pioneer education {the best way to success} …...which one of the following is a correct...

Pioneer Education {The Best Way To Success} Medical and Non - Medical Classes

www.pioneermathematics.com S.C.O.- 320, Sector 40–D, CHD. Phone: 9815527721, 4617721

1

P-SAT 2014 (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: MATHEMATICS (1 to 30)

Section-B: PHYSICS (31 to 60),

Section-C: CHEMISTRY (61 to 90).

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:__________________________________

Solution Visits: www.pioneermathematics.com/latest_updates.com

http://www.pioneermathematics.com/latest_updates.com



2

Section-A {Biology}

1. Which one of the following is a correct statements?

(a) In pteridophyte, gametophyte has a protonemal and leafy stage

(b) In gymnosperms, female gametophyte is free living

(c) Antheridiophores and archegoniophores are present in pteridophytes

(d) Origin of seed habit can be traced in pteridophytes

Ans. (d)

2. The pollination of two flowers on different plants is known as

(a) xenogamy (b) geitonogamy (c) cleistogamy (d) dichogamy

Ans.(a)

3. The angiospermic endosperm except in special cases is a triploid (3n) tissue as it a

product of triple fusion involving double fertilization. It is thus, distinct from the

endosperm of gymnosperms and heterosporous pteridophytes, where the

endosperm is

(a) diploid before fertilization

(b) simple haploid (n) tissue of the gametophyte not involving any complication like

fusion or fertilization

(c) polyploid formed after simple fertilization

(d) haploid formed after fertilization

Ans. (b)

4. Coffee plant has chromosome number of 2n in its somatic cells, what is the

chromosome number in the edible part of coffee seed?

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

Ans. (b)

5. In a normal pregnant woman, the amount of total gonadotropin activity was

assessed. The result expected was

(a) high level of circulating FSH and LH in the uterus to stimulate implementation of

the embryo

(b) high level of circulating HCG to stimulate endometrial thickening

(c) high levels of FSH and LH in uterus to simulate endometrial thickening



3

(d) high level of circulating HCG to simulate oestrogen and progesterone synthesis

Ans. (d)

6. Which of the following is not correct?

(a) Fertilisin is present in the surface of egg

(b) Antifertilisin is present in the surface of spermatozoa

(c) Fertilisation between different species is prevented due to species reaction

between fertilisin and antifertilisin

(d) Cleavage involves cell division with increase in growth

Ans. (d)

7. A cross section at the midpoint of the middle pieces of a human sperm will show

(a) centriole, mitochondria and 9 + 2 arrangement of microtubules

(b) centriole and mitochondria

(c) mitochondria and 9 + 2 arrangement

(d) 9 + 2 arrangement of microtubules only

Ans. (c)

8. Based on cellular mechanism there are the two types of regeneration found in the

animals. Which one of the following is the correct example of the type mentioned?

(a) Morphollaxis – Regeneration of two transversely cut equal pieces of a Hydra

into two small Hydra

(b) Epimorphosis – Replacement of old and dead erythrocytes by the new ones

(c) Morphollaxis – Healing up of a wound in the skin

(d) Epimorphosis –Regeneration of crushed and filtered out pieces of a planaria into

as many new planaria

Ans. (a)

9. Drosophila files with XXY genotype are females, but human beings with such

genotype are abnormal males. It shows that

(a) Y-chromosome is essential for sex determination in Drosophila

(b) Y-chromosome is female determining in Drosophila

(c)Y-chromosome is male determining in human beings.

(d) Y-chromosome has no role in sex determination either in Drosophila or in human



4

beings.

Ans. (c)

10. In a chromosome map, B and C give crossing over 3% and A and B 8%. What will be

the percentage cross over between A and C?

(a) 12% (b) 24% (c) 7% (d) 11%

Ans. (d)

11. A male human is heterozygous for autosomal genes A and B; and is also hemizygous

for haemophilic gene h. What proportion of his sperms will be abh ?

