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: In the lac operon:
i gene — Repressor
z gene — $\beta$-galactosidase
y gene — Permease
a gene — Transacetylase
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: Crypts of Lieberkuhn are present in the small intestine.
Glisson's capsule is present in the liver.
Islets of Langerhans constitute the endocrine portion of the pancreas.
Brunner's glands are found in the submucosa of the duodenum.
Q3.
What is the direction of movement of sugars in phloem?
- A. Bi-directional ✓
- B. Non-multidirectional
- C. Upward
- D. Downward
Solution: The direction of movement of sugar in phloem is bi-directional, as it depends on the source-sink relationship, which is variable in plants.
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 are lined with ciliated epithelium to move particles or mucus in a specific 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 is the most important cause driving animals and plants to extinction.
For example, loss of tropical rainforest has reduced the forest cover from 14% to 6%.
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: In lactational amenorrhoea, due to a high prolactin level, the gonadotropin level decreases.
Oral pills are either progestogens or progestogen-estrogen combinations used by females.
Emergency contraceptives include the administration of progestogens or a progestogen-estrogen combination, or IUDs, within 72 hours of coitus.
So lactational amenorrhoea, oral pills and emergency contraceptives all involve a role of hormone.
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 do not have DNA.
Mitochondria and chloroplasts are semi-autonomous organelles and 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: In parietal placentation the ovules develop on the inner wall of the ovary or in the peripheral part, e.g. mustard, $Argemone$.
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: The Earth Summit (Rio Summit) of 1992 called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits.
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: The purines found both in DNA and RNA are adenine and guanine.
Cytosine is a pyrimidine found in both, thymine is a pyrimidine found 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: Insulin deficiency leads to diabetes mellitus.
Hypersecretion or hyposecretion of thyroxine can be associated with enlargement of the thyroid gland, called goitre.
Deficiency of corticoids (glucocorticoid + mineralocorticoid) leads to Addison's disease.
Growth hormone hypersecretion in adults leads to acromegaly.
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: The correct sequence of phases of the cell cycle is
G$_1$ $\to$ S $\to$ G$_2$ $\to$ M
G$_1$, S and G$_2$ together form the interphase, which is followed by the M phase.
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 is caused by type-II herpes simplex virus.
At present there is no cure for type-II herpes simplex virus and therefore for the disease it causes, genital herpes.
The other non-curable STIs are hepatitis-B and HIV.
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: Polyblend is a fine powder of recycled modified plastic waste.
The mixture is mixed with 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: A sub-metacentric chromosome is heterobrachial, i.e. its two arms are unequal.
The short arm is designated as the 'p' arm (p = petite, i.e. short).
The long arm is designated as the 'q' arm.
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: Restriction enzymes cut DNA molecules at a particular point by recognising a specific sequence.
Each restriction endonuclease functions by inspecting the length of a DNA sequence. Once it finds its specific recognition sequence, it binds to the DNA and cuts EACH of the two strands of the double helix at specific points in their sugar-phosphate backbone.
So the statement that it cuts only one of the two strands is incorrect.
Q17.
Persistent nucellus in the seed is known as
- A. Tegmen
- B. Chalaza
- C. Perisperm ✓
- D. Hilum
Solution: Persistent nucellus is called perisperm, e.g. black pepper, 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 secrete mucus and bicarbonates present in the gastric juice.
This plays an important role in lubrication and protection of the mucosal epithelium from excoriation by 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 bud off from the trans face of the Golgi bodies, not from the endoplasmic reticulum, so statement (1) is not correct.
Precursors of lysosomal enzymes are synthesised by RER and then sent to the Golgi bodies for further processing.
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: Microbes are used in the production of several household and industrial products:
$Lactobacillus$ – production of curd
$Saccharomyces$ $cerevisiae$ – bread making
$Aspergillus$ $niger$ – citric acid production
$Acetobacter$ $aceti$ – acetic acid
Q21.
Which part of the brain is responsible for thermoregulation?
- A. Medulla oblongata
- B. Cerebrum
- C. Hypothalamus ✓
- D. Corpus callosum
Solution: Hypothalamus is the thermoregulatory centre of our brain.
It is responsible for maintaining constant 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: Genes for flower colour in snapdragon show incomplete dominance, which is an exception to Mendel's principle of dominance.
Whereas the law of segregation is universally applicable, so the statement that it does not apply here is the incorrect one.
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: The fungus $Trichoderma$ is a biological control agent being developed for use in the treatment of plant diseases.
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: The fungus $Trichoderma$, Baculoviruses (NPV) and $Bacillus$ $thuringiensis$ are used as biocontrol agents.
$Rhizobium$, $Nostoc$, $Azospirillum$ and $Oscillatoria$ are used as biofertilisers, whereas TMV is a pathogen and aphids are pests that harm crop plants.
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 explained chromosomal mapping on the basis of recombination frequency, which is directly proportional to the distance between two genes on the same chromosome.
Q26.
Respiratory Quotient (RQ) value of tripalmitin is
- A. 0.09
- B. 0.9
- C. 0.7 ✓
- D. 0.07
Solution: $$\text{RQ} = \frac{\text{Amount of CO}_2\ \text{released}}{\text{Amount of O}_2\ \text{consumed}}$$
For tripalmitin,
$$2(\text{C}_{51}\text{H}_{98}\text{O}_6) + 145\,\text{O}_2 \rightarrow 102\,\text{CO}_2 + 98\,\text{H}_2\text{O} + \text{Energy}$$
$$\text{RQ} = \frac{102}{145} = 0.7$$
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: Cardiac output $=$ stroke volume $\times$ heart rate
Cardiac output $= 5$ L $= 5000$ mL
Stroke volume $= 100 - 50 = 50$ mL
$$5000 = 50 \times \text{Heart rate}$$
Heart rate $= 100$ 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: In pteridophytes the megaspore is retained for some time in the female gametophyte on the parent sporophyte; permanent retention is required for seed formation, which occurs in gymnosperms.
That is why pteridophytes exhibit only the precursor to the seed habit, and this retention is FIRST observed in them.
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: In an aquatic ecosystem the pyramid of biomass is generally inverted.
The small standing crop of phytoplankton supports a larger biomass of zooplankton, then small fishes and finally large fishes, so the pyramid widens upwards.
The pyramid of energy is always 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: Colostrum, the yellowish fluid secreted by the mother during the initial days of lactation, contains Immunoglobulin A.
It imparts naturally acquired passive immunity to the newborn.
