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P-SAT 2014 - 15

(PIONEER’S SCHOLARSHIP - ADMISSION TEST) {+1 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 (135 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. A species inhabiting different geographical area is known as

(a) allopatric (b) sympatric (c) biospecies (d) sibling species

Sol: (a)

2. ‘Ordines Anomali’ of Bentham and Hooker includes

(a) seed plants showing abnormal forms of growth and development

(b) plants represented only in fossil state

(c) plants described in the literature but which Bentham and Hooker did not see in

original

(d) a few order, which could not be placed satisfactorily in the classification

Sol: (d)

3. Auxospores and Hormocysts are formed, respectively by

(a) several diatoms and a cyanobacteria

(b) several cyanobacteria several diatoms

(c) some diatoms and several cyanobacteria

(d) some cyanobacteria and many diatoms

Sol: (a)

4. Myxomycetes are

(a) saprobes or parasites. having mycelia, asexual reproduction by fragmentation, and

sexual reproduction by fusion gametes.

(b) Slimy mass of multinucleate protoplasm having pseudopodia like structures for

engulfing food, reproduction through fragmentation or zoospores.

(c) Prokaryotic organism, cellular or acellular, saprobes or autotrophs, reproduce by

binary fission.

(d) Eukaryotic, single-cellular or filamentous saprobes or autotrophic, asexual

reproduction by division of haploid individual, sexual reproduction by fusion of two

cells or their nuclei.

Sol: (b)

5. All of the following statements concerning the actinomycetous filamentous soil

bacterium Frankia are correct except that Frankia

(a) can induce root nodules on many plant species



(b) can fix nitrogen in the free living state

(c) like Rhizobium, it usually infects its host plant through root hair deformation and

stimulates cell proliferation in the host’s cortex.

(d) forms specialized vesicles, in which the nitrogenase is protected from oxygen by a

chemical barrier involving triterpene hopanoids

Sol: (b)

6. In which of the following groups are seeds present?

(a) Psilophyta (b) Ginkgopsida (c) Lycopodiophyta (d)

Bryophyta

Sol: (b)

7. The floral formula 2 2 4 2 4 (2)O K C A G

(a) Solanum nigrum (b) Hibiscus rosa sinensis

(c) Citrus aurantum (d) Brassica compestris

Sol: (d)

8. Which class of algae have chlorophyll-a, d, phycoerythrin and lack flagella

(a) Cyanophyceae (b) Rhodophyceae

(c) Phaeophyceae (d) Chlorophyceae

Sol: (b)

9. Which of the following is resurrection plant?

(a) Adiantum capillus-veneris (b) Dryopteris fillix – mas

(c) Selaginella lepidophylla (d) Adiantum caudatum

Sol: (c)

10. Which of the following pairs are correctly matched ?

Animals Morphological features

I. Crocodile – Four-chambered heart

II. Sea urchin – Parapodia

III. Obelia – Metagenesis

IV. Limur – Thecodont

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

Sol: (a)



11. Polyp phase does not occur in

(a) Obelia (b) Aurelia (c) Hydra (d) Physalia

Sol: (b)

12. Ink gland occurs in

(a) Asterias (b) Sepia (c) Pila (d) Fasciola

Sol: (b)

13. The Desulfovibrio bacterium breaks down organic matter (which it most have) and

uses sulphate (not oxygen) as an electron acceptor. As a result, it produces hydrogen

sulphate (H2S), accounting for the ‘rotten egg’ smell of swamp muck. Oxygen is a deadly

poison to Desulfovibrio. We would call Desulfovibrio a (n)

(a) facultatively aerobic chemoheterotroph

(b) Obligately anaerobic chemoheterotroph

(c) facultatively anaerobic chemoautotroph

(d) Obligately anaerobic chemoautotroph

Sol: (b)

14. Histogens capping root apical, meristem is

(a) dermatogens (b) calyptogen (c) periblem (d) plerome

Sol: (b)

15. Cellulose, the most important constituent of plant cell wall is made up of

(a) Branched chain of glucose molecules linked by , 1, 6-glycosidic bond at the site

branching

(b) unbranched chain of glucose molecules linked by , 1, 4-glycosidic bond

(c) branched chain of glucose molecules linked by , 1, 4-glycosidic bond in straight

chain and ,1,6-glycosidic bond at the site of branching

(d) unbranched chain of glucose molecules linked by , 1, 4-glycosidic bond

Sol: (d)

16. Three of the following statements regarding cell organelles are correct, while one is

wrong. Which one is wrong?

(a) Lysosomes are double membraned vesicles budded off from Golgi apparatus and

contain digestive enzymes.

(b) ER consists of a network of membranous tubules and helps in transport, synthesis



and secretion.

(c) Leucoplasts are bounded by two membranes, lack pigment but contain their own

DNA and protein synthesising machinery

(d) Spherosomes are single membrane bound and are associates with synthesis and

storage of lipids

Sol: (a)

17. In the somatic cell cycle

(a) In G1-phase, DNA content is double the amount of DNA present in the original cell

(b) DNA replication takes place in S-phase

(c) a short interphase is followed by a long mitotic phase

(d) G2-phase follows mitotic phase.

Sol: (b)

18. Mark the correct statement

(a) The value of TP becomes zero at the time of limiting plasmolysis and below zero

during incipient and evident plasmolysis

(b) The value of TP becomes below zero at the time of limiting plasmolysis and zero at

the time incipient and evident plasmolysis

(c) The value of TP remains same at the time of limiting, incipient and evident

plasmolysis

(d) The value of TP becomes negative in all the stages of plasmolysis

Sol: (a)

19. The OP and TP of two pairs of cells A-B and X-Y are as under ,

Cell A Cell B

OP = 10 atm OP = 10 atm

TP = 4 atm TP = 6 atm

Cell X Cell Y

OP = 10 atm OP = 8 atm

TP = 4 atm TP = 4 atm

The net movement of water shall be from

(a) A to B and X to Y (b) A to B and Y to X



(c) B to A and X to Y (d) B to A and Y to X

Sol: (d)

20. In the leaves of C4-plants, malic acid formation during CO2 fixation occurs in the cells of

(a) mesophyll (b) bundle sheath (c) phloem (d) epidermis

Sol: (a)

21. In photosystem-I, the first electron acceptor is

(a) ferredoxin (b) cytochrome

(c) plastocyanin (d) an iron-sulphur protein

Sol: (d)

22. A patient is generally advised to specially, consume more meat, lentils, milk and eggs in

diet only when he suffers from

(a) kwashiorkor (b) rickets (c) anaemia (d) scurvy

Sol: (a)

23. Which group of three of the following five statements (I-V) contains all three correct

statements regarding beri-beri?

I. A crippling disease prevalent among the native population of sub-Sahara Africa.

II. A deficiency disease caused by lack of thiamine (vitamin-B1)

III. A nutritional disorder in infants and young children, when the diet is persistently

deficient in essential protein.

IV. Occurs in those countries, where the staple diet is polished rice.

V. The symptoms are pain from neuritis, paralysis , muscle wasting, progressive

oedema, mental deterioration and finally heart failure.

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

IV and V

Sol: (d)

24. The richest sources of vitamin-B12 is

(a) goat’s liver and Spirulina (b) chocolate and green gram

(c) rice and hen’s egg (d) carrot and chicken’s breast

Sol: (a)

25. Spider web is formed by a fluid secreted by its

(a) abdominal gland (b) salivary gland (c) cephalothorax (d)



None of these

Sol: (a)

26. Isinglass is prepared from

(a) lchthyophis (b) lungs (c) fishes (d) lizards

Sol: (c)

27. One of the following is a connecting link between protozoans and poriferans

(a) Cliona (b) Leucosolenia (c) Oscarella (d)

Proterospongia

Sol: (d)

28. Which pair of the following belongs to Basidiomycetes?

(a) Birds nest fungi and puff-balls (b) Puff-balls and Claviceps

(c) Peziza and stink horns (d) Morchella and mushrooms

Sol: (a)

29. Which one of the following is a matching set of a phylum and its three examples?

(a) Cnidaria — Bonellia, Physalia and Aurelia

(b) Platyhelminthes — Planaria, Schistosoma and Enterobius

(c) Mollusca — Loligo, Teredo and Octopus

(d) Porifera — Spongilla, Euplectella and Pennatula

Sol: (c)

30. All mammals without any exception are characterized by

(a) viviparity and biconcave red blood cells

(b) extra-abdominal testes and a four-chambered heart

(c) heterodont teeth and 12 –pairs of cranial nerves

(d) a muscular diaphragm and milk producing glands

Sol: (c)

31. Which one of the following features is common in silver fish, scorpion, dragonfly and

prawn?

