Q1.
Which of the following refer to correct example(s) of organisms which have evolved due to changes in environment brought about by anthropogenic action?
(a) Darwin's Finches of Galapagos islands.
(b) Herbicide resistant weeds.
(c) Drug resistant eukaryotes.
(d) Man-created breeds of domesticated animals like dogs.
- A. (a) and (c)
- B. (b), (c) and (d) ✓
- C. only (d)
- D. only (a)
Solution: Herbicide resistant weeds, drug resistant eukaryotes and man-created breeds of domesticated animals like dogs are examples of evolution by anthropogenic action.
Darwin's Finches of Galapagos islands are an example of natural selection, adaptive radiation and founder's effect, not of human interference.
Q2.
Meiotic division of the secondary oocyte is completed
- A. At the time of copulation
- B. After zygote formation
- C. At the time of fusion of a sperm with an ovum ✓
- D. Prior to ovulation
Solution: Meiotic division of the secondary oocyte is completed after the entry of the sperm into the secondary oocyte, which leads to the formation of a large ovum and a tiny II$^{\text{nd}}$ polar body.
Q3.
Which of the following is correct about viroids?
- A. They have free RNA without protein coat ✓
- B. They have DNA with protein coat
- C. They have free DNA without protein coat
- D. They have RNA with protein coat
Solution: Viroids have free RNA without a protein coat.
T.O. Diener discovered a new infectious agent that was smaller than viruses and caused potato spindle tuber disease. It was found to be a free RNA; it lacked the protein coat that is found in viruses, hence the name viroid.
Q4.
The plant parts which consist of two generations - one within the other
(a) Pollen grains inside the anther
(b) Germinated pollen grain with two male gametes
(c) Seed inside the fruit
(d) Embryo sac inside the ovule
- A. (a), (b) and (c)
- B. (c) and (d)
- C. (a) and (d) ✓
- D. (a) only
Solution: The plant parts which consist of two generations one within the other are pollen grains inside the anther and embryo sac inside the ovule.
Pollen grain is haploid inside the diploid anther.
Embryo sac is haploid inside the diploid ovule.
Q5.
Experimental verification of the chromosomal theory of inheritance was done by
- A. Sutton
- B. Boveri
- C. Morgan ✓
- D. Mendel
Solution: Experimental verification of the chromosomal theory of inheritance was done by Morgan.
Sutton and Boveri proposed the chromosomal theory of inheritance, but it was experimentally verified by T.H. Morgan.
Q6.
Which of the following pairs is of unicellular algae?
- A. $Gelidium$ and $Gracilaria$
- B. $Anabaena$ and $Volvox$
- C. $Chlorella$ and $Spirulina$ ✓
- D. $Laminaria$ and $Sargassum$
Solution: $Chlorella$ and $Spirulina$ are unicellular algae.
$Gelidium$, $Gracilaria$, $Laminaria$ and $Sargassum$ are multicellular. $Volvox$ is colonial.
Q7.
Secondary metabolites such as nicotine, strychnine and caffeine are produced by plants for their
- A. Growth response
- B. Defence action ✓
- C. Effect on reproduction
- D. Nutritive value
Solution: A wide variety of chemical substances that we extract from plants on a commercial scale — nicotine, caffeine, quinine, strychnine, opium, etc. — are produced by them as defences against grazers and browsers.
Q8.
By which method was a new breed 'Hisardale' of sheep formed by using Bikaneri ewes and Marino rams?
- A. Mutational breeding
- B. Cross breeding ✓
- C. Inbreeding
- D. Out crossing
Solution: Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams.
In cross-breeding, superior males of one breed are mated with superior females of another breed.
Q9.
The infectious stage of $Plasmodium$ that enters the human body is
- A. Sporozoites ✓
- B. Female gametocytes
- C. Male gametocytes
- D. Trophozoites
Solution: $Plasmodium$ enters the human body as sporozoites (the infectious stage) through the bite of an infected female $Anopheles$ mosquito.
Q10.
The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is
- A. Root pressure ✓
- B. Imbibition
- C. Plasmolysis
- D. Transpiration
Solution: Root pressure is a positive hydrostatic pressure. It develops in tracheary elements at night and in early morning.
This pressure forces water out of the special openings at the tip of grass blades, and the loss of water in liquid form is called guttation.
Q11.
From his experiments, S.L. Miller produced amino acids by mixing the following in a closed flask
- A. CH$_3$, H$_2$, NH$_4$ and water vapor at 800$^{\circ}$C
- B. CH$_4$, H$_2$, NH$_3$ and water vapor at 600$^{\circ}$C
- C. CH$_3$, H$_2$, NH$_3$ and water vapor at 600$^{\circ}$C
- D. CH$_4$, H$_2$, NH$_3$ and water vapor at 800$^{\circ}$C ✓
Solution: In 1953, S.L. Miller, an American scientist, created an electric discharge in a closed flask containing CH$_4$, H$_2$, NH$_3$ and water vapor at 800$^{\circ}$C.
After running the discharge for a week he observed the formation of amino acids.
Q12.
In relation to Gross primary productivity and Net primary productivity of an ecosystem, which one of the following statements is correct?
- A. Gross primary productivity is always more than net primary productivity ✓
- B. Gross primary productivity and Net primary productivity are one and same
- C. There is no relationship between Gross primary productivity and Net primary productivity
- D. Gross primary productivity is always less than net primary productivity
Solution: Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis.
Net primary productivity is GPP minus respiration.
Hence gross primary productivity is always more than NPP.
Q13.
The sequence that controls the copy number of the linked DNA in the vector, is termed
- A. Ori site ✓
- B. Palindromic sequence
- C. Recognition site
- D. Selectable marker
Solution: The $Ori$ sequence is responsible for controlling the copy number of the linked DNA in the vector.
$Ori$, i.e. origin of replication, is responsible for initiation of replication.
Q14.
Cuboidal epithelium with brush border of microvilli is found in
- A. Ducts of salivary gland
- B. Proximal convoluted tubule of nephron ✓
- C. Eustachian tube
- D. Lining of intestine
Solution: Cuboidal epithelium with brush border of microvilli is found in the proximal convoluted tubule of the nephron (PCT).
The microvilli greatly increase the surface area available for reabsorption.
Q15.
The body of the ovule is fused within the funicle at
- A. Micropyle
- B. Nucellus
- C. Chalaza
- D. Hilum ✓
Solution: The attachment point of the funicle and the body of the ovule is known as the hilum.
Thus, hilum represents the junction between ovule and funicle.
Q16.
In light reaction, plastoquinone facilitates the transfer of electrons from
- A. Cytb$_6$f complex to PS-I
- B. PS-I to NADP$^{+}$
- C. PS-I to ATP synthase
- D. PS-II to Cytb$_6$f complex ✓
Solution: After excitement, e$^{-}$ is passed from PS-II (P$_{680}$) to the primary electron acceptor (pheophytin). From the primary e$^{-}$ acceptor, e$^{-}$ is passed to plastoquinone.
Plastoquinone (PQ) in turn transfers its e$^{-}$ to the Cyt b$_6$f complex.
Therefore plastoquinone facilitates the transfer of electrons from PS-II to the Cyt b$_6$f complex.
Q17.
Match the following diseases with the causative organism and select the correct option.
Column-I:
(a) Typhoid
(b) Pneumonia
(c) Filariasis
(d) Malaria
Column-II:
(i) $Wuchereria$
(ii) $Plasmodium$
(iii) $Salmonella$
(iv) $Haemophilus$
- A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) ✓
- B. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
- C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
- D. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
Solution: Typhoid fever in humans is caused by the pathogenic bacterium $Salmonella$ $typhi$.
Pneumonia is caused by $Streptococcus$ $pneumoniae$ and $Haemophilus$ $influenzae$.
Filariasis or elephantiasis is caused by the filarial worms $Wuchereria$ $bancrofti$ and $Wuchereria$ $malayi$.
Malaria is caused by different species of $Plasmodium$.
So (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii).
Q18.
Match the following columns and select the correct option.
Column-I:
(a) $Clostridium$ $butylicum$
(b) $Trichoderma$ $polysporum$
(c) $Monascus$ $purpureus$
(d) $Aspergillus$ $niger$
Column-II:
(i) Cyclosporin-A
(ii) Butyric Acid
(iii) Citric Acid
(iv) Blood cholesterol lowering agent
- A. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) ✓
- B. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
- C. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
- D. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Solution: $Clostridium$ $butylicum$ produces butyric acid.
$Trichoderma$ $polysporum$ produces Cyclosporin-A.
$Monascus$ $purpureus$ produces statins, used as a blood cholesterol lowering agent.
$Aspergillus$ $niger$ produces citric acid.
So (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii).
Q19.
Which of the following statements are true for the phylum-Chordata?
(a) In Urochordata notochord extends from head to tail and it is present throughout their life.
(b) In Vertebrata notochord is present during the embryonic period only.
(c) Central nervous system is dorsal and hollow.
(d) Chordata is divided into 3 subphyla : Hemichordata, Tunicata and Cephalochordata.
- A. (c) and (a)
- B. (a) and (b)
- C. (b) and (c) ✓
- D. (d) and (c)
Solution: In Vertebrata, notochord is present during the embryonic period only, as it is replaced by the vertebral column.
In chordates, the central nervous system is dorsal and hollow.
In Urochordata the notochord is present only in the larval tail, not throughout life, so (a) is wrong.
Chordata is divided into three subphyla — Urochordata (Tunicata), Cephalochordata and Vertebrata; Hemichordata is a separate phylum, so (d) is wrong.
Q20.
Goblet cells of alimentary canal are modified from
- A. Columnar epithelial cells ✓
- B. Chondrocytes
- C. Compound epithelial cells
- D. Squamous epithelial cells
Solution: Goblet cells of the alimentary canal are modified from columnar epithelial cells, which secrete mucus.
They are unicellular glands formed within the epithelium itself.
Q21.
Which of the following is not an inhibitory substance governing seed dormancy?
- A. Abscisic acid
- B. Phenolic acid
- C. Para-ascorbic acid
- D. Gibberellic acid ✓
Solution: Gibberellic acid breaks seed dormancy rather than imposing it.
It activates synthesis of $\alpha$-amylase, which breaks down starch into simple sugar.
Abscisic acid, phenolic acid and para-ascorbic acid are inhibitory substances governing seed dormancy.
Q22.
Name the enzyme that facilitates opening of DNA helix during transcription.
- A. DNA helicase
- B. DNA polymerase
- C. RNA polymerase ✓
- D. DNA ligase
Solution: RNA polymerase facilitates opening of the DNA helix during transcription.
It binds to the promoter, unwinds a short stretch of the duplex and continues polymerisation as it moves along the template.
Q23.
Match the following
Column-I:
(a) Inhibitor of catalytic activity
(b) Possess peptide bonds
(c) Cell wall material in fungi
(d) Secondary metabolite
Column-II:
(i) Ricin
(ii) Malonate
(iii) Chitin
(iv) Collagen
Choose the correct option from the following
- A. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
- B. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- C. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
- D. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) ✓
Solution: Malonate is the competitive inhibitor of the catalytic activity of succinic dehydrogenase, so (a) matches with (ii).
Collagen is proteinaceous in nature and possesses peptide bonds, so (b) matches with (iv).
Chitin is a homopolymer present in the cell wall of fungi and the exoskeleton of arthropods, so (c) matches with (iii).
Abrin and ricin are toxins, secondary metabolites, so (d) matches with (i).
Q24.
Bilaterally symmetrical and acoelomate animals are exemplified by
- A. Platyhelminthes ✓
- B. Aschelminthes
- C. Annelida
- D. Ctenophora
Solution: Platyhelminthes are bilaterally symmetrical, triploblastic and acoelomate animals with organ level of organisation.
Aschelminthes are pseudocoelomate, Annelida are coelomate, and Ctenophora are radially symmetrical.
Q25.
Presence of which of the following conditions in urine are indicative of Diabetes Mellitus?