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: Phloem in gymnosperms lacks both sieve tubes and companion cells.
Instead it has sieve cells and albuminous 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) $Pila$ is a mollusc; the mouth contains a file-like rasping organ for feeding called the radula.
(b) $Bombyx$ is an arthropod; excretion takes place through Malpighian tubules.
(c) $Pleurobrachia$ is a ctenophore; the body bears eight external rows of ciliated comb plates, which help in locomotion.
(d) $Taenia$ is a platyhelminth; specialised cells called flame cells help in osmoregulation and excretion.
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: Statements (a) and (b) are incorrect because dialysis eliminates urea and potassium from the body.
Statements (c) and (d) are correct. As phosphate ions are eliminated during dialysis, calcium ions are also eliminated along with them, so there will be reduced absorption of calcium ions from the gastrointestinal tract.
RBC production will be reduced, due to reduced erythropoietin hormone.
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 a dense matrix of collagen and corneal epithelium.
It is the most sensitive part of the eye, and it is avascular, not highly vascularised.
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 birds female heterogamety is found, so the sex of the progeny depends on the type of egg rather than the type of sperm.
In fowls the male produces only A + Z sperms (100%), while the female produces A + Z eggs (50%) and A + W eggs (50%).
Hence statement (4) is incorrect.
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: The concept of "Omnis cellula-e cellula", i.e. that cells arise from pre-existing cells, was proposed by Rudolf Virchow.
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: Growth rings are formed by the seasonal activity of the cambium. In plants of temperate regions the cambium is more active in spring and less active in autumn, so annual rings ARE prominent there.
In temperate regions climatic conditions are not uniform throughout the year, whereas in the tropics climatic conditions are uniform throughout the year.
Hence statement (1) is not true.
Q38.
$Thiobacillus$ is a group of bacteria helpful in carrying out
- A. Denitrification ✓
- B. Nitrogen fixation
- C. Chemoautotrophic fixation
- D. Nitrification
Solution: $Thiobacillus$ $denitrificans$ causes denitrification, i.e. conversion of oxides of nitrogen to free N$_2$.
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: Asthma is a difficulty in breathing causing wheezing due to inflammation of the bronchi and bronchioles.
It can be due to increasing air-borne allergens and pollutants. Asthma is an allergic condition, and many people in urban areas suffer from this respiratory disorder.
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: The phenomenon in which the female gamete develops into an embryo without getting fused with a male gamete (fertilisation) is called parthenogenesis.
Parthenocarpy is the development of fruit without fertilisation; autogamy is self-pollination within the same flower.
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: Vertebrosternal ribs are true ribs; dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage. The first seven pairs of ribs are called true ribs.
The 8$^{\text{th}}$, 9$^{\text{th}}$ and 10$^{\text{th}}$ pairs of ribs do not articulate directly with the sternum but join the seventh ribs with the help of hyaline cartilage. These are vertebrochondral or false ribs.
The last 2 pairs (11 and 12) of ribs are not connected ventrally and are therefore called floating ribs.
So only the first seven pairs of ribs are ventrally connected 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: Steroid hormones directly enter into the cell and bind with intracellular receptors in the nucleus to form a hormone-receptor complex.
The hormone-receptor complex then interacts with the genome, regulating gene expression.
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: According to the rules of binomial nomenclature, the correctly written scientific name of mango is $Mangifera$ $indica$ Linn.
The genus begins with a capital letter, the species with a small letter, and the abbreviated name of the author who first described it follows at the end.
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: 1 map unit represents 1% cross over.
Map unit is used to measure genetic distance, and this genetic distance is based on the average number of cross over frequency.
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: Cells in G$_0$ phase are said to exit the cell cycle.
These are at the quiescent stage and do not proliferate unless called upon 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: Following are the post-fertilisation changes:
Ovule $\rightarrow$ Seed
Ovary $\rightarrow$ Fruit
Zygote $\rightarrow$ Embryo
Central cell $\rightarrow$ Endosperm
The embryo sac develops inside the ovule BEFORE fertilisation, so statement (1) is incorrect.
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: In recombinant DNA technology bacteria are able to produce human insulin because the genetic code is nearly universal.
The same codons specify the same amino acids in bacteria as in humans, so a human gene is read correctly by the bacterial machinery.
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 is insulin dependent and is responsible for the majority of glucose transport into muscle and adipose cells in anabolic conditions.
GLUT-I is insulin independent and is widely distributed in different tissues.
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: The mRNA is read in triplets, so only the addition or deletion of a multiple of three bases leaves the reading frame intact.
Deleting GGU from the 7$^{\text{th}}$, 8$^{\text{th}}$ and 9$^{\text{th}}$ positions removes exactly three bases, giving
5$'$ AAC AGC GCU AUU 3$'$
so there is no change in the reading frame of the mRNA.
Insertion or deletion of one or two bases would shift 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 and LNG-20 are hormone releasing IUDs, which make the uterus unsuitable for implantation and the cervix hostile to sperms.
Lippes loop is a non-medicated IUD and Multiload 375 is a copper releasing IUD.
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: According to Hugo de Vries, mutations are random and directionless.
De Vries believed mutation caused speciation and hence called it saltation (single step large mutation).
Q52.
Expressed Sequence Tags (ESTs) refers to :
- A. Novel DNA sequences
- B. Genes expressed as RNA ✓
- C. Polypeptide expression
- D. DNA polymorphism
Solution: Expressed Sequence Tags (ESTs) are DNA sequences (genes) that are expressed as mRNA for protein synthesis.
These are used 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: $Bacillus$ $thuringiensis$ forms protein crystals during a particular phase of its growth. These crystals contain a toxic insecticidal protein.
These proteins exist as inactive protoxins, but once an insect ingests the inactive toxin it is converted into an active form of toxin due to the alkaline pH of the gut, which solubilises the crystals.
The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis, and eventually cause the death of the insect.
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: The correct match of hominids and their brain sizes is:
$Homo$ $habilis$ — 650-800 cc
$Homo$ $neanderthalensis$ — 1400 cc
$Homo$ $erectus$ — 900 cc
$Homo$ $sapiens$ — 1350 cc
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: Relative contribution of various greenhouse gases to total global warming is
CO$_2$ = 60%, CH$_4$ = 20%, CFC = 14%, N$_2$O = 6%
Therefore CO$_2$ and CH$_4$ are the major greenhouse gases.
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: Saprophytes – decomposition of dead organic materials.