(a) Three pairs of legs and segmented body

(b) Chitinous cuticle and two pairs of antennae

(c) Jointed appendages and chitinous exoskeleton

(d) Cephalothorax and tracheae



Sol: (c)

32. In Arthropoda, head and thorax are often fused to form cephalothorax but, in which,

one of the following classes, is the body divided into head, thorax and abdomen?

(a) Insecta (b) Myriapoda

(c) Crustacea (d) Arachnida and Crustacea

Sol: (d)

33. Deuterostomate.and enterocoelomate invertebrate is;

(a) Pila (b) Ascaris (c) Aphrodite (d)

Asterias

Sol: (d)

34. Larva of Balanoglossus is

(a) Muller's larva (b) tadpole (c) tornaria iarva (d) kentrogen larva

Sol: (c)

35. Which one of the following is a matching pair of a certain body feature and its

value/count in a normal human adult?

(a) Urea : 5-10 mg/100 mL of blood

(b) Blood sugar (fasting) : 70 -100mg/100 mL

(c) Total blood volume : 5-6

(d) ESR in Wintrobe method : 9-5 mm in males and 20-34 mm in females

Sol: (b)

36. Examination of blood of a person suspected of having anaemia, shows large, immature,

nucleated erythrocytes without haemoglobin. Supplementing his diet with which of the

following is likely to alleviate his symptoms?

(a) Thiamine (b) Folic acid and cobalamine

(c) Riboflavin (d) Iron compounds

Sol: (d)

37. Intrinsic factor is required for

(a) production of gastric juice (b) absorption of B12

(c) Peristalsis (d) feeling of hunger

Sol: (b)

38. A patient, who has undergone a gastrectomy (removal of the stomach) may suffer from



(a) cirrhosis (b) pernicious anaemia

(c) gastric ulcer (d) inability to digest fats

Sol: (b)

39. Which of the following statements is false concerning the hepatic palates?

(a) Each is only one or two cell layers thick

(b) They are considered the functional units of the liver

(c) A sinusoid separates adjacent hepatic palates

(d) Bile flows through the sinusoids of the hepatic palates

Sol: (d)

40. Reduced growth, hair loss and vomiting may result from deficiency of

(a) iron (b) copper (c) potassium (d) zinc

Sol: (d)

41. Gaseous exchange in submerged hydrophytes takes place through

(a) lenticles (b) stomata

(c) hydathodes (d) general surface of plants

Sol: (d)

42. Respiration is controlled by

(a) cerebellum (b) medulla oblongata (c) olfactory lobes (d)

hypothalamus

Sol: (b)

43. Oxygen content reduction makes the glycolyse (glycogenesis) intensity increased due

to

(a) increase of ADP concentration in cell

(b) increase of NAD+ concentration in cell

(c) increase of ATP concentration in cell

(d) increase of concentration of peroxides and free radicals

Sol: (a)

44. Inflammation of the lung covering causing severe chest pain is

(a) emphysema (b) pleurisy (c) asphyxia (d) hypoxia

Sol: (b)

45. Pneumatic and inhibitory centres are associated with



(a) respiration (b) breathing (c) inspiration (d)

expiration

Sol: (b)

46. Compared to those of humans, the erythrocytes in frog are

(a) without nucleus but with haemoglobin

(b) nucleated and with haemoglobin

(c) very much smaller and fewer

(d) nucleated and without haemoglobin

Sol: (d)

47. People who have migrated from the planes to an area adjoining Rohtang Pass about six

months back

(a) have more RBCs and their haemoglobin has a lower binding affinity to O2

(b) are not physically fit to play games like football.

(c) suffer from altitude sickness with symptoms like nausea, fatigue, etc.

(d) have the usual RBC count but their haemoglobin has very high binding affinity to O2

Sol: (a)

48. Blood clotting corpuscles is

(a) thrombocyte (b) monocyte (c) lymphocyte (d)

erythrocyte

Sol: (a)

49. Which of the following is not correct?

(a) Neutrophils are most abundant RBCs

(b) Rh– person, if exposed to Rh+ blood will form specific antibodies against the Rh

antigens

(c) Fishes have1

32

chambered heart

(d) Calcium ions play a very important role in clotting

Sol: (c)

50. An artificial pace-maker is implanted subcutaneously and connected to the heart in

patients

(a) having 90% blockage of the three main coronary arteries



(b) having a high blood pressure

(c) with irregularity in the heart rhythm

(d) suffering from arteriosclerosis

Sol: (c)

51. Angiotensinogen is a protein produced and secreted by

(a) macula densa cells (b) endothelial cells (cells living the blood vessels)

(c) liver cells (d) juxtaglomerular (JG) cells

Sol: (c)

52. Leucopenia is the condition, in which

(a) the total number of leucocytes decreases below 5000/cu mm

(b) the total number of leucocytes increases above 6000/cu mm

(c) bone marrow is destroyed

(d) formation of leucocytes stops

Sol: (a)

53. The chief function of the serum albumin in the blood is to

(a) produce antibodies (b) form fibrinogen

(c) maintain colloidal osmotic pressure (d) remove waste products

Sol: (c)

54. Which of the following is not a condition of late diastole?

(a) The atria and ventricles are relaxed (b) The AV valves are open

(c) The aortic semilunar valve is open (d) Both (a) and (c)

Sol: (d)

55. Consider the following four statements (I-IV) regarding kidney transplant and select

the two correct ones out of these

I. Even if a kidney transplant is proper the recipient may need to take immuno-

suppresants for a long time.

II. The cell-mediated immune response is responsible for the graft rejection.

III. The B-lymphocytes are responsible for rejection of the

graft.

IV. The acceptance of rejection of a. kidney transplant

depends on specific interferons.



The two correct statements are

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

Sol: (d)

56. Hyperosmotic urine secretion depends upon

(a) width of Bowman's capsule (b) length of loop of Henle

(c) length of proximal convoluted tubules (d) length of distal convoluted tubules

Sol: (b)

57. Marine teleosts, undergoing putrefaction, emit sharp characteristic foul odour, which is

due to the production of

(a) trimethylamine (b) hydrogen sulphide (c) ammonia (d) lactic acid

Sol: (c)

58. Rods and cones of eyes are modified

(a) multipolar neuron (b) unipolar neuron (c) bipolar neuron (d) None of these

Sol: (c)

59. Parkinson's disease (characterized - by tremors and progressive rigidity of limbs) is

caused by degeneration of brain neurons that are involved in movement control and

make use of neurotransmitter

(a) acetylcholine (b) norepinephrine (c) dopamine (d) GABA

Sol: (c)

60. Which one of the following is a living fossil?

(a) Cycas (b) Moss (c) Saccharomyces (d) Spirogyra

Sol: (a)

61. Which of the following groups represents the all members of Phaeophyceae?

(a) Chara, Spirogyra and Fucus (b) Ectocarpus, Laminaria and Sargassum

(c) Dictyota, Polysiphonia and Porphyra (d) Gracilaria, Gelidium and Volvox

Sol: (b)

62. Ergot of rye is caused by a species of

(a) Phytophthora (b) Uncinula (c) Ustilago (d) Claviceps

Sol: (d)

63. How many plants in the list given below have composite fruits that develop from an

inflorescence?



