NEET Previous Year Question Paper 2020 with Solutions

 

1. Name the enzyme that facilitates opening of DNA helix during transcription.
(1) DNA polymerase
(2) RNA polymerase
(3) DNA ligase
(4) DNA helicase
Answer (2)
Sol. RNA polymerase facilitates opening of DNA helix during transcription.

2. Which of the following would help in prevention of diuresis?
(1) Atrial natriuretic factor causes vasoconstriction
(2) Decrease in secretion of renin by JG cells
(3) More water reabsorption due to undersecretion of ADH
(4) Reabsorption of Na+ and water from renal tubules due to aldosterone
Answer (4)
Sol. Adrenal cortex secretes mineralocorticoids like aldosterone which increase the reabsorption of Na+ and water from renal tubule that prevent diuresis.

3. Meiotic division of the secondary oocyte is completed
(1) After zygote formation
(2) At the time of fusion of a sperm with an ovum
(3) Prior to ovulation
(4) At the time of copulation
Answer (2)
Sol. Meiotic division of secondary oocyte is completed after the entry of sperm in secondary oocyte which lead to the formation of a large ovum and a tiny IInd polar body.

4. Match the following concerning essential elements and their functions in plants
(a) Iron (i) Photolysis of water
(b) Zinc (ii) Pollen germination
(c) Boron (iii) Required for chlorophyll biosynthesis
(d) Manganese (iv) IAA biosynthesis
Select the correct option
(a) (b) (c) (d)
(1) (iii) (iv) (ii) (i)
(2) (iv) (i) (ii) (iii)
(3) (ii) (i) (iv) (iii)
(4) (iv) (iii) (ii) (i)
Answer (1)
Sol. (a) Iron - Essential for the formation of chlorophyll
(b) Zinc - Needed for synthesis of auxin
(c) Boron - Have a role in pollen grain germination
(d) Manganese - Is involved in the splitting of water to liberate O2 during photosynthesis

5. Which of the following pairs is of unicellular algae?
(1) Anabaena and Volvox
(2) Chlorella and Spirulina
(3) Laminaria and Sargassum
(4) Gelidium and Gracilaria
Answer (2)
Sol. Chlorella and Spirulina are unicellular algae. Gelidium, Gracilaria, Laminaria and Sargassum are multicellular. Volvox is colonial.

6. The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of
(1) 1 molecule of 6-C compound
(2) 1 molecule of 4-C compound and 1 molecule of 2-C compound
(3) 2 molecules of 3-C compound
(4) 1 molecule of 3-C compound
Answer (4)
Sol. In photorespiration, O2 binds to RubisCo. As a result RuBP instead to being converted to 2 molecules of PGA bind with O2 to form one molecule each of phosphoglycerate (3 carbon compound) and phosphoglycolate (2 carbon compound).

7. Match the following columns and select the correct option.
Column-I Column-II
(a) Eosinophils (i) Immune response
(b) Basophils (ii) Phagocytosis
(c) Neutrophils (iii) Release histaminase, destructive enzymes
(d) Lymphocytes (iv) Release granules containing histamine
(a) (b) (c) (d)
(1) (i) (ii) (iv) (iii)
(2) (ii) (i) (iii) (iv)
(3) (iii) (iv) (ii) (i)
(4) (iv) (i) (ii) (iii)
Answer (3)
Sol. Option (3) is the correct answer because Eosinophils are associated with allergic reactions and release histaminase, destructive enzymes, so (a) in column I matches with (iii) in column II.
Basophils secrete histamine, serotonin, heparin etc. and are involved in inflammatory reactions, so (b) matches with (iv).
Neutrophils are phagocytic cells; so (c) matches with (ii). Both B and T lymphocytes are responsible for immune responses of the body, so, (d) in column I matches with (i) in column II.

8. Match the following columns and select the correct option.
Column-I Column-II
(a) Placenta (i) Androgens
(b) Zona pellucida (ii) Human Chorionic Gonadotropin (hCG)
(c) Bulbo-urethral glands (iii) Layer of the ovum
(d) Leydig cells (iv) Lubrication of the Penis
(a) (b) (c) (d)
(1) (iii) (ii) (iv) (i)
(2) (ii) (iii) (iv) (i)
(3) (iv) (iii) (i) (ii)
(4) (i) (iv) (ii) (iii)
Answer (2)
Sol. The correct option is (2) because
(a) Placenta secretes human chorionic gonadotropin (hCG)
(b) Zona pellucida is a primary egg membrane secreted by the secondary oocyte
(c) The secretions of bulbourethral glands help in lubrication of the penis
(d) Leydig cells synthesise and secrete testicular hormones called androgens

9. 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
(1) (c) and (d)
(2) (a) and (d)
(3) (a) only
(4) (a), (b) and (c)
Answer (2)
Sol. 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.

10. Which of the following statements about inclusion bodies is incorrect?
(1) They lie free in the cytoplasm
(2) These represent reserve material in cytoplasm
(3) They are not bound by any membrane
(4) These are involved in ingestion of food particles
Answer (4)
Sol. These are not involved in ingestion of food particles.

11. Strobili or cones are found in
(1) Marchantia
(2) Equisetum
(3) Salvinia
(4) Pteris
Answer (2)
Sol. Strobili or cones are found in Equisetum.

12. Montreal protocol was signed in 1987 for control of
(1) Release of Green House gases
(2) Disposal of e-wastes
(3) Transport of Genetically modified organisms from one country to another
(4) Emission of ozone depleting substances
Answer (4)
Sol. Montreal protocol - Signed in 16 Sep, 1987 (Ozone day)
Came into force - 1 Jan, 1989.
It was aimed at stopping the production and import of ODS and reduce their concentration in the atmosphere.

13. Which of the following statements is correct?
(1) Adenine pairs with thymine through three H-bonds
(2) Adenine does not pair with thymine
(3) Adenine pairs with thymine through two H-bonds
(4) Adenine pairs with thymine through one H-bond
Answer (3)
Sol. Adenine pairs with thymine through two H-bonds i.e., A = T.

14. The body of the ovule is fused within the funicle at
(1) Nucleolus
(2) Chalaza
(3) Hilum
(4) Micropyle
Answer (3)
Sol. The attachment point of funicle and body of ovule is known as hilum.

15. The sequence that controls the copy number of the linked DNA in the vector, is termed
(1) Palindromic sequence
(2) Recognition site
(3) Selectable marker
(4) Ori site
Answer (4)
Sol. The correct option is (4) because 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.

16. Identify the wrong statement with regard to Restriction Enzymes.
(1) They are useful in genetic engineering.
(2) Sticky ends can be joined by using DNA ligases.
(3) Each restriction enzyme functions by inspecting the length of a DNA sequence.
(4) They cut the strand of DNA at palindromic sites.
Answer (2)
Sol. Restriction endonucleases make cuts at specific positions within the DNA.
They function by inspecting the length of a DNA sequence.
Restriction endonuclease bind to the DNA and cut the two strands of double helix at specific points in their sugar-phosphate backbones.
They are used in genetic engineering to form recombinant molecules of DNA.
DNA ligases join the DNA fragments.

17. The product(s) of reaction catalyzed by nitrogenase in root nodules of leguminous plants is/are
(1) Ammonia and oxygen
(2) Ammonia and hydrogen
(3) Ammonia alone
(4) Nitrate alone
Answer (2)
Sol. N2 + 8e- + 8H+ + 16ATP --Mg2+--> 2NH3 + H2 + 16ADP + 16Pi
Ammonia and Hydrogen.

18. In light reaction, plastoquinone facilitates the transfer of electrons from
(1) PS-I to NADP+
(2) PS-I to ATP synthase
(3) PS-II to Cytb f complex
(4) Cytb f complex to PS-I
Answer (3)
Sol. After excitement, e- is passed from PS-II (P680) to primary electron acceptor (Pheophytin). From primary e- acceptor, e- is passed to plastoquinone. Plastoquinone (PQ) in turn transfer its e- to Cyt b6f complex. Therefore plastoquinone facilitates the transfer of electrons from PS-II to Cyt b6f complex.

19. Which of the following hormone levels will cause release of ovum (ovulation) from the graffian follicle?
(1) Low concentration of LH
(2) Low concentration of FSH
(3) High concentration of Estrogen
(4) High concentration of Progesterone
Answer (3)
Sol. High level of estrogen will send positive feedback to anterior pituitary for release of LH. FSH, LH and estrogen are at peak level during mid of menstrual cycle (28 day cycle). LH surge leads to ovulation.

20. The first phase of translation is
(1) Aminoacylation of tRNA
(2) Recognition of an anti-codon
(3) Binding of mRNA to ribosome
(4) Recognition of DNA molecule
Answer (1)
Sol. The first phase of translation involves activation of amino acid in the presence of ATP and linked to their cognate tRNA - a process commonly called as charging of tRNA or aminoacylation of tRNA.

21. The roots that originate from the base of the stem are
(1) Prop roots
(2) Lateral roots
(3) Fibrous roots
(4) Primary roots
Answer (3)
Sol. The roots that originate from the base of the stem are fibrous roots.