Parasites – grow on/in living plants and animals.
Lichens – symbiotic association of algae and fungi.
Mycorrhiza – symbiotic association of fungi with plant roots.
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: Golden rice is vitamin A enriched rice, with a gene from daffodil, and is rich in carotene.
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: Individuals with Klinefelter's syndrome have trisomy of the sex chromosomes as 44 + XXY (47).
They show overall masculine development, gynaecomastia, and 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: Extrusion of the second polar body from the egg nucleus occurs after entry of the sperm but before fertilisation.
The entry of sperm into the ovum induces completion of the meiotic division of the secondary oocyte.
Entry of sperm causes breakdown of metaphase promoting factor (MPF) and turns on the anaphase promoting complex (APC).
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: Frequency of dominant allele $p = 0.4$
Frequency of recessive allele $q = 1 - 0.4 = 0.6$
Frequency of homozygous dominant individuals (AA) $= p^{2} = (0.4)^{2} = 0.16$
Frequency of heterozygous individuals (Aa) $= 2pq = 2(0.4)(0.6) = 0.48$
Frequency of homozygous recessive individuals (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: In flowering plants, out of the two male gametes discharged in the synergid, one fuses with the egg and the other fuses with the secondary or definitive nucleus present in the central cell.
Egg (n) + 1$^{\text{st}}$ male gamete (n) $\longrightarrow$ Zygote (2n)
Secondary nucleus (2n) + 2$^{\text{nd}}$ male gamete (n) $\longrightarrow$ PEN (3n)
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: The given data shows stabilising selection, as most of the newborns having average weight between 3 to 3.3 kg survive, and babies with less and more weight have a low survival rate.
Q63.
Which of the following muscular disorders is inherited?
- A. Botulism
- B. Tetany
- C. Muscular dystrophy ✓
- D. Myasthenia gravis
Solution: Progressive degeneration of skeletal muscle mostly due to a genetic disorder is muscular dystrophy.
Tetany is muscular spasm due to low calcium in body fluid.
Myasthenia gravis is an autoimmune disorder leading to paralysis of skeletal muscles.
Botulism is a rare and dangerous type of food poisoning caused by the bacterium $Clostridium$ $botulinum$.
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: To control the deleterious effect of stratospheric ozone depletion, an international treaty was signed at Montreal, Canada in 1987.
It is popularly known as the Montreal protocol.
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: Coenzymes, metal ions and prosthetic groups that are bound to enzyme protein are together called cofactors.
PROSTHETIC GROUPS are tightly bound to the enzyme, whereas metal ions and coenzymes are not tightly bound to the enzyme — so statement (A) is false.
A complete catalytically active enzyme with its bound prosthetic group is called a HOLOENZYME, not apoenzyme — so statement (B) is also false.
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: True segmentation is present in Annelida, Arthropoda and Chordata.
They also have organ system level of organisation, bilateral symmetry and are true coelomates.
Mollusca is a true coelomate but is 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: The fungus associated with the roots of $Pinus$ increases mineral and water absorption for the plant by increasing surface area, and in turn the fungus gets food from the plant.
Therefore the mycorrhizal association is obligatory for $Pinus$ seed germination.
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: The correct sequence of organs in the alimentary canal of cockroach starting from the mouth is:
Pharynx $\to$ Oesophagus $\to$ Crop $\to$ Gizzard $\to$ Ileum $\to$ Colon $\to$ 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: In mitochondria, the enzymes for electron transport are present in the INNER membrane, not the outer membrane.
So statement (3) is incorrect; the other three are correct.
Q70.
Drug called 'Heroin' is synthesized by
- A. nitration of morphine
- B. methylation of morphine
- C. acetylation of morphine ✓
- D. glycosylation of morphine
Solution: Heroin, commonly called smack, is chemically diacetylmorphine, which is synthesized by acetylation of morphine.
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 catalyses the conversion of glucose to glucose-6-phosphate.
It is the first step of the activation phase 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: During the isolation of a desired gene, chilled ethanol is used for the precipitation of DNA.
The DNA appears as a collection of fine threads in suspension.
Q73.
Which of the following is a commercial blood cholesterol lowering agent?
- A. Lipases
- B. Cyclosporin A
- C. Statin ✓
- D. Streptokinase
Solution: Statin is obtained from a yeast (fungi) called $Monascus$ $purpureus$.
It acts by competitively inhibiting the enzyme responsible for synthesis of 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: To produce enzymes in large quantity the equipment required is a bioreactor.
Large scale production involves the use of 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: The infective constituent in viruses is either DNA or RNA, not the protein coat.
So statement (4) is incorrect; the other three are correct.
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: Bulliform cells become flaccid due to water loss.
This makes the leaves curl inward to minimise 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: Xylem is associated with translocation of mainly water, mineral salts, some organic nitrogen and hormones.
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: The correct sequence for transport of sperm cells in the male reproductive system is
Seminiferous tubules $\to$ Rete testis $\to$ Vasa efferentia $\to$ Epididymis $\to$ Vas deferens $\to$ Ejaculatory duct $\to$ Urethra $\to$ 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: Storage of nuclear waste should be done in suitably shielded containers and buried within rocks deep below the earth's surface (about 500 m deep).
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: The body is able to differentiate self and non-self, and the cell-mediated immune response is responsible for graft rejection.
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: During flowering, the photoperiodic stimulus is perceived by the leaves of plants.
A hormonal substance is then translocated from the leaves to the shoot apices, where flowers are induced.
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: $Salmonella$ $typhi$ is the causative agent of typhoid fever.
The confirmatory test is the Widal test, which is based on an antigen-antibody reaction.
Q83.
Concanavalin A is
- A. a pigment
- B. an alkaloid
- C. an essential oil
- D. a lectin ✓
Solution: Concanavalin A is a secondary metabolite; it is a lectin, and has the property of agglutinating RBCs.
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: The plant hormone auxin induces flowering in pineapple.
Ethylene also helps in synchronisation of flowering and 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 EXPOSES harmful recessive genes so that they can be eliminated by selection; it does not select for them.
It also helps in accumulation of superior genes and elimination of less desirable genes. Therefore, with selection at each step, it increases the productivity of the inbred population.
Close and continued inbreeding usually reduces fertility and even productivity.
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: Yeast is a unicellular sac fungus. It lacks a filamentous structure or hyphae.
So statement (1) is incorrect; the other three statements are correct.
