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MEDICAL

Mega Test - III {+2-Medical}

General Instructions:- The question paper contains 90 objective multiple choice questions.

There are three parts in the question paper consisting of MATHS (1 to 30),

Physics (31 to 60) and Chemistry (61 to 90).

Each right answer carries (4 marks) and wrong (–1marks)

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

Maximum Time 3Hrs.

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

Note: Actual pattern of AIPMT consists of 45 questions each from

Zoology, Botany, Physics & Chemistry.

Which will be applicable in our next Mega Test.

Name: _______________________________Father Name:______________________________

Mobile: ______________________________School Name:______________________________

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

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



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Section – A {Biology}

1. On fertilization, what structure develops from carpel ?

(a) Testa (b) Tegmen (c) Pericarp (d) Perisperm

Ans. (c)

2. The constant feature of embryo sac is :

(a) Synergids (b) Antipodals (c) Egg (d) Polar nuclei

Ans. (c)

3. Endosperm in coconut is :

(a) Cellular (b) Nuclear

(c) Both cellular and nuclear (d) Neither cellular nor nuclear

Ans. (c)

4. In anther culture, some diploid plants are produced along with haploid plants, what

may be the source of diploid plants ?

(a) Generative cell of pollen (b) Cells of anther wall

(c) Vegetative cell of pollen (d) Exine of pollen wall

Ans. (b)

5. The production of gametophyte directly from sporophyte without meiosis is called :

(a) Apospory (b) Apogamy

(c) Parthenogenesis (d) Apomixis

Ans. (a)

6. Dyad stage is formed in which type of cytokinesis during meiotic division of

microspore mother cell ?

(a) Successive type (b) Simultaneous type

(c) Both (a) and (b) (d) None of these

Ans. (a)

7. Which is the most logical sequence with reference to life cycle of angiosperms ?

(a) Germination, endosperm formation, seed dispersal, double fertilization

(b) Cleavage, fertilization, grafting, fruit formation



3

(c) Pollination, fertilization, seed formation, germination

(d) Maturation, mitosis, differentiation, fertilization

Ans. (c)

8. Match the following:

I Seminiferous

II Chasmogamous

III Herkogamy

IV Homogamy

A. Flower exposing their reproductive parts

B. Speed producing plant

C. Both sex organs maturing at sametime

D. Physical barrier between male and female organs

(a) I-A, II-B, III-C, IV-D (b) I-B, II-A, III-C, IV-D

(c) I-B, II-A, III-D, IV-C (d) I-A, II-B, III-D, IV-C

Ans. (c)

9. 12 : 3 : 1, F2 ratio is obtained in interaction of ………………..factors.

(a) Complementary (b) Supplementary (c) Epistatic (d) Inhibitory

Ans. (c)

10. If haploid chromosome no. in an organism is 12, what is the chromosome no. present

in its trisomic ?

(a) 11 (b) 13 (c) 25 (d) 23

Ans. (c)

11. Test cross of a trihybrid will yield a ratio of :

(a) 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 (b) 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1

(c) 9 : 9 : 3 : 3 : 3 : 1 : 1 : 1 (d) None of the above

Ans. (b)

12. A cross between red-flowered and white-flowered plants results in pink-flowered

plants in F1; when the F1 hybrids were self-fertilised, the resulting F2 generation



4

revealed 1 red : 2 Pink : 1 white progeny. This type of inheritance is the result of :

(a) Incomplete dominance (b) Co-dominance

(c) Pleiotropic genes (d) Complementary genes

Ans. (a)

13. In maize, coloured endosperm (C) is dominant over colour-less (c) and full endosperm

(R) is dominant over shrunken (r). When a dihydrid of F1 gen. was test crossed, it

produced four phenotypes in the following percentage:

Coloured full 48 %

Coloured shrunken 5%

Colourless full 7%

Colourless shrunken 48%

From this data, what will be the distance between two non-allelic genes :

(a) 48 units (b) 5 units (c) 7 units (d) 12 units

Ans. (d)

14. A haemophilic man marries a normal homozygous woman. What is the probability

that their son will be haemophilic ?

