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P-SAT 2014 - 15

(PIONEER’S SCHOLARSHIP - ADMISSION TEST) {+2 MEDICAL}

General Instructions:-

The question paper contains 90 objective multiple choice questions.

There are three parts in the question paper consisting of

Section-A: BIOLOGY (1 to 90)

Section-B: PHYSICS (91 to 135),

Section-C: CHEMISTRY (136 to 180).

Each right answer carries 4 marks and wrong –1mark

The paper consists of 90 questions. The maximum marks are 360.

Maximum Time 3Hrs.

Give your response in the OMR Sheet provided with the Question Paper.

Name: _______________________________Father Name:__________________________

Mobile: ______________________________School:__________________________________

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Section - A {Biology}

1. Which of the leucocytes in mammals stores histamine ?

(a) Basophils (b) Monocytes (c) Eosinophils (d) Neutrophils

Ans. : (a)

2. Haversian canal in the bone of mammals are connected by small blood vessel canals

called

(a) Schlemm’s canal (b) Volkmann’s canal (c) portal caplillaries (d) sinuses

Ans. : (b)

3. Which of the following pairs is correctly matched ?

(a) Elastic fibres – Fibrous cartilage

(b) Vertebrate heart – Neurogenic

(c) Limulus – Neurogenic heart

(d) Tunica albuginea – Uterus and vagina

Ans. : (c)

4. Which types of placentation is found in pea?

(a) Marginal (b) Axile (c) Parietal (d) Basal

Ans. : (a)

5. Gynobasic style is found in

(a) Ocimum (b) Ranunculus (c) Nymphaea (d) Citrus

Ans. : (a)

6. Seminal plasma in human males is rich in

(a) fructose and calcium (b) glucose and calcium

(c) DNA and testosterone (d) ribose and potassium

Ans. : (a)

7. Which of the following stages of spermatogenesis involves meiotic division?

(a) Multiplication stage (b) Growth stage

(c) Maturation stage (d) None of these

Ans. : (a)

8. Which of the following signaling molecules is responsible for erectile function?

(a) Acetylcholine (b) Nitric oxide (c) Dopamine (d) Carbon dioxide

Ans. : (b)



9. In Lathyrus odoratus, cross between two purple flowered plants gives a pink/white

progeny. It is due to

(a) epistasis (b) multiple allelism (c) segregation (d) co-

dominance

Ans. : (c)

10. Two genes R and Y are located very close on the chromosomal linkage map of maize

plant. When RRYY and rryy genotypes are hybridized, then F2 segregation will show

(a) higher number of the recombinant types

(b) segregation in the expected 9 : 3: 3:1 ratio

(c) segregation in 3:1 ratio

(d) higher number of the parental types

Ans. : (d)

11. Extra-nuclear inheritance is a consequence of presence of genes in

(a) mitochondria and chloroplasts (b) endoplasmic reticulum and

mitochondria

(c) ribosomes and chloroplasts (d) lysosomes and ribosomes

Ans. : (a)

12. Amino acid sequence in protein synthesis is decided by the sequence of

(a) tRNA (b) mRNA (c) cDNA (d) rRNA

Ans. : (b)

13. E. coli cell with a mutated Z gene of the lac operon cannot grow in medium containing

only lactose as the source of energy because

(a) in the presence of glucose, E. coli cells do not utilize lactose

(b) they cannot transport lactose from the medium in the cells

(c) the lac operon is constitutively active in these cells

(d) they cannot synthesise functional -galactosidase

Ans. : (d)

14. Fossils of an ancient reptile called Lystrosaurus have been found in Africa, India and

Antarctica. Which of the following best explains this distribution?

(a) They were able to move between continents before the oceans filled

(b) Movement of India due to continental drift carried them from place to place



(c) These areas were once next to each other and have since drifted apart

(d) Changes in climate forced them to migrate from place to place

Ans. : (c)

15. Which one is a true statement regarding DNA polymerase used in PCR?

(a) It is used to ligate introduced DNA in recipient cells

(b) It serves as a selectable marker

(c) It is isolated from a virus

(d) It remains active at high temperature

Ans. : (d)

16. Bt gene, which produces protein toxic to insect larvae is

(a) Cry (b) cry (c) Try (d) trp

Ans. : (b)

17. Which of the following is expected to have the highest value (gm/m2/yr) in a grassland

ecosystem?

(a) Secondary Production (SP) (b) Tertiary Production (TP)

(c) Gross Production (GP) (d) Net Production (NP)

Ans. : (c)

18. Which of the following is not correct?

(a) Humification and mineralization occur during decomposition in soil

(b) Humification leads to accumulation of dark coloured humus

(c) Decomposition is an anoxygenic process

(d) The humus is further degraded by some microbes and release of inorganic

nutrients occur by the process mineralization

Ans. : (c)

19. Minamata disease was the result of

(a) mercury poisoning (b) cadmium poisoning

(c) phosphate pollution (d) eutrophication

Ans. : (a)

20. Formation of non-functional methaemoglobin causes blue-baby syndrome. This is due

to

(a) excess of arsenic concentration in drinking water



(b) excess of nitrates in drinking water

(c) deficiency of iron in food

(d) increased methane content in the atmosphere

Ans. : (b)

21. Limit of BOD prescribed by Central pollution Control Board for the discharge of

industrial and municipal waste water into natural surface water, is

(a) < 3.0 ppm (b) < 10 ppm (c) < 100 ppm (d) < 30ppm

Ans. : (b)

22. Sponteneous generation of flies from rotting meat was disproved by

(a) Francesco Reddi (b) Louis Pasteur

(c) Lazzaro Spallanzani (d) Charles Darwin.

Ans. : (a)

23. The ephemeral structure, which anchors the embryo and pushed it into the nutritional

zone of the embryo sac.

(a) haustorium (b) suspensor (c) coleorhiza (d) radical

Ans. : (b)

24. According to Geological records, presence of reduced material like uranite, pyrite etc.

in sediments indicates

(a) the presence of oxidising atmosphere on early earth

(b) the presence of moderate reducing atmosphere on early earth

(c) the presence of strongly reducing (non-oxygenic) conditions on early earth

(d) absence of atmosphere

Ans. : (c)

25. E. coli about to replicate was placed in a medium containing radioactive thymidine for

five minutes. Then it was made to replicate in a normal medium. Which of the

following observations will be correct?

(a) Both the strand of DNA will be radioactive

(b) One strand will be radioactive

(c) Each strand will be half radioactive

(d) None of the strands will be radioactive

Ans. : (b)



26. Which one of the following statements about human sperm is correct ?

(a) Acrosome has a conical pointed structure used for piercing and penetrating the

egg, resulting in fertilization

(b) The sperm lysins in the acrosome dissolves the egg envelope facilitating

fertilisation

(c) Acrosome serves as a sensory structure leading the sperm towards the ovum

(d) Acrosome serves no particular function

Ans. : (d)

27. Withdrawal of which of the following hormones is the immediate cause of

menstruation ?

(a) Oestrogens (b) FSH (c) FSH-RH (d) Progesterone

Ans. : (d)

28. In which one of the following sets of animals do all the four give birth to young ones ?

(a) Lion, bat, whale and ostrich

(b) Platypus, penguin, bat and Hippopotamus

(c) Shrew, bat, cat and kiwi

(d) Kangaroo, hedgehog, dolphin and Loris

Ans. : (d)

29. Superficial meroblastic cleavage occurs in

(a) reptiles (b) birds (c) mammals (d) insects

Ans. : (d)

30. Viagra (sildenafil citrate), which is used for the treatment of erectile dysfunction acts

by blocking the enzyme

(a) cGMP phosphodiesterase, which breakdowns cGMP

(b) cGMP phosphodiesterase, which synthesizes cGMP

(c) ATPase, which breakdowns ATP

(d) ATPase, which synthesizes ATP

Ans. : (a)

31. A sequential expression of a set of human genes occurs when a steroid molecule binds

to the

(a) transfer RNA (b) messenger RNA (c) DNA sequence (d)



ribosome

Ans. : (c)

32. Phenotype of an organisms is the result of

(a) mutations and linkages

(b) cytoplasmic effects and nutrition

(c) environmental changes and sexual dimorphism

(d) genotype and environment interactions

Ans. : (d)

33. Which one of the following pairs of features is a good example of polygenic inheritance

?