Walnut, poppy, radish, fig, pineapple, apple, tomato, mulberry

(a) four (b) five (c) two (d) three

Sol: (d)

64. Which one of the following statements is correct?

(a) Seeds or orchids have oil-rich endosperm

(b) Placentation in primrose is basal

(c) Flower of tulip is a modified shoot

(d) In tomato and fruit is a capsule

Sol: (c)

65. The correct statement is

(a) spike, spadix or catkin inflorescence give rise to sorosis type fruit

(b) in pineapple (ananas) and jack fruit (Artocarpus), rachis, - bracts and

perianth are used for eating

(c) in mulberry, fleshy perianth is eaten

(d) All of the above

Sol: (d)

66. Why is vivipary an undesirable character for annual crop plants?

(a) It reduces the vigour of plant

(b) The seeds cannot be stored .under normal conditions for the next season

(c) The seeds exhibit long dormancy

(d) It adversely affects the fertility of the plant

Sol: (b)

67. The family containing mustard and its main characters are

(a) Brassicaceae — Tetramerous flowers, six stamens, bicarpellary gynoecium and

siliqua type fruit

(b) Brassicaceae — Pentamerous flowers, many stamens, pentacarpellary

gynoecuim and capsule type fruit

(c) Solanaceae — Pentamerous flowers, five stamens, bicarpellary gynoecium and

berry type fruit

(d) Poaceae — Trimerous flowers, three stamens, monocarpellary gynoecium

and caryopsis type of fruit



Sol: (a)

68. Bicollateral vascular bundles are found in family

(a) Cucurbitaceae (b) Solanaceae (c) Myrtaceae (d) All of

these

Sol: (d)

69. In which one of the following would you expect to find glyoxysomes?

(a) Endosperm of wheat (b) Endosperm of castor

(c) Palisade cells in leaf (d) Root hairs

Sol: (b)

70. A common structural feature of vessel elements and sieve tube element is

(a) thick secondary walls (b) pores on lateral walls

(c) presence of P-protein (d) enucleate condition

Sol: (b)

71. The correct ascending order with their size is

(a) PPLO < Virus < Bacteria < Plant cell (b) Virus < PPLO < Bacteria< Plant cell

(c) Plant cell > Virus > Bacteria > PPLO (d) Bacteria > Plant cell> Virus > PPLO

Sol: (b)

72. Which of the following statements regarding mitochondrial membrane is not correct?

(a) The outer membrane is permeable to all kinds of molecules

(b) The enzymes of the electron transfer chain are embedded in the outer membrane

(c) The inner membrane is highly convoluted forming a series of infoldings

(d) The outer membrane resembles a sieve

Sol: (b)

73. Mechanical support, enzyme circulation, protein synthesis and detoxification of drugs

are functions of

(a) ER (b) ribosomes (c) dictyosomes (d) chloroplast

Sol: (a)

74. Genes present in the cytoplasm of eukaryotic cells an found in

(a) mitochondria and inherited via egg cytoplasm

(b) lysosomes and peroxisomes

(c) Golgi bodies and smooth endoplasmic reticulum



(d) plastids and inherited via male gamete

Sol: (a)

75. GLUT-4 is a protein, which

(a) fights against infectious agents (b) enables glucose transport into cells

(c) acts as enzymes (d) acts as hormones

Sol: (b)

76. As per rule of thumb, rate of chemical reaction doubles or decreases by half for every

(a) 10oC change in temperature (b) 20oC change in temperature

(c) 30oC change in temperature (d) 40oC change in temperature

Sol: (a)

77. In which of the following enzymes, copper is necessarily associated as an activator?

(a) carbonic anhydrase (b) Tryptophanase

(c) Lactic dehydrogenase (d) Tyrosinase

Sol: (d)

78. Telomere repetitive DNA sequences control the function of eukaryotic chromosomes

because they

(a) act as replicons (b) are RNA transcription initiator

(c) help chromosome pairing (d) prevent chromosomes loss

Sol: (d)

79. Which one of the following preceds reformation of the nuclear envelope during M-

phase of the cell cycle?

(a) Decondensation from chromosomes and reassembly of the nuclear lamina

(b) Transcription from chromosomes and reassembly of the nuclear lamina

(c) Formation of the contractile ring and Formation of the phragmoplast

(d) Formation of the contractile ring and transcription from chromosomes

Sol: (a)

80. What would be the consequence of inhabiting the enzyme that degrades mitotic cyclin?

(a) The cells would immediately enter-S-phase (b) Nothing would happen

(c) The cell would arrest in mitosis (d) the cells would immediately re-

enter M-phase

Sol: (c)



81. Cell would normally processed to mitosis without interruption

(a) once it has started the S period (b) once it had entered the G2-phase

(c) at any time during cell division activity (d) none of these

Sol: (a)

82. Which of the following cells secretes a glycoprotein hormone called inhibin, which is

involved in the negative feed back control of sperm production ?

(a) Leydig cells (b) Sertoli cells (c) Prostate gland (d)

Interstitial cells

Sol: (b)

83. Which one of the following pairs is correctly matches with regard to the codon and the

amino acid coded by it ?

(a) UUA-valine (b) AAA-lysine (c) AUG-cysteine (d) CCC-alanine

Sol: (b)

84. In raccoons, a dark face mask is dominant over a bleached face mask Several crosses

were made between raccoons that were heterozygous for dark face mask and raccoons

that were homozygous for bleached face mask. What percentage of the offsprings

would be expected to have a dark face mask ?

(a) 0% (b) 50% (c) 75% (d) 100%

Sol: (b)

85. What is the strongest evidence that protobionts may have formed spontaneously ?

(a) The fossil record found in the stromatolites

(b) The discovery of ribozymes showing that prebiotic RNA molecules may have been

autocatalytic

(c) The production of organic compounds within a laboratory apparatus simulating

conditions on early earth

(d) The relative ease with which liposomes can be synthesized in laboratories

Sol: (d)

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.

86. Assertion : Our body secretes adrenalin in intense cold

Reason : Adrenalin raises metabolic rate.

Sol: (a)

87. Assertion: Too much carbon dioxide is carried in the blood, yet blood does not become

acidic

Reason: this is because in carbon dioxide transport, the blood buffers play a very

important role.

Sol: (a)

88. Assertion: Asphyxia occurs due to the lack of oxygen and retention of carbon dioxide.

Reason: It is produced by occlusion of the air way.

Sol: (b)

89. Assertion: Ginger has a prostrate growing rhizome.

Reason: Shoot growth is not effected by gravity.

Sol: (b)

90. Assertion: Pneumatophores contain lenticels that bring exchange of gases.

Reason: Pneumatophores possess stomata on their surface to regulate transpiration

from their surface.

Sol: (c)



Section – B {Physics}

91. The amplitude of a damped oscillator of mass m varies with time t as:at/m

0A A e The

dimensions of a are

(a) [ML0T–1] (b) [M0LT–1] (c) [MLT–1] (d) [ML–1T ]

Sol : (a)

The exponent is a dimensionless number. Hence at/m is dimensionless. Therefore,

Dimension of a = 0 1dimension of m MML T

dimension of t T

92. If L, R, C and V respectively represent inductance, resistance, capacitance and potential

difference, then the dimensions of L

RCV are the same as those of

(a) current (b) 1

current (c) charge (d)

1

charge

Sol : (b)

RC has the dimensions of time (T), V has the dimensions of emf which has the

dimensions of L dI

dt. The correct choice is (b).

93. A particle initially (i.e. at t = 0) moving with a velocity u is subjected to a retarding force

which decelerates is at a rate a = k v where v is the instantaneous velocity and k is a

positive constant. Find the time taken by the particle to come to rest.

(a) u

tk

(b) 2 u

tk

(c) u

t2k

(d) 3 u

t2k

Sol : (b)

dv dv dva k v k dt

dt dt v

Integrating

v t

u 0

dvk dt

v 1/2 1/22 v u kt

Putting v = 0, we get 2 u

tk



94. The speed of projection of projectile increased by 5%, without charging the angle of

projection. The percentage increase in the range will be

(a) 2.5% (b) 5% (c) 7.5% (d)

10%

Sol : (d)

We have seen above that R 2 u

2 0.05R u

= u

0.1 0.05u

or R 0.1 R . Hence, R will increase by 10%.

Thus, the correct choice is (d).