- A. Uremia and Renal Calculi
- B. Ketonuria and Glycosuria ✓
- C. Renal calculi and Hyperglycaemia
- D. Uremia and Ketonuria
Solution: Presence of ketone bodies in urine (ketonuria) and presence of glucose in urine (glycosuria) are indicative of Diabetes mellitus.
Q26.
Ray florets have
- A. Superior ovary
- B. Hypogynous ovary
- C. Half inferior ovary
- D. Inferior ovary ✓
Solution: Ray florets have an inferior ovary.
Epigynous flowers are found in family Asteraceae, e.g. sunflower, and in an epigynous flower the ovary is inferior.
Q27.
Identify the substances having glycosidic bond and peptide bond, respectively in their structure
- A. Glycerol, trypsin
- B. Cellulose, lecithin
- C. Inulin, insulin ✓
- D. Chitin, cholesterol
Solution: Inulin is a fructan (polysaccharide of fructose). Adjacent fructose units are linked through a glycosidic bond.
Insulin is a protein composed of 51 amino acids. Adjacent amino acids are attached through a peptide bond.
Q28.
Which of the following statements is not correct?
- A. The proinsulin has an extra peptide called C-peptide.
- B. The functional insulin has A and B chains linked together by hydrogen bonds. ✓
- C. Genetically engineered insulin is produced in $E.$ $Coli$.
- D. In man insulin is synthesised as a proinsulin
Solution: Functional insulin has A and B chains linked together by DISULPHIDE bridges, not hydrogen bonds — so that statement is not correct.
The other three statements are correct.
Q29.
Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (G$_0$). This process occurs at the end of
- A. G$_1$ phase ✓
- B. S phase
- C. G$_2$ phase
- D. M phase
Solution: Some dividing cells exit the cell cycle and enter the vegetative inactive stage, called quiescent stage (G$_0$).
This process occurs when the cell is unable to proceed past G$_1$ phase.
Q30.
Identify the correct statement with regard to G$_1$ phase (Gap 1) of interphase.
- A. Reorganisation of all cell components takes place.
- B. Cell is metabolically active, grows but does not replicate its DNA. ✓
- C. Nuclear Division takes place.
- D. DNA synthesis or replication takes place.
Solution: During G$_1$ phase the cell is metabolically active and continuously grows, but does not replicate its DNA.
DNA synthesis takes place in S phase. Nuclear division occurs during karyokinesis.
Reorganisation of all cell components takes place in M-phase.
Q31.
The QRS complex in a standard ECG represents
- A. Depolarisation of auricles
- B. Depolarisation of ventricles ✓
- C. Repolarisation of ventricles
- D. Repolarisation of auricles
Solution: QRS complex represents the depolarisation of the ventricles, which initiates the ventricular contraction.
Q32.
If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is $6.6 \times 10^{9}$ bp, then the length of the DNA is approximately
- A. 2.5 meters
- B. 2.2 meters ✓
- C. 2.7 meters
- D. 2.0 meters
Solution: Distance between 2 base pairs in DNA helix $= 0.34$ nm $= 0.34 \times 10^{-9}$ m
Total number of base pairs $= 6.6 \times 10^{9}$ bp
Length of DNA $= \left[0.34 \times 10^{-9}\right]\ \text{m} \times 6.6 \times 10^{9}\ \text{bp}$
$$= 2.2\ \text{m}$$
Q33.
Which of the following regions of the globe exhibits highest species diversity?
- A. Madagascar
- B. Himalayas
- C. Amazon forests ✓
- D. Western Ghats of India
Solution: The largely tropical Amazonian rain forest in South America has the greatest biodiversity on earth.
Q34.
Which of the following is put into Anaerobic sludge digester for further sewage treatment?
- A. Floating debris
- B. Effluents of primary treatment
- C. Activated sludge ✓
- D. Primary sludge
Solution: The sediment in the settlement tank is called activated sludge.
A small part of the activated sludge is pumped back into the aeration tank.
The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters.
Q35.
Dissolution of the synaptonemal complex occurs during
- A. Zygotene
- B. Diplotene ✓
- C. Leptotene
- D. Pachytene
Solution: Dissolution of the synaptonemal complex occurs during the diplotene stage of Prophase-I of Meiosis-I.
Q36.
Select the option including all sexually transmitted diseases.
- A. Gonorrhoea, Malaria, Genital herpes
- B. AIDS, Malaria, Filaria
- C. Cancer, AIDS, Syphilis
- D. Gonorrhoea, Syphilis, Genital herpes ✓
Solution: Gonorrhoea, Syphilis and Genital herpes are sexually transmitted diseases.
Gonorrhoea is caused by a bacterium $Neisseria$ $gonorrhoeae$.
Syphilis is caused by a bacterium $Treponema$ $pallidum$.
Genital herpes is caused by a virus, Type-II Herpes simplex virus.
Malaria, filaria and cancer are not sexually transmitted.
Q37.
Select the correct statement.
- A. Glucagon is associated with hypoglycemia.
- B. Insulin acts on pancreatic cells and adipocytes.
- C. Insulin is associated with hyperglycemia.
- D. Glucocorticoids stimulate gluconeogenesis. ✓
Solution: Glucagon is associated with hyperglycemia.
Insulin acts on hepatocytes and adipocytes and is associated with hypoglycemia.
Glucocorticoids stimulate gluconeogenesis, so they increase blood sugar level — that is the correct statement.
Q38.
The product(s) of reaction catalyzed by nitrogenase in root nodules of leguminous plants is/are
- A. Nitrate alone
- B. Ammonia and oxygen
- C. Ammonia and hydrogen ✓
- D. Ammonia alone
Solution: The nitrogenase-catalysed reaction is
$$\text{N}_2 + 8\text{e}^- + 8\text{H}^+ + 16\text{ATP} \xrightarrow{\ \text{Mg}^{++}\ } 2\text{NH}_3 + \text{H}_2 + 16\text{ADP} + 16\text{Pi}$$
So the products are ammonia and hydrogen.
Q39.
In gel electrophoresis, separated DNA fragments can be visualized with the help of
- A. Ethidium bromide in UV radiation ✓
- B. Acetocarmine in UV radiation
- C. Ethidium bromide in infrared radiation
- D. Acetocarmine in bright blue light
Solution: The separated DNA fragments can be visualised only after staining the DNA with ethidium bromide followed by exposure to UV radiation.
Q40.
In which of the following techniques, the embryos are transferred to assist those females who cannot conceive?
- A. GIFT and ZIFT
- B. ICSI and ZIFT
- C. GIFT and ICSI
- D. ZIFT and IUT ✓
Solution: Option (4) is the answer because the ARTs in which embryos are transferred include ZIFT and IUT, i.e. Zygote Intrafallopian Transfer and Intra Uterine Transfer respectively; both are embryo transfer (ET) methods.
Options (1), (2) and (3) are incorrect because in GIFT (Gamete Intrafallopian Transfer) a gamete is transferred into the fallopian tube of a female who cannot produce ova. ICSI is Intra Cytoplasmic Sperm Injection, in which a sperm is directly injected into the ovum.
Q41.
Select the correct match
- A. Phenylketonuria – Autosomal dominant trait
- B. Sickle cell anaemia – Autosomal recessive trait, chromosome-11 ✓
- C. Thalassemia – X linked
- D. Haemophilia – Y linked
Solution: Phenylketonuria – autosomal recessive disorder.
Thalassemia – autosomal recessive disorder.
Haemophilia – X linked recessive disorder.
Sickle cell anaemia – autosomal recessive trait, caused due to mutation in a gene present on chromosome no. 11. This is the correct match.
Q42.
Which of the following is not an attribute of a population?
- A. Natality
- B. Mortality
- C. Species interaction ✓
- D. Sex ratio
Solution: Natality – population attribute.
Mortality – population attribute.
Sex ratio – population attribute.
Species interaction is a population interaction, not an attribute of a population.
Q43.
The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of
- A. 1 molecule of 3-C compound ✓
- B. 1 molecule of 6-C compound
- C. 1 molecule of 4-C compound and 1 molecule of 2-C compound
- D. 2 molecules of 3-C compound
Solution: In photorespiration, O$_2$ binds to RuBisCo. As a result RuBP, instead of being converted to 2 molecules of PGA, binds with O$_2$ to form one molecule each of phosphoglycerate (3 carbon compound) and phosphoglycolate (2 carbon compound).
So only one molecule of a 3-C compound is formed.
Q44.
Match the following concerning essential elements and their functions in plants
(a) Iron
(b) Zinc
(c) Boron
(d) Manganese
(i) Photolysis of water
(ii) Pollen germination
(iii) Required for chlorophyll biosynthesis
(iv) IAA biosynthesis
Select the correct option
- A. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
- B. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) ✓
- C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
- D. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Solution: (a) Iron – essential for the formation of chlorophyll, so (iii).
(b) Zinc – needed for synthesis of auxin (IAA), so (iv).
(c) Boron – has a role in pollen grain germination, so (ii).
(d) Manganese – is involved in the splitting of water to liberate O$_2$ during photosynthesis, so (i).
Q45.
Which is the important site of formation of glycoproteins and glycolipids in eukaryotic cells?
- A. Peroxisomes
- B. Golgi bodies ✓
- C. Polysomes
- D. Endoplasmic reticulum
Solution: Golgi apparatus is the important site of formation of glycoproteins and glycolipids in eukaryotic cells.
Carbohydrates are added to proteins and lipids in the Golgi cisternae before the finished products are packaged and dispatched.
Q46.
Select the correct events that occur during inspiration.
(a) Contraction of diaphragm
(b) Contraction of external inter-costal muscles
(c) Pulmonary volume decreases
(d) Intra pulmonary pressure increases
- A. (c) and (d)
- B. (a), (b) and (d)
- C. only (d)
- D. (a) and (b) ✓
Solution: Inspiration is initiated by the contraction of the diaphragm, which increases the volume of the thoracic chamber in the antero-posterior axis.
The contraction of the external intercostal muscles increases the volume of the thoracic chamber in the dorsoventral axis.
Both raise the pulmonary volume and LOWER the intrapulmonary pressure, so (c) and (d) are wrong.
Q47.
The roots that originate from the base of the stem are
- A. Primary roots
- B. Prop roots
- C. Lateral roots
- D. Fibrous roots ✓
Solution: The roots that originate from the base of the stem are fibrous roots.
Q48.
The ovary is half inferior in :
- A. Mustard
- B. Sunflower
- C. Plum ✓
- D. Brinjal
Solution: The ovary is half inferior in Plum, i.e. the flower is perigynous.
Mustard and brinjal are hypogynous (superior ovary); sunflower is epigynous (inferior ovary).
Q49.
Match the following columns and select the correct option.
Column-I:
(a) Floating Ribs
(b) Acromion
(c) Scapula
(d) Glenoid cavity
Column-II:
(i) Located between second and seventh ribs
(ii) Head of the Humerus
(iii) Clavicle
(iv) Do not connect with the sternum
- A. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
- B. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
- C. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) ✓
- D. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Solution: (a) The 11$^{\text{th}}$ and 12$^{\text{th}}$ pairs of ribs are not connected ventrally and are therefore called floating ribs.
(b) Acromion is a flat expanded process of the spine of the scapula. The lateral end of the clavicle articulates with the acromion process.
(c) Scapula is a flat triangular bone in the dorsal part of the thorax between the 2$^{\text{nd}}$ and the 7$^{\text{th}}$ rib.
(d) Glenoid cavity of the scapula articulates with the head of the humerus to form the shoulder joint.
Q50.
If the head of cockroach is removed, it may live for few days because
- A. the cockroach does not have nervous system.
- B. the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body. ✓
- C. the head holds a 1/3$^{\text{rd}}$ of a nervous system while the rest is situated along the dorsal part of its body.
- D. the supra-oesophageal ganglia of the cockroach are situated in ventral part of abdomen.
Solution: The head holds a small proportion of the nervous system while the rest is situated along the ventral part of the body.
Hence a decapitated cockroach can still survive for a few days.
Q51.
Identify the incorrect statement.