22. Identify the wrong statement with reference to transport of oxygen
(1) Higher H+ conc. in alveoli favours the formation of oxyhaemoglobin
(2) Low pCO2 in alveoli favours the formation of oxyhaemoglobin
(3) Binding of oxygen with haemoglobin is mainly related to partial pressure of O2
(4) Partial pressure of CO2 can interfere with O2 binding with haemoglobin
Answer (1)
Sol. The correct option is (1) because higher H+ concentration favours the dissociation of oxygen from oxyhaemoglobin in tissues.
In the alveoli, high pO2, low pCO2, lesser H+ concentration and lower temperature favour formation of oxyhaemoglobin.

23. In gel electrophoresis, separated DNA fragments can be visualized with the help of
(1) Acetocarmine in UV radiation
(2) Ethidium bromide in infrared radiation
(3) Acetocarmine in bright blue light
(4) Ethidium bromide in UV radiation
Answer (4)
Sol. The separated DNA fragments can be visualised only after staining the DNA with Ethidium bromide followed by exposure to UV radiation.

24. The enzyme enterokinase helps in conversion of
(1) caseinogen into casein
(2) pepsinogen into pepsin
(3) protein into polypeptides
(4) trypsinogen into trypsin
Answer (4)
Sol. The correct option is (4) because trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin. Trypsinogen is a zymogen from pancreas.

25. Experimental verification of the chromosomal theory of inheritance was done by
(1) Boveri
(2) Morgan
(3) Mendel
(4) Sutton
Answer (2)
Sol. Experimental verification of the chromosomal theory of inheritance was done by Morgan.
Note: Sutton and Boveri proposed chromosomal theory of inheritance but it was experimentally verified by T.H. Morgan.

26. According to Robert May, the global species diversity is about
(1) 50 million
(2) 7 million
(3) 1.5 million
(4) 20 million
Answer (2)
Sol. Robert May estimated global species diversity at about 7 million. Although some extreme estimates range from 20 to 50 million.

27. Match the organism with its use in biotechnology.
(a) Bacillus thuringiensis (i) Cloning vector
(b) Thermus aquaticus (ii) Construction of first rDNA molecule
(c) Agrobacterium tumefaciens (iii) DNA polymerase
(d) Salmonella typhimurium (iv) Cry proteins
Select the correct option from the following:
(a) (b) (c) (d)
(1) (iii) (ii) (iv) (i)
(2) (iii) (iv) (i) (iii)
(3) (ii) (iv) (iii) (i)
(4) (iv) (iii) (i) (ii)
Answer (4)
Sol. (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 1st recombinant DNA molecule was performed using native plasmid of Salmonella typhimurium.

28. Identify the correct statement with regard to G1 phase (Gap 1) of interphase.
(1) Cell is metabolically active, grows but does not replicate its DNA.
(2) Nuclear Division takes place.
(3) DNA synthesis or replication takes place.
(4) Reorganisation of all cell components takes place.
Answer (1)
Sol. During G1 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.

29. Which of the following is correct about viroids?
(1) They have DNA with protein coat
(2) They have free DNA without protein coat
(3) They have RNA with protein coat
(4) They have free RNA without protein coat
Answer (4)
Sol. Viroids have free RNA without protein coat.

30. 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 :
(1) Dicotyledonous stem
(2) Dicotyledonous root
(3) Monocotyledonous stem
(4) Monocotyledonous root
Answer (3)
Sol. All features are related to monocotyledonous stems.

31. From his experiments, S.L. Miller produced amino acids by mixing the following in a closed flask
(1) CH4, H2, NH3 and water vapor at 600°C
(2) CH3, H2, NH3 and water vapor at 600°C
(3) CH4, H2, NH3 and water vapor at 800°C
(4) CH3, H2, NH4 and water vapor at 800°C
Answer (3)
Sol. In 1953, S.L. Miller, an American scientist created electric discharge in a closed flask containing CH4, H2, NH3 and water vapor at 800°C.

32. Identify the basic amino acid from the following.
(1) Lysine
(2) Valine
(3) Tyrosine
(4) Glutamic Acid
Answer (1)
Sol. Option (1) is the correct answer because 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.

33. Snow-blindness in Antarctic region is due to
(1) High reflection of light from snow
(2) Damage to retina caused by infra-red rays
(3) Freezing of fluids in the eye by low temperature
(4) Inflammation of cornea due to high dose of UV-B radiation
Answer (4)
Sol. UV-B radiations damage DNA and mutations may occur. In human eye, cornea absorbs UV-B radiations, and a high dose of UV-B causes inflammation of cornea called snow blindness, cataract, etc.

34. Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (G0). This process occurs at the end of
(1) S phase
(2) G2 phase
(3) M phase
(4) G1 phase
Answer (3)
Sol. Some dividing cells exit the cell cycle and enter vegetative inactive stage, called quiescent stage (G0). This process occurs at the end of M-phase and beginning of G1 phase.

35. Which of the following regions of the globe exhibits highest species diversity?
(1) Himalayas
(2) Amazon forests
(3) Western Ghats of India
(4) Madagascar
Answer (2)
Sol. The largely tropical Amazonian rain forest in South America has the greatest biodiversity on earth.

36. Identify the incorrect statement.
(1) Sapwood is the innermost secondary xylem and is lighter in colour
(2) Due to deposition of tannins, resins, oils etc., heart wood is dark in colour
(3) Heart wood does not conduct water but gives mechanical support
(4) Sapwood is involved in conduction of water and minerals from root to leaf
Answer (1)
Sol. Incorrect statement: Sapwood is the innermost secondary xylem and is lighter in colour.
Correct statement: Sapwood is outermost secondary xylem.

37. Floridean starch has structure similar to
(1) Mannitol and algin
(2) Laminarin and cellulose
(3) Starch and cellulose
(4) Amylopectin and glycogen
Answer (4)
Sol. Floridean starch is stored food material in red algae. Its structure is similar to Amylopectin and Glycogen.

38. Which of the following is not an attribute of a population?
(1) Mortality
(2) Species interaction
(3) Sex ratio
(4) Natality
Answer (2)
Sol. Natality Population attribute
Mortality Population attribute
Species interaction Population interaction
Sex ratio Population attribute

39. The number of substrate level phosphorylations in one turn of citric acid cycle is
(1) Two
(2) Three
(3) Zero
(4) One
Answer (4)
Sol. One substrate level phosphorylation in one turn of citric acid cycle as per following reaction:
Succinyl Co-A -> Succinate
GDP -> GTP
ATP <- ADP

40. Identify the correct statement with reference to human digestive system.
(1) Ileum is a highly coiled part
(2) Vermiform appendix arises from duodenum
(3) Ileum opens into small intestine
(4) Serosa is the innermost layer of the alimentary canal
Answer (1)
Sol. Option (1) is correct as ileum is a highly coiled tube. Serosa is the outermost layer of the alimentary canal, thus, option (4) is an incorrect statement.
A narrow finger-like tubular projection, the vermiform appendix arises from caecum part of large intestine thus, option (2) is incorrect statement. Ileum opens into the large intestine, thus option (3) is also an incorrect statement.

41. In which of the following techniques, the embryos are transferred to assist those females who cannot conceive?
(1) ICSI and ZIFT
(2) GIFT and ICSI
(3) ZIFT and IUT
(4) GIFT and ZIFT
Answer (3)
Sol. Option (3) is the answer because ART 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.
Option (1), (2) and (4) are incorrect because in GIFT (Gamete Intrafallopian Transfer), gamete is transferred into the fallopian tube of female who cannot produce ova. ICSI is Intra cytoplasmic sperm injection in which sperm is directly injected into the ovum.

42. In relation to Gross primary productivity and Net primary productivity of an ecosystem, which one of the following statements is correct?
(1) Gross primary productivity and Net primary productivity are one and same
(2) There is no relationship between Gross primary productivity and Net primary productivity
(3) Gross primary productivity is always less than net primary productivity
(4) Gross primary productivity is always more than net primary productivity
Answer (4)
Sol. Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. Net primary productivity is GPP - respiration.
Hence gross primary productivity is always more than NPP.

43. Name the plant growth regulator which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.
(1) Ethylene
(2) Abscisic acid
(3) Cytokinin
(4) Gibberellin
Answer (4)
Sol. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.

44. Secondary metabolites such as nicotine, strychnine and caffeine are produced by plants for their
(1) Defence action
(2) Effect on reproduction
(3) Nutritive value
(4) Growth response
Answer (1)
Sol. 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 (plants) as defences against grazers and browsers.

45. Select the correct match
(1) Sickle cell anaemia - Autosomal recessive trait, chromosome-11
(2) Thalassemia - X linked
(3) Haemophilia - Y linked
(4) Phenylketonuria - Autosomal dominant trait
Answer (1)
Sol. Phenylketonuria - Autosomal recessive disorder
Thalassemia - Autosomal recessive disorder
Haemophilia - X linked recessive disorder
Sickle cell anaemia - Autosomal recessive trait, caused due to mutation in gene present on chromosome no. 11

46. Select the correct statement.
(1) Insulin acts on pancreatic cells and adipocytes.
(2) Insulin is associated with hyperglycemia.
(3) Glucocorticoids stimulate gluconeogenesis.
(4) Glucagon is associated with hypoglycemia.
Answer (3)
Sol. Glucagon is associated with hyperglycemia. Insulin acts on hepatocytes and adipocytes and is associated with hypoglycemia. Glucocorticoids stimulate gluconeogenesis, so increase blood sugar level.