95. A particle moves along with x-axis. The position x of particle with respect to time t from

origin is given by x = b0 + b1t + b2t2. The acceleration of particle is

(a) b0 (b) b1 (c) b2 (d) 2 b2

Sol : (d)

Given, x = b0 + b1t + b2t2

Also, dx

vdt

20 1 2

dx dv b b t b t

dt dt

Using n n 1dx nx ,

dx we have

1 2v b 2b t

Acceleration a = 2

dv2b

dt

96. For a particle in uniform circular motion the acceleration a

at a point P R, on the

circle of radius R is (here is measured from the x-axis)

(a) 2 2v v

cos i sin jR R

(b) 2 2v v

sin i cos jR R

(c) 2 2v v

cos i sin jR R

(d) 2 2v v

i jR R

Sol : (c)

For a particle in uniform circular motion,



2v

aR

towards centre of circle

2v

a cos i sin jR

or 2 2v v

a cos i sin jR R

97. Figure shows the variation of velocity (v) of a body with position (x) from the origin O.

Which of the graphs shown in fig, correctly represents the variation of the acceleration

(a) with position (x) ?

Sol : (d)



The slope of the given v versus x graph is m = 0

0

v

x and intercept is c = + v0. Hence v

varies with x as

00

0

vv x v

x (i)

where v0 and x0 are constants of motion. Differentiating with respect to time t, we have

0

0

vdv dx

dt x dt

or 0

0

va v

x (ii)

Using Eq. (i) in Eq. (ii), we get

0 00

0 0

v va x v

x x

or

22

0 0

0 0

v va x

x x

Thus, the graph of a versus x is a straight line having a positive slope =

2

0

0

v

x and

negative intercept = – 20

0

v

x. Hence the correct choice is (d).

98. A liquid of mass M and density p is filled in a tube AB of length L. The tube is rotated

about end A with angular velocity . Obtain the expression for the force exerted at the

other end B.

(a) 21ML

2 (b) 23

ML2

(c) 27ML

2 (d) 25

ML2

Sol : (a)



Mass of element of length dx is m = M

L dx

Centripetal force on element = L

2

0

m x

= L2

0

Mxdx

L

= 21ML

2

99. A small sphere of mass m = 500 g moving on the inner surface of a large hemispherical

bowl of radius

R = 5 m describes a horizontal circle at a distance OC = 2.5 m below the centre O of the

bowl as shown in fig. Find the force exerted by the sphere on the bowl and the time

period of revolution of the sphere around the circle. Take g = 10 ms–2.

(a) 3.16 s (b) 3.12 s (c) 3.14 s (d) 3.18 s

Sol : (c)

Given OP = 5 m and OC = 2.5 m. Therefore 0OC 1cos 60 .

OP 2

Radius of circle is r = CP = OP sin 5 sin 060 5 3

m2



Figure shows the forces acting on the sphere, N is the normal reaction.

Net force towards centre C of the circle = N sin

2mr N sin ..(i)

Also mg N cos ..(ii)

From (ii), mg 0.5 10

N 10Ncos cos

From (i), 2m Rsin N sin

1N 102 rad s

mR 0.5 5

Time period 2

T second = 3.14 s

100. Figure shows the position-time (x–t) graph of one-dimensional motion of a body of

mass 0.4 kg. What is the time interval between consecutive impulses received by the

body ?

(a) 2s (b) 4 s (c) 8 s (d) 16 s



Sol : (b)

The slope of the graph of between t = 0 and t = 2 s is constant and positive. Therefore,

the body moves from position x = 0 to x = 2 m during the time interval from t = 0 to 2s.

Between t = 2 s and t = 4 s, the slope of the graph is constant but negative. This implies

that at t = 2s, the velocity of the body is reversed and it retraces its path and returns to

x = 0 at t = 4 s; and so on. Thus, the body receives impulses at t = 0, 2 s, 4 s, …., etc.

Therefore, the internal between two consecutive impulses is 2 s. Hence, the correct

choice is (a).

101. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by

friction in 10 s. Then the coefficient of friction is

(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.06

Sol : (d)

From equation of motion

v u at 0 u gt

u 6

0.06gt 10 10

102. A particle, which is constrained to move along the x-axis, is subjected to a force in the

same direction which varies with the distance x of the particle from the origin as

F(x) = –kx + ax3. Here k and a are positive constants. For x 0 , the functional form of

the potential energy U(x) of the particle is (see figure).



Sol : (d)

The potential energy of the particle is given by

3U Fdx kx ax dx

or 2 4 2

2x x xU k a 2k ax

2 4 4 (i)

From Eq. (i) it follows that U = 0 at two values of x which are x = 0 and x = 2k / a .

Hence graphs (b) and (c) are not possible. Also U is maximum or minimum at a value of

x given by dU

0dx

, i.e.

2 4d kx ax0

dx 2 4

= 3 2kx ax x k ax

or x k / a.

At this value of x, U is maximum if 2

2

d U0

dx,

Now 2

3 2

2

d U dkx ax k 3ax .

dx dx

At x k / a ,

2

2

d U kk 3a k 3k 2k,

dx a which is negative.

Hence U is maximum at x = k / a .



Hence graph (a) is also not possible. Also U is negative for x 2k / a . Therefore, the

correct graph is (d).

103. A box is moved along a straight line by a machine delivering constant power. The

distance moved by the body in time t is proportional to

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

Sol : (c)

Power P = Fv = mav = m dv

dt. V or v dv =

P

mdt. Integrating, we have

Pvdv dt

m ( P = constant)

or 2v Pt

2 m

or 1/22Pv t

m

or 1/2dx 2Pt

dt m

or 1/22Pdx t dt

m

Integrating again, we have

1/22Pdx t dt

m

or 3/22 2Px t

3 m

i.e. 3/2x t . Hence the correct choice is (c).

104. A body of mass m, having momentum p, is moving on a rough horizontal surface. If it is

stopped in a distance x, the coefficient of friction between the body and the surface is

given by

(a) 2

2

p

2gm x (b)

2p

2mgx (c)

p

3mgx

(d) 2

p

2gm x

Sol : (a)



Force of friction = mg . Therefore, retardation a = mg / m g . Also

2 2 2 22ax v or 2am x m v . But p = mv . Therefore,

2 22 am x p

But a g . Therefore, 2

2 2

2

p2 g m x p or .

2gm x

Hence the correct choice is (a).

105. A block of mass 10 kg is moving in x-direction with a constant speed of 10 ms–1. It is

subjected to a retarding force F = –0.1 × Jm–1, during its travel from x = 20m to x = 30 m.

Its final kinetic energy will be

(a) 475 J (b) 450 J (c) 275 J (d) none of these

Sol : (a)

Work done = 2 2f i

1 1mv mu K K

2 2

2f i

1F.dx K mv

2

2

f

1F.dx K 10 10

2

fF.dx K 500

x 30

fx 200.1 dx K 500

or

x 302

f

x 20

x0.1 K 500

2

2 2

f

30 200.1 K 500

2 2

fK 500 25

fK 500 25 475J

106. The particle of mass 50 kg is at rest. The work done to accelerates it by 20 m/s in 10 s is

(a) 103 J (b) 104 J (c) 2 × 103 J (d) 4 × 104 J

Sol : (b)

v = u + at

Given, u = 0, t = 10s, v = 20 m/s



v = 0 + a × 10

20 = a × 10

a = 2 m/s2

Also, 21s ut at

2

Given, 1

u 0, s 0 2 10 102

s = 100 m

Work done = force × displacement where, f = ma we have

W = ma × s

W = 50 × 2 × 100 = 104 J

107. A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal

floor. Initially it stands vertically. It is allowed to fall freely in a vertical plane. What is

the angular acceleration of the rod when it is at an angle with the vertical?