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: Tidal Volume $= 500$ mL, Expiratory Reserve Volume $= 1000$ mL
Expiratory Capacity $=$ TV $+$ ERV
$$= 500 + 1000 = 1500\ \text{mL}$$
The residual volume 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 is ex-situ conservation (off-site conservation), i.e. living plants (flora) are conserved in a human managed system.
Sacred groves, biosphere reserves and wildlife sanctuaries are in-situ methods.
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: The proximity between the loop of Henle and the vasa recta, as well as the counter current in them, helps in maintaining an increasing osmolarity towards the inner medullary interstitium.
This mechanism helps to maintain a concentration gradient in the medullary interstitium, so human urine is nearly four times more concentrated than the initial filtrate formed.
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: In an ECG, the P-wave represents depolarisation of the atria.
The QRS complex represents depolarisation of the ventricles.
The T-wave represents repolarisation of the ventricles, i.e. return from the excited to the normal state.
Reduction in the size of the T-wave indicates insufficient supply of oxygen, i.e. coronary ischaemia.
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: On a smooth incline the retardation is $g\sin\theta$, so the stopping distance is
$$x = \frac{u^{2}}{2g\sin\theta}$$
$$x_1 = \frac{u^{2}}{2g\sin 60^{\circ}},\qquad x_2 = \frac{u^{2}}{2g\sin 30^{\circ}}$$
$$\frac{x_1}{x_2} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}} = \frac{1 \times 2}{2 \times \sqrt{3}} = 1 : \sqrt{3}$$
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: Excess pressure inside a soap bubble $= \dfrac{4T}{R}$, and the gauge pressure at depth Z$_0$ is $\rho g Z_0$.
$$P_0 + \frac{4T}{R} = P_0 + \rho g Z_0$$
$$Z_0 = \frac{4T}{R\rho g} = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 1000 \times 10}\ \text{m}$$
$$Z_0 = 1\ \text{cm}$$
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: Equivalent focal length in air:
$$\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$$
When glycerine is filled in between, the glycerine layer behaves like a diverging lens of focal length $(-f)$:
$$\frac{1}{F_2} = \frac{1}{f} + \frac{1}{f} - \frac{1}{f} = \frac{1}{f}$$
$$\frac{F_1}{F_2} = \frac{f/2}{f} = \frac{1}{2}$$
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: An $\alpha$-particle is the nucleus of helium, which has two protons and two neutrons.
Q95.
Which of the following acts as a circuit protection device?
- A. Fuse ✓
- B. Conductor
- C. Inductor
- D. Switch
Solution: A fuse wire has a low melting point, so when excess current flows, the heat produced in it melts it and the circuit breaks.
Hence a fuse acts as a circuit protection device.
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: At the critical angle $i = i_c$, the refracted ray grazes along the interface.
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: For the shortest path the resultant velocity must be along the north, so
$$\vec{V}_{SG} = \vec{V}_{SR} + \vec{V}_{RG}$$
$$\sin\theta = \frac{|\vec{V}_{RG}|}{|\vec{V}_{SR}|} = \frac{10}{20} = \frac{1}{2}$$
$$\theta = 30^{\circ}\ \text{west of north}$$
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: Capacitance $C = 20\ \mu$F $= 20 \times 10^{-6}$ F, and $\dfrac{dV}{dt} = 3$ V/s.
$$q = CV \Rightarrow \frac{dq}{dt} = C\frac{dV}{dt}$$
$$i_c = 20 \times 10^{-6} \times 3 = 60 \times 10^{-6}\ \text{A} = 60\ \mu\text{A}$$
The displacement current equals the conduction current, so $i_d = i_c = 60\ \mu$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: In Bohr's model of the H atom,
$$\text{K.E.} = |\text{TE}| = \frac{|U|}{2}$$
So K.E. $= 3.4$ eV and $U = -6.8$ 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: For $X = \dfrac{A^{2}B^{1/2}}{C^{1/3}D^{3}}$, the fractional errors add with the magnitudes of the powers:
$$\frac{\Delta X}{X} \times 100 = 2\frac{\Delta A}{A}\times 100 + \frac{1}{2}\frac{\Delta B}{B}\times 100 + \frac{1}{3}\frac{\Delta C}{C}\times 100 + 3\frac{\Delta D}{D}\times 100$$
$$= 2 \times 1\% + \frac{1}{2}\times 2\% + \frac{1}{3}\times 3\% + 3 \times 4\%$$
$$= 2\% + 1\% + 1\% + 12\% = 16\%$$
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: Charge Q will be distributed over the surface of the hollow metal sphere.
For $r < R$ (inside), by Gauss's law $\oint \vec{E}_{in}\cdot d\vec{S} = \dfrac{q_{en}}{\epsilon_0} = 0$ since $q_{en} = 0$, so $E_{in} = 0$.
For $r > R$ (outside), $q_{en} = Q$, so
$$E_0\,4\pi r^{2} = \frac{Q}{\epsilon_0} \Rightarrow E_0 \propto \frac{1}{r^{2}}$$
So the field is zero inside and decreases as r increases outside.
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: At the mid-point each line charge is at a distance R, and the two fields point in the same direction (away from the positive line and towards the negative line).
Electric field due to line charge 1: $\vec{E}_1 = \dfrac{\lambda}{2\pi\epsilon_0 R}\hat{i}$ N/C
Electric field due to line charge 2: $\vec{E}_2 = \dfrac{\lambda}{2\pi\epsilon_0 R}\hat{i}$ N/C
$$\vec{E}_{net} = \frac{\lambda}{2\pi\epsilon_0 R}\hat{i} + \frac{\lambda}{2\pi\epsilon_0 R}\hat{i} = \frac{\lambda}{\pi\epsilon_0 R}\hat{i}\ \text{N/C}$$
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: The heat current related to a difference of temperature across the length $\ell$ of a conductor of area A is
$$\frac{dH}{dt} = \frac{KA}{\ell}\Delta T$$
where K is the coefficient of thermal conductivity, so
$$K = \frac{\ell}{A}\frac{dH}{dt}\frac{1}{\Delta T}$$
Hence the unit of K is W m$^{-1}$ 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: A$_0$ only shifts the mean position, so the oscillation itself is
$$y' = y - A_0 = A\sin\omega t + B\cos\omega t$$
The two terms differ in phase by 90$^{\circ}$, so the resultant amplitude is
$$R = \sqrt{A^{2} + B^{2} + 2AB\cos 90^{\circ}} = \sqrt{A^{2} + B^{2}}$$
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: In air the angular fringe width is $\theta_0 = \dfrac{\beta}{D}$.