(a) 1

8 (b)

1

32 (c)

1

16 (d)

1

4

Ans. (a)

12. In guinea pigs, black fur (B) is dominant over white fur (b) and rough fur (R) is

dominant over smooth fur (r). A cross between two guinea pigs hybrid for both traits

(BbRr × BbRr) produces some offsprings that have rough, black fur and some that

have smooth, black fur. The genotypes of these offsprings illustrate the genetic

concept of

(a) intermediate inheritance (b) multiple alleles

(c) independent assortment (d) codominance

Ans. (c)

13. Given below is a sample of portion of DNA strand giving the base sequence on the

opposite strands. What is so, special shown in it?

5’ _______GAATTC___3’

3’ _______CTTAAG____5’

(a) Deletion mutation

(b) Start codon at the 5’ end

(c) Palindromic sequence of base pairs



5

(d) Replication completed

Ans. (c)

14. The following ratio is generally constant for a given species

(a) A + G/C+ T (b) T + C/G + A (c) G + C/A + T (d) A + C/T + G

Ans. (c)

15. When genes resemble functional genes but do not produce functional substance are

called

(a) jumping genes (b) overlapping gens

(c) pseudogenes (d) non-functional genes

Ans. (c)

16. During splicing the exons are joined and the enzyme, which catalyses this reaction si

(a) RNA ligase (b) RNA catalase (c) RNA polymerase (d) RNA permease

Ans. (a)

17. Which one of the following options gives one correct example each of convergent

evolution and divergent evolution ?

Convergent Evolution Divergent Evolution

(a) Eyes of Octopus and mammals Bones of forelimbs of

vertebrates

(b) Thorns of Bougainvillia and tendrils of

Cucurbita

Wings of butterflies and birds

(c) Bones of forelimbs of vertebrates Wings of butterfly and birds

(d) Thorns of Bougainvillia and tendrils of

Cucurbita

Eyes of Octopus and

mammals

Ans. (a)

18. During embryonic development in mammals, heart is first 2-chambered as in fishes,

then 3-chambered as in amphibians and finally becomes 4-chambered. This fact is

related with

(a) Hardy-Weinberg law (b) Mendelism

(c) Biogenetic law (d) Lamarckism

Ans. (a)



6

19. 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 revision 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 Sping Resonance (ESR) and fossil DNA

(d) study of carbohydrates/proteins in rocks

Ans. (c)

20. In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are

(a) n = 7 and x = 21 (b) n = 21 and x = 2 1

(c) n = 21 and x = 14 (d) n = 21 and x = 7

Ans. (d)

21. Parthenocarpic tomato fruits can be produced by

(a) removing androecium of flowers before pollen grains are released

(b) treating the plants with low concentrations of gibberellic acid and auxins

(c) raising the plants from vernalised seeds

(d) treating the plants with phenylmercuric acetate

Ans. (b)

22. Treatment of seed at low temperature under moist conditions to break its dormancy

is called

(a) scarification (b) vernalization (c) chelation (d) stratification

Ans. (d)

23. The technique used for detecting specific target DNA sequences in genes, is called

(a) Eastern blotting (b) Western blotting

(c) Southern blotting (d) Northern blotting

Ans. (c)

24. The figure below is the diagrammatic representation of the E. coli vector pBR 322.

Which one of the given options correctly identifies its certain components (s) ?



7

(a) Ori-original restriction enzyme (b) rop-reduced osmotic pressure

(c) Hind III, Eco RI-selectable markers (d) ampR, tetR–antibiotic resistance genes

Ans. (d)

25. Golden rice is a

(a) vitamin-A rich transgenic rice (b) vitamin-B1 rich transgenic rice

(c) vitamin-C rich transgenic rice (d) vitamin-D rich transgenic rice

Ans. (a)

26. The population of an insect species shows an explosive increase in number during

rainy season followed by its disappearance at the end of the season. What does this

show ?