(a) 3g sin

L (b)

2g sin

L (c)

3g sin

2L (d)

2g sin

3L

Sol : (c)

The entire mass of the rod acts at its centre of mass C. AC = L/2. The magnitude of the

torque due to weight Mg is

Mg r

= L

Mg AD Mg sin2



Moment of inertia of the rod about A is I = 2ML

3

Angular acceleration 2

MgLsin 3

I 2 ML=

3g sin

2L

108. A thin wire of length L and uniform linear mass density p is bent into a circular loop

with centre at O as shown in fig. The moment of inertia of the loop about the axis XX’ is

(a) 3

2

L

8 (b)

3

2

L

16 (c)

3

2

5 L

16 (d)

3

2

3 L

8

Sol : (d)

Let m be the mass of the loop and r its radius. The moment of inertia of the loop about

an axis passing through the centre O is fig.

2O

1I mr

2

From the parallel axes theorem, the moment of inertia about XX’ is

2 2 2 20

1 3I I mr mr mr mr

2 2

The mass of the loop, m = pL and radius r = L/2 .

Hence

2 3

2

3 L 3pLI pL

2 2 8

Thus the correct choice is (d).



109. A circular portion of diameter R is cut out from a uniform circular disc of mass M and

radius R as shown in fig. 5.43. The moment of inertia of the remaining (shaded) portion

of the disc about an axis passing through the centre O of the disc and perpendicular to

its plane is

(a) 15

32MR2 (b) 27

MR16

(c) 213MR

32 (d) 23

MR8

Sol : (c)

Moment of inertia of complete disc about O is I = 1

2 MR2. Mass of the cut-out part is

m = M

4. The moment of inertia of the cut-out portion about its own centre

2

20

1 1 M R 1I mr

2 2 4 2 32 MR2 because r =R/2. From the parallel axes theorem, the

moment of inertia of the cut out portion about O is

2

2 2c 0

1 M RI I m MR

32 4 2

= 23MR

32

Moment of inertia of the shaded portion about O is

2 2 2s c

1 3 13I I I MR MR MR

2 32 32, which is choice (c).

110. The moment of inertia of a circular loop of radius R, at a distance of R

2 around a

rotating axis parallel to horizontal diameter of loop is

(a) MR2 (b) 21MR

2 (c) 2MR2 (d) 23

MR4



Sol : (d)

From theorem of parallel axis,

2

CM

Rl l M

2

221 MR

l MR2 4

23l MR

4

111. A uniform cylinder has a radius R and length L. If the moment of inertia of this cylinder

about an axis passing through its centre and normal to its circular face is equal to the

moment of inertia of the same cylinder about an axis passing through its centre and

perpendicular to its length, then

(a) L = R (b) L = 3 R (c)R

L3

(d)

3L R

2

Sol : (b)

Moment of inertia of a cylinder about its centre and parallel to its length = 2MR

2

Moment of inertia about its centre and perpendicular to its length = M 2 2L L

12 4

2 2 2ML MR MR

12 4 2

L 3 R

112. The centres of a ring of mass m and a sphere of mass M of equal radius R, are at a

distance 8 R apart as shown in fig. The force of attraction between the ring and the



sphere is

(a) 2

2 2 GmM

27 R (b)

2

GmM

8R (c)

2

GmM

9R (d)

2

2 GmM

9 9R

Sol: (a)

Refer to fig. Let be the mass per unit length of the ring. L = 2 R is the length of the

ring. Consider a small element of length dx of the ring located at C. Then

Force along BC is 2

GM dxf

3R. Therefore, force along BA is dF = f cos

2

GM dx

9R

8 R

3R

= 2

8 GM dx

27 R

Total force = 2 2

8 GM 8 GMmdx

27 R 27 R because dx L m, the mass of the

ring. Hence the correct choice is (a).

113. The change in the gravitational potential energy when a body of mass m is raised to a

height nR above the surface of the earth is (here R is the radius of the earth)

(a) n

mgRn 1

(b) n

n 1mgR (c) nmgR (d)

mgR

n

Sol : (a)



Change in PE = GMm GMm n

mgRR n 1 R n 1

Hence the correct choice is (a).

114. A binary start system consists of two stars of masses M1 and M2 revolving in circular

orbits of radii R1 and R2 respectively. If their respective time periods are T1 and T2, then

(a) T1 > T2 if R1 > R2 (b) T1 > T2 if M1 > M2 (c) T1 = T2 (d)

3/2

1 1

2 2

T R

T R

Sol : (c)

In a binary start system, the two starts move under their mutual gravitational force.

Therefore, their angular velocities and hence their time periods are equal. Thus the

correct choice is (c).

115. Two spheres of same size, one of mass 2 kg and another of mass 4 kg are dropped

simultaneously from the top of Qutub Minar (height = 72 m). When they are 1 m above

the ground, the two spheres have the same

(a) momentum (b) kinetic energy (c) potential energy

(d) acceleration

Sol : (d)

116. The condition for a uniform spherical mass m of radius r to be a black hole is

(G = gravitational constant and g = acceleration due to gravity]

(a) 1/2

2Gmc

r (b)

1/22Gm

cr

(c) 1/2

2Gmc

r (d)None of these

Sol : (b, c) any body answered either b or c will get the mark

117. A rubber cord of mass M, length L and cross-sectional area A is hung vertically from a

ceiling. The Young’s modulus of rubber is Y. If the change in the diameter of the cord

due to its own weight is neglected, the increase in its length due to its own weight is

(a) MgL

AY (b)

MgL

2AY (c)

2MgL

AY (d)

MgL

2AY

Sol : (b)

Let M be the mass of the rubber cord, L its length, A its cross-sectional area (assumed

constant). Let us first find the elongation dl of an element AB of length dy at a distance y



from the fixed end fig. The force due to the weight of the cord is maximum at P(y = 0)

and zero at Q(y = L). Therefore, the force acting at the element is

MgF L y

L (1)

This force is responsible for the elongation of element AB. Now

stress F / A FdyY

strain dl / dy Adl

or Fdy

dlAY

(2)

Using Eq. (1) in Eq. (2), we get

Mgdl L y dy

LAY

To obtain the total elongation l of the cord, we integrate from y = 0 to y = L. Thus

L

0

Mgl dl L y dt

LAY

=

L2

0

Mg yLy

LAY 2=

22Mg L MgL

LLAY 2 2AY

The correct choice is (b).

118. A large container (with open top) of negligible mass and uniform cross-sectional area A

has a small hole of cross-sectional area a in its side wall near the bottom. The container

is keep on a smooth horizontal platform and contains a liquid of density p and mass m.

If the liquid starts flowing out of the hole at time t = 0, the initial acceleration of the



container is

(a) ga

A (b)

gA

a (c)

2ga

A (d)

gA

2a

Sol : (c)

Let h be the initial height of the liquid of density p in the container of cross-sectional

area A. The mass of the liquid in the container initially is fig.

m = Ahp

From Torricelli’s theorem, the velocity of the liquid flowing out of the hole is

v 2gh

Volume of liquid flowing out per unit time = av.

Hence the mass of liquid flowing out per unit time = pav. Therefore, the momentum

carried per unit time by the liquid flowing out is = (mass per unit time) × velocity =

(pav)v = pav2.

This is the rate of change of momentum of the liquid flowing out which is the force with

which the liquid flows out at t = 0.

Initial acceleration = 2 2force pav av

mass Ahp Ah

= a 2gh

v 2ghAh

= 2ga

A, which is choice (c).

119. A force F is applied on the wire of radius r and length L and change in the length of wire

is l. If the same force F is applied on the wire of the

(a) l

2 (b)

2l

2 (c)

3l

2 (d) None of

these

Sol : (a)



Elongation, FL

lAY

or 2

Ll

r

2

1 2 1

1 1 2

l L r

l L r

= 2

1 12

2 2

or 12

ll

2

The change in the length of other wire is l

2.

120. In steel, the Young’s modulus and the strain at the breaking point are 2 × 1011 N/m2 and

0.15 respectively. The stress at the breaking point for steel is therefore

(a) 2 × 108 N/m2 (b) 3 × 1010 N/m2 (c) 3 × 1012 N/m2 (d) None of these

Sol : (b)

Breaking stress = Strain × Young’s modulus

= 0.15 × 2 × 1011

= 3 × 1010 N/m2.