In water the wavelength becomes $\lambda/\mu$, so the angular fringe width becomes
$$\theta_w = \frac{\beta}{\mu D} = \frac{\theta_0}{\mu} = \frac{0.2^{\circ}}{4/3} = 0.15^{\circ}$$
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: Acceleration due to gravity at a depth d from the surface of the earth is
$$g' = g\left(1 - \frac{d}{R}\right)$$
Multiplying by mass m on both sides, with $d = \dfrac{R}{2}$,
$$mg' = mg\left(1 - \frac{R}{2R}\right) = 200\left(\frac{1}{2}\right) = 100\ \text{N}$$
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: As the three forces form a closed loop taken in the same order, their vector sum is zero.
$$\vec{F}_{net} = 0 \Rightarrow m\frac{d\vec{v}}{dt} = 0 \Rightarrow \vec{v} = \text{constant}$$
So the velocity of the particle remains constant.
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: Since the time periods are the same, $T_A = T_B = T$.
$$\omega_A = \frac{2\pi}{T_A},\qquad \omega_B = \frac{2\pi}{T_B}$$
$$\frac{\omega_A}{\omega_B} = \frac{T_B}{T_A} = \frac{T}{T} = 1$$
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: $B = 5 \times 10^{-5}$ T, $N = 800$, $A = 0.05$ m$^{2}$, $\Delta t = 0.1$ s, $\theta_1 = 0^{\circ}$, $\theta_2 = 90^{\circ}$.
$$\Delta\phi = NBA\cos 90^{\circ} - NBA\cos 0^{\circ} = -NBA$$
$$\Delta\phi = -800 \times 5 \times 10^{-5} \times 0.05 = -2 \times 10^{-3}\ \text{weber}$$
$$e = -\frac{\Delta\phi}{\Delta t} = \frac{2 \times 10^{-3}}{0.1} = 0.02\ \text{V}$$
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: The normal reaction supplies the centripetal force, $N = mr\omega^{2}$, and limiting friction must balance the weight.
$$f_L \geq mg \Rightarrow \mu N \geq mg \Rightarrow \mu\, m r\omega^{2} \geq mg$$
$$\omega \geq \sqrt{\frac{g}{r\mu}}$$
$$\omega_{min} = \sqrt{\frac{10}{0.1 \times 1}} = 10\ \text{rad/s}$$
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: The load is applied gradually, so the average force acting through the extension l is $\dfrac{Mg}{2}$.
Elastic potential energy stored $=$ half the work done by gravity
$$U = \frac{1}{2}Mgl$$
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: An increase in temperature leads to an increase in the kinetic energy of the gas (assuming it to be ideal), since
$$U = \frac{F}{2}nRT$$
where F is the degree of freedom.
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: Inside the conductor $(d < R)$:
$$B = \frac{\mu_0}{2\pi}\frac{i}{R^{2}}d \quad\text{i.e.}\quad B = Kd$$
so B is a straight line passing through the origin, maximum at the surface where
$$B = \frac{\mu_0}{2\pi}\frac{i}{R}$$
Outside the conductor $(d > R)$:
$$B = \frac{\mu_0}{2\pi}\frac{i}{d} \quad\text{i.e.}\quad B \propto \frac{1}{d}$$
which is hyperbolic. So 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: Fractional loss of KE of the colliding body in a head-on elastic collision is
$$\frac{\Delta KE}{KE} = \frac{4m_1m_2}{(m_1 + m_2)^{2}}$$
$$= \frac{4(4m)(2m)}{(4m + 2m)^{2}} = \frac{32m^{2}}{36m^{2}} = \frac{8}{9}$$
Q115.
Which colour of the light has the longest wavelength?
- A. Violet
- B. Red ✓
- C. Blue
- D. Green
Solution: In the visible spectrum the wavelength increases from violet to red.
So red has the longest wavelength among the given options.
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: For the difference in lengths to be independent of temperature, the two rods must expand equally:
$$\alpha_{Cu}L_{Cu} = \alpha_{Al}L_{Al}$$
$$1.7 \times 10^{-5} \times 88 = 2.2 \times 10^{-5} \times L_{Al}$$
$$L_{Al} = \frac{1.7 \times 88}{2.2} = 68\ \text{cm}$$
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: In a p-type semiconductor an intrinsic semiconductor is doped with trivalent impurities.
This creates deficiencies of valence electrons called holes, which are the majority charge carriers.
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: At $t = 0$ the y displacement is maximum, so the equation will be a cosine function.
With $T = 4$ s,
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\ \text{rad/s}$$
$$y = a\cos\omega t = 3\cos\frac{\pi}{2}t$$
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: Work done by a variable force is
$$W = \int_{y_i}^{y_f} F\,dy$$
Here $y_i = 0$ and $y_f = 1$ m, so
$$W = \int_0^1 (20 + 10y)\,dy = \left[20y + \frac{10y^{2}}{2}\right]_0^1 = 25\ \text{J}$$
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: At the lowest point,
$$T - mg = \frac{mu^{2}}{l} \Rightarrow T = mg + \frac{mu^{2}}{l}$$
The tension is maximum at the lowest position of the mass, so the chance of breaking is maximum there.
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: In one complete vibration the displacement is zero. So the average velocity in one complete vibration is
$$= \frac{\text{Displacement}}{\text{Time interval}} = \frac{y_f - y_i}{T} = 0$$
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: A rainbow cannot be observed when the observer faces towards the sun.
The observer must stand with the sun behind them and the rain in front, so statement (4) is wrong.
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: For an electron accelerated through a potential V,
$$\lambda = \frac{12.27}{\sqrt{V}}\ \text{Å} = \frac{12.27 \times 10^{-10}}{\sqrt{10000}} = 12.27 \times 10^{-12}\ \text{m}$$
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: Work required $=$ change in kinetic energy, and the final KE is zero.
For a rolling disc,
$$\text{Initial KE} = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} = \frac{3}{4}mv^{2}$$
$$= \frac{3}{4} \times 100 \times (20 \times 10^{-2})^{2} = 3\ \text{J}$$
$$|\Delta KE| = 3\ \text{J}$$
Q125.
The correct Boolean operation represented by the circuit diagram drawn is :
- A. NOR
- B. AND
- C. OR
- D. NAND ✓
Solution: From the given logic circuit the LED will glow when the voltage across the LED is high.