- A. Sapwood is involved in conduction of water and minerals from root to leaf
- B. Sapwood is the innermost secondary xylem and is lighter in colour ✓
- C. Due to deposition of tannins, resins, oils etc., heart wood is dark in colour
- D. Heart wood does not conduct water but gives mechanical support
Solution: Incorrect statement: Sapwood is the innermost secondary xylem and is lighter in colour.
Correct statement: Sapwood is the OUTERMOST secondary xylem.
The other three statements are correct.
Q52.
Bt cotton variety that was developed by the introduction of toxin gene of $Bacillus$ $thuringiensis$ (Bt) is resistant to
- A. Fungal diseases
- B. Plant nematodes
- C. Insect predators
- D. Insect pests ✓
Solution: Bt cotton is resistant to cotton bollworm, which is an insect pest.
$cry$ I Ac and $cry$ II Ab genes have been introduced in cotton to protect it from cotton bollworm. This makes Bt cotton a biopesticide.
Q53.
The number of substrate level phosphorylations in one turn of citric acid cycle is
- A. One ✓
- B. Two
- C. Three
- D. Zero
Solution: There is one substrate level phosphorylation in one turn of the citric acid cycle, in the reaction
Succinyl Co-A $\xrightarrow{\text{Succinate thiokinase}}$ Succinate
where GDP is converted to GTP, and GTP then transfers its phosphate to ADP to give ATP.
Q54.
Identify the wrong statement with regard to Restriction Enzymes.
- A. They cut the strand of DNA at palindromic sites.
- B. They are useful in genetic engineering.
- C. Sticky ends can be joined by using DNA ligases. ✓
- D. Each restriction enzyme functions by inspecting the length of a DNA sequence.
Solution: Restriction endonucleases make cuts at specific positions within the DNA. They function by inspecting the length of a DNA sequence, bind to the DNA and cut the two strands of the double helix at specific points in their sugar-phosphate backbones. They are used in genetic engineering to form recombinant DNA molecules.
Joining of the sticky ends is done by DNA ligases, not by restriction enzymes — so statement (3) does not describe a restriction enzyme.
Q55.
Flippers of Penguins and Dolphins are examples of
- A. Convergent evolution ✓
- B. Industrial melanism
- C. Natural selection
- D. Adaptive radiation
Solution: Flippers of Penguins and Dolphins are an example of analogous organs.
Analogous structures are a result of convergent evolution.
Q56.
Identify the wrong statement with reference to transport of oxygen
- A. Partial pressure of CO$_2$ can interfere with O$_2$ binding with haemoglobin
- B. Higher H$^{+}$ conc. in alveoli favours the formation of oxyhaemoglobin ✓
- C. Low pCO$_2$ in alveoli favours the formation of oxyhaemoglobin
- D. Binding of oxygen with haemoglobin is mainly related to partial pressure of O$_2$
Solution: Higher H$^{+}$ concentration favours the DISSOCIATION of oxygen from oxyhaemoglobin in tissues, so statement (2) is wrong.
In the alveoli, high pO$_2$, low pCO$_2$, lesser H$^{+}$ concentration and lower temperature favour formation of oxyhaemoglobin.
Q57.
Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.
- A. A person will have only two of the three alleles.
- B. When I$^{A}$ and I$^{B}$ are present together, they express same type of sugar. ✓
- C. Allele 'i' does not produce any sugar.
- D. The gene (I) has three alleles.
Solution: ABO blood groups are controlled by the gene I. The gene I has three alleles I$^{A}$, I$^{B}$ and i.
The alleles I$^{A}$ and I$^{B}$ produce a slightly DIFFERENT form of the sugar, so statement (2) is wrong.
Allele i does not produce any sugar. Because humans are diploid organisms, each person can possess at the most any two of the three I gene alleles.
Q58.
Identify the basic amino acid from the following.
- A. Glutamic Acid
- B. Lysine ✓
- C. Valine
- D. Tyrosine
Solution: Lysine is a basic amino acid.
Valine is a neutral amino acid.
Glutamic acid is an acidic amino acid, while tyrosine is an aromatic amino acid.
Q59.
Name the plant growth regulator which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.
- A. Gibberellin ✓
- B. Ethylene
- C. Abscisic acid
- D. Cytokinin
Solution: Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
Q60.
Match the organism with its use in biotechnology.
(a) $Bacillus$ $thuringiensis$
(b) $Thermus$ $aquaticus$
(c) $Agrobacterium$ $tumefaciens$
(d) $Salmonella$ $typhimurium$
(i) Cloning vector
(ii) Construction of first rDNA molecule
(iii) DNA polymerase
(iv) Cry proteins
Select the correct option from the following:
- A. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) ✓
- B. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
- C. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- D. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
Solution: (a) $Bacillus$ $thuringiensis$ is a source of Cry-proteins.
(b) $Thermus$ $aquaticus$ is a source of thermostable DNA polymerase (Taq polymerase) used in PCR.
(c) $Agrobacterium$ $tumefaciens$ is a cloning vector.
(d) The construction of the 1$^{\text{st}}$ recombinant DNA molecule was performed using the native plasmid of $Salmonella$ $typhimurium$.
Q61.
Which of the following statements is correct?
- A. Adenine pairs with thymine through one H-bond
- B. Adenine pairs with thymine through three H-bonds
- C. Adenine does not pair with thymine
- D. Adenine pairs with thymine through two H-bonds ✓
Solution: Adenine pairs with thymine through two H-bonds, i.e. A $=$ T.
Guanine pairs with cytosine through three H-bonds.
Q62.
Match the following columns and select the correct option.
Column-I:
(a) Gregarious, polyphagous pest
(b) Adult with radial symmetry and larva with bilateral symmetry
(c) Book lungs
(d) Bioluminescence
Column-II:
(i) $Asterias$
(ii) Scorpion
(iii) $Ctenoplana$
(iv) $Locusta$
- A. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓
- B. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- C. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
- D. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
Solution: (a) $Locusta$ is a gregarious pest.
(b) In Echinoderms such as $Asterias$, adults are radially symmetrical but larvae are bilaterally symmetrical.
(c) Scorpions respire through book lungs.
(d) Bioluminescence is well marked in ctenophores such as $Ctenoplana$.
Q63.
Which of the following would help in prevention of diuresis?
- A. Reabsorption of Na$^{+}$ and water from renal tubules due to aldosterone ✓
- B. Atrial natriuretic factor causes vasoconstriction
- C. Decrease in secretion of renin by JG cells
- D. More water reabsorption due to undersecretion of ADH
Solution: Adrenal cortex secretes mineralocorticoids like aldosterone, which increase the reabsorption of Na$^{+}$ and water from the renal tubule, and that prevents diuresis.
Atrial natriuretic factor causes vasodilation, ADH undersecretion causes more water loss, and decreased renin lowers aldosterone — none of these prevent diuresis.
Q64.
Choose the correct pair from the following
- A. Polymerases - Break the DNA into fragments
- B. Nucleases - Separate the two strands of DNA
- C. Exonucleases - Make cuts at specific positions within DNA
- D. Ligases - Join the two DNA molecules ✓
Solution: Ligases join the two DNA molecules.
Polymerases synthesise DNA, nucleases cut DNA, and it is the endonucleases (not exonucleases) that make cuts at specific positions within the DNA.
Q65.
Identify the correct statement with reference to human digestive system.
- A. Serosa is the innermost layer of the alimentary canal
- B. Ileum is a highly coiled part ✓
- C. Vermiform appendix arises from duodenum
- D. Ileum opens into small intestine
Solution: Option (2) is correct as the ileum is a highly coiled tube.
Serosa is the outermost layer of the alimentary canal, so option (1) is incorrect.
A narrow finger-like tubular projection, the vermiform appendix, arises from the caecum part of the large intestine, so option (3) is incorrect.
Ileum opens into the large intestine, so option (4) is also incorrect.
Q66.
Embryological support for evolution was disapproved by
- A. Alfred Wallace
- B. Charles Darwin
- C. Oparin
- D. Karl Ernst von Baer ✓
Solution: Embryological support for evolution was disapproved by Karl Ernst von Baer.
He noted that embryos never pass through the adult stages of other animals during embryonic development.
Q67.
Which of the following hormone levels will cause release of ovum (ovulation) from the graffian follicle?
- A. High concentration of Progesterone
- B. Low concentration of LH
- C. Low concentration of FSH
- D. High concentration of Estrogen ✓
Solution: A high level of estrogen sends positive feedback to the anterior pituitary for release of LH.
FSH, LH and estrogen are at peak level during the middle of the menstrual cycle (28 day cycle).
The LH surge leads to ovulation.
Q68.
The specific palindromic sequence which is recognized by EcoRI is
- A. 5$'$ - GGAACC - 3$'$ / 3$'$ - CCTTGG - 5$'$
- B. 5$'$ - CTTAAG - 3$'$ / 3$'$ - GAATTC - 5$'$
- C. 5$'$ - GGATCC - 3$'$ / 3$'$ - CCTAGG - 5$'$
- D. 5$'$ - GAATTC - 3$'$ / 3$'$ - CTTAAG - 5$'$ ✓
Solution: EcoRI recognises the palindromic sequence
5$'$ - GAATTC - 3$'$
3$'$ - CTTAAG - 5$'$
and cuts between G and A on both strands, leaving sticky ends.
BamHI recognises GGATCC, so option (3) is not EcoRI.
Q69.
The first phase of translation is
- A. Recognition of DNA molecule
- B. Aminoacylation of tRNA ✓
- C. Recognition of an anti-codon
- D. Binding of mRNA to ribosome
Solution: The first phase of translation involves activation of an amino acid in the presence of ATP and its linkage to the cognate tRNA.
This process is commonly called charging of tRNA or aminoacylation of tRNA.
Q70.
Floridean starch has structure similar to
- A. Amylopectin and glycogen ✓
- B. Mannitol and algin
- C. Laminarin and cellulose
- D. Starch and cellulose
Solution: Floridean starch is the stored food material in red algae.
Its structure is similar to amylopectin and glycogen.
Q71.
Strobili or cones are found in
- A. $Pteris$
- B. $Marchantia$
- C. $Equisetum$ ✓
- D. $Salvinia$
Solution: Strobili or cones are found in $Equisetum$.
In $Equisetum$ the sporangia bearing structures are arranged into compact strobili.
Q72.
How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?
Solution: Mendel selected 14 true breeding pea plant varieties, as 7 pairs, each pair differing in only one character with contrasting traits.
Q73.
Snow-blindness in Antarctic region is due to
- A. Inflammation of cornea due to high dose of UV-B radiation ✓
- B. High reflection of light from snow
- C. Damage to retina caused by infra-red rays
- D. Freezing of fluids in the eye by low temperature
Solution: UV-B radiations damage DNA and mutations may occur.
In the human eye, the cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of the cornea, called snow blindness, cataract, etc.
Q74.
The enzyme enterokinase helps in conversion of
- A. trypsinogen into trypsin ✓
- B. caseinogen into casein
- C. pepsinogen into pepsin
- D. protein into polypeptides
Solution: Trypsinogen is activated by the enzyme enterokinase, secreted by the intestinal mucosa, into active trypsin.
Trypsinogen is a zymogen from the pancreas.
Q75.
Match the following with respect to meiosis
(a) Zygotene
(b) Pachytene
(c) Diplotene
(d) Diakinesis
(i) Terminalization
(ii) Chiasmata
(iii) Crossing over
(iv) Synapsis
Select the correct option from the following
- A. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) ✓
- B. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
- C. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
- D. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Solution: Zygotene $\rightarrow$ Synapsis
Pachytene $\rightarrow$ Crossing over
Diplotene $\rightarrow$ Chiasmata formation
Diakinesis $\rightarrow$ Terminalisation
Q76.
Which of the following statements about inclusion bodies is incorrect?
- A. These are involved in ingestion of food particles ✓
- B. They lie free in the cytoplasm
- C. These represent reserve material in cytoplasm
- D. They are not bound by any membrane
Solution: Inclusion bodies are not involved in ingestion of food particles, so statement (1) is incorrect.
They represent reserve material in the cytoplasm, lie free in the cytoplasm and are not bound by any membrane.
Q77.
Match the following columns and select the correct option.