121. The physical quantity which does not have the same dimensions as the other three is

(a) spring constant (b) surface tension

(c) surface energy (d) acceleration due to gravity

Sol : (d)

Spring constant, surface tension and surface energy have the same dimensions namely

[ML0T–2], acceleration due, to gravity has dimensions [LT–2].

122. Velocity of sound in a gas is given by p

v . Dimensional formula for is

(a) [MLT] (b) [M0L0T0] (c) [M0LT0] (d) [ML0T0]

Sol : (b)

is the ratio of Cp to Cv. It has no unit. Its dimensional formula is [M0L0T0].



pv

p

2 2 2 3 00 0 0

0 0 0

v p [L T ][ML T ][M L T ]

p [M L T ]

123. A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed

vd. The average speed for this round trip is

(a) d u

d u

2v v

v v (b) u dv v (c) d u

d u

v v

v v (d) u dv v

2.

Sol : (a)

Average speed = Distance travelled

Time taken

Let t1 and t2 be times taken by the car to go from X to Y and then from Y to X

respectively.

Then, 1 2

u d

XY XYt t

V V

= u d

u d

v vXY

v v

Total distance travelled = XY + XY = 2XY

u dav

u du d

u d

2v v2XYv

v vv vXY

v v

124. A particle moves along a straight line OX. At a time t (in second) the distance x (in

metre) of the particle form O is given by x = 40 + 12t t3. How long would the particle

travel before coming to rest?

(a) 24 m (b) 40 m (c) 56 m (d) 16m.

Sol : (c)

Distance travelled by the particle is

x = 40 + 12t – t3

v = dx

dt (x = distance)

3 n n 1d dv 40 12t t x nx

dt dx



But final velocity v = 0

12 – 3t2 = 0

or 2 12t 4

3

or t = 2 s

Hence, distance travelled by the particle before coming to rest is given by

x = 40 + 12 (2) – (2)3

= 40 + 24 – 8

= 64 – 8 = 56 m

125. When a ball is thrown up vertically with velocity v0. It reaches a maximum height of h. If

on e wishes to triple the maximum height, then the ball should be thrown with velocity

(a) 03v (b) 03v (c) 9 v0 (d) 3/2 v0

Sol : (a)

v2 = u2 – 2gh

Here, v = 0, u = v0

200 v 2gh

0v 2gh

When h’ = 3h, then

'0 0v 2g 3h 3 2gh 3 v

126. A stone of mass m is tied to a string of length l and rotated in a circle with a constant

speed v. If the string is released, the stone flies

(a) Tangential outward (b) Radically outward

(c) Radially inward (d) With an acceleration 2mv

l

Sol : (a)

127. A rough vertical board has an acceleration ‘a’ so that a 2 kg block pressing against it

does not fall. The coefficient of friction between the block and the board should be



(a) <a

g (b)

g

a (c)

a

g (d)

g

a

Sol : (d)

The block will not fall if friction > weight

R Mg

Here R = Ma

Ma Mg

g

a

128. Which of the following statements is true

(a) Total kinetic energy is conserved in elastic collisions but momentum is not

conserved in elastic collisions

(b) Total kinetic energy is not conserved but momentum is conserved in inelastic

collisions

(c) In elastic collisions, the momentum is conserved but not in inelastic collisions

(d) Both kinetic energy and momentum are conserved in elastic as well as inelastic

collisions

Sol : (b)

129. A bomb of mass 3.0 kg explodes in air into two places of masses 2.0 kg and 1.0 kg. The

smaller mass goes at a speed of 80 ms/. The total energy imparted to the two fragments

is :

(a) 1.07 kJ (b) 2.14 kJ (c) 2.4 kJ (d) 4.8 kJ

Sol : (d)

Centre of mass doesnot change

2 × v + 1(80) = 0

2v = – 80



v = 80

2 = –40m/sec

Now E = 2 21 1 2 2

1 1m v m v

2 2

= 2 21 1

2 40 1 802 2

= 1600 + 3200

= 4800 = 4.8 kJ

130. For inelastic collision between two spherical rigid bodies :

(a) The total kinetic energy is conserved

(b) The total potential energy is conserved

(c) The linear momentum is not conserved

(d) The linear momentum is conserved

Sol : (d)

131. A horizontal platform is rotating with uniform angular velocity around the vertical axis

passing through its centre. At some instant of time a viscous fluid of mass m is dropped

at the centre and is allowed to spread out and finally fall, the angular velocity during

this period

(a) decreases continuously (b) decreases initially and increases again

(c) remains unaltered (d) increases continuously

Sol : (b)

132. Two spheres of masses M and 2 M are initially at rest at a distance R apart. Due to

mutual force of attraction, they approach each other. When they are at separation R/2,

the acceleration of their centre of mass would be

(a) 0 (b) g (c) 3 g (d) 12 g

Sol : (a)

Since no external force is acting on the system

Position of com doesnot change.

Acceleration of com doesnot change

Option (a) is correct.

133. In an orbital motion, the angular momentum vector is

(a) parallel to the linear momentum. (b) in the orbital plane.



(c) along the radius vector. (d) perpendicular to the orbital plane.

Sol : (d)

134. The mass of moon is 1/81 of earth’s mass and its radius 1/4 of that of earth. If the

escape velocity from the earth’s surface is 11.2 km s–1, its value for the moon is

(a) 0.14 km s–1 (b) 0.76 km s–1 (c) 2.45 km s–1 (d) 5.28 km s–1

Sol : (c)

Given moon earth

1M M

81

moon earth

1R R

4

Now = moon

2moon

gmg'

r

earth2earth

GM16g'

81 r

earth

16g' g

81

Now eV 2gr

moon moon

16 rV 2g'R 2 g

81 4

= 2 2 22.4

2gr 11.2 2.48 km / s9 9 9

135. A satellite is in an orbit around the earth. If its kinetic energy is doubled, then

(a) it will escape out of earth’s gravitational field.

(b) it will maintain its path.

(c) it will fall on the earth.

(d) it will rotate with a great speed.

Sol : (a)



Section-C Chemistry

136. Amount of oxygen require for combustion of 1 kg of mixture of butane and isobutane is

(a) 35.86 (b) 3.586 (c) 3.23 (d) none of these

Sol : (b)

4 1O 2 2 2

13C H O 4CO 5H O

2

58 g of C4H10 require O2 = 13

322

1000g of C4H10 require O2 = 13 32 1000

3586.2g2 58

= 3.586 kg

137. In a compound C, H and N are present in 9 : 1 : 3.5 by weight. If molecular weight of

compound is 108, the molecular formula of compound is

(a) C2H6N2 (b) C3H4N (c) C6H8N2 (d) C9H12N3

Sol : (c)

Element Ratio (by weight) Atomic weight Mol Ratio Simplest Mol

Ratio

C 9 12 9

0.7512

0.75

30.25

H 1 1 1

1.001

1.00

40.25

N 3.5 14 3.5

0.2514

0.25

10.25

Empirical formula = C3H4N

and empirical formula weight = 3×12 + 4×1 +14 = 54

Molecular weight = 108

Molecular weightn

Empirical formula weight

108

n 254

Molecular formula = (Empirical formula)n



= (C3H4N)2

= C6H8N2

138. Which of the following is non-permissible ?

(a) n = 4, l = 3, m = 0 (b) n = 4, l = 2, m = 1

(c) n = 4, l = 4, m = 1 (d) n = 4, l = 0, m = 0

Sol : (c)

According to the rules for quantum number the possible values of n, l, m and s are

n=1 to , any whole number ,

l = 0 to (n–1) for every value of n.

m = –l to 0 +l, for every value of l.

and s = 1 1

or2 2

(a) n = 4, l = 2, m = 1

All the values are according to rules.

(b) n = 4, l =2, m = 1

All the values are according to rules

(c) n = 4, l = 4, m = 1

This value l can have maximum (n–1)

value i.e., …3 in this case.

This set of quantum numbers us non-permissible.

(d) n = 4, l = 0, m = 0

All the values are according to rules.

Choice (a), (b) and (d) are permissible.