(a) 1/2 (b) 1/4 (c) 1/6 (d) 0

Ans. (d)

15. In a nucleotide, N-base is attached to sugar at :

(a) C-1 (b) C-2 (c) C-5 (d) C-3

Ans. (a)

16. Which sequence is according to increasing molecular weight ?

(a) t-RNA, DNA, m-RNA, r-RNA (b) r-RNA, DNA, t-RNA, m-RNA

(c) DNA, m-RNA, r-RNA, t-RNA (d) t-RNA, r-RNA, m-RNA, DNA

Ans. (d)

17. Ambiguous codon in nature is :

(a) UAG (b) UGA (c) UAA (d) UUU

Ans. (b)



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18. If one strand of DNA has the base sequence CAT, GAC, TAG, what would be the base

sequence in the other strand ?

(a) GTA, CTG, ATC (b) GAT, GTC, ACT

(c) TAC, ACT, GCT (d) CAT, TAG, GAC

Ans. (a)

19. In polymerase chain reaction, DNA polymerase enzyme used is :

(a) DNA polymerase I (b) DNA polymerase II

(c) Taq polymerase (d) DNA polymerase III

Ans. (c)

20. Match the List-I with List-II :

List-I List-II

A. Diameter to the helix 1. 34o

A

B. Distance between two adjacent 2. 20o

A

C. Distance between two complete turns 3. 3o

A

D. Length of the hydrogen bonds 4. 4.4o

A

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

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

Ans. (b)

21. From the following, identify DNA fragment showing palindromic sequence :

(a)

1 1

1 1

5 A C T C T G 3

| | | | | |

3 T G A G A C 5

(b)

1 1

1 1

5 A A C C A G 3

| | | | | |

3 T T G G T C 5

(c)

1 1

1 1

5 T T G A G T 3

| | | | | |

3 A A C T C A 5

(d)

1 1

1 1

5 G T G C A C 3

| | | | | |

3 C A C G T G 5

Ans. (d)



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22. Control of aphids (homopterous insects) is done by introduction of :

(a) Praying mantis (lady bug) (b) Bt

(c) Nematodes (d) None of the above

Ans. (a)

23. Cylinder acting as death trap, having pheromones, have :

(a) Male pheromones (b) Female pheromones

(c) Sterile males (d) All of these

Ans. (b)

24. Cloves are :

(a) Fruits (b) Seeds (c) Flower buds (d) Flowers

Ans. (c)

25. Genes for antibiotic resistance are located in :

(a) Chromosomes (b) Nucleus (c) Cell wall (d) Plasmid

Ans. (d)

26. Single cell protein refers to :

(a) A specific protein extracted from pure culture of single type of cells

(b) Sources of mixed proteins extracted from pure or mixed culture of organisms or

cells

(c) Proteins extracted from a single cell

(d) A specific protein extracted from a single cell

Ans. (b)

27. Which of the following is not achievement of gene transfer technology ?

(a) Golden rice (b) Bt cotton

(c) Decaffeinated coffee (d) None of these

Ans. (d)

In the questions given below, two statements-an Assertion (A) and a Reason (R) are

given. Given the appreciate response as.

(a) If both A and R are true and R is the correct explanation of A.



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(b) If both A and R are true but R is not the correct explanation of A.

(c) If A is true but R is false.

(d) If both A and R are false.

28. A : Endosperm is a characteristic storage tissue, which supplies nutrition to the

developing embryo.

R : Development of endosperm is from primary endosperm nucleus.

Ans. (b)

29. A: In a DNA molecule A-T rich portions melt before G-C rich portions.

R: In between A and T, there are 2 H-bonds whereas in between G-C, there are 3 H-

bonds.

Ans. (a)

30. A : Cross over frequencies are used in formation of linkage maps.

R: Recombination frequencies are directly proportional to distances between genes.