47. 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.
(1) (b), (c) and (d)
(2) only (d)
(3) only (a)
(4) (a) and (c)
Answer (1)
Sol. The correct option is (1) because :
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 example of natural selection, adaptive radiation and founder's effect.

48. Choose the correct pair from the following
(1) Nucleases - Separate the two strands of DNA
(2) Exonucleases - Make cuts at specific positions within DNA
(3) Ligases - Join the two DNA molecules
(4) Polymerases - Break the DNA into fragments
Answer (3)
Sol. Ligases join the two DNA molecules.

49. Embryological support for evolution was disapproved by
(1) Charles Darwin
(2) Oparin
(3) Karl Ernst von Baer
(4) Alfred Wallace
Answer (3)
Sol. 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.

50. Goblet cells of alimentary canal are modified from
(1) Chondrocytes
(2) Compound epithelial cells
(3) Squamous epithelial cells
(4) Columnar epithelial cells
Answer (4)
Sol. Goblet cells of alimentary canal are modified from columnar epithelial cells which secrete mucus.

51. Bt cotton variety that was developed by the introduction of toxin gene of Bacillus thuringiensis (Bt) is resistant to
(1) Plant nematodes
(2) Insect predators
(3) Insect pests
(4) Fungal diseases
Answer (3)
Sol. Bt cotton is resistant to cotton bollworm (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 as biopesticide.

52. 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.
(1) (a) and (b)
(2) (b) and (c)
(3) (d) and (c)
(4) (c) and (a)
Answer (2)
Sol. In vertebrates, notochord is present during embryonic period only as it is replaced by vertebral column.
In chordates, central nervous system is dorsal and hollow.

53. Which of the following is put into Anaerobic sludge digester for further sewage treatment?
(1) Effluents of primary treatment
(2) Activated sludge
(3) Primary sludge
(4) Floating debris
Answer (2)
Sol. The sediment in settlement tank is called activated sludge.
A small part of the activated sludge is pumped back into aeration tank Remaining major part of the sludge is pumped into large tank called anaerobic sludge digesters.

54. Identify the substances having glycosidic bond and peptide bond, respectively in their structure
(1) Cellulose, lecithin
(2) Inulin, insulin
(3) Chitin, cholesterol
(4) Glycerol, trypsin
Answer (2)
Sol. Inulin is a fructan (polysaccharide of fructose). Adjacent fructose units are linked through glycosidic bond.
Insulin is a protein composed of 51 aminoacids. Adjacent aminoacids are attached through peptide bond.

55. Match the following diseases with the causative organism and select the correct option.
Column-I Column-II
(a) Typhoid (i) Wuchereria
(b) Pneumonia (ii) Plasmodium
(c) Filariasis (iii) Salmonella
(d) Malaria (iv) Haemophilus
(a) (b) (c) (d)
(1) (ii) (i) (iii) (iv)
(2) (iv) (i) (ii) (iii)
(3) (i) (iii) (ii) (iv)
(4) (iii) (iv) (i) (ii)
Answer (4)
Sol. Typhoid fever in humans is caused by pathogenic bacterium Salmonella typhi.
Pneumonia is caused by Streptococcus Pneumoniae and Haemophilus influenzae.
Filariasis or elephantiasis is caused by the filarial worm, Wuchereria bancrofti and Wuchereria malayi.
Malaria is caused by different species of Plasmodium.

56. Match the following columns and select the correct option.
Column-I Column-II
(a) Clostridium butylicum (i) Cyclosporin-A
(b) Trichoderma polysporum (ii) Butyric Acid
(c) Monascus purpureus (iii) Citric Acid
(d) Aspergillus niger (iv) Blood cholesterol lowering agent
(a) (b) (c) (d)
(1) (i) (ii) (iv) (iii)
(2) (iv) (iii) (ii) (i)
(3) (iii) (iv) (ii) (i)
(4) (ii) (i) (iv) (iii)
Answer (4)
Sol. Column-I Column-II
(a) Clostridium butylicum (ii) Butyric acid
(b) Trichoderma polysporum (i) Cyclosporin-A
(c) Monascus purpureus (iv) Blood cholesterol lowering agent
(d) Aspergillus niger (iii) Citric acid

57. By which method was a new breed 'Hisardale' of sheep formed by using Bikaneri ewes and Marino rams?
(1) Cross breeding
(2) Inbreeding
(3) Out crossing
(4) Mutational breeding
Answer (1)
Sol. Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri-ewe and Marino rams. In cross-breeding, superior male of one breed are mated with superior females of another breed.

58. 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
(1) (a), (b) and (d)
(2) only (d)
(3) (a) and (b)
(4) (c) and (d)
Answer (3)
Sol. Inspiration is initiated by the contraction of diaphragm, which increases the volume of thoracic chamber in the anterio-posterior axis.
The contraction of external intercostal muscles increase the volume of the thoracic chamber in the dorsoventral axis.

59. Match the following columns and select the correct option.
Column-I Column-II
(a) Gregarious, polyphagous pest (i) Asterias
(b) Adult with radial symmetry and larva with bilateral symmetry (ii) Scorpion
(c) Book lungs (iii) Ctenoplana
(d) Bioluminescence (iv) Locusta
(a) (b) (c) (d)
(1) (iii) (ii) (i) (iv)
(2) (ii) (i) (iii) (iv)
(3) (i) (iii) (ii) (iv)
(4) (iv) (i) (ii) (iii)
Answer (4)
Sol. (a) Locusta is a gregarious pest.
(b) In Echinoderms, adults are radially symmetrical but larvae are bilaterally symmetrical.
(c) Scorpions respire through book lungs.
(d) Bioluminescence is well marked in ctenophores.

60. Which is the important site of formation of glycoproteins and glycolipids in eukaryotic cells?
(1) Golgi bodies
(2) Polysomes
(3) Endoplasmic reticulum
(4) Peroxisomes
Answer (1)
Sol. Golgi bodies are site of formation of glycoproteins and glycolipids in eukaryotic cells.

61. The specific palindromic sequence which is recognized by EcoRI is
(1) 5'-CTTAAG-3'
3'-GAATTC-5'
(2) 5'-GGATCC-3'
3'-CCTAGG-5'
(3) 5'-GAATTC-3'
3'-CTTAAG-5'
(4) 5'-GGAACC-3'
3'-CCTTGG-5'
Answer (3)
Sol. The correct option is (3) because the specific palindromic sequence which is recognised by EcoRI is
5'-GAATTC-3'
3'-CTTAAG-5'

62. Dissolution of the synaptonemal complex occurs during
(1) Diplotene
(2) Leptotene
(3) Pachytene
(4) Zygotene
Answer (1)
Sol. Dissolution of the synaptonemal complex occurs During Diplotene stage of Prophase-I of Meiosis-I.

63. Match the trophic levels with their correct species examples in grassland ecosystem.
(a) Fourth trophic level (i) Crow
(b) Second trophic level (ii) Vulture
(c) First trophic level (iii) Rabbit
(d) Third trophic level (iv) Grass
Select the correct option
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (i) (ii) (iii) (iv)
(3) (ii) (iii) (iv) (i)
(4) (iii) (ii) (i) (iv)
Answer (3)
Sol. Grassland ecosystem is a terrestrial ecosystem. It includes various trophic levels
First trophic level (T1) - Grass
Second trophic level (T2) - Rabbit
Third trophic level (T3) - Crow
Fourth trophic level (T4) - Vulture

64. How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?
(1) 14
(2) 8
(3) 4
(4) 2
Answer (1)
Sol. Mendel selected 14 True breeding plant varieties.

65. Match the following columns and select the correct option.
Column-I Column-II
(a) Bt cotton (i) Gene therapy
(b) Adenosine deaminase deficiency (ii) Cellular defence
(c) RNAi (iii) Detection of HIV infection
(d) PCR (iv) Bacillus thuringiensis
(a) (b) (c) (d)
(1) (ii) (iii) (iv) (i)
(2) (i) (ii) (iii) (iv)
(3) (iv) (i) (ii) (iii)
(4) (iii) (ii) (i) (iv)
Answer (3)
Sol. The correct option is (3) because
(a) In Bt cotton the specific Bt toxin gene was isolated from Bacillus thuringiensis.
(b) The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency.
(c) RNAi (RNA interference) takes place in all eukaryotic organisms as a method of cellular defense.
(d) PCR is now routinely used to detect HIV in suspected AIDS patients.

66. Match the following columns and select the correct option.
Column-I Column-II
(a) 6-15 pairs of gill slits (i) Trygon
(b) Heterocercal caudal fin (ii) Cyclostomes
(c) Air Bladder (iii) Chondrichthyes
(d) Poison sting (iv) Osteichthyes
(a) (b) (c) (d)
(1) (iv) (ii) (iii) (i)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iii) (iv) (i)
(4) (iii) (iv) (i) (ii)
Answer (3)
Sol. Cyclostomes have an elongated body bearing 6-15 pairs of gill slits for respiration, so (a) matches with (ii) in column-II.
Air bladder is present in bony fishes belonging to class Osteichthyes which regulates buoyancy, so (c) matches with (iv) in column-II.
Trygon, a cartilaginous fish, possesses poison sting, so, (d) matches with (i) in column-II.
Heterocercal caudal fin is present in members of class Chondrichthyes, so (b) in column-I matches with (iii) in column-II.

