ✓
- D. Li < B < Be < C < N < O < F < Ne
Solution: Li < B < Be < C < O < N < F < Ne First ionisation enthalpy generally increases across period 2, with two exceptions. B < Be: the 2p electron of B is more shielded than the 2s electron of Be, so it is easier to remove. O < N: N has a stable half-filled 2p³, while in O two 2p electrons share an orbital and repel each other. ∴ Li < B < Be < C < O < N < F < Ne
Q177.
4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is
- A. 5f > 6p > 4d > 5p
- B. 5f > 6p > 5p > 4d ✓
- C. 6p > 5f > 5p > 4d
- D. 6p > 5f > 4d > 5p
Solution: 5f > 6p > 5p > 4d By the (n + l) rule, a larger n + l means higher energy; for equal n + l, the larger n is higher. 4d: 4 + 2 = 6; 5p: 5 + 1 = 6; 5f: 5 + 3 = 8; 6p: 6 + 1 = 7 5p and 4d both have n + l = 6, so 5p (larger n) is higher. ∴ 5f > 6p > 5p > 4d
Q178.
For the cell reaction
2Fe$^{3+}$(aq) + 2I$^{-}$(aq) $\to$ 2Fe$^{2+}$(aq) + I$_2$(aq)
E$^{\ominus}_{cell}$ = 0.24 V at 298 K. The standard Gibbs energy ($\Delta_r$G$^{\ominus}$) of the cell reaction is : [Given that Faraday constant F = 96500 C mol$^{-1}$]
- A. 23.16 kJ mol$^{-1}$
- B. – 46.32 kJ mol$^{-1}$ ✓
- C. – 23.16 kJ mol$^{-1}$
- D. 46.32 kJ mol$^{-1}$
Solution: $-46.32\ \text{kJ mol}^{-1}$ $\Delta_r G^\ominus = -nFE^\ominus_{cell}$, with $n = 2$ (two Fe³⁺ ions each gain one electron). $\Delta_r G^\ominus = -2 \times 96500\ \text{C mol}^{-1} \times 0.24\ \text{V} = -46320\ \text{J mol}^{-1}$ ∴ $\Delta_r G^\ominus = -46.32\ \text{kJ mol}^{-1}$
Q179.
Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region?
- A. Brackett series
- B. Lyman series
- C. Balmer series ✓
- D. Paschen series
Solution: Balmer series Balmer series lines (transitions ending at n = 2) fall in the visible region; Lyman lies in the ultraviolet, Paschen and Brackett in the infrared.
Q180.
The biodegradable polymer is:
- A. Buna-S
- B. Nylon-6,6
- C. Nylon-2-Nylon 6 ✓
- D. Nylon-6
Solution: Nylon-2–nylon-6 Nylon-2–nylon-6, an alternating polyamide copolymer of glycine and aminocaproic acid, is biodegradable; Buna-S, nylon-6,6 and nylon-6 are not.