The LED stays lit for the input pairs (0, 0), (0, 1) and (1, 0), and goes off only when both switches A and B are closed, i.e. (1, 1).
So the truth table gives Y = 1, 1, 1, 0 for A, B = 00, 01, 10, 11, which is the output 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: $$r = \frac{p}{qB}$$
For an ionized hydrogen atom $q = e$, so $r_H = \dfrac{p}{eB}$.
For an $\alpha$-particle $q = 2e$, so $r_\alpha = \dfrac{p}{2eB}$.
$$\frac{r_H}{r_\alpha} = \frac{p/eB}{p/2eB} = \frac{2}{1}$$
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: $$F = \frac{kQ^{2}}{r^{2}}$$
If 25% of the charge of A is transferred to B, then
$$q_A = Q - \frac{Q}{4} = \frac{3Q}{4},\qquad q_B = -Q + \frac{Q}{4} = \frac{-3Q}{4}$$
$$F_1 = \frac{k q_A q_B}{r^{2}} = \frac{k\left(\frac{3Q}{4}\right)^{2}}{r^{2}} = \frac{9}{16}\frac{kQ^{2}}{r^{2}} = \frac{9F}{16}$$
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: An electric heater does not involve eddy currents. It uses Joule's heating effect in a resistance wire.
The induction furnace and magnetic braking in trains both use 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: Angle of dip is the angle of the earth's resultant magnetic field from the horizontal.
Dip is zero at the equator and positive in the northern hemisphere.
In the southern hemisphere the dip angle is considered as negative.
So A ($\delta = +25^{\circ}$) lies in the northern hemisphere and B ($\delta = -25^{\circ}$) lies 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: (i) All bulbs glowing: section A is three bulbs in parallel and so is section B, and the two sections are in series.
$$R_{eq} = \frac{R}{3} + \frac{R}{3} = \frac{2R}{3}$$
$$P_i = \frac{E^{2}}{R_{eq}} = \frac{3E^{2}}{2R}$$
(ii) Two from section A and one from section B glowing:
$$R_{eq} = \frac{R}{2} + R = \frac{3R}{2}$$
$$P_f = \frac{2E^{2}}{3R}$$
$$\frac{P_i}{P_f} = \frac{3E^{2}}{2R} \times \frac{3R}{2E^{2}} = \frac{9}{4}$$
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: By Torricelli's theorem the efflux speed is $u = \sqrt{2gh}$, so the rate of flow is
$$Q = au = a\sqrt{2gh}$$
$$= 2 \times 10^{-6}\ \text{m}^{2} \times \sqrt{2 \times 10 \times 2}\ \text{m/s}$$
$$= 2 \times 10^{-6} \times 6.28 = 12.56 \times 10^{-6}$$
$$\approx 12.6 \times 10^{-6}\ \text{m}^{3}/\text{s}$$
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: For an ideal voltmeter the resistance is infinite and for an ideal ammeter the resistance is zero.
Circuit 1: $V_1 = i_1 \times 10 = \dfrac{10}{10} \times 10 = 10$ volt
Circuit 2: $V_2 = i_2 \times 10 = \dfrac{10}{10} \times 10 = 10$ volt
So $V_1 = V_2$, and
$$i_1 = i_2 = \frac{10\ \text{V}}{10\ \Omega} = 1\ \text{A}$$
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: Initial potential energy at the earth's surface is
$$U_i = \frac{-GMm}{R}$$
Final potential energy at height $h = R$ is
$$U_f = \frac{-GMm}{2R}$$
Work done $=$ change in PE,
$$W = U_f - U_i = \frac{GMm}{2R} = \frac{gR^{2}m}{2R} = \frac{mgR}{2}$$
using $GM = gR^{2}$.
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: In an adiabatic process there is no exchange of heat between the system and its surroundings.
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: By the work-energy theorem for rotation,
$$W = \frac{1}{2}I(\omega_f^{2} - \omega_i^{2})$$
with $\theta = 2\pi$ revolutions $= 2\pi \times 2\pi = 4\pi^{2}$ rad and $\omega_i = 3 \times \dfrac{2\pi}{60}$ rad/s.
$$-\tau\theta = \frac{1}{2}\times\frac{1}{2}mr^{2}(0^{2} - \omega_i^{2})$$
$$\tau = \frac{\frac{1}{2}\times\frac{1}{2}\times 2 \times (4 \times 10^{-2})^{2}\left(3 \times \frac{2\pi}{60}\right)^{2}}{4\pi^{2}}$$
$$\tau = 2 \times 10^{-6}\ \text{N m}$$
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: Malachite is CuCO$_3\cdot$Cu(OH)$_2$, a green coloured ore of copper.
CuFeS$_2$ is copper pyrites and Fe$_3$O$_4$ is magnetite.
Q137.
Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is :
Solution: All enzymes that utilise ATP in phosphate transfer require magnesium (Mg) as the co-factor.
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: For an ideal solution:
$\Delta_{mix}$H $= 0$
$\Delta_{mix}$S $> 0$
$\Delta_{mix}$G $< 0$
$\Delta_{mix}$V $= 0$
So the correct option is $\Delta_{mix}$H $= 0$ at constant T and P.
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: In K$_4$[Fe(CN)$_6$] the oxidation state of Fe is $+2$.
Fe ground state: [Ar]3d$^{6}$4s$^{2}$, so Fe$^{2+}$ is 3d$^{6}$4s$^{0}$.
CN$^{-}$ is a strong field ligand, so pairing occurs in an octahedral field and all six d electrons occupy the lower t$_{2g}$ set.
Hence the configuration is $t_{2g}^{6}e_g^{0}$.
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: Pent-2-en-4-yne is CH$_3$–CH=CH–C$\equiv$CH.
Counting the bonds: 4 C–C skeletal bonds and 6 C–H bonds give 10 $\sigma$ bonds.
The C=C double bond contributes 1 $\pi$ bond and the C$\equiv$C triple bond contributes 2 $\pi$ bonds, so there are 3 $\pi$ bonds.
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: Anions (A) are in hcp, so the number of anions per unit cell is 6.
The number of octahedral voids equals the number of anions, i.e. 6. Cations occupy 75% of them:
$$\text{Number of cations} = 6 \times \frac{3}{4} = \frac{18}{4} = \frac{9}{2}$$
$$\text{C} : \text{A} = \frac{9}{2} : 6 = 9 : 12 = 3 : 4$$
So the formula is C$_3$A$_4$.
Q142.