67. The process of growth is maximum during
(1) Senescence
(2) Dormancy
(3) Log phase
(4) Lag phase
Answer (3)
Sol. In exponential growth, the initial growth is slow (lag phase) and it increases rapidly thereafter at an exponential rate in log or exponential phase.

68. Identify the wrong statement with reference to immunity.
(1) Active immunity is quick and gives full response.
(2) Foetus receives some antibodies from mother, it is an example for passive immunity.
(3) When exposed to antigen (living or dead) antibodies are produced in the host's body. It is called "Active immunity".
(4) When ready-made antibodies are directly given, it is called "Passive immunity".
Answer (1)
Sol. The correct option is (1) because active immunity is slow and takes time to give its full effective response in comparison to passive immunity where pre-formed antibodies are administered.

69. Match the following columns and select the correct option.
Column-I Column-II
(a) Floating Ribs (i) Located between second and seventh ribs
(b) Acromion (ii) Head of the Humerus
(c) Scapula (iii) Clavicle
(d) Glenoid cavity (iv) Do not connect with the sternum
(a) (b) (c) (d)
(1) (iii) (ii) (iv) (i)
(2) (iv) (iii) (i) (ii)
(3) (ii) (iv) (i) (iii)
(4) (i) (iii) (ii) (iv)
Answer (2)
Sol. (a) 11th and 12th pairs of ribs are not connected ventrally and are therefore, called floating ribs.
(b) Acromion is a flat expanded process of spine of scapula. The lateral end of clavicle articulates with acromion process.
(c) Scapula is a flat triangular bone in the dorsal part of the thorax between 2nd and the 7th rib.
(d) Glenoid cavity of scapula articulates with head of the humerus to form the shoulder joint.

70. 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 x 10^9 bp, then the length of the DNA is approximately
(1) 2.2 meters
(2) 2.7 meters
(3) 2.0 meters
(4) 2.5 meters
Answer (1)
Sol. Length of DNA = [0.34 x 10^-9] m x 6.6 x 10^9 bp = 2.2 m
Distance between 2 base pair in DNA helix = 0.34 nm = 0.34 x 10^-9 m
Total number of base pair = 6.6 x 10^9 bp

71. Presence of which of the following conditions in urine are indicative of Diabetes Mellitus?
(1) Ketonuria and Glycosuria
(2) Renal calculi and Hyperglycaemia
(3) Uremia and Ketonuria
(4) Uremia and Renal Calculi
Answer (1)
Sol. Presence of Ketone bodies in urine (Ketonuria) and presence of glucose in urine (Glycosuria) are indicative of Diabetes mellitus.

72. Bilaterally symmetrical and acoelomate animals are exemplified by
(1) Aschelminthes
(2) Annelida
(3) Ctenophora
(4) Platyhelminthes
Answer (4)
Sol. Platyhelminthes are bilaterally symmetrical, triploblastic and acoelomate animals with organ level of organisation.

73. Ray florets have
(1) Hypogynous ovary
(2) Half inferior ovary
(3) Inferior ovary
(4) Superior ovary
Answer (3)
Sol. Ray florets have inferior ovary. Epigynous flower are formed in family Asteraceae (e.g., Sunflower)

74. The infectious stage of Plasmodium that enters the human body is
(1) Female gametocytes
(2) Male gametocytes
(3) Trophozoites
(4) Sporozoites
Answer (4)
Sol. Plasmodium enters the human body as sporozoites (Infectious stage) through the bite of Infected Female Anopheles mosquito.

75. Which of the following statements is not correct?
(1) The functional insulin has A and B chains linked together by hydrogen bonds.
(2) Genetically engineered insulin is produced in E.Coli.
(3) In man insulin is synthesised as a proinsulin
(4) The proinsulin has an extra peptide called C-peptide.
Answer (1)
Sol. The correct option is (1) because functional insulin has A and B chains linked together by disulphide bridges.

76. In water hyacinth and water lily, pollination takes place by :
(1) Wind and water
(2) Insects and water
(3) Insects or wind
(4) Water currents only
Answer (3)
Sol. In majority of aquatic plants, the flowers emerge above the level of water. These may be pollinated by insects or wind eg.: Water hyacinth and water lily.

77. Cuboidal epithelium with brush border of microvilli is found in
(1) Proximal convoluted tubule of nephron
(2) Eustachian tube
(3) Lining of intestine
(4) Ducts of salivary gland
Answer (1)
Sol. Cuboidal epithelium with brush border of microvilli is found in proximal convoluted tubule of nephron (PCT).

78. Match the following columns and select the correct option.
Column-I Column-II
(a) Pituitary gland (i) Grave's disease
(b) Thyroid gland (ii) Diabetes mellitus
(c) Adrenal gland (iii) Diabetes insipidus
(d) Pancreas (iv) Addison's disease
(a) (b) (c) (d)
(1) (iii) (i) (iv) (ii)
(2) (ii) (i) (iv) (iii)
(3) (iv) (iii) (i) (ii)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. Graves' disease is due to excess secretion of thyroid hormones (T3 & T4).
Diabetes mellitus is due to hyposecretion of insulin from beta-cells of pancreas.
Diabetes insipidus is due to hyporelease of ADH from posterior pituitary.
Addison's disease is due to hyposecretion of hormone from adrenal cortex.

79. Which one of the following is the most abundant protein in the animals?
(1) Lectin
(2) Insulin
(3) Haemoglobin
(4) Collagen
Answer (4)
Sol. Collagen is the most abundant protein in animal world and RuBisCO is the most abundant protein in the whole of the Biosphere.

80. If the head of cockroach is removed, it may live for few days because
(1) the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.
(2) the head holds a 1/3rd of a nervous system while the rest is situated along the dorsal part of its body.
(3) the supra-oesophageal ganglia of the cockroach are situated in ventral part of abdomen.
(4) the cockroach does not have nervous system.
Answer (1)
Sol. The head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.

81. Flippers of Penguins and Dolphins are examples of
(1) Industrial melanism
(2) Natural selection
(3) Adaptive radiation
(4) Convergent evolution
Answer (4)
Sol. The correct option is (4) because flippers of Penguins and Dolphins are an example of analogous organs. Analogous structures are a result of convergent evolution.

82. The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is
(1) Imbibition
(2) Plasmolysis
(3) Transpiration
(4) Root pressure
Answer (4)
Sol. Root pressure is positive hydrostatic pressure.
It develops in tracheary element at night and in early morning.

83. Match the following with respect to meiosis
(a) Zygotene (i) Terminalization
(b) Pachytene (ii) Chiasmata
(c) Diplotene (iii) Crossing over
(d) Diakinesis (iv) Synapsis
Select the correct option from the following
(a) (b) (c) (d)
(1) (i) (ii) (iv) (iii)
(2) (ii) (iv) (iii) (i)
(3) (iii) (iv) (i) (ii)
(4) (iv) (iii) (ii) (i)
Answer (4)
Sol. Zygotene -> Synapsis
Pachytene -> Crossing over
Diplotene -> Chiasmata formation
Diakinesis -> Terminalisation

84. The QRS complex in a standard ECG represents
(1) Depolarisation of ventricles
(2) Repolarisation of ventricles
(3) Repolarisation of auricles
(4) Depolarisation of auricles
Answer (1)
Sol. QRS complex represents the depolarisation of ventricles.

85. Select the option including all sexually transmitted diseases.
(1) AIDS, Malaria, Filaria
(2) Cancer, AIDS, Syphilis
(3) Gonorrhoea, Syphilis, Genital herpes
(4) Gonorrhoea, Malaria, Genital herpes
Answer (3)
Sol. Gonorrhoea, Syphilis, 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.

86. Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.
(1) When IA and IB are present together, they express same type of sugar.
(2) Allele 'i' does not produce any sugar.
(3) The gene (I) has three alleles.
(4) A person will have only two of the three alleles.
Answer (1)
Sol. ABO blood groups are controlled by the gene I. The gene I has three alleles IA, IB and i. The alleles IA and IB produce a slightly different form of the sugar while 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.

87. Which of the following is not an inhibitory substance governing seed dormancy?
(1) Phenolic acid
(2) Para-ascorbic acid
(3) Gibberellic acid
(4) Abscisic acid
Answer (3)
Sol. Gibberellic acid break seed dormancy.
It activate synthesis of alpha-amylase which breakdown starch into simple sugar.

88. Match the following columns and select the correct option.
Column-I Column-II
(a) Organ of Corti (i) Connects middle ear and pharynx
(b) Cochlea (ii) Coiled part of the labyrinth
(c) Eustachian tube (iii) Attached to the oval window
(d) Stapes (iv) Located on the basilar membrane
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (i) (ii) (iv) (iii)
(3) (ii) (iii) (i) (iv)
(4) (iii) (i) (iv) (ii)
Answer (1)
Sol. Option (1) is correct because organ of Corti is located on the Basilar membrane, thus (a) in column-I matches with (iv) in column-II.
The coiled portion of the labyrinth is called cochlea, so (b) matches with (ii) in column II.
The eustachian tube connects the middle ear cavity with the pharynx, thus (c) matches with (i) in column-II.
The middle ear contains ossicle called Stapes that is attached to the oval window of the cochlea, so (d) matches with (iii) in column II.