139. The wavelength of a spectral line emitted by hydrogen atom in the Lyman series is 16

15R

cm. What is the value of n2 ? (where, R = Rydberg constant)

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

Sol : (c)

for Lyman series ,



2 22

2 22

22

22

22

22

1 1 1R

1 n

15R 1 1R

16 1 n

15R n 1

16R n

15 n 1

16 n

2 22 2

22

2

15n 16n 16

n 16

n 4

140. Which one of the following ions has electronic configuration [Ar] 3d6 ?

(At. no : Mn = 25, Fe = 26, Co = 27, Ni = 28)

(a) Ni3+ (b) Mn3+ (c) Fe3+ (d) Co3+

Sol : (d)

Ni3+ (28) = [Ar] 3d7

Mn3+ (25) = [Ar] = 3d4

Fe3+ (26) = [Ar] = 3d5

Co3+ (27) = [Ar] = 3d6

141. Among the alkali metal halides, the covalent character decreases in order

(a) MF > MCl > MBr > MI (b) MF > MCl > MI > MBr

(c) MI > MBr > MCl > MF (d) MCl > MI > MBr > MF.

Sol : (c)

The covalent character decreases with the decrease in the size of halide ion. Hence,

covalent character decreases MI > MBr > MCl > MF

142. Identify the wrong statement in the following.

(a) Amongst isoelectronic species, smaller the positive charge on the cation, smaller is

the ionic radius

(b) Amongst isoelectronic species, greater the negative charge on the anion, larger is

the ionic radius

(c) Atomic radius of the elements increases as one moves down the first group of the



Periodic Table

(d) Atomic radius of the elements decreases as one moves across from left to right in

the second period of the Periodic Table

Sol : (a)

Atomic radius of the elements decreases across a period from left to right due to

increase in effective nuclear charge, On moving down a group, since number of shells

increases , so atomic radius increases. Amongst isoelectronic species, ionic radius

increases with increases negative charge or decrease in positive charge.

143. In which one of the following species, the central atom has the type of hybridisation

which is not the same as that present in the other three ?

(a) SF4 (b) 3I (c) 25SbCl (d) 5PCl

Sol : (a)

Molecules having the same number of hybrid orbitals, have same hybridization and

number of hybrid orbitals.

1H V X C A

2

where, V = number of valence electron of centraql atom

X= number of monovalent atoms

C= charge on cation

A = Charge on anion.

(a) In SF4, 1

H 6 4 0 0 52

(b) In I 3 , 1

H 7 2 1 52

(c) In SbCl 25 ,

1H 5 5 2 6

2

(d) In PCl2, 1

H 5 5 0 0 52

Since, only SbCl 25 has different number of hybrid orbitals (i.e., 6) from the other given

species, its hybridization is different from the others, i.e., sp3d2. (The hybridization of

other species is sp3d)



144. Which of the following molecules has trigonal planar geometry ?

(a) BF3 (b) NH3 (c) PH3 (d) IF3

Sol : (a)

(a) BF3 3bp +0lp sp3-hybridisation and trigonal planar geometry.

(b) NH3 3bp +1/p sp3-hybridisation and pyramidal geometry.

(c) PH3 3bp +1/p sp3-hybridisation pyramidal geometry.

(d) IF3 3bp +2/p sp3d-hybridisation and T shape.

145. The root mean square velocity of one mole of a monoatomic gas having molar mass M is

Urms. The relation between the average kinetic energy (E) of the gas and Urms is

(a) rms

3EU

2M (b) rms

2EU

3M (c) rms

2EU

M (d) rms

EU

3M

Sol : (c)

Average kinetic energy, 3

E RT2

………..(1)

Root mean square velocity, rms

3RTU

M

From equ.(i),

On putting the value of T in the expression of Urms, we get

rms

3 2E 2EU

3M M

146. Which of the following statement is not true about the effect of an increase in

temperature on the distribution of molecular speeds

(a) Area under distribution curve remains the same as that under lower temperature

(b) The distribution becomes broader

(c) The fraction of molecules, with the most probable speeds increases

(d) The most probable speed increases.

Sol: (c)

147. Which of the following has the minimum bond length?

(a) 2O (b) 22O (c) 2O (d) 2O

Sol : (d)



2

2

22

2

10 5Bond order of O 2.5

2

10 7Bond order of O 1.5

2

10 8Bond order of O 1

2

10 6Bond order of O 2

2

Maximum bond order = minimum bond length

Bond length is minimum for O 2 .

148. The vapour pressure of two liquids P and Q are 80 and 60 torr respectively. The total

vapour pressure of solution obtained by mixing 3 moles of P and 2 moles of Q would be

(a) 140 torr (b) 20 torr (c) 68 torr (d) 72 torr

Sol : (d)

Mole fraction of P = 3 3

3 2 5

Mole fraction of Q = 2 2

3 2 5

Hence, total vapour pressure = (Mole fraction of P × Vapour pressure of P)+(Mole

fraction of Q × vapour pressure of Q)

=3 2

80 60 48 24 72torr5 5

149. A gaseous mixture was prepared by taking equal moles of CO and N2. If the total

pressure of the mixture was found 1 atm, the partial pressure of the nitrogen (N2) in the

mixture is

(a) 0.8 atm (b) 0.9 atm (c) 1 atm (d) 0.5 atm

Sol : (d)

2

2

2

2

2

CO N

CO N

CO N

N

N

n N

p P

Given, p p 1 atm

or 2p 1atm

or p 0.5atm



150. Which of the following is an intensive property ?

(a) Temperature (b) Viscosity (c) Surface tension (d) All of the above

Sol : (d)

The properties of the system, value of which is independent of the amount of substance

in the system are called intensive properties e.g., viscosity, surface tension

temperature, pressure etc.

151. The enthalpy change of reaction does not depend upon

(a) state of reactants and products

(b) nature of reactants and products

(c) various of intermediate reactions

(d) enthalpy of reactants and products.

Sol : (c)

According to Hess’s law, the enthalpy change for a reaction does not depend on the

nature of intermediate reaction steps.

152. If 1 mole of an ideal gas expands isothermally at 370C from 15 L to 25 L, the maximum

work obtained is

(a) 12.87 J (b) 6.43 J (c) 8.57 J (d) 2.92 J

Sol : (a)

2rev

1

VW 2.303 nRT log

V

=25

2.303 1 0.0821 (273 37)log15

=5

2.303 0.0821 310log3

=13J 12.87 J

153. The change in entropy for the fusion of 1 mole of ice is (melting point of ice = 273 K,

molar enthalpy of fusion for ice = 6.0 kJ mol–1)

(a) 11.73 JK–1mol–1 (b) 18.84 JK–1mol–1 (c) 21.97 JK–1mol–1 (d)

24.47 JK–1mol–1

Sol : (c)

Entropy change of fusion o

o tf

HS

T



3

f

60 10S

273

= 1 121.97 JK mol

154. In which of the following equilibrium Kc and Kp are not equal ?

(a) 2 22C s O g 2CO g

(b) 2 2 3SO g NO g SO g NO g

(c) 2 2H g I g 2HI g

(d) 2 22NO g N g O g

Sol : (a)

The reaction for which the number of moles of gaseous products (np) is not equal to the

number of moles of gaseous reactants (nR), has different value of Kc and Kp.

(a) np = nR = 2, thus, Kp = Kc.

(b) np = nR = 2, thus, Kp = Kc.

(c) np = nR = 2, thus, Kp = Kc.

(d) np = 2, nR=1 thus, Kp Kc.

155. A reaction is, A + B C + D; initially we start with equal concentrations of A and B.

At equilibrium, we find the moles of C is two times of A. What is the equilibrium

constant of the reaction ?

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

2 (d)

1

4

Sol : (b)

c

A B C D

At equ. a a 2a 2a

C C 2a 2aK 4

A B a a

156. If the concentration of 24CrO ions in a saturated solution of 2 4Ag CrO is 2 × 10–4 M, the

solubility product of 2 4Ag CrO would be

(a) 32 × 10–12 (b) 8 × 10–12 (c) 16 × 10–12 (d) 8 × 10–8

Sol : (a)



22 4 4

2s ss

Ag CrO s 2Ag 2Ag CrO

Given, 2 44CrO s 2 10

4

4

2 2sp 4

24 4

12

Ag 2s 2 2 10

4 10

K Ag CrO

4 10 2 10

32 10

157. In which of the following, the oxidation number of oxygen has been arranged in

increasing order ?