The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is :
Solution: Haber's process:
$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$
2 moles of NH$_3$ require 3 moles of H$_2$.
Hence 20 moles of NH$_3$ require $\dfrac{3 \times 20}{2} = 30$ moles of H$_2$.
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: PbF$_4$ and SnF$_4$ are both ionic in nature, because fluorine is small and highly electronegative.
So the statement that PbF$_4$ is covalent is incorrect.
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)$_2$ is amphoteric in nature, since it can react both with acid and base:
Be(OH)$_2$ + 2HCl $\longrightarrow$ BeCl$_2$ + 2H$_2$O
Be(OH)$_2$ + 2NaOH $\longrightarrow$ Na$_2$[Be(OH)$_4$]
The hydroxides of the other alkaline earth metals listed 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: In manganate (MnO$_4^{2-}$) and permanganate (MnO$_4^{-}$), the $\pi$-bonds are of the d$\pi$–p$\pi$ type.
That is, the $\pi$-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese.
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: Ca(OH)$_2$ $\rightleftharpoons$ Ca$^{2+}$ + 2OH$^{-}$
pH $= 9$, hence pOH $= 14 - 9 = 5$, so [OH$^{-}$] $= 10^{-5}$ M.
Since two OH$^{-}$ come from each formula unit, [Ca$^{2+}$] $= \dfrac{10^{-5}}{2}$.
$$K_{sp} = [\text{Ca}^{2+}][\text{OH}^{-}]^{2} = \left(\frac{10^{-5}}{2}\right)(10^{-5})^{2} = 0.5 \times 10^{-15}$$
Q147.
The mixture that forms maximum boiling azeotrope is:
- A. Heptane + Octane
- B. Water + Nitric acid ✓
- C. Ethanol + Water
- D. Acetone + Carbon disulphide
Solution: Solutions showing negative deviation from Raoult's law form a maximum boiling azeotrope.
Water and nitric acid show negative deviation and form a maximum boiling azeotrope.
Ethanol + water and acetone + carbon disulphide show positive deviation, 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: XeF$_4$ has two lone pairs on Xe and is square planar.
XeF$_6$ has one lone pair and is distorted octahedral.
XeOF$_4$ has one lone pair and is square pyramidal.
XeO$_3$ has one lone pair and is pyramidal.
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: In (a) Cu is in $+1$ and goes to $+2$ and $0$, so it is a disproportionation.
In (b) Mn is in $+6$ and goes to $+7$ and $+4$, so it is a disproportionation.
In (c) Mn goes from $+7$ to $+6$ and $+4$ but oxygen is also oxidised from $-2$ to 0, so it is not a simple disproportionation of one species.
In (d) Mn goes from $+7$ and $+2$ to a single state $+4$, which is comproportionation, not disproportionation.
Hence only (a) and (b) are disproportionation reactions.
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: A conjugate base is what remains after an acid loses an H$^{+}$ ion.
H$_2$O on loss of H$^{+}$ gives OH$^{-}$, its conjugate base (gaining H$^{+}$ would give H$_3$O$^{+}$, its conjugate acid).
HF on loss of H$^{+}$ becomes F$^{-}$, the conjugate base of HF.
For example: HF + H$_2$O $\rightleftharpoons$ F$^{-}$ + H$_3$O$^{+}$
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: Chlorination of benzene in the presence of anhydrous AlCl$_3$ is an electrophilic substitution.
Generation of the electrophile:
Cl–Cl + AlCl$_3$ $\longrightarrow$ Cl$^{+}$ + AlCl$_4^{-}$
The electrophile Cl$^{+}$ attacks the $\pi$ cloud of benzene to give the arenium ion, and loss of H$^{+}$ from the sp$^{3}$ carbon restores aromaticity, giving chlorobenzene.
The UV-light reaction is a free radical addition, while the other two are not substitutions on the ring.
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: Ozonolysis giving propanone and ethanal in equimolar ratio means the alkene A is
(CH$_3$)$_2$C=CH–CH$_3$, i.e. 2-methylbut-2-ene.
Addition of HCl follows Markovnikov's rule, so H adds to the carbon bearing more hydrogens and Cl to the more substituted carbon, giving the more stable tertiary carbocation.
So B is (CH$_3$)$_2$CCl–CH$_2$–CH$_3$, i.e. 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: $$Z = \frac{V_{real}}{V_{ideal}}$$
Since $V_{real} < V_{ideal}$, we get $Z < 1$.
If $Z < 1$, attractive forces are dominant among gaseous molecules and liquefaction of the gas will be easy.
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: SO$_2$(g) is not a greenhouse gas.
Methane, nitrous oxide and ozone all act as 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: H$_2$O($\ell$) $\rightleftharpoons$ H$_2$O(v), $\Delta$S > 0
Expansion of gas at constant temperature, $\Delta$S > 0
Sublimation of solid to gas, $\Delta$S > 0
2H(g) $\longrightarrow$ H$_2$(g), $\Delta$S < 0, because $\Delta n_g < 0$ and randomness decreases.
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: $$W_{irr} = -P_{ext}\Delta V$$
$$= -2\ \text{bar} \times (0.25 - 0.1)\ \text{L} = -2 \times 0.15\ \text{L bar}$$
$$= -0.30\ \text{L bar} = -0.30 \times 100\ \text{J} = -30\ \text{J}$$
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: Due to the presence of d-orbitals in Si, Ge and Sn they can form species like [SiF$_6$]$^{2-}$, [GeCl$_6$]$^{2-}$ and [Sn(OH)$_6$]$^{2-}$.
[SiCl$_6$]$^{2-}$ does not exist, because six large chloride ions cannot be accommodated around Si$^{4+}$ due to the limitation of its size.
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: $$E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n}\log Q$$
At equilibrium $Q = K_{eq}$ and $E_{cell} = 0$, with $n = 1$:
$$0 = E^{\circ}_{cell} - \frac{0.059}{1}\log K_{eq}$$
$$\log K_{eq} = \frac{E^{\circ}_{cell}}{0.059} = \frac{0.59}{0.059} = 10$$
$$K_{eq} = 10^{10} = 1.0 \times 10^{10}$$
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: Temporary hardness of water is removed by Clark's method, in which a calculated amount of lime is added to the hard water.
This precipitates the bicarbonates of calcium and magnesium as insoluble carbonates.
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: Option (1): 10 mmol HCl and 10 mmol NaOH react completely to give a neutral solution, so no buffer.