(a) S-shaped or sigmoid growth of this insect

(b) The food plants mature and die at the end of the rainy season

(c) Its population growth curve is of J-type

(d) The population of its predators increases enormously

Ans. (c)

27. Aquatic photo diffraction produces zones

(a) euphotic, disphotic and aphotic (b) aphotic, euphotic and disphotic

(c) euphotic, aphotic and disphotic (d) disphotic, aphotic and euphotic

Ans. (a)

28. Rare endangered and endemic taxa can be found intact and flourishing in

(a) sacred groves (b) buffer zone (c) tropical forests (d)

temperate forests

Ans. (a)



8

29. What is the correct ascending order in respect of complexity of the following ?

I Ecosphere II Species

III Population IV Community

V Ecosystem

(a) I, II, III, IV, V (b) III, II, IV, V, I

(c) I, III, II, IV, V (d) II, IV, III, I, V

Ans. (d)

30. Match the following columns.

Column A Column B

A. Fertilisers

B. Fly ash

C. Scrubbers

D. Soot

1. Incomplete burning of carbohydrates

2. Dust and smoke

3. Coal based thermal plants

4. Gaseous pollutants

5. Eutrophication

Codes

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

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

Ans. (b)



9

Section – B {Chemistry}

31. The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and

c = 0.504 nm and 090 and 0120 is :

(a) cubic (b) Hexagonal (c) Orthorhombic (d) Rhombohedral.

Sol : (b)

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

radius)

(a) 320

r13

b (b) 324

r13

(c) 312

r3

(d) 316

r3

.

Sol: (d)

Number of atoms per unit cell = 8×1 1

6 48 2

Volume of 4 atoms = 4×3 34 16

r r3 3

33. Which of the following statements is not correct ?

(a) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.

(b) Molecular solids are generally volatile.

(c) The number of carbon atoms in a unit cell of diamond is 8.

(d) The number of Bravais lattices in which a crystal can be categorized is 14.

Sol: (a)

The fraction of the total volume occupied by the atoms in a primitive cell is 0.524.

34. Percentage of free space in cubic close packed structure and in body centred cubic

structure are respectively

(a) 48% and 26% (b) 30% and 26% (c) 26% and 32% (d) 32% and

48%.

Sol: (c)

Packing fractions of fcc and bcc lattice are 74% and 68%

Vacancies are 26% and 32%.

35. At 800C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid

‘B’ is 1000 mm Hg. If a mixture of solution of ‘A’ and ‘B’ boils at 800C and 1 atm



10

pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg) :

(a) 50 mol percent (b) 52 mol percent (c) 34 mol percent

(d) 48 mol percent.

Sol: (a)

PTotal = o oA A B BP x P x

760 = 520xA + 1000(1–xA)

= 520xA + 1000–1000xA

or 480xA = 240 or xA =0.5

It means moles of A = 50%.

36. A solution of Sucrose (Molar mass 342 g mol–1) has been prepared by dissolving 6.85

g of sucrose in 100 g of water. The freezing print of solution obtained will be

(Kf = 1.86 K kg mol–1)

(a) – 0.5700C (b) – 0.3720C (c) – 0.5200C (d) +0.3720C

Sol: (b)

0f

6.85 1000T 1.86 0.372 C.

342 100

0Freezing point 0 0.372 0.372 C.

37. Which of the following compounds is used as antifreeze in automobile radiators ?

(a) Methyl alcohol (b) Glycol (c) Nitrophenol (d) Ethyl

alcohol

Sol:(b)

38. If 2Fe /FeE =– 0.441 V and 3 2Fe /Fe

E 0.771V The standard emf. of reaction

3 2Fe 2Fe 3Fe will be approximately

(a) 0.111 V (b) 0.330 V (c) 1.653 V (d) 1.212 V.

Sol: (d)

21Fe Fe 2e; E 0441 V

3 222Fe 2Fe 2e; E 0.771 V

The approximate value of standard emf of given reaction can be obtained

substanding equation (i) from equation (ii)



11

2Fe3+ – Fe2+ 2Fe2+ – Fe; E 0.771 ( 0.441)

or 2Fe3+ + Fe 3Fe2+; E 1.212V

39. The efficiency of fuel cell is given by

(a) S / G (b) H/ G (c) G / S (d) G/ H .