(a) Human height and skin colour

(b) ABO blood group in humans and flower colour of Mirabilis jalapa

(c) Hair pigment of mouse and tongue rolling in humans

(d) Human eye colour and sickle-cell anaemia

Ans. : (a)

34. Mating of an organism to a double recessive in order to determine whether it is

homozygous or heterozygous for a character under consideration is called

(a) reciprocal cross (b) test cross (c) dihybrid cross (d) back cross

Ans. : (b)

35. In polytene chromosomes, dark bands are visible. These bands are formed by the

aposition of

(a) protein particles (b) chromomeres on chromonemata

(c) nucleosomes (d) None of the above

Ans. : (b)

36. At a particular locus, frequency of ‘A’ allele is 0.6 and that of ‘a’ is 0.4. What would be

the frequency of heterozygotes in a random mating population at equilibrium ?

(a) 0.16 (b) 0.48 (c) 0.36 (d) 0.24

Ans. : (b)

37. A woman with normal vision, but whose father was colourblind marries a colourblind

man. Suppose that the fourth child of this couple was a boy. This boy

(a) must have normal colour vision



(b) will be partially colourblind, since he is heterozygous for the colourblind mutant

allele

(c) must be colourblind

(d) may be colourblind or may be of normal vision

Ans. : (d)

38. A pregnant woman underwent amniocentesis. An extra Barr body is present in the

embryo. The syndrome likely to occur in the child is

(a) Patau’s syndrome (b) Edward’s syndrome

(c) Down’s syndrome (d) Klinefelter’s syndrome

Ans. : (d)

39. Given below is a representation of a kind of chromosomal mutation. What is the kind

of mutation represented ?

(a) Deletion (b) Duplication

(c) Inversion (d) Reciprocal translocation

Ans. : (c)

40. Cri-du-chat syndrome in humans is caused by the

(a) fertilisation of an XX egg by a normal Y-bearing sperm

(b) loss of half of the short arm of chromosome 5

(c) loss of half of the long arm of chromosome 5

(d) trisomy of 21st chromosome

Ans. : (b)



41. Match the following columns.

Column A Column B

A.

B.

C.

D.

Secondary sex organ

Sertoli cells

Serosa

Blastocyst stage

1.

2.

3.

4.

5.

Uterus

Fallopian tube

Ovary

Inhibin

Epididymis

Codes

(a) A–1, B–2, C–3, D–4 (b) A–1, B–2, C–3, D–5

(c) A–5, B–4, C–1, D–2 (d) A–5, B–2, C–1, D–3

Ans. : (c)

42. Which of the following is not correctly matched ?

(a) Equal holoblastic cleavage – Marsupials

(b) Unequal holoblastic cleavage – Amphibians

(c) Meroblastic cleavage – Humans

(d) Radial cleavage – Synapata paracentrotums

Ans. : (c)

43. At menopause there is rise in urinary excretion of

(a) STH (b) FSH (c) LTH (d) MSH

Ans. : (b)

44. Which of the following is not correct ?

(a) The embryo develops up to blastocyst stage in Fallopian tube

(b) Fallopian tube conveys ovum from the ovary to the uterus through peristalsis

(c) Implantation takes place after 25 days of fertilisation

(d) Cowper’s glands are present in male mammals

Ans. : (a)

45. Which of the following is not correct about Leydig cells ?

(a) Leydig cells are exocrine in nature

(b) Leydig cells are characteristic of testes of mammal

(c) Leydig cells remain absent in testis of frog

(d) Leydig cells secrete male hormone testosterone, which influences secondary sexual



characters in male

Ans. : (a)

46. Which one of the following statements is false in respect of viability of mammalians

sperm ?

(a) Sperm is viable for only up to 24 h

(b) Survival of sperm depends on the pH of the medium and is more active in alkaline

medium

(c) Viability of sperm is determined by its motility

(d) Sperms must be concentrated in a thick suspension

Ans. : (a)

47. Women, who consumed the drug thalidomide for relief from vomiting during early

months of pregnancy gave birth to children with

(a) no spleen (b) hare-lip

(c) extra fingers and toes (d) under developed limbs

Ans. : (d)

48. Which of the following are diploid cells ?

(a) Primary oocyte and primary spermatocyte

(b) Secondary oocyte and secondary spermatocyte

(c) Spermatid and egg

(d) Sperm and fertilised egg

Ans. : (a)

49. Grey crescent is the area

(a) at the point of entry of sperm into ovum

(b) just opposite to the sire of entry of sperm into ovum

(c) at the animal pole

(d) at the vegetal pole

Ans. : (b)

50. The mRNA sequences in eukaryotes, which do not code for anything are called

(a) introns (b) interferons (c) exons (d) endons

Ans. : (a)



51. Removal of introns and joining of exons in a defined order during transcription is

called

(a) looping (b) inducing (c) slicing (d) splicing

Ans. : (d)

52. During translation initiation in prokaryotes, a GTP molecule is needed in

(a) association of 30 S , mRNA with formyl-met-tRNA

(b) association of 50 S subunit of ribosome with initiation complex

(c) formation of formyl-met-tRNA

(d) binding of 30 subunit of ribosome with mRNA

Ans. : (a)

53. In E. coli, an operator gene combines with

(a) inducer gene to switch on structural gene action

(b) inducer gene to switch off structural gene action

(c) regulator protein (repressor) to switch off structural gene action

(d) regulator protein to switch on gene action

Ans. : (c)

54. Catabolite Activator Protein (CAP) and repressor protein control the

(a) mRNA (b) rRNA (c) lac operon (d) trp operon

Ans. : (c)

55. Coacervates containing nucleoprotein, surrounded by several nutritive substances and

covered by a surface membrane represent

(a) pre-cell (b) post-cell (c) microsphere (d) liposome

Ans. : (a)

56. Which of the following is not evidence for the role of endosymbiosis in the origin of

eukaryotes ?

(a) Chloroplasts have their own DNA

(b) The inner membrane of a chloroplast is similar to prokaryotic membranes

(c) Mitochondria reproduced by binary fission

(d) The DNA in the eukaryotic nucleus codes for some enzymes in mitochondria

Ans. : (d)



57. The extinct human who lived 100000-40000 years ago, in Europe, Asia and parts of

Africa, with short stature, heavy eye brows, retreating fore heads, large jaws with

heavy teeth, stocky bodies, a lumbering gait and stopped posture was

(a) Homo habilis (b) Neanderthal man (c) Cro-magnon man (d) Ramapithecus

Ans. : (b)

58. Which of the following discoveries would force scientists to revise their present

theories regarding the origin of life on earth ?

(a) The earth is found to be billion years ago, rather than 4.6 billion

(b) Polypeptides can catalyse replication of small RNA molecules

(c) There was a log of oxygen gas in the atmosphere 4 billion years ago

(d) Minerals in lava catalyse formation of polypeptides from amino acids

Ans. : (c)

59. Age of fossils in the past was generally determined by radio-carbon method and other

methods involving radioactive elements found in the rocks. More precise methods,

which were used recently and led to the revisions of the evolutionary period for

different groups of organisms, includes

(a) study of carbohydrates/proteins in fossils

(b) study of the conditions of fossilisation

(c) Electron Spin Resonance (ESR) and fossil DNA

(d) study of carbohydrates/proteins in rocks

Ans. : (c)

60. In the developmental history of mammalian heart, it is observed that it passes through

a two-chambered fish-like heart, three chambered frog-like heart and finally four-

chambered stage. To which hypothesis can this above cited statement by

approximated ?