Option (2): 2.5 mmol CH$_3$COOH and 5 mmol NaOH leave 2.5 mmol excess NaOH, so the solution is basic due to a strong base, not a basic buffer.
Option (3): 10 mmol CH$_3$COOH and 10 mmol NaOH give only the salt CH$_3$COONa, which hydrolyses; this is not a basic buffer.
Option (4): 10 mmol HCl and 20 mmol NH$_4$OH leave 10 mmol NH$_4$OH together with 10 mmol NH$_4$Cl, i.e. a weak base with its salt — this is a basic 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: Partial hydrogenation of an alkyne over Lindlar's catalyst (H$_2$, Pd/C poisoned with quinoline) gives syn addition of hydrogen.
So but-2-yne gives cis-2-butene.
Na/liquid NH$_3$ would give the trans (anti addition) product.
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: In phenol the lone pair of electrons on oxygen is involved in resonance with the ring.
Hence the oxygen carries a partial positive charge, and the incoming proton will not be able to attack easily.
So phenol is the most difficult to protonate.
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: On going down the group the thermal stability order for H$_2$E decreases, because the H–E bond energy decreases as the size of E increases.
Therefore the order of stability is
H$_2$Po < H$_2$Te < H$_2$Se < H$_2$S < H$_2$O
Q164.
The correct structure of tribromooctaoxide is
- A. Structure (1)
- B. Structure (2) ✓
- C. Structure (3)
- D. Structure (4)
Solution: The correct structure of tribromooctaoxide, Br$_3$O$_8$, is a chain of three bromine atoms
O=Br–Br–Br=O
with the central bromine carrying two doubly bonded oxygens and each terminal bromine carrying two doubly bonded oxygens in addition to its terminal Br=O.
There are no formal negative charges on any of the oxygen atoms.
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: Phthalic acid first reacts with NH$_3$ to give the diammonium salt.
On heating, loss of 2H$_2$O gives the diamide, benzene-1,2-dicarboxamide.
On strong heating, loss of NH$_3$ from the diamide closes the ring 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: Pure nitrogen is obtained by thermal decomposition of sodium azide or barium azide.
Haber process gives ammonia.
Contact process gives sulphuric acid.
Deacon's process gives chlorine.
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: The rate of reaction is given as
$$-\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt} = +\frac{1}{2}\frac{d[\text{NH}_3]}{dt}$$
So the correct relation among the options is
$$-\frac{d[\text{N}_2]}{dt} = \frac{1}{2}\frac{d[\text{NH}_3]}{dt}$$
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: Cumene on oxidation with air gives cumene hydroperoxide, C$_6$H$_5$–C(CH$_3$)$_2$–O–O–H, which is the intermediate A.
Cumene hydroperoxide on treatment 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 is a non-essential amino acid, i.e. it can be synthesised in the body.
Lysine, valine and leucine are essential amino acids that must be supplied through 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: MO configuration of C$_2$ is
$\sigma 1s^{2}$, $\sigma^{*}1s^{2}$, $\sigma 2s^{2}$, $\sigma^{*}2s^{2}$, $\pi 2p_x^{2} = \pi 2p_y^{2}$
The bonding electrons that remain after cancelling the $\sigma$ and $\sigma^{*}$ pairs are only in the $\pi$ orbitals, so C$_2$ has only $\pi$ bonds.
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: In aqueous solution the electron donating inductive effect, the solvation effect (H-bonding) and steric hindrance all together affect the basic strength of substituted amines.
Hence the basic character is
(CH$_3$)$_2$NH (2$^{\circ}$) > CH$_3$NH$_2$ (1$^{\circ}$) > (CH$_3$)$_3$N (3$^{\circ}$)
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: The charge present on a colloid is due to adsorption of the common ion from the dispersion medium.
AgNO$_3$ + KI $\longrightarrow$ AgI + KNO$_3$
For the sol to be negatively charged, KI must be in excess so that I$^{-}$ is adsorbed on the AgI formed.
In option (3), AgNO$_3$ is $50 \times 1 = 50$ mmol and KI is $50 \times 2 = 100$ mmol, so KI is in largest excess and I$^{-}$ is adsorbed, giving the negatively charged [AgI]I$^{-}$ sol.
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: PCl$_5$ has a trigonal bipyramidal shape: the three equatorial bonds are 120$^{\circ}$ apart, the two axial bonds are 180$^{\circ}$ apart, and the axial bonds (240 pm) are longer than the equatorial bonds (202 pm).
Due to the longer and hence weaker axial bonds, PCl$_5$ is a REACTIVE molecule, so statement (1) is incorrect.
Q174.
Among the following, the narrow spectrum antibiotic is :
- A. Chloramphenicol
- B. Penicillin G ✓
- C. Ampicillin
- D. Amoxycillin
Solution: Penicillin G is a narrow spectrum antibiotic.
Chloramphenicol is broad spectrum, while ampicillin and amoxycillin are broad spectrum antibiotics.
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: First order rate constant is given as
$$k = \frac{2.303}{t}\log\frac{[A_0]}{[A]_t}$$
For 99% completion, $[A_0] = 100$ and $[A]_t = 1$:
$$k = \frac{2.303}{t}\log\frac{100}{1} = \frac{2.303}{t}\times 2\log 10$$
$$t = \frac{2.303}{k}\times 2 = \frac{4.606}{k}$$
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: 'Be' and 'N' have comparatively more stable valence sub-shells than 'B' and 'O', so their ionisation enthalpies are higher than expected from the general trend.
Hence the correct order of first ionisation enthalpy is
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: $(n + l)$ values are:
4d $= 4 + 2 = 6$
5p $= 5 + 1 = 6$
5f $= 5 + 3 = 8$
6p $= 6 + 1 = 7$
For equal $(n + l)$, the orbital with higher n has higher energy, so 5p > 4d.
Hence the correct order of decreasing energy is 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: $$\Delta G^{\ominus} = -nFE^{\ominus}_{cell}$$
Here $n = 2$, so
$$\Delta G^{\ominus} = -2 \times 96500 \times 0.24\ \text{J mol}^{-1} = -46320\ \text{J mol}^{-1}$$
$$= -46.32\ \text{kJ mol}^{-1}$$
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: In the hydrogen spectrum, the Balmer series transitions (those ending at n = 2) fall in the visible region.
Lyman lies in the ultraviolet, while Paschen and Brackett lie 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 is a biodegradable polymer.
It is an alternating polyamide copolymer of glycine and aminocaproic acid.