Sol: (d)

40. For the reaction 2A + B 3C + D

which of the following does not express the reaction rate ?

(a) d[A]

2dt (b)

d[B]

dt (c)

d[C]

3dt (d)

d[D]

dt.

Sol: (c)

The rate in terms of C is given as + 1 d[C]

.3 dt

41. Consider the reaction

2 2Cl aq H S aq S s 2H aq 2Cl aq

The rate equation for this reaction is rate = k [Cl2] [H2S]

Which of these mechanisms is/are consistent with this rate equation ?

2 2A. Cl H S H Cl Cl HS slow

Cl HS H Cl S fast

2B. H S H HS fast equilibrium

2Cl HS 2Cl H S slow

(a) A only (b) B only (c) Both (A) and (B) (d) Neither (A) nor (B).

Sol : (a)

Rate depends only on slow step

42. Among the electrolytes, Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effect

coagulating agent for Sb2S3 sol is

(a) Na2SO4 (b) CaCl2 (c) Al2(SO4)3 (d) Mn4Cl.

Sol : (c)

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

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



12

43. The correct order of acidic strength is

(a) Cl2O7 > SO2 > P4O10 (b) CO2 > N2O4 > SO3

(c) Na2O > MgO > Al2O3 (d) K2O > CaO > MgO.

Sol : (a)

44. Which of the following does not have S–S bond?

(a) 2

2 4S O (b) 2

2 3S O (c) 2

2 5S O (d) 2

2 7S O .

Sol : (d)

22 7S O has no S S linkage :

All others has at least one S-S linkage.

45. How many EDTA (ethylenediamine-tetraacetic acid) molecules are required to make

an octahedral complex with a Ca2+ ion ?

(a) Six (b) Three (c) One (d) Two.

Sol: (c)

ethylenediaminetetraacetate ion is hexadentate. It binds with one Ca2+ ion to form

octahedral complex.

46. Which type of isomerism is shown by the octahedral complex Co(NH3)4Br2Cl?

(a) Geometrical and ionization (b) Geometrical and optical

(c) Optical and ionization (d) Geometrical only.

Sol : (a)

[Co(NH3)4 Br2]Cl and [Co(NH3)4 BrCl] Br are ionization isomers.

47. Which among the following will be named as dibromidobis (ethylene diamine)

chromium (III) bromide?



13

(a) [Cr(en)3]Br3 (b) [Cr(en)2 Br2]Br

(c) [Cr(en)Br4]– (d) [Cr(en)Br2]Br

Sol : (b)

As per IUPAC rule.

48. The organic chloro compound, which shows complete stereochemical inversion

during of SN2 reaction, is

(a) CH3Cl (b) (C2H5)2CHCl (c) (CH3)3CCl (d)

(CH3)2CHCl.

Sol: (a)

In SN2 reaction, the order of reactivities of alkyl halides is 10 > 20 > 30

49. For reaction:

3 2

3

H C CH CH CH HBr A

|

CH

A (predominantly) is

Sol : (d) The product is formed through rearrangement of the intermediate

carbocation.

HBr 1,2 hydride shift3 2 3 3

3 3

CH CH CH CH CH CH CH CH

| |

CH CH



14

Br3 2 3 3 2 3

3 3

Br

|

CH C CH CH CH C CH CH

| |

CH CH

50. The major product of the following reaction is

Sol : (a)

In the presence of dimethyl formamide which is an aprotic solvent, the reaction

follows SN2 mechanism resulting in inversion.

51. The product of acid catalyzed hydration of 2-phenylpropene is

(a) 3-phenylpropene is (b) 1-phenyl-2-prppanol

(c) 2-phenyl-2-propanol (d) 2-phenyl-1-propanol.

Sol : (c)



15

The reaction proceeds via more stable benzylic carbocation.