Ans. (a)



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Section B {Physics}

31. Two point charges + 4e and e are kept at distance x apart. At what distance a charge q

must be placed from charge +e, so that q is in equilibrium :

(a) x

2 (b)

2x

3 (c)

x

3 (d)

x

6

Sol : (c)

The situation is show n in figure.

Let charge q is placed at a distance y from charge +e.

For equilibrium of charge q,

2 20 0

1 4e q 1 q e

4 4 yx y

or 2 2

4qe qe

yx y

or 2 2

4 1

yx y

or 2 1

x y y

or 2y = x – y

x

y3



9

32. In the given figure, two point charge q1 and q2 are placed at distances a and b from

centre of a metallic sphere having charge Q. The electric field due to the metallic

sphere at the point P is :

(a)

2 2

1 22 2

0

q q1

4 a b (b)

20

1 Q

4 R

(c)

22

1 22 2 2

0

q q1 Q

4 R a b (d) none of a above

Sol : (a)

The net electric field at P is zero.

p 1 2 sphereE E E E

or 1 2 sphere0 E E E

sphere 1 2E E E

2 2sphere 1 2E E E

=

2 22

12 2

0 0

q q

4 a 4 b

=

2 2

1 22 2

0

q q1

4 a b



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33. If two electric charges q and –2q are placed at distance 6a, apart, then the locus of

point in the plane of charge, where the field potential is zero, is :

(a) x2 + 2y2 – 4ax – 12a2 = 0 (b) 2x2 + y2 + 4ax – 12a2 = 0

(c) x2 + y2 + 4ax – 12a2 = 0 (d) x2 + y2 + 8ax + 12a2 = 0

Sol : (c)

Let at P the field potential is zero. Let r1 and r2 are the distances of P from charges +q

and – 2q respectively.

2 21r x y

2 22r 6a x y

Net potential at P

0 1 0 2

2q1 qV

4 r 4 r

= 2 2 2 2

0 0

1 q 1 2q

4 4x y 6a x y

Given, V = 0

2 2 2 2

1 2

x y 6a x y

2 2 2x y 4ax 12a 0



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34. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0… upto and a charge –

q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0, … upto .Here x0 is a positive

constant. Take the electric potential at a point due to a charge Q at a distance r from it

to be 0

Q

4 r. Then, the potential at the origin due to the above system of charges is :

(a) 0 (b) 0 0

q

8 x ln 2 (c) (d)

0

q ln 2

4 x.

Sol: (d)

0 0 0 0 0 0 0 0

q q q1 1 q q 1V .... ....

4 x 3x 5x 4 2x 4x 6x

= 0 0

1 q 1 1 1 1 11 ....

4 x 2 3 4 5 6

= ee

0 0 0

q log 21 qlog 1 1

4 x 4 x.

35. A radioactive source in the form of a metal sphere of radius 10–2 m, emits beta paricles

(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) 600 s . (b) 700 s . (c) 800 s . (d) 900 s .

Sol : (b)

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

= 5 1010 t.

The deficiency of electrons from a conductor 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.



12

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.

36. In the given circuit the value of resistance R is :

(a) 42 (b) 62 (c) 72 (d) 82

Sol : (a)

Since 10 , 20 & 60 are parallel to each other

p

1 1 1 1

R 10 20 60

= 6 3 1 10

60 60

pR 6

Now R & Rp are in series

According to ohm’s Law.

12 = 0.25 (R + Rp)

R 42



13

37. In the circuit shown in figure, the current through :

(a) the 3 resistor is 0.50 A (b) the 3 resistor is 0.25 A

(c) the 4 resistor is 0.50 A (d) the 4 resistor is 0.25 A.

Sol : (d)

The equivalent resistance between A and D (of right circuit) is 4

Net resistance = 3 + 4 + 2 = 9

Current I = 9

1 A9

.

This current will be divided in 8 between A and D and remainder 8 equally; so

current in AB = 0.5 A. Similarly the current 0.5 A in AB is equally divided between BC

and to the right of BC.

Current in4 0.25A .

38. Two heater wires of equal length are first connected in series and then the parallel.

The ratio of heat produced in the two cases will be :

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

Sol : (c)

In series 2

1

VQ

2R.