89. The ovary is half inferior in :
(1) Sunflower
(2) Plum
(3) Brinjal
(4) Mustard
Answer (2)
Sol. The ovary is half inferior in Plum.

90. Match the following
(a) Inhibitor of catalytic activity (i) Ricin
(b) Possess peptide bonds (ii) Malonate
(c) Cell wall material in fungi (iii) Chitin
(d) Secondary metabolite (iv) Collagen
Choose the correct option from the following
(a) (b) (c) (d)
(1) (iii) (iv) (i) (ii)
(2) (ii) (iii) (i) (iv)
(3) (ii) (iv) (iii) (i)
(4) (iii) (i) (iv) (ii)
Answer (3)
Sol. Option (3) is the correct answer because Malonate is the competitive inhibitor of catalytic activity of succinic dehydrogenase, so (a) matches with (ii) in column II.
Collagen is proteinaceous in nature and possesses peptide bonds, so (b) matches with (iv) in column II.
Chitin is a homopolymer present in the cell wall of fungi and exoskeleton of arthropods, so, (c) matches with (iii) in column II.
Abrin and Ricin are toxins, secondary metabolites, so (d) in column I matches with (i) in column II.

91. Which of the following oxoacid of sulphur has -O-O-linkage?
(1) H2S2O8, peroxodisulphuric acid
(2) H2S2O7, pyrosulphuric acid
(3) H2SO3, sulphurous acid
(4) H2SO4, sulphuric acid
Answer (1)
Sol. HO-S(=O)2-O-O-S(=O)2-OH
Peroxodisulphuric acid

92. An increase in the concentration of the reactants of a reaction leads to change in
(1) threshold energy
(2) collision frequency
(3) activation energy
(4) heat of reaction
Answer (4)
Sol. Heat of reaction is an extensive property. Hence, on change of amount/concentration of reactants heat of reaction changes.

93. Identify the incorrect match.
Name IUPAC Official Name
(a) Unnilunium (i) Mendelevium
(b) Unniltrium (ii) Lawrencium
(c) Unnilhexium (iii) Seaborgium
(d) Unununium (iv) Darmstadtium
(1) (c), (iii)
(2) (d), (iv)
(3) (a), (i)
(4) (b), (ii)
Answer (2)
Sol. Unununium
Atomic number = 111
IUPAC official name : Roentgenium

94. A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 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 N2 is :
[Use atomic masses (in g mol-1) : N = 14, Ar = 40]
(1) 15 bar
(2) 18 bar
(3) 9 bar
(4) 12 bar
Answer (1)
Sol. nN2 = 7/28 = 1/4 = 0.25
nAr = 8/40 = 1/5 = 0.20
Now, Applying Dalton's law of partial pressure,
pN2 = (chiN2) PTotal
= (0.25/0.45) x 27 bar
= (5/9) x 27 = 15 bar

95. Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as
(1) Cross Cannizzaro's reaction
(2) Cross Aldol condensation
(3) Aldol condensation
(4) Cannizzaro's reaction
Answer (2)
Sol. In the presence of dil. OH(-), benzaldehyde and acetophenone will react to undergo cross-aldol condensation.

96. Which one of the following has maximum number of atoms?
(1) 1 g of O2(g) [Atomic mass of O = 16]
(2) 1 g of Li(s) [Atomic mass of Li = 7]
(3) 1 g of Ag(s) [Atomic mass of Ag = 108]
(4) 1 g of Mg(s) [Atomic mass of Mg = 24]
Answer (2)
Sol. Number of Mg atoms = (1/24) x NA
Number of O atoms = (1/32) x 2 x NA
Number of Li atoms = (1/7) x NA
Number of Ag atoms = (1/108) x NA

97. Anisole on cleavage with HI gives
Answer (3)
Sol. Anisole + HI -> Phenol + CH3I

98. Which of the following amine will give the carbylamine test?
Answer (3)
Sol. Aliphatic and aromatic primary amines give carbylamine reaction.

99. Identify the incorrect statement.
(1) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
(2) The oxidation states of chromium in CrO4^2- and Cr2O7^2- are not the same.
(3) Cr2+ (d4) is a stronger reducing agent than Fe2+ (d6) in water.
(4) 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.
Answer (2)
Sol. Oxidation state of Cr in CrO4^2- and Cr2O7^2- is +6.

100. Which of the following is a basic amino acid?
(1) Tyrosine
(2) Lysine
(3) Serine
(4) Alanine
Answer (2)
Sol. H2N-CH2-CH2-CH2-CH2-CH(NH2)-COOH
(Structure of Lysine)
Lysine is a basic amino acid.

101. Which of the following is a natural polymer?
(1) polybutadiene
(2) poly (Butadiene-acrylonitrile)
(3) cis-1,4-polyisoprene
(4) poly (Butadiene-styrene)
Answer (3)
Sol. Naturally occuring polymer, natural rubber is cis-1,4-polyisoprene.

102. Match the following and identify the correct option.
(a) CO(g) + H2(g) (i) Mg(HCO3)2 + Ca(HCO3)2
(b) Temporary hardness of water (ii) An electron deficient hydride
(c) B2H6 (iii) Synthesis gas
(d) H2O2 (iv) Non-planar structure
(a) (b) (c) (d)
(1) (iii) (iv) (ii) (i)
(2) (i) (iii) (ii) (iv)
(3) (iii) (i) (ii) (iv)
(4) (iii) (ii) (i) (iv)
Answer (3)
Sol. Mixture of CO and H2 gases is known as water gas or synthesis gas.
Temporary hardness of water is due to bicarbonates of calcium and magnesium.
Diborane (B2H6) is an electron deficient hydride.
H2O2 is non-planar molecule having open book like structure.

103. An alkene on ozonolysis gives methanal as one of the product. Its structure is
Answer (1)
Sol. Alkene with terminal =CH2 gives methanal.

104. The rate constant for a first order reaction is 4.606 x 10^-3 s^-1. The time required to reduce 2.0 g of the reactant to 0.2 g is :
(1) 500 s
(2) 1000 s
(3) 100 s
(4) 200 s
Answer (1)
Sol. k = (2.303/t) log(A0/A) (First order rate equation)
4.606 x 10^-3 = (2.303/t) log(2/0.2)
t = (2.303/(4.606 x 10^-3)) x log 10
= 10^3/2 = 500 sec

105. On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be
(1) H2S gas
(2) SO2 gas
(3) Hydrogen gas
(4) Oxygen gas
Answer (4)
Sol. During the electrolysis of dil. sulphuric acid using Pt electrodes following reaction will take place.
At cathode :
4H+(aq) + 4e- -> 2H2(g)
At anode :
2H2O(l) -> O2(g) + 4H+(aq) + 4e-

106. An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is
(1) (4/sqrt(3)) x 288 pm
(2) (4/sqrt(2)) x 288 pm
(3) (sqrt(3)/4) x 288 pm
(4) (sqrt(2)/4) x 288 pm
Answer (3)
Sol. For BCC,
sqrt(3)a = 4r
r = sqrt(3)a/4
Given, a = 288 pm
r = (sqrt(3)/4) x 288

107. Sucrose on hydrolysis gives
(1) α-D-Glucose + β-D-Fructose
(2) α-D-Fructose + β-D-Fructose
(3) β-D-Glucose + α-D-Fructose
(4) α-D-Glucose + β-D-Glucose
Answer (1)
Sol. Sucrose --Hydrolysis--> α-D-Glucose + β-D-Fructose

108. Which of the following is not correct about carbon monoxide?
(1) The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
(2) It is produced due to incomplete combustion.
(3) It forms carboxyhaemoglobin
(4) It reduces oxygen carrying ability of blood.
Answer (1)
Sol. The carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin.

109. The mixture which shows positive deviation from Raoult's law is
(1) Acetone + Chloroform
(2) Chloroethane + Bromoethane
(3) Ethanol + Acetone
(4) Benzene + Toluene
Answer (3)
Sol. 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.

110. Identify compound X in the following sequence of reactions
Answer (1)
Sol. Toluene --Cl2/hv--> Benzyl chloride (X) --H2O/373K--> Benzyl alcohol --(oxidation)--> Benzaldehyde

111. The freezing point depression constant (Kf) 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):
(1) 0.40 K
(2) 0.60 K
(3) 0.20 K
(4) 0.80 K
Answer (1)
Sol. ΔTf = kf m
= 5.12 (K kg mol-1) x 0.078 (mol kg-1)
= 0.399 K
≈ 0.40 K

112. Which of the following is a cationic detergent?
(1) Cetyltrimethyl ammonium bromide
(2) Sodium dodecylbenzene sulphonate
(3) Sodium lauryl sulphate
(4) Sodium stearate
Answer (1)
Sol. CH3-(CH2)15-N+(CH3)3 Br-
Cetyltrimethyl ammonium bromide

113. Paper chromatography is an example of
(1) Thin layer chromatography
(2) Column chromatography
(3) Adsorption chromatography
(4) Partition chromatography
Answer (4)
Sol. Paper chromatography is a type of partition chromatography in which a special quality paper known as chromatography paper is used.