2H O3 2 3 3H

6 5 6 5

OH

|

CH C CH CH C CH

| |

C H C H

2 phenyl 2 propanol

52. In the following sequence of reactions,

2 2P I H OMg HCHO3 2 ether

CH CH OH A B C D

the compound ‘D’ is

(a) n-propyl alcohol (b) propanal (c) butanal (d) n-

butyl alcohol.

Sol : (a)

(A) is CH3CH2I (B) is CH3CH2MgI (c) is CH3CH2CH2OmgI (d)is CH3CH2CH2OH

53. Among the following four compounds

I. Phenol II. Methylphenol

III. meta-Nitrophenol IV. para-Nitrophenol

The acidity order is

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

Sol : (a)

Electron withdrawing group increases the acidic character of phenol. The effect is

more when the electron withdrawing group is at para position.

54. The correct order of reactivity of PhMgBr with

3 3 31 II III

O O O

|| || ||

Ph C PH CH C H CH C CH , is

(a) (I) > (II) > (III) (b) (III) > (II) > (I) (c) (II) > (III) > (I) (d) (I) > (III) > (II).

Sol : (c)

The reactivity of carbonyl compound towards nucleophilic addition of Grignard’s

reagent depends on extent of steric hindrance at α – carbon. Greater the steric



16

hindrance smaller the reactivity. Hence, reactivity order is:

3 3 3CH CHO CH CO CH Ph CO Ph

(II) (III) (I)

55.

What is X?

(a) CH3COOH (b) BrCH2COOH (c) (CH3CO)2O (d) CHO–COOH.

Sol : (c)

X is (CH3CO)2O it is an example of Perkin’s reaction.

56. 3

dil. NaOH HCN3 heat H O

CH CHO HCHO A B

The structure of compound B is

(a)

2CH CH CH COOH

|

OH

(b)

2CH CH CH OH

|

CN

(c)

2 2CH CH CH COOH

|

CN

(d)

3CH CH COOH

|

OH

Sol : (A)

3

2

H Odil.NaOH HCN3 2 2 2 2H O

OH CN COOH

| | |

HCHO CH CHO H C CH CHO CH CHCHO CH CH CHOH CH CH CH OH

| (A) (B)

OH

57. Which will not undergo diazotisation ?

(a) C6H5NH2 (b) C6H5CH2NH2



17

Sol : (b)

A reaction in which – NH2 group is converted into diazo group ( N N) at low

temperature (00-50C) is diazotisation. Only 10 (primary) aromatic amine form

diazonium salts. Diazotised salts are stable in cold aqueous solution.

58. CH3NH2+CHCl3+3KOH (?) + 3KCl + H2O The missing product is

(a) CH3–CN (b) CH3NHCl (c) 3CH N C (d) 3CH N C.

Sol : (d)

The product formed during the reaction is carbylamines having structure as given

3CH N C

59. Amongst the compounds given, the one that would form a brilliant coloured dye on

treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of -

naphthol is.

Sol : (c)

60. The two forms of D-glucopyranose are called

(a) Enantiomers (b) Epimers (c) Anomers (d) Diastereomers.



18

Sol : (c)

Two forms of D-glucopyranose differ in the orientation of groups around C-1. Such

isomers are called anomers.

Section – C {Physics}

61. A radioactive source in the form of a metal sphere of radius 10–2 m, emits beat

particles (electrons) at the rate of 5 × 1010 particles per second. The source is

electrically insulated. How long will it take for its potential to be raised by 2 volts

assuming that 40% of the emitted beta particles escape the source.

(a) 200 s (b) 500 s (c) 700 s (d) 900 s

Sol : (c)

Beta particles (i.e., electrons) emitted in t seconds



19

= 5 × 1010 t.

The deficiency of electrons from a conductors results in positive charge.

As 40% beta-particles escape from the source; positive charge gained by source

(metal sphere),

1040q ne 5 10 t.e

100

10 19 92 10 t 1.6 10 3.2 10 t coul.

Due to this charge the potential acquired by sphere

0

1 qV

4 R

Substituting given values

99

2

3.2 10 t2 9.0 10

10.