Column-I:
(a) Eosinophils
(b) Basophils
(c) Neutrophils
(d) Lymphocytes
Column-II:
(i) Immune response
(ii) Phagocytosis
(iii) Release histaminase, destructive enzymes
(iv) Release granules containing histamine
- A. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
- B. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
- C. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
- D. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) ✓
Solution: Eosinophils are associated with allergic reactions and release histaminase and destructive enzymes, so (a) matches (iii).
Basophils secrete histamine, serotonin, heparin etc. and are involved in inflammatory reactions, so (b) matches (iv).
Neutrophils are phagocytic cells, so (c) matches (ii).
Both B and T lymphocytes are responsible for immune responses of the body, so (d) matches (i).
Q78.
The transverse section of a plant shows following anatomical features :
(a) Large number of scattered vascular bundles surrounded by bundle sheath
(b) Large conspicuous parenchymatous ground tissue
(c) Vascular bundles conjoint and closed
(d) Phloem parenchyma absent
Identify the category of plant and its part :
- A. Monocotyledonous root
- B. Dicotyledonous stem
- C. Dicotyledonous root
- D. Monocotyledonous stem ✓
Solution: All the listed features — scattered vascular bundles with a bundle sheath, conspicuous parenchymatous ground tissue, conjoint and closed vascular bundles and absence of phloem parenchyma — are features of a monocotyledonous stem.
Q79.
Match the following columns and select the correct option.
Column-I:
(a) Pituitary gland
(b) Thyroid gland
(c) Adrenal gland
(d) Pancreas
Column-II:
(i) Grave's disease
(ii) Diabetes mellitus
(iii) Diabetes insipidus
(iv) Addison's disease
- A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- B. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) ✓
- C. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
- D. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Solution: Diabetes insipidus is due to hyporelease of ADH from the posterior pituitary, so (a) matches (iii).
Graves' disease is due to excess secretion of thyroid hormones (T$_3$ and T$_4$), so (b) matches (i).
Addison's disease is due to hyposecretion of hormone from the adrenal cortex, so (c) matches (iv).
Diabetes mellitus is due to hyposecretion of insulin from $\beta$-cells of the pancreas, so (d) matches (ii).
Q80.
Match the following columns and select the correct option.
Column-I:
(a) Placenta
(b) Zona pellucida
(c) Bulbo-urethral glands
(d) Leydig cells
Column-II:
(i) Androgens
(ii) Human Chorionic Gonadotropin (hCG)
(iii) Layer of the ovum
(iv) Lubrication of the Penis
- A. (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
- B. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
- C. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓
- D. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Solution: (a) Placenta secretes human chorionic gonadotropin (hCG).
(b) Zona pellucida is a primary egg membrane secreted by the secondary oocyte, i.e. a layer of the ovum.
(c) The secretions of the bulbourethral glands help in lubrication of the penis.
(d) Leydig cells synthesise and secrete testicular hormones called androgens.
Q81.
In water hyacinth and water lily, pollination takes place by :
- A. Water currents only
- B. Wind and water
- C. Insects and water
- D. Insects or wind ✓
Solution: In the majority of aquatic plants, the flowers emerge above the level of water.
These may be pollinated by insects or wind, e.g. water hyacinth and water lily.
Q82.
According to Robert May, the global species diversity is about
- A. 20 million
- B. 50 million
- C. 7 million ✓
- D. 1.5 million
Solution: Robert May estimated global species diversity at about 7 million.
Some extreme estimates range from 20 to 50 million, and about 1.5 million species have actually been described so far.
Q83.
Match the following columns and select the correct option.
Column-I:
(a) 6-15 pairs of gill slits
(b) Heterocercal caudal fin
(c) Air Bladder
(d) Poison sting
Column-II:
(i) $Trygon$
(ii) Cyclostomes
(iii) Chondrichthyes
(iv) Osteichthyes
- A. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- B. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
- C. (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
- D. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓
Solution: Cyclostomes have an elongated body bearing 6-15 pairs of gill slits for respiration, so (a) matches (ii).
Heterocercal caudal fin is present in members of class Chondrichthyes, so (b) matches (iii).
Air bladder is present in bony fishes belonging to class Osteichthyes, where it regulates buoyancy, so (c) matches (iv).
$Trygon$, a cartilaginous fish, possesses a poison sting, so (d) matches (i).
Q84.
The process of growth is maximum during
- A. Lag phase
- B. Senescence
- C. Dormancy
- D. Log phase ✓
Solution: In exponential growth the initial growth is slow (lag phase) and it increases rapidly thereafter at an exponential rate in the log or exponential phase.
So growth is maximum during the log phase.
Q85.
Match the following columns and select the correct option.
Column-I:
(a) Bt cotton
(b) Adenosine deaminase deficiency
(c) RNAi
(d) PCR
Column-II:
(i) Gene therapy
(ii) Cellular defence
(iii) Detection of HIV infection
(iv) $Bacillus$ $thuringiensis$
- A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- B. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
- C. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
- D. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓
Solution: Bt cotton carries the toxin gene of $Bacillus$ $thuringiensis$, so (a) matches (iv).
Adenosine deaminase deficiency is treated by gene therapy, so (b) matches (i).
RNA interference is a method of cellular defence, so (c) matches (ii).
PCR is used for detection of HIV infection in suspected AIDS patients, so (d) matches (iii).
Q86.
Match the following columns and select the correct option.
Column-I:
(a) Organ of Corti
(b) Cochlea
(c) Eustachian tube
(d) Stapes
Column-II:
(i) Connects middle ear and pharynx
(ii) Coiled part of the labyrinth
(iii) Attached to the oval window
(iv) Located on the basilar membrane
- A. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
- B. (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) ✓
- C. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
- D. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
Solution: Organ of Corti is located on the basilar membrane, so (a) matches (iv).
The coiled portion of the labyrinth is called the cochlea, so (b) matches (ii).
The eustachian tube connects the middle ear cavity with the pharynx, so (c) matches (i).
The middle ear contains the ossicle called stapes, which is attached to the oval window of the cochlea, so (d) matches (iii).
Q87.
Which one of the following is the most abundant protein in the animals?
- A. Collagen ✓
- B. Lectin
- C. Insulin
- D. Haemoglobin
Solution: Collagen is the most abundant protein in the animal world.
RuBisCO is the most abundant protein in the whole of the biosphere.
Q88.
Identify the wrong statement with reference to immunity.
- A. When ready-made antibodies are directly given, it is called "Passive immunity".
- B. Active immunity is quick and gives full response. ✓
- C. Foetus receives some antibodies from mother, it is an example for passive immunity.
- D. When exposed to antigen (living or dead) antibodies are produced in the host's body. It is called "Active immunity".
Solution: Active immunity is SLOW and takes time to give its full effective response, in comparison to passive immunity where pre-formed antibodies are administered.
So statement (2) is the wrong one; the other three are correct.
Q89.
Montreal protocol was signed in 1987 for control of
- A. Emission of ozone depleting substances ✓
- B. Release of Green House gases
- C. Disposal of e-wastes
- D. Transport of Genetically modified organisms from one country to another
Solution: The Montreal protocol was signed on 16 September 1987 (Ozone day) and came into force on 1 January 1989.
It was aimed at stopping the production and import of ozone depleting substances and reducing their concentration in the atmosphere.
Q90.
Match the trophic levels with their correct species examples in grassland ecosystem.
(a) Fourth trophic level
(b) Second trophic level
(c) First trophic level
(d) Third trophic level
(i) Crow
(ii) Vulture
(iii) Rabbit
(iv) Grass
Select the correct option
- A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- B. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
- C. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
- D. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) ✓
Solution: Grassland ecosystem is a terrestrial ecosystem. It includes various trophic levels:
First trophic level (T$_1$) – Grass
Second trophic level (T$_2$) – Rabbit
Third trophic level (T$_3$) – Crow
Fourth trophic level (T$_4$) – Vulture
Q91.
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitch of the screw gauge is :
- A. 0.25 mm
- B. 0.5 mm ✓
- C. 1.0 mm
- D. 0.01 mm
Solution: Least count $= \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
$$0.01\ \text{mm} = \frac{\text{Pitch}}{50}$$
$$\text{Pitch} = 0.5\ \text{mm}$$
Q92.
The mean free path for a gas, with molecular diameter d and number density n can be expressed as :
- A. $\dfrac{1}{\sqrt{2}\,n\pi d^{2}}$ ✓
- B. $\dfrac{1}{\sqrt{2}\,n^{2}\pi d^{2}}$
- C. $\dfrac{1}{\sqrt{2}\,n^{2}\pi^{2} d^{2}}$
- D. $\dfrac{1}{\sqrt{2}\,n\pi d}$
Solution: The mean free path of a gas molecule is given by
$$\lambda = \frac{1}{\sqrt{2}\,n\pi d^{2}}$$
where $n$ is the number density and $d$ the molecular diameter.
Q93.
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
- A. four times
- B. one-fourth
- C. zero ✓
- D. doubled
Solution: Given $\nu = \dfrac{3}{2}\nu_0$
On halving, $\nu' = \dfrac{\nu}{2} = \dfrac{3}{4}\nu_0$
Since $\nu' < \nu_0$, the incident frequency is below the threshold frequency.
Therefore no photoelectric emission will take place, whatever the intensity, and the photoelectric current is zero.
Q94.
In a certain region of space with volume 0.2 m$^{3}$, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is :
- A. 0.5 N/C
- B. 1 N/C
- C. 5 N/C
- D. zero ✓
Solution: Since the electric potential is found to be constant throughout the region,
$$E = -\frac{dV}{dr} = 0$$
Q95.
Which of the following graph represents the variation of resistivity ($\rho$) with temperature (T) for copper?
- A. Graph (1)
- B. Graph (2) ✓
- C. Graph (3)
- D. Graph (4)
Solution: For a metal such as copper the resistivity increases with temperature.
At temperatures much lower than 0$^{\circ}$C the graph deviates considerably from a straight line, curving upward as the temperature rises.
Hence option (2) is correct.
Q96.
A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L$_1$ when mass M is suspended from its free end. The expression for Young's modulus is :
- A. $\dfrac{Mg(L_1 - L)}{AL}$
- B. $\dfrac{MgL}{AL_1}$
- C. $\dfrac{MgL}{A(L_1 - L)}$ ✓
- D. $\dfrac{MgL_1}{AL}$
Solution: Young's modulus is the ratio of longitudinal stress to longitudinal strain.
Stress $= \dfrac{Mg}{A}$ and strain $= \dfrac{L_1 - L}{L}$
$$Y = \frac{Mg/A}{(L_1 - L)/L} = \frac{MgL}{A(L_1 - L)}$$
Q97.
In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be:
- A. 524 Hz ✓
- B. 536 Hz
- C. 537 Hz
- D. 523 Hz
Solution: The difference of $f_A$ and $f_B$ is 6 Hz.
If the tension decreases, $f_B$ decreases and becomes $f_B'$.
Now the difference of $f_A$ and $f_B'$ is 7 Hz, i.e. it increases. So $f_A > f_B$.
$$f_A - f_B = 6\ \text{Hz},\qquad f_A = 530\ \text{Hz}$$
$$f_B = 524\ \text{Hz (original)}$$
Q98.
A 40 $\mu$F capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly :
- A. 2.05 A
- B. 2.5 A ✓
- C. 25.1 A
- D. 1.7 A
Solution: $$i_{rms} = C\omega\,\varepsilon_{rms}$$
$C = 40 \times 10^{-6}$ F, $\omega = 2\pi f = 100\pi$, $\varepsilon_{rms} = 200$ V
$$i_{rms} = 200 \times 40 \times 10^{-6} \times 2\pi \times 50 = 2.5\ \text{A}$$
Q99.
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g = 10 m/s$^{2}$)
- A. 340 m
- B. 320 m
- C. 300 m ✓
- D. 360 m
Solution: $$v^{2} = u^{2} + 2gh$$
with $v = 80$ m/s and $u = 20$ m/s,
$$h = \frac{v^{2} - u^{2}}{2g} = \frac{6400 - 400}{20} = 300\ \text{m}$$
Q100.