In parallel2 2

2

V 2VQ

R / 2 R.

2

22

1

Q 2V / R4.

Q V /2R



14

39. Two electric bulbs one of 200 V-40 watt and other of 200V–100 watt are connected in

series to a 200 volt line, then:

(a) The potential drop across two bulbs is same

(b) The potential drop across 40 watt bulb is greater than the potential drop across

100 watt bulb

(c) The potential drop across 100 watt bulb is greater than the potential drop across

40 watt bulb

(d) the potential drop across both the bulbs is 200 volt.

Sol :(b)

2

40

100

VR

P

200 200R 1000

40

200 200R 400

100

Since they are connected in series thereforeR V

40 100V V

40. Two particles X and Y having equal charges after being accelerated through the same

potential difference, enter a region of uniform magnetic field and describe circular

paths of radii R1 and R2 respectively. The ratio of the mass of X to than of Y is :

(a)

1/2

1

2

R

R (b) 2

1

R

R (c)

2

1

2

R

R (d) 1

2

R

R

Sol : (c)

Radius of circular path, R = mv

qB.

Also kmv 2mE 2mqV .



15

2mqV 1 2mVR m

qB B q.

2

2 1 1

2 2

m Rm R

m R.

41. An electron having a charge 191.6 10 C and mass 319 10 kg is moving with a speed

64 10 m/s in a magnetic field 12 10 T in a circular orbit. The force acting on the

electron and the radius of circular orbit will be :

(a) 12.8 10–13 N, 1.1 10–4 m (b) 1.28 10–14 N, 1.1 10–3 m

(c) 1.28 10–13 N, 1.1 10–3 m (d) 1.28 10–13 N, 1.1 10–4 m.

Sol : (d)

By using formula r = 2mv mv

& fqB r

Option (d) is correct

42. Two straight long conductors AOB and COD are perpendicular to each other and carry

current I1 and I2. The magnitude of magnetic induction at a point P at a distance a from

O in a direction perpendicular the plane ABCD is :

(a) 0 1 2I I

2 a (b) 0

1 2I I2 a

(c) 1/22 20

1 2I I2 a

(d) 0 1 2

1 2

I I

2 a I I.

Sol : (c)

Here B1 =2

0 1I,

2 a



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B2 = 2

0 2I,

2 a

Since they are perpendicular to each other as shown in figure therefore resultant

magnetic field B’ = 2 21 2B B =

1/22 201 2I I

2 a

Therefore option (c) is correct.

43. 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) 0

e

NI blog

2 (b a) a (d) 0

e

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

0

0

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

44. The resistance of a galvanometer is 90 . If only 10% of the main current passes

through the galvanometer, in what way and of what value a resistor is to be used ?

(a) 10 in series (b) 10 in parallel

(c) 810 in series (d) 810 in parallel.



17

Sol : (b)

The distribution of current is shown in figure.

Since potential is same

I 9I90 R

10 10

R 10

45. A wire carrying current i is in the shape of a plane curve r f , where r, being

polar co-ordinates. The magnetic field at pole is :

(a) 0I d

4 r (b) 0I d

2 r (c) 0

2

I d

4 r (d)

20

2

I d

4 r

Sol : (a)

The magnitude field at the point C due to considered element is

002 2

I dl sinIdl sindB

4 r 4 r

or 0 02

Ird IddB

4 r4 r

dl sin rd

0I dB

4 r



18

46. In a series resonant L-C-R circuit, the voltage across R is 100V and R = 1 k with

C = 2 F . The resonant frequency is 200 rad/s At resonance the voltage across L is :

(a) 2.5 10–2 V (b) 40 V (c) 250 V (d) 4 10–3 V

Sol: (c)

At resonance, 1

LC

Current flowing through the circuit,

RV 100I 0.1A

R 1000

So, voltage across L is given by

L LV I X I L

but 1

LC

L 6

I 0.1V 250 V

C 200 2 10

47. A conducting ring of radius r is rolling without slipping with a constant angular

velocity . If the uniform magnetic field strength is B and is direction into the page

then emf induced across PQ is :