2

9 9

2 10t

9 10 3.2 10

47 10 s 700 s.

62. Three point charges each +q, placed at three of vertices of an equilateral triangle and

a charge – Q placed at the centre are in equilibrium. The ratio q

Qis :

(a) 3 (b) 3 (c) 1

3 (d) 3 3

Sol : (b)

The system will be in equilibrium if the net force on charge q at one vertex due to

charges q at the other two vertices is equal and opposite to the force due to charge Q

at the centroid, i.e. (here a is the side of the triangle)

2

220

0

3q Qq

4 a a4

3

Therefore, q

Q= 3 .



20

63. A conducting spherical bubble of radius a and thickness t (t << a) is charged to a

potential V. Now it collapses to form a spherical droplet. Find the potential of the

droplet.

(a)

2/3a

V3t

(b)

1/33a

Vt

(c)

2/33a

Vt

(d)

1/3a

V3t

Sol : (d)

The charge and mass both are conserved. If r is radius of droplet and v its potential,

then from conservation of mass 3 24

r 4 a t3

1/32r 3a t . …(1)

For same charge q, the potential of bubble,

0

1 qV

4 a ..(2)

and the potential of droplet,

0

1 qv .

4 r ..(3)

From (2), 0q 4 Va

From (3), 0q 4 ve

0 04 Va 4 vr

1/32

a aVv V

r 3a t [Using (1)]

1/3a

or v V3t

64. Two identical capacitor have the same capacitance C. One of them is charged to a

potential V1 and other to V2. The negative ends of the capacitors are connected

together. When the positive ends are also connected, the decrease in the energy of

the combined system is :



21

(a) 2 2

1 2

1C V V

4 (b)

2 21 2

1C V V

4 (c)

2

1 2

1C V V

4 (d)

2

1 2

1C V V

4

Sol : (c)

Decrease in energy, initial finalU U U .

When positive ends are connected, the common potential is

1 2 1 2CV CV V VU

C C 2.

2

2 2 1 21 2

1 1 V VU C V V 2C

2 2 2

2

1 2

1C V V .

4

65. A heater coil is cut into two equal parts and only one part is now used in the heater,

the heat generated will now be :

(a) four times (b) doubled (c) halved (d) one-

fourth.

Sol:(b)

Resistance of each first will be R

2 ,

2

1

VH t

R

2 2

2 1

V 2VH t t 2H

R /2 R

so heat produced in one will be doubled.

66. A circuit is shown below. Then :

(a) The currents through three branches are equal



22

(b) The currents through three branches are not related

(c) The currents through AB, AC and AD will be in the ratio 2 : 1 : 2

(d) The currents through AB, AC and AD will be in the ratio 1 : 2 : 1.

Sol : (a)

67. The Kirchhoff’s first law i 0 and second law iR E where the symbols

have usual meanings, are respectively based on:

(a) conservation of charge, conservation of momentum

(b) conservation of energy, conservation of charge

(c) conservation of momentum, conservation of charge

(d) conservation of charge, conservation of energy.

Sol : (d)

68. A coil having N-turns is wound tightly in the form of a spiral with inner and outer

radii a and b respectively. When the current a passes through the coil, the magnetic

field at the centre is :

(a) 0NI

b (b)

02 NI

a (c)

0e

NI blog

2 b a a (d)

0e

2 NI blog .

b a a

Sol : (c)

Consider an element of thickness dx at a distance x from centre. Number of turns in

the element

NdN dx.

b a

00

Ndx i

dNi b adB

2x 2x

Ndx i

b a

2x

Net magnetic field B = b

0

a

Ni dxdB

2 b a x

= 0e

Ni blog .

2 b a a



23

69. A long straight wire along Z-axis carries a current i in the negative Z-direction. The

magnetic vector field B at a point having co-ordinates (x, y) on z = 0 plane is :

(a) 0

2 2

i yi xj

2 x y

(b)

0

2 2

i xi yj

2 x y

(c)

0

2 2

.i xj yi

2 x y

(d)

0

2 2

i xi yj

x y

.