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2}$ nm, the potential difference is :
- A. $10^{2}$ V
- B. $10^{3}$ V
- C. $10^{4}$ V ✓
- D. 10 V
Solution: $$\lambda = \frac{12.27}{\sqrt{V}}\ \text{Å}$$
$$\sqrt{V} = \frac{12.27 \times 10^{-10}}{1.227 \times 10^{-11}} = 10^{2}$$
$$V = 10^{4}\ \text{volts}$$
Q101.
For the logic circuit shown, the truth table is:
- A. (A, B) $\to$ Y : (0, 0) $\to$ 0, (0, 1) $\to$ 1, (1, 0) $\to$ 1, (1, 1) $\to$ 1
- B. (A, B) $\to$ Y : (0, 0) $\to$ 1, (0, 1) $\to$ 1, (1, 0) $\to$ 1, (1, 1) $\to$ 0
- C. (A, B) $\to$ Y : (0, 0) $\to$ 1, (0, 1) $\to$ 0, (1, 0) $\to$ 0, (1, 1) $\to$ 0
- D. (A, B) $\to$ Y : (0, 0) $\to$ 0, (0, 1) $\to$ 0, (1, 0) $\to$ 0, (1, 1) $\to$ 1 ✓
Solution: Input A passes through a NOT gate giving $\overline{A}$, and input B through a NOT gate giving $\overline{B}$. Both feed a NOR gate whose output is Y.
$$Y = \overline{\overline{A} + \overline{B}} = \overline{\overline{A \cdot B}} = A \cdot B$$
So the circuit behaves as an AND gate, and Y is 1 only when both A and B are 1.
Q102.
A short electric dipole has a dipole moment of $16 \times 10^{-9}$ C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60$^{\circ}$ with the dipole axis is : $\left(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \text{N m}^{2}/\text{C}^{2}\right)$
- A. 200 V ✓
- B. 400 V
- C. zero
- D. 50 V
Solution: $$V = \frac{k\,p\cos\theta}{r^{2}}$$
$$V = \frac{9 \times 10^{9} \times 16 \times 10^{-9} \times \cos 60^{\circ}}{0.36}$$
$$V = 200\ \text{V}$$
Q103.
An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m$^{-1}$. The permeability of the material of the rod is : ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)
- A. $8.0 \times 10^{-5}$ T m A$^{-1}$
- B. $2.4\pi \times 10^{-5}$ T m A$^{-1}$
- C. $2.4\pi \times 10^{-7}$ T m A$^{-1}$
- D. $2.4\pi \times 10^{-4}$ T m A$^{-1}$ ✓
Solution: $$\chi_m = 599$$
$$\mu_r = 1 + \chi_m = 600$$
$$\mu = \mu_r\mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7}$$
$$\mu = 2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$$
Q104.
The increase in the width of the depletion region in a p-n junction diode is due to :
- A. reverse bias only ✓
- B. both forward bias and reverse bias
- C. increase in forward current
- D. forward bias only
Solution: Due to reverse biasing, the width of the depletion region increases.
Forward bias reduces the depletion width.
Q105.
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is :
- A. 5.0 g
- B. 10.0 g ✓
- C. 20.0 g
- D. 2.5 g
Solution: The force of surface tension balances the weight of water in the capillary tube.
$$F_S = 2\pi r T\cos\theta = mg$$
Here T and $\theta$ are constant, so $m \propto r$.
$$\frac{m_2}{5.0} = \frac{2r}{r}$$
$$m_2 = 10.0\ \text{g}$$
Q106.
The energy equivalent of 0.5 g of a substance is :
- A. $4.5 \times 10^{13}$ J ✓
- B. $1.5 \times 10^{13}$ J
- C. $0.5 \times 10^{13}$ J
- D. $4.5 \times 10^{16}$ J
Solution: From mass-energy equivalence,
$$E = mc^{2}$$
$$E = 0.5 \times 10^{-3} \times (3 \times 10^{8})^{2}$$
$$E = 4.5 \times 10^{13}\ \text{J}$$
Q107.
The solids which have the negative temperature coefficient of resistance are:
- A. insulators only
- B. semiconductors only
- C. insulators and semiconductors ✓
- D. metals
Solution: In insulators and semiconductors the number of free charge carriers increases rapidly as the temperature rises, so the resistance falls with rising temperature.
Hence both insulators and semiconductors have a negative temperature coefficient of resistance, while metals have a positive one.
Q108.
A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$, then the angle of incidence is nearly equal to :
- A. $\dfrac{2A}{\mu}$
- B. $\mu A$ ✓
- C. $\dfrac{\mu A}{2}$
- D. $\dfrac{A}{2\mu}$
Solution: The light ray emerges normally from the second surface, hence the angle of emergence $e = 0$ and so $r_2 = 0$.
Since $r_1 + r_2 = A$, we get $r_1 = A$.
Applying Snell's law on the first surface,
$$1 \cdot \sin i = \mu \sin r_1 = \mu \sin A$$
For small angles $\sin\theta \approx \theta$, hence $i = \mu A$.
Q109.
For which one of the following, Bohr model is not valid ?
- A. Singly ionised helium atom (He$^{+}$)
- B. Deuteron atom
- C. Singly ionised neon atom (Ne$^{+}$) ✓
- D. Hydrogen atom
Solution: Bohr model is only valid for single electron species.
A singly ionised neon atom has more than one electron in orbit, hence the Bohr model is not valid for it.
Q110.
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is :
- A. $1.83 \times 10^{-7}$ rad
- B. $7.32 \times 10^{-7}$ rad
- C. $6.00 \times 10^{-7}$ rad
- D. $3.66 \times 10^{-7}$ rad ✓
Solution: $$\theta_R = 1.22\frac{\lambda}{d}$$
with $\lambda = 600 \times 10^{-9}$ m and $d = 2$ m,
$$\theta_R = \frac{1.22 \times 600 \times 10^{-9}}{2} = 3.66 \times 10^{-7}\ \text{rad}$$
Q111.
A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
- A. 32 N ✓
- B. 30 N
- C. 24 N
- D. 48 N
Solution: $$mg_h = \frac{mg_0}{\left(1 + \dfrac{h}{R}\right)^{2}}$$
$$W = \frac{72}{\left(1 + \dfrac{R/2}{R}\right)^{2}} = \frac{72}{(3/2)^{2}} = \frac{4}{9} \times 72 = 32\ \text{N}$$
Q112.
A charged particle having drift velocity of $7.5 \times 10^{-4}$ m s$^{-1}$ in an electric field of $3 \times 10^{-10}$ Vm$^{-1}$, has a mobility in m$^{2}$ V$^{-1}$ s$^{-1}$ of :
- A. $2.5 \times 10^{6}$ ✓
- B. $2.5 \times 10^{-6}$
- C. $2.25 \times 10^{-15}$
- D. $2.25 \times 10^{15}$
Solution: Mobility is defined as
$$\mu = \frac{v_d}{E}$$
$$\mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^{6}\ \text{m}^{2}\text{V}^{-1}\text{s}^{-1}$$
Q113.
For transistor action, which of the following statements is correct?
- A. Base, emitter and collector regions should have same size.
- B. Both emitter junction as well as the collector junction are forward biased.
- C. The base region must be very thin and lightly doped. ✓
- D. Base, emitter and collector regions should have same doping concentrations.
Solution: For a bipolar junction transistor the length profile is $L_C > L_E > L_B$ and the doping profile is E $>$ C $>$ B.
So the base region must be very thin and lightly doped.
For transistor action the base-emitter junction is forward biased and the base-collector junction is reverse biased.
Q114.
The capacitance of a parallel plate capacitor with air as medium is 6 $\mu$F. With the introduction of a dielectric medium, the capacitance becomes 30 $\mu$F. The permittivity of the medium is : ($\epsilon_0 = 8.85 \times 10^{-12}$ C$^{2}$ N$^{-1}$ m$^{-2}$)
- A. $1.77 \times 10^{-12}$ C$^{2}$ N$^{-1}$ m$^{-2}$
- B. $0.44 \times 10^{-10}$ C$^{2}$ N$^{-1}$ m$^{-2}$ ✓
- C. $5.00$ C$^{2}$ N$^{-1}$ m$^{-2}$
- D. $0.44 \times 10^{-13}$ C$^{2}$ N$^{-1}$ m$^{-2}$
Solution: $$C = KC_0 \Rightarrow K = \frac{C}{C_0} = \frac{30}{6} = 5$$
Since $K = \dfrac{\epsilon}{\epsilon_0}$,
$$\epsilon = K\epsilon_0 = 5 \times 8.85 \times 10^{-12} = 0.44 \times 10^{-10}\ \text{C}^{2}\text{N}^{-1}\text{m}^{-2}$$
Q115.
Taking into account of the significant figures, what is the value of 9.99 m – 0.0099 m?
- A. 9.98 m ✓
- B. 9.980 m
- C. 9.9 m
- D. 9.9801 m
Solution: $$9.99 - 0.0099 = 9.9801\ \text{m}$$
In subtraction the answer should be reported to the least number of decimal places among the terms, i.e. two decimal places.
So the answer should be 9.98 m.
Q116.
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :
- A. g/2
- B. g/5 ✓
- C. g/10
- D. g
Solution: For an Atwood machine,
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2}\quad\text{where } m_1 > m_2$$
$$a = \frac{(6 - 4)g}{6 + 4} = \frac{g}{5}$$
Q117.
A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27$^{\circ}$C. Its density is : (R = 8.3 J mol$^{-1}$ K$^{-1}$)
- A. 0.2 kg/m$^{3}$ ✓
- B. 0.1 kg/m$^{3}$
- C. 0.02 kg/m$^{3}$
- D. 0.5 kg/m$^{3}$
Solution: $$PM = \rho RT \Rightarrow \rho = \frac{PM}{RT}$$
with $P = 249 \times 10^{3}$ N/m$^{2}$, $M = 2 \times 10^{-3}$ kg and $T = 300$ K,
$$\rho = \frac{(249 \times 10^{3})(2 \times 10^{-3})}{8.3 \times 300} = 0.2\ \text{kg/m}^{3}$$
Q118.
The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is : (c = speed of electromagnetic waves)
- A. 1 : 1 ✓
- B. 1 : c
- C. 1 : c$^{2}$
- D. c : 1
Solution: In an electromagnetic wave the electric and magnetic field components carry equal energy.
The average energy density of the electric field equals that of the magnetic field, so the two contributions to the intensity are in the ratio 1 : 1.
Q119.
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)
- A. $3.14 \times 10^{-4}$ T
- B. $6.28 \times 10^{-5}$ T
- C. $3.14 \times 10^{-5}$ T
- D. $6.28 \times 10^{-4}$ T ✓
Solution: Magnetic field at the centre of a solenoid $= \mu_0 n I$
$$n = \frac{N}{L} = \frac{100}{50 \times 10^{-2}} = 200\ \text{turns/m},\qquad I = 2.5\ \text{A}$$
$$B = 4\pi \times 10^{-7} \times 200 \times 2.5 = 6.28 \times 10^{-4}\ \text{T}$$
Q120.
In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :
- A. half
- B. four times ✓
- C. one-fourth
- D. double
Solution: Fringe width $\beta = \dfrac{\lambda D}{d}$
Now $d' = \dfrac{d}{2}$ and $D' = 2D$, so
$$\beta' = \frac{\lambda(2D)}{d/2} = \frac{4\lambda D}{d} = 4\beta$$
Q121.
A resistance wire connected in the left gap of a metre bridge balances a 10 $\Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 $\Omega$ of the resistance wire is :
- A. $1.0 \times 10^{-1}$ m ✓
- B. $1.5 \times 10^{-1}$ m
- C. $1.5 \times 10^{-2}$ m
- D. $1.0 \times 10^{-2}$ m
Solution: For a balanced metre bridge,
$$\frac{P}{10} = \frac{l_1}{l_2} = \frac{3}{2} \Rightarrow P = \frac{30}{2} = 15\ \Omega$$
So 1.5 m of the resistance wire has a resistance of 15 $\Omega$.