(a) Biogenetic law (b) Hardy-Weinberg law

(c) Lamarck’s principle (d) Mendelian principles

Ans. : (a)

61. Motile zygote of Plasmodium occurs in

(a) gut of female Anopheles (b) salivary glands of Anopheles



(c) human RBCs (d) human liver

Ans. : (a)

62. The gynoecium consists of many free pistils in flowers of

(a) Aloe (b) Tomato (c) Papaver (d)

Michelia

Ans. : (d)

63. The ovary is half inferior in flowers of

(a) cucumber (b) cotton (c) guava (d) peach

Ans. : (d)

64. In a type of apomixis known as adventives embryony, embryos develop, directly from

the

(a) nucellus or integuments (b) synergids or antipodals in an

embryo sac

(c) accessory embryo sacs in the ovule (d) zygote

Ans. : (a)

65. Genes for cytoplasmic male sterility in plants are generally located in

(a) mitochondrial genome (b) cytosol

(c) chloroplast genome (d) nuclear genome

Ans. : (a)

66. Eight nucleate embryo sacs are

(a) always bisporic

(b) always tetrasporic

(c) always monosporic

(d) sometimes monosporic, sometimes bisporic and sometimes tetrasporic

Ans. : (d)

67. Tetradynamous condition is found in

(a) Hibiscus rosa-sinensis (b) Petunia hybrida

(c) Helianthus annuus (d) Brassica campestris

Ans. : (d)

68. Triple fusion in angiosperm is the fusion of second sperm with

(a) antipodal cell and one synergid cell (b) two antipodal cells



(c) two synergid cells (d) two polar nuclei

Ans. : (d)

69. The pollen tube usually enters the embryo sac

(a) between the egg cell and synergid (b) by directly penetrating the

egg

(c) between one synergid and antipodal cell (d) by knocking off the

antipodal cells

Ans. : (a)

70. In oogamy, fertilisation involves

(a) a small non-motile female gamete and a large motile male gamete

(b) a large non-motile female gamete and a small motile male gamete

(c) a large non-motile female gamete and a small non-motile male gamete

(d) a large motile female gamete and a small non-motile male gamete

Ans. : (b)

71. If ovule is inverted, its body is parallel to funicle and micropyle is near helium, then it

is called

(a) orthotropous (b) anatropous (c) amphitropous (d)

campylotropous

Ans. : (b)

72. An ovule, which becomes curved so that the nucellus and embryo sac lie at right angles

to the funicle is

(a) hemitropous (b) campylotropous (c) anatropous (d)

orthotropous

Ans. : (a)

73. Select the characters, which are not applicable to the family-Solanaceae.

I. Epipetalous and syngenesious anthers.

II. Bicarpellary and syncarpous ovary.

III. Oblique ovary with axile placentation.

IV. Stamen six, arranged in two whorls.

V. Bicarpellary, syncarpous and inferior ovary.



(a) II and III (b) I, IV and V (c) II, IV and V (d) I and III

Ans. : (b)

74. What is the amino acid sequence encoded by the base sequence: UGA UUU UCC GGG

AGU of a mRNA segment ?

(a) Methionine–Phenylalanine–Serine–Glycine–Serine

(b) Glycine–Serine–Phenylalanine–Serine–Glycine

(c) Serine–Phenylalanine–Serine–Glycine–Serine

(d) Serine–Phenylalanine–Glycine–Serine–Glycine

Ans. : (c)

75. Cultivation of Bt cotton has been much in the news. The prefix Bt means

(a) barium-treated cotton seeds

(b) bigger thread variety of cotton with better tensile strength

(c) produced by biotechnology using restriction enzymes and ligases

(d) carrying an endotoxin gene from Bacillus thuringiensis

Ans. : (d)

76. Which of the following aquatic weeds is not used in production of biogas ?

(a) Eichhornia crassipes (b) Hydrilla

(c) Pistia stratiotes (d) Spirulina

Ans. : (d)

77. The formula of growth rate for population in a given time is

(a) dt/dN = rN (b) dt/rN = dN (c) rN/dN = dt (d) dN/dt =

rN

Ans. : (d)

78. The correct sequence of plants in a hydrosere is

(a) Oak Lantana Scirpus Pistia Hydrilla Volvox

(b) Volvox Hydrilla Pistia Scirpus Lantana Oak

(c) Pistia Volvox Scirpus Hydrilla Oak Lantana

(d) Oak Lantana Volvox Hydrilla Pistia Scirpus

Ans. : (b)



Directions : (Q. Nos. 79–90) In each of the following questions a statement of

Assertion is given followed by a corresponding statement of Reason just below it. Of

the statements, mark the correct answer as

(a) If both Assertion and Reason are true and Reason is the correct explanation of the

Assertion.

(b) If both Assertion and Reason are true, but Reason is not the correct explanation of

Assertion.

(c) If Assertion is true, but Reason is false.

(d) If both Assertion and Reason are false.

79. Assertion: The quiescent centre acts as a reservoir of relatively resistant cells, which

constitute a permanent source of active initials.

Reason: The cells of the inactive region of quiescent centre become active when the

previous active initials get damaged.

Ans. : (a)

80. Assertion: In humans, the gamete contributed by the male determines whether the

child produced will be male or female

Reason: Sex in humans is a polygenic trait depending upon a cumulative effect of some

genes on X-chromosome and some on Y- chromosome

Ans. : (c)

81. Assertion : Organochlorine pesticides are organic compounds that have been

chlorinated

Reason : Fenitrothion is one of the organochlorine pesticides

Ans. : (c)

82. Assertion : In recombinant DNA technology, human genes are often transferred into

bacteria (prokaryotes) or yeast (eukaryote).

Reason : Both bacteria and yeast multiply very fast to form huge population, wich

expresses the desired gene.

Ans. : (a)

83. Assertion : Transgenic plant production is an application of plant tissue culture

Reason : An organism that contains and expresses a transgene is called transgenic



organism

Ans. : (b)

84. Assertion : A Suspended Particulate Matter (SPM) is an important pollutant released

by diesel vehicles

Reason : Catalytic converters greatly reduce pollution caused by automobiles

Ans. : (b)

85. Assertion An organism with lethal mutation may not even develop beyond the zygote

stage.

Reason : All types of gene mutations are lethal.

Ans. : (c)

86. Assertion : Gene expression is a molecular mechanism by which a gene expression a

phenotype.

Reason : Structural genes are controlled by control genes.

Ans. : (c)

87. Assertion : Amber codon is a termination codon.

Reason : If in a mRNA, a termination codon is present, the protein synthesis stops

abruptly whether, the protein synthesis is complete or not.

Ans. : (a)

88. Assertion : Polytene chromosomes have a high amount of DNA.

Reason : Polytene chromosomes are formed by repeated replication of chromosomal

DNA without separation of chromatids.

Ans. : (a)

89. Assertion : 7-celled, 8 nucleate and monosporic embryo sac is called Polygonum type

of embryo sac.

Reason : It was discovered by Hofmeister for the first time in Polygonum.

Ans. : (c)

90. Assertion : Insects visit flower of gather honey.

Reason : Attraction of flowers prevents insects from damaging other parts of the plant.

Ans. : (d)



Section – B {Physics}

91. The capacitance of a sphere of radius R1 is increased 3 times when it enclosed by an

earthed sphere of radius R2. The ratio R2/R1 is

(a) 2 (b) 3

2 (c)

4

3 (d) 3

Sol: (b)

1 0 1C 4 R

0 1 22

2 1

4 R RC

R R

Given C2 = 3C1 Hence

0 1 20 1

2 1

4 R R3 4 e R

R R

which gives 2

1

R 3

R 2, which is choice (b).

92. A parallel plate capacitor of plate area A and plate separation d is charged by a

battery or voltage V. The battery is then disconnected. The work needed to pull the

plates to a separation 2d is

(a) 2

0AV

d (b)

202AV

d (c)

20AV

2d (d)

203AV

2d

Sol :(c)

Energy stored initially is 2

i

QU

2C. If d is doubled, C becomes C/2. Hence, energy

stored when d is doubled is 2

f

QU

C

Work needed is

2 2 22

f i

Q Q Q 1W U U CV

C 2C 2C 2 Q CV

Now, 0AC

d. Hence

20A V

W2d

, which is choice (c)



93. Two equal negative charges –q are fixed at points (0, a) and (0, – a) on the y-axis. A

positive charge Q is released from rest at a point (2a, 0) on the x-axis. The charge Q

will

(a) execute simple harmonic motion about the origin

(b) move to the origin and remain at rest there

(c) move to infinity

(d) execute oscillatory but not simple harmonic motion.

Sol :(d)

Let the charge Q be a P, with OP = x. The resultant force F is along the x-axis directed

towards the origin. The charge Q moves to O, and acquires kinetic energy. It will

cross O and move to –ve x-axis until it comes to rest. It is again attracted towards O

and cross it and this process continues. Therefore charge Q executes periodic motion

(see fig).

Let AP = BP = r. Then

1 2 20

qQF F

4 r

The resultant force on Q is

1 2 20

2qQF F cos F cos cos

4 r

3/23 2 20 0

2qQx 2qQ xF

4 r 4 a x

Thus F is not of the form F = kx (where k = constant) and hence the motion is not

simple harmonic. Hence the correct choice is (d).

94. A capacitor of capacitance 5 F is connected as shown in the fig. The internal

resistance of the cell is 0.5 . The amount of charge on the capacitor plate is



(a) zero (b) 5 C (c) 10 C (d) 25 C

Sol : (c)

We know that capacitor offers infinite resistance for DC source. Battery is DC source)

Current taken from the cell

E 25i 1A

R r 1 1 0.5

Potential drop across capacitor

V = E –ir

= 2.5 –1×0.5 = 2 V

The current in 2 resister is zero

Charge on capacitor

Q = CV = 5×2 = 10 C

95. The reading of the ammeter in the circuit in fig, is

(a) 3

A5

(b) 4

A5

(c) 6

A5

(d) 7

A5

Sol :(b)

Since the seven resistances are in parallel, the effective resistance is R = 70/7 = 10 .