(a) OF2 < KO2 < BaO2 < O3 (b) BaO2 < KO2 < O3 < OF2

(c) BaO2 < O3 < OF2 < KO2 (d) KO2 < OF2 < O2 < BaO2

Sol : (b)

Let the oxidation number of oxygen in the following compounds be x.

In OF2

x + (–1)2 = 0

In KO2

+1+(x +2) =0

2x = –1

x = 1

2

In BaO2

+2+(x ×2) =0

2x = –2

x = –1.

InO3, oxidation number of oxygen is zero because oxidation number of an element in

free state or in any of its allotropic form is always zero.

Thus, the increasing order of oxidation number is

2 2 3 211 202

BaO KO O OF



158. Given 2 2

o o

Cr /Cr Fe /FeE 0.72 V, E 0.42 V. The EMF for the cell Cr

2 2Cr 0.1 M Fe 0.01 M Fe is

(a) 0.26 V (b) –1.14 V (c) 0.30 V (d) – 0.30 V.

Sol: (c)

As 2 2

o o

Cr /Cr Fe /FeE 0.72 V and E 0.42 V

EMF of cell = o 0cathode anodeE E

=–0.42 – (–0.72) = 0.30 V

159. The reduction potential values of M, N, and O are +2.46, –1.13 and – 3.13 V respectively.

Which of the following order is correct regarding their reducing property?

(a) O > N > M (b) O > M > N (c) M > N > O (d) M > O > N.

Sol: (a)

Lower is the value of Ered better is the reducing property.

160. Chemical ‘A’ is used for water softening to remove temporary hardness. ‘A’ reacts with

sodium carbonate to generate caustic soda. When CO2 is bubbled through a solution of

‘A’, it turns cloudy. What is the chemical formula of A?

(a) CaO (b) Ca(OH)2 (c) CaCO3 (d) Ca(HCO3)2

Sol : (b)

Ca(OH)2 is the used for the softening of temporary hard water,

161. The bond that undergoes heterolytic cleavage most easily is

(a) C-O (b) C-C (c) C-H (d) O-H

Sol: (d)

Greater the difference in Electronegativity of bonded atom easier will be heterolytic

cleavage.

162. In ethane and cyclohexane, which one of the following pairs of conformations are more

stable ?

(a) eclipsed and chair conformation

(b) staggered and chair conformation

(c) staggered and boat conformation

(d) eclipsed and boat conformation

Sol : (b)



In ethane and cyclohexane, staggered and chair forms are more stable respectively.

163. Dehydrohalogenation of an alkyl halide is

(a) nucleophilic substitution reaction

(b) elimination reaction

(c) rearrangement reaction

(d) Both (a) and (b)

Sol : (b)

Dehydrohalogenation of alkyl halide in the presence of alc. KOH is the example of

elimination reaction.

2 2 2 2alc.

RCH CH Cl KOH RCH CH KCl H O

164. Acetylene on treatment with alkaline KMnO4 yields

(a) Oxalic acid (b) Ethanol (c) Glyoxal (d) 1, 2-Ethanediol.

Sol : (a)

165. Which amongst the following is the most stable carbocation ?

(a) 3CH (b) 3 2CH CH (c)

3

3

CH C

|

CH

(d)

3

3

3

CH

|

CH C

|

CH

Sol : (d)



More is the number of + I groups (r) greater is the dispersal of positive charged and

hence more is the stability of carbocation.

166.

(a) 1-methyl spiro [221] hept-2-one (b) 1-methyl bicyclo [221] heptane

(c) 1- methyl bicyclo [122] heptane (d) Bicyclo [122] heptanes

Sol (b)

167. Which among the following is likely to show geometrical isomerism?

(a) CH2=CH–CH=CCl2 (b) CH3C(Cl)=C(CH3)2

(c) CH3CH=CH2 (d) CH3CH=NOH

Sol : (d)

168. For (i) I–, (ii) Cl–, (iii) Br–, the increasing order of nucleophilicity would be

(a) Cl– < Br– < I– (b) I– < Cl– < Br–

(c) Br– <Cl– < I– (d) I– < Br– < Cl–



Sol : (a)

For the nucleophilic atoms, down the group of the periodic table, nucleophilicity

increases whereas basicity decreases.

169. In reaction

3 2

3

H C CH CH CH HBr A

|

CH

A (Predominantly) is :

Sol : (c)

170. 2

4 2 4

H O Tautomerism2 2 HgSO /H SO

C H A B

The compounds B is

(a) An acid (b) Acetaldehyde (c) Acetone (d) Ethanol.

Sol : (b)

4

2 4

HgSO Tautomerism2 2 2 3H SO

C H [CH CHOH] CH CHO.

171. Aromatisation of n-heptane by passing over (Al2O3 + Cr2O3) catalyst at 773 K gives

(a) benzene (b) toluene (c) mixture of both (d) heptylene

Sol : (b)

n-heptane 2 3 3 2Al O /Cr O toluene



172. The reaction of toluence with Cl2 in presence of Halogen carriers gives predominantly

(a) o- and p- chlorotoluence (b) m-chlorotoluene

(c) benzoyl chloride (d) benzyl chloride.

Sol : (a)

CH3–group in toluene is ortho-para directing. Hence, chlorination of toluene takes place

at ortho para positions.

173. Violet coloured complex obtained in the detection of sulphur is

(a) Na2[Fe(NO)(CN)5] (b) Na3[Fe(ONS)(CN)5]

(c) Na4[Fe(CN)5NOS] (d) Both (b) and (c)

Sol : (c)

Sulphur of organic compounds is detected b y using sodium nitroprusside. With it

sulphur forms violet colour complex, called thionitroprusside.

2 2 5 4 5sodium nitroprusside sodium thionitroprusside

Na S Na [Fe(CN) NO] Na [Fe(CN) NOS]

174. Structural formula of Lewisite is

Sol : (c)

Lewisite is obtained when acetylene reacts with arsenic chloride.

3AsCl

2Lewisite

CH CH CHCl CHAsCl

175. In the following reaction, the product ‘R’ is

32

3

CH ClH O Hot iron2 tube AlCl

CaC P Q R

(a) benzene (b) ethyl benzene

(c) toluene (d) n-propylbenzene



Sol : (c)

176. Which one of the following alkenes will react faster with H2 under catalytic

hydrogenation conditions ?

Sol : (a)

1Stablility of alkene

Heat of hydrogenation of alkene

Greater the number of alkyl groups attached to the doubly bounded carbon atoms,

more stable is the alkene.

177. The function of AlCl3 in Friedel-Crafts reaction is

(a) to absorb HCl (b) to absorb water

(c) to produce nucleophile (d) to produce electrophile

Sol : (d)

The function of AlCl3, in Friedel-Crafts reaction, is to produce electrophile, which later

add to benzene nucleus.

178. X is heated with sodalime and gives ethane. X is

(a)sodium ethanoate (b) sodium methanoate

(c) sodium propanoate (d) either (a) or (c)

Sol : (c)

NaOH/CaO2 5 2 6 2 3C H COOH C H Na CO



179. In triplets carbenes, the two electrons

(a) Are paired in one orbital (b) Are present in different orbitals

(c) Have the same spin (d) Bothe (b) and (c)

Sol : (d)

Carbenes is neutral divalent carbon intermediate in which carbon is covalently bonded

to two monovalent atoms or groups and has two non-bonded electrons. In triplets

carbene two electrons are present in different orbitals with same spin( ).

180. Which is following statements is false:

(a) Beilstein test is reliable test for halogens in organic compounds

(b) The purpose of boiling Lassaigne’s extract with concentrated HNO3 before testing

for halogen is to destroy any Na2S or NaCN if formed in the reaction

(c) Presence of sulphur in Lassaigne’s test can be shown with the help of violet colour

(d) Halogen cannot be estimated by duma’s method

Sol : (a)

Beilstein test is not reliable test for halogen.

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