Since $R = \dfrac{\rho l}{A}$, length is proportional to resistance:
$$\frac{15}{1} = \frac{1.5}{l}$$
$$l = 0.1\ \text{m} = 1.0 \times 10^{-1}\ \text{m}$$
Q122.
The energy required to break one bond in DNA is $10^{-20}$ J. This value in eV is nearly:
- A. 0.6
- B. 0.06 ✓
- C. 0.006
- D. 6
Solution: $$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$$
$$1\ \text{J} = \frac{1}{1.6 \times 10^{-19}}\ \text{eV}$$
$$10^{-20}\ \text{J} = \frac{10^{-20}}{1.6 \times 10^{-19}}\ \text{eV} = 0.06\ \text{eV}$$
Q123.
When a uranium isotope $^{235}_{92}$U is bombarded with a neutron, it generates $^{89}_{36}$Kr, three neutrons and :
- A. $^{91}_{40}$Zr
- B. $^{101}_{36}$Kr
- C. $^{103}_{36}$Kr
- D. $^{144}_{56}$Ba ✓
Solution: $$^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{89}_{36}\text{Kr} + 3\,^{1}_{0}\text{n} + ^{A}_{Z}\text{X}$$
Conserving charge: $92 + 0 = 36 + Z \Rightarrow Z = 56$
Conserving mass number: $235 + 1 = 89 + 3 + A \Rightarrow A = 144$
So $^{144}_{56}$Ba is generated.
Q124.
Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is :
- A. adiabatic ✓
- B. isochoric
- C. isobaric
- D. isothermal
Solution: The entire system is thermally insulated, so no heat exchange takes place with the surroundings.
Hence the process is adiabatic.
Q125.
Light with an average flux of 20 W/cm$^{2}$ falls on a non-reflecting surface at normal incidence having surface area 20 cm$^{2}$. The energy received by the surface during time span of 1 minute is :
- A. $12 \times 10^{3}$ J
- B. $24 \times 10^{3}$ J ✓
- C. $48 \times 10^{3}$ J
- D. $10 \times 10^{3}$ J
Solution: Energy received $=$ Intensity $\times$ Area $\times$ Time
$$= 20 \times 20 \times 60 = 24 \times 10^{3}\ \text{J}$$
Q126.
The quantities of heat required to raise the temperature of two solid copper spheres of radii r$_1$ and r$_2$ (r$_1$ = 1.5 r$_2$) through 1 K are in the ratio :
- A. $\dfrac{9}{4}$
- B. $\dfrac{3}{2}$
- C. $\dfrac{5}{3}$
- D. $\dfrac{27}{8}$ ✓
Solution: $$\Delta Q = ms\Delta T = \frac{4}{3}\pi r^{3}\rho\, s\,\Delta T$$
$$\frac{\Delta Q_1}{\Delta Q_2} = \left(\frac{r_1}{r_2}\right)^{3} = (1.5)^{3} = \frac{27}{8}$$
Q127.
The average thermal energy for a mono-atomic gas is : (k$_B$ is Boltzmann constant and T, absolute temperature)
- A. $\dfrac{3}{2}k_BT$ ✓
- B. $\dfrac{5}{2}k_BT$
- C. $\dfrac{7}{2}k_BT$
- D. $\dfrac{1}{2}k_BT$
Solution: For monoatomic gases the degree of freedom is 3.
Hence the average thermal energy per molecule is
$$KE_{avg} = \frac{3}{2}k_BT$$
Q128.
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\dfrac{\pi}{3}$. If instead C is removed from the circuit, the phase difference is again $\dfrac{\pi}{3}$ between current and voltage. The power factor of the circuit is :
- A. 0.5
- B. 1.0 ✓
- C. –1.0
- D. zero
Solution: When L is removed,
$$\tan\phi = \frac{|X_C|}{R} \Rightarrow \tan\frac{\pi}{3} = \frac{X_C}{R}\qquad \ldots(i)$$
When C is removed,
$$\tan\phi = \frac{|X_L|}{R} \Rightarrow \tan\frac{\pi}{3} = \frac{X_L}{R}\qquad \ldots(ii)$$
From (i) and (ii), $X_L = X_C$, so the circuit is in resonance and $Z = R$.
$$\text{Power factor} = \cos\phi = \frac{R}{Z} = 1$$
Q129.
Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass.
The centre of mass of the system from the 5 kg particle is nearly at a distance of :
- A. 50 cm
- B. 67 cm ✓
- C. 80 cm
- D. 33 cm
Solution: Taking the 5 kg particle at the origin and the 10 kg particle at $x = 1$ m,
$$x_{cm} = \frac{5(0) + 10(1)}{5 + 10} = \frac{10}{15} = 0.67\ \text{m}$$
So the centre of mass is nearly 67 cm from the 5 kg particle.
Q130.
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :
- A. $\dfrac{3\pi}{2}$ rad
- B. $\dfrac{\pi}{2}$ rad
- C. zero
- D. $\pi$ rad ✓
Solution: If $y = A\sin\omega t$, then
$$v = \frac{dy}{dt} = A\omega\cos\omega t$$
$$a = \frac{dv}{dt} = -A\omega^{2}\sin\omega t = A\omega^{2}\sin(\omega t + \pi)$$
So the phase difference between displacement and acceleration is $\pi$.
Q131.
The Brewsters angle i$_b$ for an interface should be
- A. 30$^{\circ}$ < i$_b$ < 45$^{\circ}$
- B. 45$^{\circ}$ < i$_b$ < 90$^{\circ}$ ✓
- C. i$_b$ = 90$^{\circ}$
- D. 0$^{\circ}$ < i$_b$ < 30$^{\circ}$
Solution: By Brewster's law, $\mu = \tan i_b$.
For any interface $1 < \mu < \infty$, so $1 < \tan i_b < \infty$.
$$\tan^{-1}(1) < i_b < \tan^{-1}(\infty)$$
$$45^{\circ} < i_b < 90^{\circ}$$
Q132.
Dimensions of stress are :
- A. [ML$^{2}$T$^{-2}$]
- B. [ML$^{0}$T$^{-2}$]
- C. [ML$^{-1}$T$^{-2}$] ✓
- D. [MLT$^{-2}$]
Solution: $$\text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^{2}]} = [ML^{-1}T^{-2}]$$
Q133.
The color code of a resistance is given below: the four bands, read in order, are Yellow, Violet, Brown and Gold.
The values of resistance and tolerance, respectively, are
- A. 47 k$\Omega$, 10%
- B. 4.7 k$\Omega$, 5%
- C. 470 $\Omega$, 5% ✓
- D. 470 k$\Omega$, 5%
Solution: According to colour coding:
Yellow $= 4$, Violet $= 7$, Brown $= 1$ (multiplier $10^{1}$), Gold $= 5\%$ tolerance.
$$R = 47 \times 10^{1} \pm 5\%$$
$$R = 470 \pm 5\%\ \Omega$$
Q134.
A spherical conductor of radius 10 cm has a charge of $3.2 \times 10^{-7}$ C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? $\left(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \text{Nm}^{2}/\text{C}^{2}\right)$
- A. $1.28 \times 10^{5}$ N/C ✓
- B. $1.28 \times 10^{6}$ N/C
- C. $1.28 \times 10^{7}$ N/C
- D. $1.28 \times 10^{4}$ N/C
Solution: For a point outside a conducting sphere the whole charge behaves as if concentrated at the centre:
$$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^{2}}$$
$$E = \frac{9 \times 10^{9} \times 3.2 \times 10^{-7}}{225 \times 10^{-4}} = 0.128 \times 10^{6}$$
$$E = 1.28 \times 10^{5}\ \text{N/C}$$
Q135.
Find the torque about the origin when a force of $3\hat{j}$ N acts on a particle whose position vector is $2\hat{k}$ m.
- A. $6\hat{j}$ Nm
- B. $-6\hat{i}$ Nm ✓
- C. $6\hat{k}$ Nm
- D. $6\hat{i}$ Nm
Solution: $$\vec{\tau} = \vec{r} \times \vec{F}$$
$$\vec{\tau} = 2\hat{k} \times 3\hat{j} = 6(\hat{k} \times \hat{j})$$
Since $\hat{k} \times \hat{j} = -\hat{i}$,
$$\vec{\tau} = -6\hat{i}\ \text{Nm}$$
Q136.
The mixture which shows positive deviation from Raoult's law is
- A. Benzene + Toluene
- B. Acetone + Chloroform
- C. Chloroethane + Bromoethane
- D. Ethanol + Acetone ✓
Solution: Pure ethanol molecules are hydrogen bonded.
On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them.
This weakens the intermolecular attractive interactions and the solution shows positive deviation from Raoult's law.
Benzene + toluene and chloroethane + bromoethane are nearly ideal, and acetone + chloroform shows negative deviation.
Q137.
Which of the following is not correct about carbon monoxide ?
- A. It reduces oxygen carrying ability of blood.
- B. The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin. ✓
- C. It is produced due to incomplete combustion.
- D. It forms carboxyhaemoglobin
Solution: Carboxyhaemoglobin is about 300 times MORE stable than oxyhaemoglobin, so statement (2) is not correct.
That greater stability is exactly why CO reduces the oxygen carrying ability of blood.
Q138.
The number of Faradays(F) required to produce 20 g of calcium from molten CaCl$_2$ (Atomic mass of Ca = 40 g mol$^{-1}$) is
Solution: 1 equivalent of any substance is deposited by 1 F of charge.
Equivalent mass of Ca $= \dfrac{40}{2} = 20$
Number of equivalents $= \dfrac{\text{Given mass}}{\text{Equivalent mass}} = \dfrac{20}{20} = 1$
So 1 faraday of charge is required.
Q139.
Hydrolysis of sucrose is given by the following reaction.
Sucrose $+$ H$_2$O $\rightleftharpoons$ Glucose $+$ Fructose
If the equilibrium constant (K$_C$) is $2 \times 10^{13}$ at 300 K, the value of $\Delta_r G^{\ominus}$ at the same temperature will be :
- A. 8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(2 $\times$ 10$^{13}$)
- B. 8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(3 $\times$ 10$^{13}$)
- C. $-$8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(4 $\times$ 10$^{13}$)
- D. $-$8.314 J mol$^{-1}$K$^{-1}$ $\times$ 300 K $\times$ ln(2 $\times$ 10$^{13}$) ✓
Solution: $$\Delta G = \Delta G^{\circ} + RT\ln Q$$
At equilibrium $\Delta G = 0$ and $Q = K_{eq}$, so
$$\Delta_r G^{\circ} = -RT\ln K_{eq}$$
$$\Delta_r G^{\circ} = -8.314\ \text{J mol}^{-1}\text{K}^{-1} \times 300\ \text{K} \times \ln(2 \times 10^{13})$$
Q140.
For the reaction, 2Cl(g) $\longrightarrow$ Cl$_2$(g), the correct option is :
- A. $\Delta_r$H > 0 and $\Delta_r$S < 0
- B. $\Delta_r$H < 0 and $\Delta_r$S > 0
- C. $\Delta_r$H < 0 and $\Delta_r$S < 0 ✓
- D. $\Delta_r$H > 0 and $\Delta_r$S > 0
Solution: Cl$_2$(g) $\longrightarrow$ 2Cl(g) is an endothermic reaction because it requires energy to break the bond.
So the reverse reaction, 2Cl(g) $\longrightarrow$ Cl$_2$(g), is exothermic and $\Delta_r$H < 0.
Two gaseous atoms combine to form one gaseous molecule, so randomness decreases and $\Delta_r$S < 0.
Q141.
Paper chromatography is an example of
- A. Partition chromatography ✓
- B. Thin layer chromatography
- C. Column chromatography
- D. Adsorption chromatography
Solution: Paper chromatography is a type of partition chromatography in which a special quality paper known as chromatography paper is used.
The water trapped in the pores of the paper acts as the stationary phase.
Q142.