Therefore, the current in the circuit is I = 14/10 = 7/5A. The given circuit can be

redrawn as shown in fig, where 1 2

70 70R and R

3 4. The current I2 is given by

2

2

RI I

R



7 10 4A

5 70 / 4 5, which is choice (b).

96. A battery of emf. E and internal resistance r is connected across a pure resistive

device (such as an electric heater) of resistance R. The power output of the device

will be maximum if

(a) R = r (b) R = 2 r (c) R = 2r (d) R = 4r

Sol :(a)

The current in the circuit is

EI

R r

Therefore the power output of the device is given by

22

2

E RP I R

R r (i)

For given values of E and r, power output P will be maximum if dP/dP = 0 and

d2P/dR2 < 0. Differentiating (i) with respect to R we get (with E and r fixed)

2

2

dP E 2R1

dR R rR r (ii)

Now dP/dR = 0 if

2R1 0

R r

which gives R = r. Thus, P will be either maximum or minimum when R = r. To decide

whether P is maximum at R = r, we find d2P/dR2 at R = r. If its value is negative, P will

be maximum. Differentiating (ii) we have

2 2

32

d P 2E 3R2

dR R rR r



2 2

2 3

at R r

d P E

dR 8r

which is negative. Hence P is maximum when R = r.

97. A current of 2 A flows through a 2 resistor when connected across a battery. The

same battery supplies a current of 0.5 A when connected across a 9 resistor. The

internal resistance of the battery is

(a) 1

3 (b)

1

4 (c) 1 (d) 0.5

Sol : (a)

Current E

iR r

E2

2 r ………..(i)

E0.5

9 r ………..(ii)

From (i) and (ii), we have

2 9 r

0.5 2 r

9 r4

2 r

3r = 1

r = 1

3

98. A uniform wire is bent into the sphere of an equilateral triangle of side a. It is

suspended from a vertex at place where a uniform magnetic field B exists parallel to

its parallel. Find the magnitude of the torque acting on the coil when a current I is

passes through it.

Sol (a)

(a) 23Ia B

4 (b) 23

Ia B2

(c) 2Ia B (c) 21Ia B

2

Area of the coil is (AB = a, BD = a/2)



A = 2 × area of triangle ABD

= 2 × 1

AD BD2

= 1 3 a

2 a2 2 2

= 23a

4

Magnetic moment of the loop is

23M IA I a

4

Since the current is clockwise the direction of vector M is perpendicular to the plane

of the coil directed inwards as shown in fig. Hence 090 . The magnitude of the

torque acting on the coil is

2 0 23 3MB sin Ia B sin90 Ia B

4 4

99. A particle of mass m and charge q, accelerated by a potential difference V enters a

region of a uniform transverse magnetic field B. If d is the thickness of the region of B,

the angle through which the particle deviates from the initial direction on leaving

the region is given by

(a)

1/2q

sin Bd2mV

(b) 1/2

qcos Bd

2mV

(c) 1/2

qtan Bd

2mV (d)

1/2q

cot Bd2mV



Sol :(a)

Refer to fig, Let v the velocity of the particle. Its kinetic energy is

1/2

21 2qVmv qV or v

2 m (1)

The particle follows a circular path from A to B of radius r which is given by

2mv mvqvB or r

r qB (2)

Using (1) and (2), we have

1/21/2m 2qV 1 2mV

rqB m B q

In triangle BCD, BD d

sinBC r

. Therefore, 1/2

qsin Bd

2mV, which is choice (a).

100. A square conducting loop of side length L carries a current l. The magnetic field at the

centre of the loop is

(a) independent L (b) proportional to L2

(c) inversely proportional to L (d) linearly proportional to L

Sol : (c)

magnetic field at the centre due to either arm



o o01

0

iB sin 45 sin45

L4

2

2 2i

4 L

Field at centre due to the four arms of the square

01

2 2iB 4B

L

1i.e., B

L

101. A metal rod PQ moves with a velocity v parallel to a very long straight wire CD

carrying a current I as shown in fig. The ends P and Q of the rod are at distances a and

b from the wire as shown. Obtain the expression for the emf induced between the

ends of the rod.

(a) 0Iv bln

a (b) 02 Iv b

lna

(c) 0Iv bln

2 a (d) 02 Iv a

lnb

Sol : (c)

Divide the rod into a large number of very small electrons, each of length dx.

Consider one such element at distance x from wire CD as shown in fig.



The magnetic field at the element due to current I in wire CD is

0IB

2 x

The direction of the field is upwards perpendicular to velocity v. The magnitude of

magnetic field is different at different points on the rod PQ. From Fleming’s L.H. rule,

the free electrons in the rod will experience force in the direction P to Q. So free

electrons move from P to Q. Hence end P acquires a positive charge (due to loss of

electrons) and end Q acquires a negative charged (due to gain of electrons).

Force on the element is

0IdF qvB qv

2 x

Therefore, electric field set up in the element is

0 0qv I IvdFdE

q 2 xq 2 x

Now dV

dEdx

0IvdV dE dx dx

2 x

Where dV is the voltage induced in the element. The voltage induced in the rod PQ is

b

0 0

a

Iv Ivdx bV ln

2 x 2 a

102. A thin non-conducting disc of radius R and mass M is held horizontally and is capable

of rotation about an axis passing through its centre and perpendicular to its plane. A

charge Q is distributed uniformly over the surface of the disc.

A time-varying magnetic field B = kt (where k is a constant and t is the time) directed



perpendicular to the plane of the disc is applied to it. If the disc is stationary initially

(i.e. at t = 0). Find the torque acting on the disc.

(a) 25kQR

4 (b)

23kQR

4 (c)

2kQR

2 (d)

2kQR

4

Sol : (d)

Area of disc = 2R

Charge per unit area = 2

Q

R

Area of a small element of width dx at a distance x from the centre of the disc =

2 xdx. Therefore, charge of the element is

2

Qdq 2 xdx

R

At time-varying magnetic field gives rise to an electric field E. Since

dVE

dl

dV Edl E 2 x

V E 2 x (1)

where V is emf induced in the element, which is given by

d dV

dt dt (BA)

= 2 2dkt x kx

dt (2)

From (1) and (2) we get

2E 2 x kx



kx

E2

(3)

Force acting on the element is

2

Q kxdF dq E 2 xdx

R 2

= 2

2

kQx dx

R

Torque acting on the disc is

R R 23

2

0 0

kQ kQRxdF x dx

R 4

103. In the capacitor of capacitance C. charge Q and energy W is stored. If charge is

increased upto 2Q, the energy stored will be

(a) W

4 (b)

W

2 (c) 2W (d) 4W

Sol : (d)

From formula

2

2

2

QW

2C

(2Q)W'

2C

QW' 4

2C

W' 4W

104. The electric and the magnetic field, associated with an electromagnetic wave,

propagating along the +z-axis can be represented by

(a) 0 0ˆ Ê E ,k,B B i (b) 0 0

ˆ Ê E , j,B B j

(c) 0 0ˆÊ E , j,E B k (d) 0 0

ˆ Ê E , j,B B j

Sol : (d)

0 0 0 0

E B

ˆˆ Ê i B j E B k

E×B points in the direction of wave propagation



105. A point source of light S is placed at the bottom of a vessel containing a liquid of

refractive index 2 . A person is viewing the source from above the surface. There is

an opaque disc of radius r floating on the surface. The centre of disc is vertically

above the source S. The liquid is gradually drained out from the vessel through a tap.

The maximum height of the liquid for which the source cannot be seen from above is

(a) r (b) 2 r (c) 3 r (d) 2r

Sol :(a)

Referring to fig, the source S cannot be seen at all from above when the water level

attains a critical maximum height x so that rays such as SA and SB suffer internal

reflection. The critical angle ic is given by

c

1 1sini

2, which gives 0

ci 45 .

In triangle OAS, we have

c 0c

r r rtan i or x r.

x tan i tan 45

Hence the correct choice is (a).

106. Light is incident at an angle an one planar end of a transparent cylindrical rod of

refractive index n. The least value of n so that the light entering the rod does not

emerge from the curved surface of the rod for any value of is

(a) 4

3 (b) 2 (c) 1.5 (d) 3

Sol :(b)

Refer to fig.