Sol:(a)

From fig.

B B sin i B cos j,

0 aB

2 r.

0 iB sin i cos j

2 r

0i y xi j

2 r r r

02 2

yi xji.

2 x y

70. The potential difference V and current i flowing through an a.c. circuit are given by V

= 5 cos t volt,

i = 2 sin t amp. the power dissipated in the circuit :

(a) 0 W (b) 10 W (c) 5 W (d) 2.5 W.

Sol : (a)



24

Phase difference between sine and cosine terms is .2

So rmsp V cos 0.2

71. A circuit has a resistance of 12 and an impedance of 15 . The power factor of the

circuit will be :

(a) 0.4 (b) 0.8 (c) 0.125 (d)

1.25.

Sol : (b)

Rcos

Z

12 4

15 5

= 0.8

72. In a LCR series circuit, the potential difference between the terminals of the

inductance is 60 V, between the terminals of the capacitor is 30 V and that across the

resistance is 40 V. Then the supply voltage will be equal to :

(a) 50 V (b) 70 V (c) 130 V (d) 10 V.

Sol : (a)

2 2R L CV V (V V )

2 2V (40) (60 30)

2 240 30

2500 50

73. An alternating current circuit consists of an inductance and a resistance in series. In

this circuit:

(a) The potential difference across and current in resistance leads the potential

difference across inductance by angle /2 .

(b) The potential difference across and current in resistance lags behind the potential

difference across inductance by an angle /2 .

(c) The potential difference across and current in resistance lags behind the potential



25

difference across inductance by an angle

(d) The potential difference across resistance leads the potential difference across

inductance by /2.

Sol : (b)

74. If i j represents refractive index when a light ray goes from medium i to medium j,

then the product 2 1 3 2 4 3 is equal to :

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

1 (d) 4 2.

Sol:(c)

2 1 3 2 4 3

1 2 2 3 3 4 1 4

1 1 1 1

75. A ray of light is incident on an equilateral glass prism placed on a horizontal table.

For minimum deviation which of the following is true :

(a) PQ is horizontal (b) QR is horizontal

(c) RS is horizontal (d) Either PQ or RS is horizontal

Sol : (b)

For minimum deviation the refracted ray within prism is always parallel to base, so

ray QR is horizontal.

76. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and

D2 be angles of minimum deviation for red and blue light respectively in a prism of

this glass. Then :

(a) D1 < D2

(b) D1 = D2

(c) D1 > D2



26

(d) D1 can be less than or greater than D2 depending upon the angle of prism.

Sol : (a)

Deviation caused by a prism D = 1 .

As blue red 1 2D D .

77. White light is incident on the interface of glass and air as shown in fig., If green light

is just totally internally reflected, then the emerging ray in air contains:

(a) yellow, orange, red (b) violet, indigo, blue

(c) all colours (d) all colours except green.

Sol: (a)

violet green redC C C .

Condition is yellow.C

Yellow, orange, red rays are transmitted, while violet, indigo and blue and totally

reflected

78. A biconvex lens of focal length f forms a circular image of sun of radius r in the focal

plane. Then :

(a) Area of image is 2r and is proportional to f

(b) Area of image is 2r and is proportional to f2

(c) If lower part is covered by black sheet, then area of image is 2r

2

(d) with increase of f, the intensity will increase.

Sol:(b)

79. Interference pattern is obtained by using light having two wavelengths 400 nm and

560 nm. The light falls normally on two narrow slits 0.1 mm apart. The screen is

placed at a distance of 1 m from the slits. The distance between the successive region



27

of total darkness is :

(a) 4 mm (b) 5.6 mm (c) 14 mm (d) 28 mm.

Sol : (d)

Let nth minima of 400 nm coincides with mth minima of 560 nm, then

400 5602n 1 2m 1

2 2

2n 1 7 14or ....

2m 1 5 10

i.e., 4th minima of 400 nm coincides with 3rd minima of 560 nm.