The rate constant for a first order reaction is $4.606 \times 10^{-3}$ s$^{-1}$. The time required to reduce 2.0 g of the reactant to 0.2 g is :
- A. 200 s
- B. 500 s ✓
- C. 1000 s
- D. 100 s
Solution: First order rate equation:
$$k = \frac{2.303}{t}\log\frac{A_0}{A}$$
$$4.606 \times 10^{-3} = \frac{2.303}{t}\log\frac{2}{0.2}$$
$$t = \frac{2.303}{4.606 \times 10^{-3}} \times \log 10 = \frac{10^{3}}{2} = 500\ \text{s}$$
Q143.
Which of the following oxoacid of sulphur has – O – O – linkage?
- A. H$_2$SO$_4$, sulphuric acid
- B. H$_2$S$_2$O$_8$, peroxodisulphuric acid ✓
- C. H$_2$S$_2$O$_7$, pyrosulphuric acid
- D. H$_2$SO$_3$, sulphurous acid
Solution: Peroxodisulphuric acid, H$_2$S$_2$O$_8$, has the structure
HO – SO$_2$ – O – O – SO$_2$ – OH
so it contains a peroxide (– O – O –) linkage.
Pyrosulphuric acid has an S – O – S bridge without a peroxide link.
Q144.
Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as
- A. Cannizzaro's reaction
- B. Cross Cannizzaro's reaction
- C. Cross Aldol condensation ✓
- D. Aldol condensation
Solution: In the presence of dilute OH$^{-}$, benzaldehyde and acetophenone react to undergo cross-aldol condensation.
Acetophenone provides the $\alpha$-hydrogen and forms the carbanion; it adds to the carbonyl carbon of benzaldehyde, and loss of water gives the $\alpha$,$\beta$-unsaturated ketone C$_6$H$_5$CH=CH–CO–C$_6$H$_5$.
Since the two carbonyl partners are different, it is a cross aldol condensation.
Q145.
An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is
- A. $\dfrac{\sqrt{2}}{4} \times 288$ pm
- B. $\dfrac{4}{\sqrt{3}} \times 288$ pm
- C. $\dfrac{4}{\sqrt{2}} \times 288$ pm
- D. $\dfrac{\sqrt{3}}{4} \times 288$ pm ✓
Solution: For a bcc lattice the body diagonal is $\sqrt{3}a$ and it holds four atomic radii:
$$\sqrt{3}a = 4r \Rightarrow r = \frac{\sqrt{3}a}{4}$$
Given $a = 288$ pm,
$$r = \frac{\sqrt{3}}{4} \times 288\ \text{pm}$$
Q146.
Which of the following is a cationic detergent?
- A. Sodium stearate
- B. Cetyltrimethyl ammonium bromide ✓
- C. Sodium dodecylbenzene sulphonate
- D. Sodium lauryl sulphate
Solution: Cetyltrimethyl ammonium bromide, CH$_3$–(CH$_2$)$_{15}$–N$^{+}$(CH$_3$)$_3$ Br$^{-}$, carries the long chain on a quaternary ammonium cation, so it is a cationic detergent.
Sodium stearate is a soap; sodium dodecylbenzene sulphonate and sodium lauryl sulphate are anionic detergents.
Q147.
The calculated spin only magnetic moment of Cr$^{2+}$ ion is
- A. 4.90 BM ✓
- B. 5.92 BM
- C. 2.84 BM
- D. 3.87 BM
Solution: Electronic configuration of Cr is [Ar] 3d$^{5}$ 4s$^{1}$.
Electronic configuration of Cr$^{2+}$ is [Ar] 3d$^{4}$, so the number of unpaired electrons $n = 4$.
$$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} = 4.9\ \text{BM}$$
Q148.
HCl was passed through a solution of CaCl$_2$, MgCl$_2$ and NaCl. Which of the following compound(s) crystallise(s)?
- A. Only NaCl ✓
- B. Only MgCl$_2$
- C. NaCl, MgCl$_2$ and CaCl$_2$
- D. Both MgCl$_2$ and CaCl$_2$
Solution: CaCl$_2$ and MgCl$_2$ are more soluble than NaCl.
So on passing HCl(g) through a solution containing CaCl$_2$, MgCl$_2$ and NaCl, the common ion effect makes only NaCl crystallise out.
Q149.
Match the following and identify the correct option.
(a) CO(g) + H$_2$(g)
(b) Temporary hardness of water
(c) B$_2$H$_6$
(d) H$_2$O$_2$
(i) Mg(HCO$_3$)$_2$ + Ca(HCO$_3$)$_2$
(ii) An electron deficient hydride
(iii) Synthesis gas
(iv) Non-planar structure
- A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
- B. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
- C. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
- D. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) ✓
Solution: A mixture of CO and H$_2$ gases is known as water gas or synthesis gas, so (a) matches (iii).
Temporary hardness of water is due to bicarbonates of calcium and magnesium, so (b) matches (i).
Diborane (B$_2$H$_6$) is an electron deficient hydride, so (c) matches (ii).
H$_2$O$_2$ is a non-planar molecule having an open book like structure, so (d) matches (iv).
Q150.
Elimination reaction of 2-Bromo-pentane to form pent-2-ene is
(a) $\beta$-Elimination reaction
(b) Follows Zaitsev rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction
- A. (a), (c), (d)
- B. (b), (c), (d)
- C. (a), (b), (d)
- D. (a), (b), (c) ✓
Solution: CH$_3$–CHBr–CH$_2$–CH$_2$–CH$_3$ on elimination gives CH$_3$–CH=CH–CH$_2$–CH$_3$.
Since the $\beta$-hydrogen is abstracted, it is a $\beta$-elimination.
Since the more substituted alkene is formed, it follows Zaitsev's rule.
Since 'H' and 'Br' are removed, it is dehydrohalogenation.
It is not a dehydration reaction, because no water is eliminated, so (d) is wrong.
Q151.
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
- A. SCN$^{-}$ < F$^{-}$ < CN$^{-}$ < C$_2$O$_4^{2-}$
- B. F$^{-}$ < SCN$^{-}$ < C$_2$O$_4^{2-}$ < CN$^{-}$
- C. CN$^{-}$ < C$_2$O$_4^{2-}$ < SCN$^{-}$ < F$^{-}$
- D. SCN$^{-}$ < F$^{-}$ < C$_2$O$_4^{2-}$ < CN$^{-}$ ✓
Solution: Spectrochemical series as given in NCERT:
I$^{-}$ < Br$^{-}$ < SCN$^{-}$ < Cl$^{-}$ < S$^{2-}$ < F$^{-}$ < OH$^{-}$ < C$_2$O$_4^{2-}$ < H$_2$O < NCS$^{-}$ < EDTA$^{4-}$ < NH$_3$ < en < CN$^{-}$ < CO
Hence the increasing order asked for is SCN$^{-}$ < F$^{-}$ < C$_2$O$_4^{2-}$ < CN$^{-}$.
Q152.
Identify the correct statement from the following :
- A. Blister copper has blistered appearance due to evolution of CO$_2$.
- B. Vapour phase refining is carried out for Nickel by Van Arkel method.
- C. Pig iron can be moulded into a variety of shapes. ✓
- D. Wrought iron is impure iron with 4% carbon.
Solution: The iron obtained from the blast furnace contains about 4% carbon and many impurities like S, P, Si and Mn in smaller amounts. This is known as pig iron and it is cast into a variety of shapes.
Blister copper blisters because of evolution of SO$_2$, not CO$_2$.
Nickel is refined by the Mond process; the Van Arkel method is used for Zr and Ti.
Wrought iron is the purest form of iron, with about 0.1 to 0.5% carbon.
Q153.
Sucrose on hydrolysis gives
- A. $\alpha$-D-Glucose + $\beta$-D-Glucose
- B. $\alpha$-D-Glucose + $\beta$-D-Fructose ✓
- C. $\alpha$-D-Fructose + $\beta$-D-Fructose
- D. $\beta$-D-Glucose + $\alpha$-D-Fructose
Solution: Sucrose on hydrolysis gives $\alpha$-D-Glucose and $\beta$-D-Fructose.
The glycosidic linkage in sucrose joins C1 of $\alpha$-D-glucose to C2 of $\beta$-D-fructose, so hydrolysis releases exactly these two units.
Q154.
What is the change in oxidation number of carbon in the following reaction?
CH$_4$(g) $+$ 4Cl$_2$(g) $\rightarrow$ CCl$_4$(l) $+$ 4HCl(g)
- A. 0 to + 4
- B. – 4 to + 4 ✓
- C. 0 to – 4
- D. + 4 to + 4
Solution: In CH$_4$: $x + 4 \times 1 = 0 \Rightarrow x = -4$
In CCl$_4$: $x + 4 \times (-1) = 0 \Rightarrow x = +4$
So the change in oxidation state of carbon is from $-4$ to $+4$.
Q155.
The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.
- A. Copper
- B. Calcium
- C. Potassium ✓
- D. Iron
Solution: Potassium (K) activates many enzymes, participates in the oxidation of glucose to produce ATP and, along with Na, helps in the transmission of nerve signals.
Q156.
Which of the following alkane cannot be made in good yield by Wurtz reaction?
- A. 2,3-Dimethylbutane
- B. n-Heptane ✓
- C. n-Butane
- D. n-Hexane
Solution: Wurtz reaction is used to prepare symmetrical alkanes of the type R$_1$–R$_1$:
R$_1$–X + 2Na + X–R$_1$ $\xrightarrow{\text{Dry ether}}$ R$_1$–R$_1$ + 2NaX
If R$_1$ and R$_2$ are different, a mixture of alkanes is obtained:
R$_1$–X + 2Na + R$_2$–X $\xrightarrow{\text{Dry ether}}$ R$_1$–R$_1$ + R$_1$–R$_2$ + R$_2$–R$_2$ + 2NaX
n-Heptane has an odd number of carbon atoms, so it cannot be made from two identical alkyl halides and cannot be obtained in good yield.
Q157.
Measuring Zeta potential is useful in determining which property of colloidal solution?
- A. Solubility
- B. Stability of the colloidal particles ✓
- C. Size of the colloidal particles
- D. Viscosity
Solution: Zeta potential is the potential difference between the fixed layer and the diffused layer of the electrical double layer around a colloidal particle.
It is a measure of the mutual repulsion between the particles, and hence of the stability of the colloidal solution.
Q158.
The freezing point depression constant (K$_f$) of benzene is 5.12 K kg mol$^{-1}$. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) :
- A. 0.80 K
- B. 0.40 K ✓
- C. 0.60 K
- D. 0.20 K
Solution: $$\Delta T_f = K_f m$$
$$\Delta T_f = 5.12\ (\text{K kg mol}^{-1}) \times 0.078\ (\text{mol kg}^{-1})$$
$$\Delta T_f = 0.399\ \text{K} \approx 0.40\ \text{K}$$
Q159.
Which of the following amine will give the carbylamine test?
- A. C$_6$H$_5$NHCH$_3$
- B. C$_6$H$_5$N(CH$_3$)$_2$
- C. C$_6$H$_5$NHC$_2$H$_5$
- D. C$_6$H$_5$NH$_2$ ✓
Solution: Only aliphatic and aromatic PRIMARY amines give the carbylamine reaction.
Aniline, C$_6$H$_5$NH$_2$, is a primary amine, so it responds to the test; the other three are secondary or tertiary amines.
Q160.
Which of the following is a natural polymer?
- A. poly (Butadiene-styrene)
- B. polybutadiene
- C. poly (Butadiene-acrylonitrile)
- D. cis-1, 4-polyisoprene ✓
Solution: The naturally occurring polymer natural rubber is cis-1,4-polyisoprene.
The other three are synthetic rubbers.
Q161.
Identify the incorrect statement.
- A. The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
- B. Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
- C. The oxidation states of chromium in CrO$_4^{2-}$ and Cr$_2$O$_7^{2-}$ are not the same. ✓
- D. Cr$^{2+}$ (d$^{4}$) is a stronger reducing agent than Fe$^{2+}$ (d$^{6}$) in water.
Solution: The oxidation state of Cr in CrO$_4^{2-}$ and in Cr$_2$O$_7^{2-}$ is $+6$ in both.
So the statement that they are not the same is incorrect; the other three statements are correct.
Q162.
Which of the following set of molecules will have zero dipole moment?