Ray OA is incident at an angle at the planar face of the cylindrical rod. Let be the

angle of refraction. From Snell’s law, we have



sin sinn or sin

sin n (1)

The ray AB is incident at point B of the curved surface of the cylinder at an angle

090 . This ray is travelling in a denser medium of refractive index n and is

incident at the cylinder-air interface at point B. The ray will not emerge from the

curved surface if it suffers total internal reflection at B. For this to happen

0c90 i , the critical angle or

0c csin 90 sin i or cos sin i

or 1/2

2c1 sin sin i

or 2 2c1 sin sin i (2)

The critical angle is given by

c

1sin i

n (3)

Using Eqs. (1) and (3) in Eq. (2) we get

22 2

2 2

sin 11 or n sin 1

n n

or 2 2n 1 sin

Since the maximum value of 2sin 1 , if follows that

2minn 2 or minn 2

This is the minimum value of refractive index of the cylindrical rod for the ray AB to

suffer total internal reflection at point B. By symmetry, ray BC will be totally reflected

along CD suffering another total internal reflection at D and so on until the ray finally

emergs from the opposite planar face of the rod.



107. A square wire of side 3.0 m is placed 25 cm from a concave mirror of focal length 10

cm. The area enclosed by the image of the wire is

(a) 1 cm2 (b) 4 cm2 (c) 16 cm2 (d) 25 cm2

Sol :(b)

u = – 25 cm and f = – 10 cm. The distance of the image is given by

1 1 1 1 1

v f u 10 25 or

3v

50 cm

Area of the object wire is 3.0 × 3.0 = 9.cm2. The area magnification is given by

area of image

area of object = (linear magnification)2

=

22

2

v 50 4

u 3 25 9

Therefore, the area enclosed by image is

2 249 cm 4 cm ,

9

which is choice (b).

108. Parallel rays from a distance object fall on a solid transparent sphere of radius R and

refractive index . The distance of the image from the sphere is

(a) R 2

2 1 (b)

R

1 (c)

R

1 (d) R 1

Sol :(a)

Refer to fig. For refraction at face I,

1 1

1 1

u v R

Since 1u , we have

1

1

v R

or 1

Rv

1 (1)



For refraction at face II, 2 1 1u v 2R 2R v .

Using Eq. (1), we get

2

R 2Ru 2R

1 1 (2)

The image distance v2 is given by

2 2

1 1

v R (3)

Using Eq. (2) in Eq. (3) and simplifying, we get

2

R 2v .

2 1

The correct choice is (a).

109. A thin prism P1 angle 4o and refractive index 154o is combined with another thin

prism P2 refractive index 1.72 to produce dispersion without deviation. The angle of

P2 is

(a) 4o (b) 5.33o (c) 2.6o (d) 3o

Sol : (d)

Angle of deviation (n 1)A

Dispersion produced by both the prism will be equal

(n1–1) A1 = (n1–1) A2

o1 12

2 2

(n 1)A (154 1) 4A 3

(n 1)A (172 1)

Hence, the angle of prism P2 is = 3o.

110. Monochromatic light of wavelength emerging from slit S illuminates slits S1 and S2

which are placed with respect to S as shown in fig. The distances x and D are large



compared to the separation d between the slits. If x = D/2, the minimum value of d so

that there is a dark fringe at the centre P of the screen is

(a) D

3 (b)

2 D

3 (c) D (d)

D2

3

Sol :(a)

Refer to fig. To reach point P, wave 1 has to travel a path (SS2 + S2P) while wave 2 has

to travel a path (SS1 + S1P). Therefore, when the waves arrive at P, the path

differences is

2 2 1 1SS S P SS S P (1)

Now, in triangle SS2S1, we have

1/221/2

2 22 2

dSS x d x 1

x

= 2

2

dx 1

2x d x

Similarly, 1/2

2 22S P D d D

2

2

d1

2D

Also 1 1SS S P x D. Using these in Eq. (1), we have

2 2

2 2

d dx 1 D 1 x D

2x 2D



= 2 2d d

x D x D2x 2D

or 2d 1 1

2 x D

In order to have a dark finger at P, 2

. Hence

2d 1 1

2 2 x D (1)

or

1/2

xDd

x D (2)

Putting x = D

2 in Eq. (2), we find that the correct choice is (a).

111. In fig, PQ represents a plane wavefront and AO and BP the corresponding extreme

rays of monochromatic light of wavelength . The value of angle for which the ray

BP and the reflected ray OP interfere ray OP interfere constructively as given by

(a) cos 2d

(b) cos4d

(c) sec3d

(d)

2sec

3d

Sol :(b)

Since P and Q are points on the same wavefront, they are in the same phase.

Therefore, the path difference at point P between the ray BP and the reflected ray OP

is

QO OP (i)



Now, in triangle POR, OP = PR d

cos cos. Also in triangle QOP, QO = OP sin

090 2 OP cos 2

OP cos 2 OP

= 2OP cos 2 1 2OP cos

= 2d2 cos 2d cos

cos

Since there is a sudden path change of 2

due to reflection, the condition of

constructive interference at P is

3,

2 2 etc.

or 3

2d cos , ,2 2

….etc.

or 3

cos ,4d 4d

, ….etc.

Hence the correct choice is (b).

112. For a photoelectric cell, the graph in fig. Showing the variation of the cut-off voltage

V0 with frequency (v) of incident light is

Sol :(d)

The correct choice is (d), because max 0 0K eV h v v



113. Two photons of energies twice and thrice the work function of a metal are incident

on the metal surface. Then the ratio of maximum velocities of the photoelectrons

emitted in the two cases respectively, is :

(a) 2 : 1 (b) 3 : 3 (c) 3 : 2 (d) 1: 2

Sol :(d)

Let v1 and v2 be the maximum velocities of photoelectrons in the two cases. The

energies or photon are given by

21 0 1

1E W mv

2 (i)

And 22 0 2

1E W mv

2 (ii)

Given E1 = 2W0 and E2 = 3W0. Using these in (i) and (ii), we have

2 20 0 1 1 0

1 12W W mv mv W

2 2 (iii)

and 2 20 0 2 2 0

1 13W W mv mv 2W

2 2 (iv)

From (iii) and (iv), we get 1

2

v 1

v 2, which is choice (d).

114. What kV potential is to be applied on X-ray tube so that minimum wavelength of

emitted

X-ray may be 10

A (h = 6.6×10–34 J-s)

(a) 12.42 kV (b) 12.84 kV

(c) 11.98 kV (d) 10.78 kV

Sol : (a)

0 0

min

12375 12375A A

V 1

12.375 kV 12.42kV

115. 22187 Ra undergoes radioactive decay with a half life of 4 days. The probability that a Ra

nucleus will disintegrate in 8 days is

(a) 1 (b) 1

2 (c)

1

4 (d)

3

4



Sol :(d)

The probability at time t that a radioactive nucleus will disintegrate is defined s

0

0

N NP

N

where N0 = number of nuclei present initially at time t = 0

N = number of nuclei left undecayed at time t.

Now 8 days = 2 half lives. After 2 half lives

0 02

N NN

42

00

0

NN 34P

N 4

Hence, the correct choice is (d).

116. The half life of a radioactive sample is 6.93 days. After how many days will only one-

twentieth of the sample be left over? Take elog 20 3.0.

(a) 20 days (b) 27 days (c) 30 days (d) 35 days

Sol :(c)

t

0

Ne .

N Thus t t1

e or 20 e20

or

et log 20 3.0

where 1/2

0.693 0.693

T 6.93 = 0.1 per day

Therefore t = 3.0 3.0

0.1 = 30 days. Thus the correct choice is (c).

117. In photocell, with exciting wavelength , the faster electron has speed v, If the

exciting wavelength is changed to 3 / 4 , the speed of the fastest electron will be

(a) 1/2

3v

4 (b)

1/24

v3

(c) less than 1/2

4v

3 (d) greater than

1/24

v3

Sol : (b)



According to Einstein’s photoelectrons.

20 1

1

hc 1W mv

2

and 20 2

2

hc 1W mv

2

These expression shown that,

2 1v

11 2

2 1

2

1

v 3 / 4

v 1

1/2

2 1

4v v

3

118. Hydrogen atom excites energy level from fundamental state to n = 3. Number of

spectrum lines according to Bohr’ is

(a) 4 (b) 3 (c) 1 (d) 2

Sol : (b)

Number of spectral lines

E

n(n 1) 3(3 1)N 3

2 2.