Location of this minima is,

6

1

2 4 1 1000 400 10Y 14 mm

2 0.1

Next 11th minima of 400 nm will coincide with 8th minima of 560 nm.

Location of this minima is,

6

2

2 11 1 1000 400 10Y 42 mm

2 0.1

Required distance = Y2 – Y1 = 28 mm

Hence, the correct option is (d).

80. According to Einstein’s photoelectric equation, the plot of the kinetic energy of the

emitted photoelectrons from a metal versus frequency of incident radiation gives a

straight line whose slope :

(a) depends on the nature of to metal

(b) depends on the intensity of radiation

(c) depends both on the intensity of radiation and the metal used

(d) is the same for all metals and independent of the intensity of radiation.

Sol : (d)

Ek = hv – W.

This is equation of a straight line of slope h.

81. The maximum kinetic energy of photoelectrons emitted from a surface when photons

of energy 6 eV fall on it is 4 eV. The stopping potential in volts is :



28

(a) 2 (b) 4 (c) 6 (d) 10.

Sol : (b)

Max KE = eVs

s4eV eV

sV 4 volt.

82. When 73Li nuclei are bombarded by protons and the resultant nuclei are

84Be, the

emitted particle will be :

(a) particles (b) particles (c) photons (d) neutrons.

Sol : (c)

7 1 B3 1 4Li H Be photon .

83. 280 days old radioactive sample has activity 6000 disintegrations/second (dps).

After 140 days to activity decreases to 3000 dps. The initial activity of the sample in

dps is :

(a) 9000 (b) 12000 (c) 16000 (d) 24000.

Sol : (d)

Number of half lives from t = 0 is

280 140n 3

140

n 3

0 0

R 1 R 1gives

R 2 R 2

3

0or R 2 R 8 3000 24000.

84. A transistor oscillator using a resonant circuit with an inductor L (of negligible

resistance) and a capacitor C in series produce oscillation, of frequency f. If L is

doubled and C is change to 4C, then frequency will be :

(a) f

2 2 (b)

f

2 (c)

f

4 (d) 8f.

Sol : (a)

1f

2 LC



29

1f '

2 2L 4C

f ' 1 1f ' f .

f 2 2 2 2

85. In common base amplifier the phase difference between the input signal voltage and

output voltage is :

(a) (b) 4

(c) 2

(d) 0.

Sol : (d)

86. When npn-transistor is used as an amplifier :

(a) electrons move from base to collector

(b) holes move from emitter to base

(c) electron move from collector to base

(d) holes move from base to emitter.

Sol : (a)

87. The energy required to remove an electron in the n = 2 state of hydrogen atom is :

(a) 27.2 eV (b) 13.6 eV (c) 6.8 eV (d) 3.4 eV

Sol:(d)

2 2

1 1E 13.6 eV

2

13.63.4 eV.

4

88. The ratio of areas with the electron orbits for first excited to the ground state for

hydrogen atom is :

(a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1.

Sol:(d)

2 2

2 2

1 1

r n 24.

r n 1

22

2 2 22

1 1 1

A r r 16.

A r r 1



30

89. The X-ray emission line of tungsten occurs at 0.021nm. The energy difference

between K and L levels in this atom is about :

(a) 0.51 MeV (b) 1.2 MeV (c) 59 keV (d) 136 eV.

Sol: (c)

o

k 0.021 nm 0.21A

Since k corresponds to the transition of an electron from L-shell to k-shell,

therefore

L K o

12375 12375E E in eV 58928eV

0.21(in A)

or E 59keV

90. The wavelength of K X ray , of an element having atomic number Z = 11 is . The

wavelength of K X ray of another element of atomic numbers Z’ is 4 . Then Z’ is :

(a) 11 (b) 44 (c) 6 (d) 4. Sol: (c)

21Z 1

2

1 2

2 1

Z 1

Z 1 or

2

21 Z 1

4 11 1 Solving this, we get Z2=6.

pioneer education {the best way to success} …...which one of the following is a correct...

Documents