114. Identify the correct statement from the following :
(1) Vapour phase refining is carried out for Nickel by Van Arkel method.
(2) Pig iron can be moulded into a variety of shapes.
(3) Wrought iron is impure iron with 4% carbon.
(4) Blister copper has blistered appearance due to evolution of CO2
Answer (2)
Sol. The iron obtained from blast furnace contains about 4% carbon and many impurities like S, P, Si, Mn in smaller amount. This is known as pig iron and cast into variety of shapes.

115. What is the change in oxidation number of carbon in the following reaction?
CH4(g) + 4Cl2(g) -> CCl4(l) + 4HCl(g)
(1) -4 to +4
(2) 0 to -4
(3) +4 to +4
(4) 0 to +4
Answer (1)
Sol. CH4 => x + 4 x 1 = 0 => x = -4
CCl4 => x + 4 x (-1) = 0 => x = +4
CH4(g) + 4Cl2(g) -> CCl4(l) + 4HCl(g)
Change in oxidation state of carbon is from -4 to +4.

116. Elimination reaction of 2-Bromo-pentane to form pent-2-ene is
(a) β-Elimination reaction
(b) Follows Zaitsev rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction
(1) (b), (c), (d)
(2) (a), (b), (d)
(3) (a), (b), (c)
(4) (a), (c), (d)
Answer (3)
Sol. Since β-hydrogen is abstracted it is β-elimination.
Since more substituted alkene is formed, it follows Zaitsev's rule.
Since H and Br are removed, it is dehydrohalogenation.

117. Hydrolysis of sucrose is given by the following reaction.
Sucrose + H2O ⇌ Glucose + Fructose
If the equilibrium constant (Kc) is 2 x 10^13 at 300 K, the value of ΔrG° at the same temperature will be :
(1) 8.314 J mol-1 K-1 x 300 K x ln(3 x 10^13)
(2) -8.314 J mol-1 K-1 x 300 K x ln(4 x 10^13)
(3) -8.314 J mol-1 K-1 x 300 K x ln(2 x 10^13)
(4) 8.314 J mol-1 K-1 x 300 K x ln(2 x 10^13)
Answer (3)
Sol. ΔG = ΔG° + RT ln Q
At equilibrium ΔG = 0, Q = Keq
So ΔrG° = -RT ln Keq
ΔrG° = -8.314 J mol-1 K-1 x 300 K x ln(2 x 10^13)

118. Match the following :
Oxide Nature
(a) CO (i) Basic
(b) BaO (ii) Neutral
(c) Al2O3 (iii) Acidic
(d) Cl2O7 (iv) Amphoteric
Which of the following is correct option?
(a) (b) (c) (d)
(1) (iii) (iv) (i) (ii)
(2) (iv) (iii) (ii) (i)
(3) (i) (ii) (iii) (iv)
(4) (ii) (i) (iv) (iii)
Answer (4)
Sol. CO : Neutral oxide
BaO : Basic oxide
Al2O3 : Amphoteric oxide
Cl2O7 : Acidic oxide

119. Identify a molecule which does not exist.
(1) C2
(2) O2
(3) He2
(4) Li2
Answer (3)
Sol. For He2 molecule
Electronic configuration is σ1s^2 σ*1s^2
so bond order = (1/2)[Nb - Na]
= (1/2)[2 - 2]
= 0
Since, bond order is zero, so He2 molecule does not exist.

120. The number of Faradays(F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca = 40 g mol-1) is
(1) 3
(2) 4
(3) 1
(4) 2
Answer (3)
Sol. 1 equivalent of any substance is deposited by 1 F of charge.
We have, 20 g calcium
Number of equivalents = Given mass/Equivalent mass
= 20/20 = 1
Equivalent mass of Ca = 40/2 = 20
So, 1 faraday of charge is required.

121. Urea reacts with water to form A which will decompose to form B. B when passed through Cu2+(aq), deep blue colour solution C is formed. What is the formula of C from the following?
(1) Cu(OH)2
(2) CuCO3.Cu(OH)2
(3) CuSO4
(4) [Cu(NH3)4]2+
Answer (4)
Sol. NH2CONH2 + H2O -> (NH4)2CO3 (A)
(NH4)2CO3 -> NH3(g) + CO2(g) + H2O(l) (B)
NH3(g) --Cu2+(aq)--> [Cu(NH3)4]2+ (C)
[Blue coloured solution]

122. Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :
(1) Tert. butyl alcohol
(2) Isobutyl alcohol
(3) Isopropyl alcohol
(4) Sec. butyl alcohol
Answer (1)
Sol. Acetone + CH3MgBr -> (after hydrolysis) tert-Butyl alcohol

123. The number of protons, neutrons and electrons in 175Lu respectively, are
(1) 71, 71 and 104
(2) 175, 104 and 71
(3) 71, 104 and 71
(4) 104, 71 and 71
Answer (3)
Sol. 175Lu
No. of Protons = 71 = No. of Electrons
No. of Neutrons = Mass no. - No. of Protons
= 175 - 71
= 104

124. Which of the following alkane cannot be made in good yield by Wurtz reaction?
(1) n-Heptane
(2) n-Butane
(3) n-Hexane
(4) 2,3-Dimethylbutane
Answer (1)
Sol. Wurtz reaction is used to prepare symmetrical alkanes like R1-R1 as
R1-X + 2Na + X-R1 --Dry ether--> R1-R1 + 2NaX
If R1 and R2 are different, then mixture of alkanes may be obtained as
R1-X + 2Na + R2-X --Dry ether-->
R1-R1 + R1-R2 + R2-R2 + 2NaX

125. HCl was passed through a solution of CaCl2, MgCl2 and NaCl. Which of the following compound(s) crystallise(s)?
(1) Only MgCl2
(2) NaCl, MgCl2 and CaCl2
(3) Both MgCl2 and CaCl2
(4) Only NaCl
Answer (4)
Sol. Since CaCl2 and MgCl2 are more soluble than NaCl, on passing HCl(g) through a solution containing CaCl2, MgCl2 and NaCl then NaCl crystallizes out.

126. Measuring Zeta potential is useful in determining which property of colloidal solution?
(1) Stability of the colloidal particles
(2) Size of the colloidal particles
(3) Viscosity
(4) Solubility
Answer (1)
Sol. In colloidal solution, the potential difference between the fixed layer and the diffused layer of opposite charge is known as Zeta potential.
The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution.

127. Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2 x 10^-15
(1) 1 x 10^-13 M
(2) 1 x 10^8 M
(3) 2 x 10^-13 M
(4) 2 x 10^-8 M
Answer (3)
Sol. Ni(OH)2 ⇌ Ni2+ + 2OH-
s s 2s
NaOH -> Na+ + OH-
0.1 0.1 0.1
Total [OH-] = 2s + 0.1 = 0.1
Ionic product = [Ni2+][OH-]^2
2 x 10^-15 = s(0.1)^2
s = 2 x 10^-13
Solubility of Ni(OH)2 = 2 x 10^-13 M

128. For the reaction, 2Cl(g) -> Cl2(g), the correct option is :
(1) ΔrH < 0 and ΔrS > 0
(2) ΔrH < 0 and ΔrS < 0
(3) ΔrH > 0 and ΔrS > 0
(4) ΔrH > 0 and ΔrS < 0
Answer (2)
Sol. Given reaction, 2Cl(g) -> Cl2(g)
We know that,
Cl2(g) -> 2Cl(g) is endothermic reaction because it requires energy to break bond.
So reverse reaction is exothermic ΔrH < 0
Also, two gaseous atom combine together to form 1 gaseous molecule.
So, randomness ΔrS < 0

129. The calculated spin only magnetic moment of Cr2+ ion is
(1) 5.92 BM
(2) 2.84 BM
(3) 3.87 BM
(4) 4.90 BM
Answer (4)
Sol. Electronic configuration of Cr - [Ar]3d^5 4s^1
Electronic configuration of Cr2+ - [Ar]3d^4
Number of unpaired e- = 4
Spin only magnetic moment = sqrt(n(n+2))
n = number of unpaired e-
Spin only magnetic moment = sqrt(4(4+2))
= sqrt(24) BM
= 4.9 BM

130. Identify the correct statements from the following :
(a) CO2(g) is used as refrigerant for ice-cream and frozen food.
(b) The structure of C60 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.
(1) (b) and (c) only
(2) (c) and (d) only
(3) (a), (b) and (c) only
(4) (a) and (c) only
Answer (2)
Sol. Dry ice, CO2(s), is used as refrigerant
C60 contains 20 six membered rings, 12 five membered rings

131. 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.
(1) Calcium
(2) Potassium
(3) Iron
(4) Copper
Answer (2)
Sol. Potassium (K) activates many enzymes participate in oxidation of glucose to produce ATP and helps in the transmission of nerve signal along with Na.