119. Application of a forward bias to a p-n junction

(a) increase the number of donors on the n-side

(b) increases the electric field in the depletion zone

(c) increases the potential difference across the depletion zone

(d) widens the depletion zone

Sol:(a)

When p-side of junction diode is connected to positive of battery and n-side to the

negative, then junction diode is forward biased



In this condition, more number of electrons enter in n- side from battery thereby

increasing the number of donors on n-side.

120. A hollow cylinder has a charge q coulomb within it. If is the electric flux in unit of V-

m associated with the curved surface B, the flux linked with the plane surface A in

unit of V-m will be

(a) 0

1 q

2 (b)

0

q

2 (c)

3 (d)

0

q

Sol: (No Option) fluke

Gauss’s law states that the net electric flux through any closed surface is equal to the

net charge the surface divided by 0 .

i.e., total

0

q

Let electric flux linked with surfaces A, B and C are ,A B Cand respectively, that is

total A B C

Since, C A

A B total

0

q2



or A B

0

1 q

2

But B (given)

Hence , A

0

1 q

2

121. Charges +q and q are placed at points A and B respectively which are a distance 2L

apart, C is the mid-point between A and B. The work done in moving a charge +Q

along the semicircle CRD is

(a) 0

qQ

4 L (b)

0

qQ

2 L (c)

0

qQ

6 L (d)

0

qQ

6 L

Sol: (d)

Electric potential energy of system

1

0 0 0

1 9( q) 1 qQ 1 qQV

4 2L 4 L 2 L

In Case-II: when charge +Q is moved from C to D

V2=0 0 0

1 (q)( q) 1 qQ 1 ( q)(Q)

4 2L 4 3L 4 L

Work done 2 1

0

u U U

qQ

6 L

122. When air is replaced by a dielectric medium of constant K. The maximum force of

attraction between two charges separated by a distance

(a) Increases K 1 times (b) increases K times

(c) decreases K times (d) remains constant

Sol: (c)

123. Two batteries, one of emf 18 V and internal resistance 2 and the other of emf 12 V

and internal resistance 1 , are connected as shown. The voltmeter V will record a

reading of



(a) 15 V (b) 30 V (c) 14 V (d) 18V.

Sol: (c)

Enet = 12 2 18 1 24 18 42

14V3 3 3

124. When a wire of uniform cross-section a, length l and resistance is bent into a

complete circle, resistance between two of diametrically opposite points will be

(a) R

4 (b)

R

8 (c) 4R (d)

R

2.

Sol: (a)

1

l RR

2a 2

2RR

2

Since they are parallel

1 2

1 1 1

Reff R R

1 2 2

Reff R R

RReff4



125. Charge passing through a conductor of cross-section area A = 0.3 m2 is given by q =

3t2 + 5t +2 in coulomb, where t is in second. What is the value of drift velocity at t =

2s? (Given, n = 2×1025/m3)

(a) 0.77 × 10 5 m/s (b) 1.77 × 10 5 m/s

(c) 2.08 × 105 m/s (d) 0.57 × 105 m/s.

Sol: (b)

dqI 6t 5

dt

at t 2

I 17A

d

d

I neV A

IV

neA

25 19

6

5

17

2 10 1.6 10 0.3

17.70 10

1.77 10

126. A charged particle (charge q ) is moving in a circle of radius R with uniform speed v.

The associated magnetic moment is given by

(a) qvR

2 (b) 2qvR (c)

2qvR

2 (d) qvR

Sol: (a)

As revolving charge is equivalent to a current

9 W

9t 9t 2

Now V

WR

qVi

2 r

Now magnet moment =i×A

= 2R i

= 2qV 1R qVR

2 R 2



127. A beam of electrons passes undeflected through mutually perpendicular electric and

magnetic fields. If the electric field is witched off and the same magnetic field is

maintained, the electrons move

(a) in an elliptical orbit (b) in a circular orbit

(c) along a parabolic path (d) along a straight line

Sol: (b)

128. When a charged particle moving with velocity v is subjected to a magnetic field of

induction B, the force on it is non-zero. This implies that

(a) angle between v and B is necessarily 900

(b) angle between v and B can have any value other than 900

(c) angle between v and B can have any value other than zero and 1800

(d) angle between v and B is either zero of 1800

Sol: (c)

129. No current is drawn from the secondary coil of an ideal transformer. What is the

power factor of the primary coil?

(a) 1/2 (b) 1 (c) (d) zero

Sol: (d)

130. A double slit experiment is performed with light of wavelength 500 nm. A thin film of

thickness 2 m and refractive index 1.5 is introduced in the path of the upper beam.

The location of the central maximum will

(a) remain unshifted (b) shift downward by nearly two fringes

(c) shift upward by nearly two fringes (d) shift downward by ten fringes

Sol: (d)

131. When a compact disc is illuminated by a source of white light, coloured lines are

observed. This is due to

(a) dispersion (b) diffraction (c) interference (d) refraction

Sol: (b)

132. Sodium lamps are used in foggy conditions because

(a) yellow light scattered less by the fog particles

(b) yellow light is scattered more by the fog particles

(c) yellow light is unaffected during its passage through the fog



(d) wavelength of yellow light is the mean of the visible part of the spectrum

Sol: (a)

133. The curve drawn between velocity and frequency of photon in vaccum will be a

(a) straight line parallel to frequency axis

(b) straight line parallel to velocity axis

(c) straight line passing through origin and making an angle of 450 with frequency

axis

(d) hyperbola

Sol: (a)

134. The work functions for metals A, B and C are respectively 1.92 eV, 2.0 and 5 eV.

According to Einstein’s equation, the metal(s) which will emit photoelectrons for a

radiation of wavelength 4100 o

A is/are

(a) A only (b) A and B (c) All of these (d) None of these

Sol: (d)

Work function for wavelength of 4100 0

A is

hcW

=34 8

10

6.62 10 3 10

4100 10

= 4.8 ×10–19J

=19

19

4.8 10eV 3eV

16 10

Now, we have

A B C

A

W 1.92eV, W 20eV, W 5eV

Since, W W

and WB < W, hence, A and B will emit photoelectrons.

Directions:

(a) If both assertion and reason are true and reason is the correct explanation of the

Assertion.

(b) If both assertion and reason are true and but reason is not correct explanation of



the Assertion.

(c) If Assertion is true, but reason is false.

(d) If both Assertion and reason is false.

135. Assertion : The pattern and position of fringes always remain same even after the

introduction of transparent medium in a path of one of the slit.

Reason : The central fringes is bright or dark depends upon the initial phase

difference between the two coherence sources.

Solution :(d)

If a transparent medium of the thickness t and refractive index is introduced in the

path of one of the slits, then effective path in air is increased by an amount( –1)t due

to introduction of plate. Therefore, the zeroth fringe shifts to a new position where

the two optical paths are equal. In such case fringe width remains unchanged.

The central fringe is bright or dark depends upon the initial phase difference

between the two coherent sources.



Section – B {Chemistry}

136. Lithium forms body centred cubic structure. The length of the side of its unit cell is

351 pm. Atomic radius of the lithium will be :

(a) 75 pm (b) 300 pm (c) 240 pm (d) 152 pm

Sol : (d)

For bcc, the body diagonal is four times the atomic radius.

4r 3 a

3 a 3 351 1.732 351r 152 pm

4 4 4

137. Total volume of atoms present in face centred cubic unit cell of a metal is (r = atomic

radius)

(a) 320r

3b (b) 324

r3

(c) 312r

3 (d) 316

r3

.

Sol: (d)

Number of atoms per unit cell = 8×1 1

6 48 2

Volume of 4 atoms = 4× 3 34 16r r

3 3

138. In a face centred cubic lattice, a unit cell is shared equally by how many unit cells ?

(a) 8 (b) 4 (c) 2 (d) 6

Sol : (d)

In a face centred cubic lattice, a unit cell is shared equally by six unit cells.

139. Which one of the following statements is false ?

(a) Raoult’s law states that the vapour pressure of a component over a solution is

proportional to its mole fraction

(b) The osmotic pressure of a solution is given by the equation = MRT, where,

M is the molarity of the solution

(c) The correct order of osmotic pressure for 0.01 M aqueous solution of each

compound is

BaCl2 > KCl > CH3COOH > sucrose

(d) Two sucrose solutions of same molality prepared in different solvents will have



the same freezing point depression

Sol : (d)

1.00 molal aqueous solution = 1.0 mol in 1000 g water

solute solvent

solvent

solute

d n 1; W 1000g

1000n 55.56

18

1n 0.0177

1 55.56

140. Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, how

many gram of ethylene glycol must you add to get freezing point of the solution

lowered to -2.80C?