- A. Boron trifluoride, hydrogen fluoride, carbon dioxide, 1,3-dichlorobenzene
- B. Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene
- C. Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene ✓
- D. Ammonia, beryllium difluoride, water, 1,4-dichlorobenzene
Solution: BF$_3$ is trigonal planar and symmetrical, so $\mu = 0$.
BeF$_2$ is linear and symmetrical, so $\mu = 0$.
CO$_2$ is linear with the two C=O dipoles opposed, so $\mu = 0$.
In 1,4-dichlorobenzene the two C–Cl dipoles point in exactly opposite directions, so $\mu = 0$.
HF, NF$_3$, H$_2$O, NH$_3$ and 1,3-dichlorobenzene all have a resultant dipole moment.
Q163.
On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be
- A. Oxygen gas ✓
- B. H$_2$S gas
- C. SO$_2$ gas
- D. Hydrogen gas
Solution: During the electrolysis of dil. sulphuric acid using Pt electrodes the following reactions take place.
At cathode:
$$4\text{H}^{+}(aq) + 4e^{-} \longrightarrow 2\text{H}_2(g)$$
At anode:
$$2\text{H}_2\text{O}(l) \longrightarrow \text{O}_2(g) + 4\text{H}^{+}(aq) + 4e^{-}$$
So oxygen gas is obtained at the anode.
Q164.
Anisole on cleavage with HI gives
- A. Iodobenzene + CH$_3$OH
- B. Phenol + C$_2$H$_5$I
- C. Iodobenzene + C$_2$H$_5$OH
- D. Phenol + CH$_3$I ✓
Solution: Anisole, C$_6$H$_5$–O–CH$_3$, is first protonated on the ether oxygen by HI.
The iodide ion then attacks by S$_N$2 at the methyl carbon, because attack at the sp$^{2}$ carbon of the ring is not possible.
So the products are phenol and CH$_3$I.
Q165.
The number of protons, neutrons and electrons in $^{175}_{71}$Lu, respectively, are
- A. 104, 71 and 71
- B. 71, 71 and 104
- C. 175, 104 and 71
- D. 71, 104 and 71 ✓
Solution: For $^{175}_{71}$Lu:
Number of protons $= 71 =$ number of electrons.
Number of neutrons $=$ mass number $-$ number of protons $= 175 - 71 = 104$.
So the answer is 71, 104 and 71.
Q166.
Match the following :
Oxide:
(a) CO
(b) BaO
(c) Al$_2$O$_3$
(d) Cl$_2$O$_7$
Nature:
(i) Basic
(ii) Neutral
(iii) Acidic
(iv) Amphoteric
Which of the following is correct option?
- A. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) ✓
- B. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- C. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
- D. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
Solution: CO : Neutral oxide
BaO : Basic oxide
Al$_2$O$_3$ : Amphoteric oxide
Cl$_2$O$_7$ : Acidic oxide
Q167.
A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following ?
- A. + R effect of – CH$_3$ groups
- B. – R effect of – CH$_3$ groups
- C. Hyperconjugation ✓
- D. – I effect of – CH$_3$ groups
Solution: Tertiary butyl carbocation, (CH$_3$)$_3$C$^{+}$, has 9 $\alpha$-H atoms.
Secondary butyl carbocation, CH$_3$–CH$^{+}$–CH$_2$–CH$_3$, has 5 $\alpha$-H atoms.
The more the number of $\alpha$-H atoms, the more will be the hyperconjugation effect, hence the more will be the stability of the carbocation.
Q168.
Which one of the followings has maximum number of atoms ?
- A. 1 g of Mg(s) [Atomic mass of Mg = 24]
- B. 1 g of O$_2$(g) [Atomic mass of O = 16]
- C. 1 g of Li(s) [Atomic mass of Li = 7] ✓
- D. 1 g of Ag(s) [Atomic mass of Ag = 108]
Solution: Number of Mg atoms $= \dfrac{1}{24} \times N_A$
Number of O atoms $= \dfrac{1}{32} \times 2 \times N_A = \dfrac{1}{16} \times N_A$
Number of Li atoms $= \dfrac{1}{7} \times N_A$
Number of Ag atoms $= \dfrac{1}{108} \times N_A$
The largest of these is for lithium, so 1 g of Li has the maximum number of atoms.
Q169.
Which of the following is a basic amino acid ?
- A. Alanine
- B. Tyrosine
- C. Lysine ✓
- D. Serine
Solution: Lysine is H$_2$N–CH$_2$–CH$_2$–CH$_2$–CH$_2$–CH(NH$_2$)–COOH.
It carries an extra –NH$_2$ group in the side chain, so it is a basic amino acid.
Q170.
The correct option for free expansion of an ideal gas under adiabatic condition is
- A. q = 0, $\Delta$T < 0 and w > 0
- B. q < 0, $\Delta$T = 0 and w = 0
- C. q > 0, $\Delta$T > 0 and w > 0
- D. q = 0, $\Delta$T = 0 and w = 0 ✓
Solution: Free expansion means $P_{ex} = 0$, so $w = -P_{ex}\Delta V = 0$.
Adiabatic process means $q = 0$.
By the first law of thermodynamics $\Delta U = q + w = 0$.
The internal energy of an ideal gas is a function of temperature only, so if the internal energy remains constant then $\Delta T = 0$.
Q171.
Identify the incorrect match.
Name:
(a) Unnilunium
(b) Unniltrium
(c) Unnilhexium
(d) Unununnium
IUPAC Official Name:
(i) Mendelevium
(ii) Lawrencium
(iii) Seaborgium
(iv) Darmstadtium
- A. (b), (ii)
- B. (c), (iii)
- C. (d), (iv) ✓
- D. (a), (i)
Solution: Unununnium has atomic number 111 and its IUPAC official name is Roentgenium, not Darmstadtium.
So the match (d) with (iv) is incorrect; the other three matches are correct.
Q172.
Identify a molecule which does not exist.
- A. Li$_2$
- B. C$_2$
- C. O$_2$
- D. He$_2$ ✓
Solution: For the He$_2$ molecule the electronic configuration is $\sigma 1s^{2}$, $\sigma^{*}1s^{2}$.
$$\text{Bond order} = \frac{1}{2}[N_b - N_a] = \frac{1}{2}[2 - 2] = 0$$
Since the bond order is zero, the He$_2$ molecule does not exist.
Q173.
Identify the correct statements from the following :
(a) CO$_2$(g) is used as refrigerant for ice-cream and frozen food.
(b) The structure of C$_{60}$ contains twelve six carbon rings and twenty five carbon rings.
(c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline.
(d) CO is colorless and odourless gas.
- A. (a) and (c) only
- B. (b) and (c) only
- C. (c) and (d) only ✓
- D. (a), (b) and (c) only
Solution: It is dry ice, CO$_2$(s), and not CO$_2$(g), that is used as a refrigerant, so (a) is wrong.
C$_{60}$ contains 20 six-membered rings and 12 five-membered rings, which is the reverse of what (b) states, so (b) is wrong.
ZSM-5, a type of zeolite, is used to convert alcohols directly into gasoline, so (c) is correct.
CO is a colourless and odourless gas, so (d) is correct.
Q174.
An alkene on ozonolysis gives methanal as one of the product. Its structure is
- A. Structure (1)
- B. Structure (2) ✓
- C. Structure (3)
- D. Structure (4)
Solution: Methanal (HCHO) is produced only from a terminal $=$CH$_2$ group.
Allyl cyclohexane, in which a cyclohexane ring carries a –CH$_2$–CH=CH$_2$ side chain, has such a terminal methylene group.
On ozonolysis followed by Zn/H$_2$O it gives cyclohexyl acetaldehyde and methanal.
Q175.
Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :
- A. Sec. butyl alcohol
- B. Tert. butyl alcohol ✓
- C. Isobutyl alcohol
- D. Isopropyl alcohol
Solution: The Grignard reagent CH$_3$MgCl adds across the polar C=O bond of acetone, the methyl carbanion attacking the electrophilic carbonyl carbon.
This gives the magnesium alkoxide (CH$_3$)$_3$C–OMgCl.
On hydrolysis it gives (CH$_3$)$_3$C–OH, tert-butyl alcohol.
Q176.
A mixture of N$_2$ and Ar gases in a cylinder contains 7 g of N$_2$ and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N$_2$ is : [Use atomic masses (in g mol$^{-1}$) : N = 14, Ar = 40]
- A. 12 bar
- B. 15 bar ✓
- C. 18 bar
- D. 9 bar
Solution: $$n_{N_2} = \frac{7}{28} = 0.25,\qquad n_{Ar} = \frac{8}{40} = 0.20$$
Applying Dalton's law of partial pressure,
$$p_{N_2} = \chi_{N_2}\,P_{Total} = \frac{0.25}{0.45} \times 27 = \frac{5}{9} \times 27 = 15\ \text{bar}$$
Q177.
An increase in the concentration of the reactants of a reaction leads to change in
- A. heat of reaction
- B. threshold energy
- C. collision frequency ✓
- D. activation energy
Solution: The number of collisions per second per unit volume of the reaction mixture is known as collision frequency.
Therefore with an increase in the concentration of reactants the collision frequency changes.
Heat of reaction, threshold energy and activation energy do not depend on concentration.
Q178.
Find out the solubility of Ni(OH)$_2$ in 0.1 M NaOH. Given that the ionic product of Ni(OH)$_2$ is $2 \times 10^{-15}$
- A. $2 \times 10^{-8}$ M
- B. $1 \times 10^{-13}$ M
- C. $1 \times 10^{8}$ M
- D. $2 \times 10^{-13}$ M ✓
Solution: Ni(OH)$_2$ $\rightleftharpoons$ Ni$^{2+}$ + 2OH$^{-}$, giving s and 2s.
NaOH $\longrightarrow$ Na$^{+}$ + OH$^{-}$, giving 0.1 M OH$^{-}$.
Total [OH$^{-}$] $= 2s + 0.1 \approx 0.1$
Ionic product $= [\text{Ni}^{2+}][\text{OH}^{-}]^{2}$
$$2 \times 10^{-15} = s(0.1)^{2}$$
$$s = 2 \times 10^{-13}$$
Solubility of Ni(OH)$_2$ $= 2 \times 10^{-13}$ M.
Q179.
Identify compound X in the following sequence of reactions
Toluene $\xrightarrow{\ \text{Cl}_2/h\nu\ }$ X $\xrightarrow[373\ \text{K}]{\ \text{H}_2\text{O}\ }$ Benzaldehyde
- A. C$_6$H$_5$CH$_2$Cl
- B. C$_6$H$_5$CHCl$_2$ ✓
- C. C$_6$H$_5$CCl$_3$
- D. C$_6$H$_5$Cl
Solution: In presence of light, chlorine attacks the side chain of toluene and gives benzal chloride, C$_6$H$_5$CHCl$_2$, which is X.
Hydrolysis of benzal chloride at 373 K first gives the unstable gem-diol C$_6$H$_5$CH(OH)$_2$.
Loss of water from the gem-diol gives benzaldehyde.
Q180.
Urea reacts with water to form A which will decompose to form B. B when passed through Cu$^{2+}$ (aq), deep blue colour solution C is formed. What is the formula of C from the following ?
- A. [Cu(NH$_3$)$_4$]$^{2+}$ ✓
- B. Cu(OH)$_2$
- C. CuCO$_3\cdot$Cu(OH)$_2$
- D. CuSO$_4$
Solution: NH$_2$CONH$_2$ + H$_2$O $\longrightarrow$ (NH$_4$)$_2$CO$_3$, which is A.
On heating, A decomposes: (NH$_4$)$_2$CO$_3$ $\xrightarrow{\Delta}$ NH$_3$(g) + CO$_2$(g) + H$_2$O(l), so B is NH$_3$.
NH$_3$ passed through Cu$^{2+}$(aq) gives the deep blue tetraamminecopper(II) ion:
$$\text{NH}_3(g) \xrightarrow{\text{Cu}^{2+}(aq)} [\text{Cu(NH}_3)_4]^{2+}$$
So C is [Cu(NH$_3$)$_4$]$^{2+}$.