132. Which of the following set of molecules will have zero dipole moment?
(1) Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene
(2) Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene
(3) Ammonia, beryllium difluoride, water, 1,4-dichlorobenzene
(4) Boron trifluoride, hydrogen fluoride, carbon dioxide, 1,3-dichlorobenzene
Answer (2)
Sol. BF3: μ = 0
BeF2: μ = 0
CO2: μ = 0
1,4-dichlorobenzene: μ = 0

133. The correct option for free expansion of an ideal gas under adiabatic condition is
(1) q < 0, ΔT = 0 and w = 0
(2) q > 0, ΔT > 0 and w > 0
(3) q = 0, ΔT = 0 and w = 0
(4) q = 0, ΔT < 0 and w > 0
Answer (3)
Sol. Free expansion => Pex = 0
therefore w = -Pex ΔV = 0
Adiabatic process => q = 0
also, ΔU = q + w [first law of thermodynamics]
therefore ΔU = 0
Internal energy of an ideal gas is a function of temperature
If internal energy remains constant
therefore ΔT = 0

134. Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(1) F- < SCN- < C2O4^2- < CN-
(2) CN- < C2O4^2- < SCN- < F-
(3) SCN- < F- < C2O4^2- < CN-
(4) SCN- < F- < CN- < C2O4^2-
Answer (3)
Sol. Spectrochemical series (as given in NCERT):
I- < Br- < SCN- < Cl- < S^2- < F- < OH- < C2O4^2- < H2O < NCS- < EDTA^4- < NH3 < en < CN- < CO

135. A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?
(1) -R effect of -CH3 groups
(2) Hyperconjugation
(3) -I effect of -CH3 groups
(4) +R effect of -CH3 groups
Answer (2)
Sol. Tertiary butyl carbocation (9 α-H atoms)
Secondary butyl carbocation (5 α-H atoms)
More the number of α-H atoms, more will be the hyperconjugation effect hence more will be the stability of carbocation.

136. 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:
(1) 536 Hz
(2) 537 Hz
(3) 523 Hz
(4) 524 Hz
Answer (4)
Sol. Difference of fA and fB is 6 Hz
If tension decreases, fB decreases and becomes fB'.
Now, difference of fA and fB' = 7 Hz (increases)
So, fA > fB
fA - fB = 6 Hz
fA = 530 Hz
fB = 524 Hz (original)

137. The increase in the width of the depletion region in a p-n junction diode is due to :
(1) both forward bias and reverse bias
(2) increase in forward current
(3) forward bias only
(4) reverse bias only
Answer (4)
Sol. Due to reverse biasing, the width of the depletion region increases.

138. The quantities of heat required to raise the temperature of two solid copper spheres of radii r1 and r2 (r1 = 1.5 r2) through 1 K are in the ratio :
(1) 3/2
(2) 27/8
(3) 27/8
(4) 9/4
Answer (2)
Sol. ΔQ = msΔT
ΔQ = (4/3) π r^3 ρ s ΔT
ΔQ1/ΔQ2 = (r1/r2)^3
= (1.5)^3
= 27/8

139. 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 π/3. If instead C is removed from the circuit, the phase difference is again π/3 between current and voltage. The power factor of the circuit is
(1) 1.0
(2) -1.0
(3) zero
(4) 0.5
Answer (1)
Sol. When L is removed,
tan φ = |XC|/R => tan π/3 = XC/R ...(i)
When C is removed,
tan φ = |XL|/R => tan π/3 = XL/R ...(ii)
From (i) and (ii), XL = XC
Since, XL = XC, the circuit is in resonance. Z = R
Power factor = cos φ = R/Z = 1

140. 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)
(1) 320 m
(2) 300 m
(3) 360 m
(4) 340 m
Answer (2)
Sol. v^2 = u^2 + 2gh
v = 80 m/s
u = 20 m/s
h = (v^2 - u^2)/(2g) = (6400 - 400)/20 = 300 m

141. 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 :
(1) four times
(2) one-fourth
(3) double
(4) half
Answer (1)
Sol. Fringe width β = λD/d
Now, d' = d/2 and D' = 2D
So, β' = λ(2D)/(d/2) = 4λD/d
β' = 4β

142. Which of the following graph represents the variation of resistivity (ρ) with temperature (T) for copper?
Answer (1)
Sol. At temperature much lower than 0°C graph deviates considerably from a straight line.
Option (1) is correct

143. 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 :
(μ0 = 4π x 10^-7 T m A^-1)
(1) 6.28 x 10^-5 T
(2) 3.14 x 10^-5 T
(3) 6.28 x 10^-4 T
(4) 3.14 x 10^-4 T
Answer (3)
Sol. Magnetic field at centre of solenoid = μ0nI
n = N/L = 100/(50 x 10^-2) = 200 turns/m
I = 2.5 A
On putting the values
B = 4π x 10^-7 x 200 x 2.5
= 6.28 x 10^-4 T

144. 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?
(1) one-fourth
(2) zero
(3) doubled
(4) four times
Answer (2)
Sol. ν = (3/2)ν0
ν' = ν/2 = (3/4)ν0
therefore ν' < ν0
No photoelectric emission will take place.

145. 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 :
(1) 0.5 mm
(2) 1.0 mm
(3) 0.01 mm
(4) 0.25 mm
Answer (1)
Sol. Least count = Pitch/Number of divisions on circular scale
0.01 mm = Pitch/50
Pitch = 0.5 mm

146. 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:
(1) g/5
(4) g/2
(2) g/10
(3) g
Answer (1)
Sol. a = (m1 - m2)g/(m1 + m2) where m1 > m2
a = (6 - 4)g/(6 + 4)
a = g/5
Note : Here no option is given according to acceleration of COM of the system.

147. For transistor action, which of the following statements is correct?
(1) Both emitter junction as well as the collector junction are forward biased.
(2) The base region must be very thin and lightly doped.
(3) Base, emitter and collector regions should have same doping concentrations.
(4) Base, emitter and collector regions should have same size.
Answer (2)
Sol. For Bi-polar junction transistor
Length Profile is LC > LE > LB
and doping profile is E > C > B
For transistor action Base-emitter junction is forward biased and Base-collector junction is reversed biased.

148. For which one of the following, Bohr model is not valid?
(1) Deuteron atom
(2) Singly ionised neon atom (Ne+)
(3) Hydrogen atom
(4) Singly ionised helium atom (He+)
Answer (2)
Sol. Bohr model is only valid for single electron species.
Singly ionised neon atom has more than one electron in orbit. Hence, Bohr model is not valid.

149. 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 :
(1) 10.0 g
(2) 20.0 g
(3) 2.5 g
(4) 5.0 g
Answer (1)
Sol. Force of surface tension balances the weight of water in capillary tube.
FS = 2πrTcosθ = mg
Here, T and θ are constant
So, m ∝ r
Hence, m2/5.0 = 2r/r
=> m2 = 10.0 g

150. 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)
(1) 1 : c
(2) 1 : c2
(3) c : 1
(4) 1 : 1
Answer (4)
Sol. In an electromagnetic wave, half of the intensity is provided by the electric field and half by the magnetic field Hence required ratio should be 1 : 1

151. 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 :
(μ0 = 4π x 10^-7 T m A^-1)
(1) 2.4π x 10^-5 T m A^-1
(2) 2.4π x 10^-7 T m A^-1
(3) 2.4π x 10^-4 T m A^-1
(4) 8.0 x 10^-5 T m A^-1
Answer (3)
Sol. χm = 599
μr = 1 + χm = 600
μ = μrμ0
μ = 600 x 4π x 10^-7
μ = 2400π x 10^-7
μ = 2.4π x 10^-4 T m A^-1

152. The Brewsters angle ib for an interface should be
(1) 45° < ib < 90°
(2) ib = 90°
(3) 0° < ib < 30°
(4) 30° < ib < 45°
Answer (1)
Sol. μ = tan ib
1 < μ < ∞
1 < tan ib < ∞
tan^-1(1) < ib < tan^-1(∞)
45° < ib < 90°

153. The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :
(1) π/2 rad
(2) zero
(3) π rad
(4) 3π/2 rad
Answer (3)
Sol. If y = A sin ωt
then v = dy/dt
v = Aω cos ωt
a = dv/dt
a = -Aω^2 sin(ωt)
a = Aω^2 sin(ωt + π)
So phase difference between displacement and acceleration is π.

154. 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 :
(1) 67 cm
(2) 80 cm
(3) 33 cm
(4) 50 cm
Answer (1)
Sol. xcm = (m1x1 + m2x2)/(m1 + m2)
= (5 x 0 + 100 x 10)/(5 + 10) = 200/3 = 66.66 cm
xcm ≈ 67 cm

155. A spherical conductor of radius 10 cm has a charge of 3.2 x 10^-7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
(1/(4πε0) = 9 x 10^9 Nm^2/C^2)
(1) 1.28 x 10^6 N/C
(2) 1.28 x 10^7 N/C
(3) 1.28 x 10^4 N/C
(4) 1.28 x 10^5 N/C
Answer (4)
Sol. Electric field outside a conducting sphere
E = (1/(4πε0)) Q/r^2
= (9 x 10^9 x 3.2 x 10^-7)/(225 x 10^-4)
= 0.128 x 10^6
= 1.28 x 10^5 N/C

156. 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 :
(1) 7.32 x 10^-7 rad
(2) 6.00 x 10^-7 rad
(3) 3.66 x 10^-7 rad
(4) 1.83 x 10^-7 rad
Answer (3)
Sol. θR = 1.22 λ/d ; λ = 600 x 10^-9 m d = 2 m
= (1.22 x 600 x 10^-9)/2
θ = 3.66 x 10^-7 rad

157. A charged particle having drift velocity of 7.5 x 10^-4 m s^-1 in an electric field of 3 x 10^-10 V m^-1, has a mobility in m^2 V^-1 s^-1 of:
(1) 2.5 x 10^-6
(2) 2.25 x 10^-15
(3) 2.25 x 10^15
(4) 2.5 x 10^6
Answer (4)
Sol. Mobility, μ = vd/E
= (7.5 x 10^-4)/(3 x 10^-10)
= 2.5 x 10^6 m^2 V^-1 s^-1

158. Taking into account of the significant figures, what is the value of 9.99 m - 0.0099 m ?
(1) 9.980 m
(2) 9.9 m
(3) 9.9801 m
(4) 9.98 m
Answer (4)
Sol. 9.99 - 0.0099 = 9.9801 m
In subtraction, answer should be reported to least number of decimal places, so answer should be 9.98 m.