(a) 72 g (b) 93 g (c) 39 g (d) 27 g

Sol : (b)

f fT i K m

or x

2.8 1 1.8662 1

or 2.8 62

x 93 g1.86

141. On mixing heptanes and octane, an ideal solution is formed. At 373 K, the vapour

pressure of two liquid components (heptanes and octane) are 105 kPa and 45 kPa

respectively the vapour pressure of solution obtained by mixing 25.0g of heptanes

and 35g of octane (molar mass of heptanes = 100 g

mol–1 and molar mass of octane = 114 g mol–1) is

(a) 72.0 kPa (b) 36.1 kPa (c) 96.2 kPa (d) 144.5

kPa

Sol: (a)

Heptane

25/100 0.25x 0.45

25/100 35/114 0.557

Xoctane=1–0.45 = 0.55

PTotal = XHeptane× 105 +Xoctane× 0.55

= (0.45×105+45×0.55) = 72.0 k Pa.



142. Equivalent conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45,

426.16 and 91 ohm–1 cm2 respectively. The equivalent conductance of CH3COOH at

infinite dilution would be :

(a) 101.38 ohm–1 cm2 (b) 253.62 ohm–1 cm2

(c) 390.71 ohm–1 cm2 (d) 678.90 ohm–1 cm2.

Sol : (c)

3 3m CH COOH m CH COONa m HCl m NaCl

= 91 + 426.16 – 126.45

= 390.71 ohm–1 cm2.

143. 4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a

certain quantity of electric charge. The volume of hydrogen produced at STP from H+

ions in solution by the same quantity of electric charge will be

(a) 44.8 L (b) 11.2 L (c) 22.4 L (d) 5.6 L.

Sol : (d)

Electrode reaction is 3Al 3e Al

Electricity required to produce 4.5 g of Al

= 3 4.5

F 0.5 F27

H2 produced by 0.5 F = 22.4 0.05

2 = 5.66 L at S.T.P.

144. When lead storage battery is charged

(a) lead dioxide dissolves

(b) sulphuric acid is regenerated

(c) the lead electrode becomes coated with lead sulphate

(d) the amount of sulphuric acid decrease

Sol : (b)

During the charging of a lead storage battery, the reaction at the anode and cathode

are

Anode PbSO4 +2e– Pb + SO 24



Cathode PbSO4 +2H2O PbO2 + 4H+ + SO 24 + e–

In both the reactions H2SO4 is regenerated.

145. The rate equation for the reaction :

2 NO + Cl2 2 NOCl is given by the rate equation : rate = k [NO]2 [Cl2]. The value of

rate constant can be increased by :

(a) increasing the temperature

(b) increasing the concentration of NO

(c) increasing the concentration of Cl2

(d) doing all these

Sol : (a)

The value of rate constant depends upon temperature and not upon the initial

concentration of reactants and products.

146. For the reaction, 2x + y z, the following kinetic data was obtained :

[x] [y] d[z]

dt

(mol L–1) (mol L–1) (mol L–1 s–1)

4.0 1.0 1.0

4.0 4.0 2.0

1.0 4.0 1.0

The rate expression is :

(a) k[x][y]1/2 (b) k[x]2[y] (c) k[x]1/2[y]1/2 (d)

k[x]1/2[y].

Sol : (c)

The rate expressions of given data are

1.0 = k[4.0]a [1.0]b ..(i)

2.0 = k[4.0]a [4.0]b ..(ii)

1.0 = k[10]a [4.0]b ..(iii)

To get value of b divide (ii) by (i)

2.0 = (4.0)b or b = 1/2

Similarly, to get value of a divide (ii) by (iii)



2.0 = (4.0)a or a = ½

rate expression = k[x]1/2[y]1/2

147. The reaction,

2 5 4 2 2

1N O in CCl solution 2NO solution O g

2is of first order in N2O5 with rate

constant 6.2 × 10–4s–1. What is the value of rate of reaction when [N2O5] = 1.25 mol L–1

(a) 5.15 × 10–5 mol L–1 s–1 (b) 6.35 × 10–3 mol L–1 s–1

(c) 7.75 × 10–4 mol L–1 s–1 (d) 3.85 × 10–4 mol L–1 s–1

Sol : (c)

For a first order reaction,

Rate = k[N2O5]

= 6.2×10–4 s–1 × 1.25 mol L–1

=7.75×10–4 mol L–1 s–1

148. The efficiency of enzyme catalysis is due to its capacity to

(a) form a strong enzyme-substrate complex

(b) change the shape of the substrate

(c) lower the activation energy of the reaction

(d) form a colloidal solution in water

Sol : (c)

The efficiency of enzyme catalysis is due to its capacity to lower the activation energy

of the reaction. Enzymes are biocatalysis which increases the rate of reaction without

being consumed in the reaction. In case of equilibrium reactions, catalyst help in

attaining the equilibrium quickly without disturbing the equilibrium.

149. A plot of log x/m vs. log P for the adsorption of a gas on a solid gives a straight line

with slope equal to

(a) log K (b) – log K (c) n (d) n–1.

Sol : (d)

The plot of log x/m vs. log P gives a straight line with slope = 1

n

150. Among the electrolytes, Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective

coagulating agent for Sb2S3 sol is



(a) Na2SO4 (b) CaCl2 (c) Al2(SO4)3 (d) NH4Cl.

Sol : (c)

Sb2S3 is a negative sol, so according to Hardy-Schulz rule, cation with highest positive

charge is best coagulating agent. Hence Al2(SO4)3 is the correct choice.

151. Which of the following pairs of metals is purified by van Arkel method ?

(a) Ni and Fe (b) Ga and I (c) Zr and Ti (d) Ag and Au.

Sol : (c)

152. During the process of electrolytic refining of copper, some metals present as impurity

settle as ‘anode mud’. These are

(a) Fe and Ni (b) Ag and Au (c) Pb and Zn (d) Se and Ag

Sol : (b)

During electrolysis, noble metals (inert metals) like Ag, Au and Pt are not affected

and separated as anode mud from the impure anode.

153. The tendency of BF3, BCl3 and BBr3 to behave as Lewis acid decreases in the sequence

(a) BCl3 > BF3 > BBr3 (b) BBr3 > BCl3 > BF3

(c) BBr3 < BF3 > BCl3 (d) BF3 > BCl3 > BBr3

Sol : (b)

As the size of halogen atom increases, the acidic strength of boron halides increases.

Thus, BF3 is the weakest Lewis acid. This is because of the p p back bonding

between the fully-filled unutilized 2p orbitals of F and vacant 2p orbitals of boron

which makes BF3 less electron deficient. Such that back donation is not possible in

case of BCl3 or BBr3 due to larger energy difference between their orbitals. Thus,

these are more electron deficient. Since on moving down the group the energy

differences increases, the Lewis acid character also increases. Thus, the tendency to

behave as Lewis acid follows the order

BBr3 > BCl3 < BF3

154. Correct order of decreasing thermal stability is as

(a) NH3 > PH3 > AsH3 > SbH3 (b) PH3 > NH3 > AsH3 > SbH3

(c) AsH3 > Ph3 > NH3 > SbH3 (d) SbH3 > AsH3 > PH3 > NH3

Sol : (a)



Thermal stability is inversely proportional to the M-H bond length. More the bond

length, lesser will be the stability.

155. Nitrogen forms a variety of compounds in all oxidation sates ranging from

(a) – 3 to + 5 (b) – 3 to + 3 (c) – 3 to + 4 (d) – 3 to + 6.

Sol : (a)

156. Which of the following statement is wrong ?

(a) The stability of hydrides increase from NH3 to BiH3 in group 15 of the periodic

table.

(b) Nitrogen cannot form d p bond.

(c) Single N-N bond is weaker than the single P-P bond.

(d) N2O4 has two resonance structure.

Sol : (a)

The stability of hydrides decreases form NH3 to BiH3 which can be observed from

their bond dissociation enthalpy. The correct order is NH3 < PH3 < AsH3 < SbH3 <

BiH3.

Property NH3 PH3 AsH3 SbH3 BiH3

1dissH E H / kJ mol 389 322 297 255 –

157. Which of the following statements is not valid for oxoacids of phosphorus ?

(a) All oxoacids contain tetrahedral four coordinated phosphorus

(b) All oxoacids contain at least one P = O units and one P–OH group

(c) Orthophosphoric acid is used in the manufacture of triple superphosphate

(d) Hypo phosphorus acid is a diprotic acid

Sol : (d)

Hypo phosphorus acid (H3PO2) is monobasic acid due to presence of one OH group

158. The main reason for larger number of oxidation states exhibited by the actinoids

than the corresponding lanthanoids, is :

(a) more energy difference between 5f-and 6d-orbtials than between 4f-and 5d-

orbitals.