159. The energy equivalent of 0.5 g of a substance is:
(1) 1.5 x 10^13 J
(2) 0.5 x 10^13 J
(3) 4.5 x 10^16 J
(4) 4.5 x 10^13 J
Answer (4)
Sol. From mass-energy equivalence.
E = mc^2
= 0.5 x 10^-3 x (3 x 10^8)^2
= 4.5 x 10^13 J

160. When a uranium isotope 235/92 U is bombarded with a neutron, it generates 89/36 Kr, three neutrons and:
(1) 101/36 Kr
(2) 103/36 Kr
(3) 144/56 Ba
(4) 91/40 Zr
Answer (3)
Sol. U(235,92) + n(1,0) -> Kr(89,36) + 3n(1,0) + X(A,Z)
92 + 0 = 36 + Z
=> Z = 56
235 + 1 = 89 + 3 + A
=> A = 144
So, Ba(144,56) is generated.

161. A short electric dipole has a dipole moment of 16 x 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° with the dipole axis is:
(1/(4πε0) = 9 x 10^9 N m^2/C^2)
(1) 400 V
(2) zero
(3) 50 V
(4) 200 V
Answer (4)
Sol. V = kp cos θ/r^2
V = (9 x 10^9 x 16 x 10^-9 x cos 60)/0.36
V = 200 V

162. A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C.
Its density is: (R = 8.3 J mol-1 K-1)
(1) 0.1 kg/m^3
(2) 0.02 kg/m^3
(3) 0.5 kg/m^3
(4) 0.2 kg/m^3
Answer (4)
Sol. PM = ρRT => ρ = PM/RT
P = 249 x 10^3 N/m^2
M = 2 x 10^-3 kg
T = 300 K
therefore ρ = (249 x 10^3)(2 x 10^-3)/(8.3 x 300) = 0.2 kg/m^3

163. The average thermal energy for a mono-atomic gas is: (kB is Boltzmann constant and T, absolute temperature)
(1) (5/2)kB T
(2) (7/2)kB T
(3) (1/2)kB T
(4) (3/2)kB T
Answer (4)
Sol. For monoatomic gases, degree of freedom is 3. Hence average thermal energy per molecule is KEavg = (3/2)kB T

164. The color code of a resistance is given below
Yellow Violet Brown Gold
The values of resistance and tolerance, respectively, are
(1) 4.7 kΩ, 5%
(2) 470 Ω, 5%
(3) 470 kΩ, 5%
(4) 47 kΩ, 10%
Answer (2)
Sol. According to colour coding
Yellow Violet Brown Gold
4 7 1 5%
So, R = 47 x 10^1 ± 5%
R = 470 ± 5% Ω

165. For the logic circuit shown, the truth table is:
(1) A B Y
0 0 1
0 1 1
1 0 1
1 1 0
(2) A B Y
0 0 1
0 1 0
1 0 0
1 1 0
(3) A B Y
0 0 0
0 1 0
1 0 0
1 1 1
(4) A B Y
0 0 0
0 1 1
1 0 1
1 1 1
Answer (3)
Sol. Y = A + B (bar)
= A(bar) B(bar)
= A . B -> AND Gate
Truth Table
A B Y
0 0 0
0 1 0
1 0 0
1 1 1

166. A resistance wire connected in the left gap of a metre bridge balances a 10 Ω 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 Ω of the resistance wire is :
(1) 1.5 x 10^-1 m
(2) 1.5 x 10^-2 m
(3) 1.0 x 10^-2 m
(4) 1.0 x 10^-1 m
Answer (4)
Sol. Initially, P/10 = l1/l2 = 3/2
=> P = 30/2 = 15 Ω
Now Resistance, R = ρl/A
R1/R2 = l1/l2
=> 15/1 = 1.5/l2
l2 = 0.1 m = 1.0 x 10^-1 m

167. Find the torque about the origin when a force of 3j N acts on a particle whose position vector is 2k m.
(1) -6i N m
(2) 6k N m
(3) 6i N m
(4) 6j N m
Answer (1)
Sol. τ = r x F
τ = 2k x 3j
τ = -6i N m

168. A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L1 when mass M is suspended from its free end. The expression for Young's modulus is :
(1) MgL/(AL1)
(2) MgL/(A(L1 - L))
(3) MgL1/(AL)
(4) Mg(L1 - L)/(AL)
Answer (2)
Sol. Stress = Mg/A
Strain = ΔL/L = (L1 - L)/L
Young's modulus = Stress/Strain = MgL/(A(L1 - L))

169. A 40 μF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly :
(1) 2.5 A
(2) 25.1 A
(3) 1.7 A
(4) 2.05 A
Answer (1)
Sol. irms = Cωεrms
C = 40 x 10^-6 F
ω = 2πf = 100π
εrms = 200 V
therefore irms = 200 x 40 x 10^-6 x 2π x 50
= 2.5 A

170. 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?
(1) 30 N
(2) 24 N
(3) 48 N
(4) 32 N
Answer (4)
Sol. mgh = mg0/(1 + h/R)^2
W = 72/(1 + R/2/R)^2
W = 72/(3/2)^2 = (4/9) x 72 = 32 N

171. An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 x 10^-2 nm, the potential difference is :
(1) 10^3 V
(2) 10^4 V
(3) 10 V
(4) 10^2 V
Answer (2)
Sol. λ = 12.27/sqrt(V) Å
sqrt(V) = (12.27 x 10^-10)/(1.227 x 10^-11) = 10^2
therefore V = 10^4 volts

172. 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 μ, then the angle of incidence is nearly equal to :
(1) μA
(2) μA/2
(3) A/(2μ)
(4) 2A/μ
Answer (1)
Sol. Light ray emerges normally from another surface, hence, e(angle of emergence) = 0
r2 = 0
r1 + r2 = A
=> r1 = A
Applying Snell's law on first surface
1. sin i = μ sin r1
=> sin i = μ sin A
For small angles (sin θ ≈ θ)
hence i = μA

173. The solids which have the negative temperature coefficient of resistance are:
(1) semiconductors only
(2) insulators and semiconductors
(3) metals
(4) insulators only
Answer (2)
Sol. For metals temperature coefficient of resistance is positive while for insulators and semiconductors, temperature coefficient of resistance is negative.

174. 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:
(1) 1 N/C
(2) 5 N/C
(3) zero
(4) 0.5 N/C
Answer (3)
Sol. Since, electric potential is found throughout constant, hence electric field, E = -dV/dr = 0

175. 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:
(1) 24 x 10^3 J
(2) 48 x 10^3 J
(3) 10 x 10^3 J
(4) 12 x 10^3 J
Answer (1)
Sol. Energy received = Intensity x Area x Time
= 20 x 20 x 60
= 24 x 10^3 J

176. The capacitance of a parallel plate capacitor with air as medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permittivity of the medium is :
(ε0 = 8.85 x 10^-12 C^2 N^-1 m^-2)
(1) 0.44 x 10^-10 C^2 N^-1 m^-2
(2) 5.00 C^2 N^-1 m^-2
(3) 0.44 x 10^-13 C^2 N^-1 m^-2
(4) 1.77 x 10^-12 C^2 N^-1 m^-2
Answer (1)
Sol. C = K C0
K = C/C0 = 30/6 = 5
K = ε/ε0
ε = Kε0
= 5 x 8.85 x 10^-12
= 0.44 x 10^-10 C^2 N^-1 m^-2

177. The energy required to break one bond in DNA is 10^-20 J. This value in eV is nearly:
(1) 0.06
(2) 0.006
(3) 6
(4) 0.6
Answer (1)
Sol. 1 eV = 1.6 x 10^-19 J
1 J = 1/(1.6 x 10^-19) eV
10^-20 J = 10^-20/(1.6 x 10^-19) eV
= 0.06 eV

178. Dimensions of stress are :
(1) [MLT-2]
(2) [ML-1 T-2]
(3) [MLT-2]
(4) [MLT-2]
Answer (2)
Sol. Stress = Force/Area
= [MLT-2]/[L2]
= [ML-1 T-2]

179. 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 :
(1) isochoric
(2) isobaric
(3) isothermal
(4) adiabatic
Answer (4)
Sol. Entire system is thermally insulated. So, no heat exchange will take place. Hence, process will be adiabatic.

180. The mean free path for a gas, with molecular diameter d and number density n can be expressed as :
(1) 1/(sqrt(2) n^2 π d^2)
(2) 1/(sqrt(2) n^2 π^2 d^2)
(3) 1/(sqrt(2) n π d)
(4) 1/(sqrt(2) n π d^2)
Answer (4)
Sol. According to the formula
λ = 1/(sqrt(2) n π d^2)

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