(b) lesser energy difference between 5f- and 6d-orbitals than between 4f- and 5d-

orbitals.



(c) large atomic size of actinoids than the lanthanoids.

(d) greater reactive nature of the actinoids than the lanthanoids.

Sol : (b)

Actinoids exhibit more number of oxidation states in general because 5f orbitals

extend further from the nucleus than the 4f orbitals. As a result, nucleus has less

control on the 5f orbitals. The electrons present are more available for bonding and

to exhibit oxidation states.

159. 2 2 7 2 4 24K Cr O 4K CrO 3O X. In the above reaction X is

(a) CrO3 (b) Cr2O7 (c) Cr2O3 (d) CrO5

Sol : (c)

4K2Cr2O7 4K2CrO4 + 3O2 + 2Cr2 O3

160. The IUPAC name of 2 4K [Ni CN ] is

(a) potassium tetracyanonickelate (II)

(b) potassium tetracyanato nickelate (III)

(c) potassium tetracyanatonickel (II)

(d) potassium tetracyanonickel (III)

Sol : (a)

The IUPAC name of K2[Ni(CN)4] is potassium tetracyanonickelate (II)

161. Of the following complex ions, which is diamagnetic in nature ?

(a) 24[NiCl ] (b) 2

4[Ni CN ] (c) 2

4[CuCl ] (d)

36[CoF ]

Sol : (b)

Ni2+ in 2

4[Ni CN ] undergoes dsp2 hybridisation. No unpaired electrons are present

and hence it is diamagnetic .

162. The complexes 3 36 6 6[Co NH ][Cr CN ] and [Cr NH ]

6[Co CN ] are the examples of

which type of isomerism ?

(a) Linkage isomerism

(b) Ionization isomerism

(c) Coordination isomerism



(d) Geometrical isomerism.

Sol : (c)

163. Freon used as refrigerant is

(a) CF2 = CF2 (b) CH2F2 (c) CCl2F2 (d) CF4

Sol : (c)

Freon used as refrigerant is CCl2F2.

2 5Sb F

4 2 2freon

CCl 2HF CF Cl 2HCl

164. The alcohol that produces turbidity immediately with ZnCl2/conc. HCl at room

temperature is

(a) 1-Butanol

(b) 2-Butanol

(c) 2-Methylpropan-2-ol

(d) 2-Methylpropanol

Sol : (c)

Mixture of anhydrous ZnCl2 and conc. HCl is known as Lucas reagent. Lucas test is

used for the distinction between primary, secondary and tertiary alcohols.

The tertiary alcohol reacts immediately with Lucas reagent producing turbidity.

The secondary alcohol gives turbidity within 5-10 min and primary alcohol does not

give turbidity at room temperature. In the given alternates, 2-hydoxy-2-methyl

propane is a 3o alcohol, so it is more reactive.

165. Identify (Z’) in the following sequence of reactions.

2KOH alc. KCN alc. BrNBS, hv3 7C H I X Y Z Z'

(a) 3 2CH CH CN (b) Br CH CH CN

(c) 2 2CH CH CH CN (d) none of these.

Sol : (d)

2

KOH alc. KCN alc.NBS, hv3 7 2 3 2 2

Br2 2 2 2

C H I CH CH CH CH CH CH Br

CH CH CH CN Br CH CH(Br) CH CN



166. In the chemical reactions,

the compounds ‘A’ and ‘B’ respectively are

(a) Nitrobenzene and chlorobenzene

(b) Nitrobenzene and flurobenzene

(c) Phenol and benzene

(d) Benzene diazonium chloride and fluorobenzene.

Sol : (d)

167. The correct decreasing order of acidity of the following. P-nitrophenol (I), o-

nitrophenol(II), phenol (III), m-nitrophenol (IV)

(a) I > II > III > IV (b) II > I > IV > III

(c) I > II > IV > III (d) II > I > III > IV.

Sol : (c)

M group substituted phenols are more acidic than phenol. Among o- and p-

nitrophenol, p-nitrophenol is more acidic because in o-nitrophenol there is

intramolecular H-bonding.

168. A compound which gives a yellow solid on adding to an alcoholic solution of 2, 4-

dinitrophenyl-hydrazine but does not reduce Fehling’s solution and ammonical silver

nitrate solution, is

(a) CH3CHOHCH3 (b) CH3COCH3 (c) CH3CHO (d) CH3COOH.

Sol : (b)



169. Formaldehyde reacts with ammonia to give urotropine, which is

(a) 2 46CH N (b) 2 34

CH N (c) 2 66CH N

(d) 2 33CH N

Sol : (a)

170. CH3CHO and C6H5CH2CHO can be distinguished chemically by

(a) Benedict test (b) Iodoform test

(c) Tollen’s reagent test (d) Fehling solution test

Sol : (b)

CH3CHO give positive iodoform test whereas C6H5CH2CHO do not give due to the

absence of

171. Which of the following statements about primary amines is false ?

(a) Alkyl amines are stronger bases than aryl amines

(b) Alkyl amines react with nitrous acid to produce alcohols

(c) Aryl amines react with nitrous acid to produce phenols

(d) Alkyl amines are stronger bases than ammonia

Sol: (c)

HNO2 converts–NH2 group of aliphatic amino acid –OH while that aromatic amino –

N=NCl. Since, phenyl group is a electron withdrawing group, it decreases the basicity.

Alkyl group, on the other hand, being electron releasing. Increases the basicity. Thus,

alkyl amines are more basic as compare to aryl amines as well as ammonia.

R–NH2 2HNO R–OH

Thus, HNO2 (nitrous acid) converts alkyl amines to alcohols.

But 2HNO6 5 2 6 5

benzenediazonium chloride

C H NH C H N NCl

Thus, HNO2 does not convert aryl amines into phenol.

172. Consider the following reactions :

3 4

3

CH Cl Alkaline KMnOZn dust

Anhdrous AlClPhenol X Y Z , the product Z is :

(a) Benzaldehyde (b) Benzoic acid (c) Benzene (d) Toluene



Sol : (b)

173. Which of the following statements is not true about the following reaction ?

4Pb/BaSO2 Boiling xylene

RCOCl H RCHO HCl

(a) All aldehydes can be obtained by reduction of the corresponding acid chlorides

(b) This reaction is called Resoenmund reduction

(c) Here BaSO4 is used to poison the catalyst so that further reduction of aldehyde

cannot occur

(d) None of the above.

Sol : (a)

Since formyl chloride is unstable and hence formaldehyde cannot be prepared by

Rosenmund reduction.

174. When benzoic acid is treated with LAH, it forms:

(a) Benzaldehyde (b) Benzyl alcohol

(c) Benzene (d) Toluene.

Sol : (b)

175. What is the major product of the following reaction ?



Sol. : (c)

With Combination of active metal/conc. HCl such as Zn/HCl ,Fe/HCl or Sn/HCl nitro

group (-NO2) is reduced to 10-amino(-NH2) group.

176. In a set of reactions, m-bromo benzoic acid gave a product ‘D’. Identify the product.

Sol : (c)

177. The presence or absence of hydroxyl group on which carbon atom of sugar

differentiates RNA and DNA

(a) 1st (b) 2nd (c) 3rd (d) 4th



Sol : (b)

RNA has ribose and DNA has deoxyribose sugar portion. Deoxyribose differs from

ribose due to absence of hydroxyl group at 2nd carbon.

178. During the process of digestion, the proteins present in food materials are

hydrolysed to amino acids. The two enzymes involved in the process

protein enzyme A enzyme B

polypeptides amino acids, are respectively

(a) Invertase and zymase (b) Amylase and maltase

(c) Diastase and lipase (d) Pepsin and trypsin.

Sol : (d)

179. Which of the following is not correctly matched ?

(a)

2 2

n

CH C CH CH

Neoprene; |

Cl

(b) Nylon-66:

2 26 4 n

O

||

NH CH NH CO CH C O

(c) Tery lene ;

(d) PMMA :

3

2

3 n

CH

|

CH C

|

COOCH

Sol : (c)

Terylene contains ester linkage,

i.e. – OCH2 – CH2 – O – CO – and not ketone linkage.

180. Which one of the following is employed as a tranquillizer ?

(a) Naproxen (b) Tetracyline (c) Chlorpheniramine (d) Equanil.

Sol : (d)

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