(a) 2B r (b) 2B r

2 (c) 24B r (d)

2 2r B

8

Sol : (d)

2B le

2



19

l

2 r

l r

2

2

B r2

e2

2 2r B

8

48. The number of turns in the primary and secondary of a transformer are 5 and 10

respectively and the mutual inductance is 25 H. If the number of turns in the primary

and secondary are made 10 and 5 respectively, then the mutual inductance of the

transformer will be :

(a) 6.25 H (b) 12.5 H (c) 25 H (d) 50 H.

Sol: (c)

1 2M N N

N1N2 in both case is same so

M2 = M1 = 25 H.

49. If vs, vx and vm are the speeds of gamma rays, X-rays and microwaves respectively in

vacuum then :

(a) vs > vx > vm (b) vs < vx < vm (c) vs > vx < vm (d) vs = vx = vm

Sol : (d)

50. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive

index n2 (n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The



20

maximum value of angle of incidence max , such that the ray comes out only from the

other surface CD, is given by :

(a) 1 11 2

2 1

n nsin cos sin

n n (b) 1 12

1 2

n 1sin cos sin

n n

(c) 1 1

2

nsin

n (d) 1 2

1

nsin

n.

Sol : (a)

The incident ray PQ will emerge from the surface CD only of it is width reflected at

surface AD. For this

2r C

As 01 2r r 90

or 02 1r 90 r

or 0190 r c

or 1r 90 C .

From Snell’s law 1

1 2

nsin

sin r n.



21

For maximum 1, r 90 C .

max 1

2

sin n

sin 90 C n.

1max

2

nsin cos C

n 1 2

1

nAs C sin

n

1 11 2max

2 1

n nsin cos sin

n n.

51. A ray of light passes through an equilateral prism such that the angle of incidence is

equal to the angle of emergence and the latter is equal to 3/4th the angle of prism. The

angle of deviation is :

(a) 450 (b) 390 (c) None of these (d) 300

Sol : (d)

1 e A

2i A

3AA

2

1A

2

60

2

030

52. In Young’ doubled slit experiment 12 fringes are observed to be formed in a certain

segment of the screen, when light of wavelength 600 nm is used. If the wavelength of

light is changed to 400 nm. The number of fringes observed in the same segment of the

screen is :

(a) 12 (b) 18 (c) 24 (d) 30.

Sol : (b)



22

Fringe width, ω = D

d

`When the wavelength is decreased from 600 nm to 400 nm, fringe width will also

decrease by a factor of 4

6 or

2

3 or the number of fringes in the same segment will

increase by a factor of 3/2.

Therefore, number of fringes observed in the same segment = 12 × 3

2= 18

53. To demonstrate the phenomenon of interference, we require :

(a) two sources which emit radiation of the same frequency,

(b) two sources which emit radiation of nearly the same frequency,

(c) two sources which emit radiation of the same frequency and have a definite phase

relationship

(d) two sources which emit the radiation of different wavelengths.

Sol : (c)

54. Two beam of light having intensities I and 4I interfere to produce a fringe pattern on

the screen. The phase difference between the beams is π/2 at point A and at point B.

Then the difference between the resultant intensities at A and B is :

(a) 2I (b) 4I (c) 5I (d) 7I.

Sol : (b)

Intensity at A,

A 1 2 1 2I I I 2 I I cos I 4I 5I2

Intensity at B,

1 2 1 2IB I I 2 I I cos I 4I 2 I 4I I.

Difference in intensities, A BI I 4I.

55. If in a cinema hall, the distance between the projector and the screen is increased by

2%, keeping everything else unchanged, then the intensity of illumination on the

screen is :



23

(a) decreased by 4% (b) decreased by 2%

(c) increased by 2% (d) increased by 4%

Sol : (a)

Let illuminance on screen = E

2

I cosE

r

Taking log on both sides

log E log I log cos 2 log r

dE dr2

E r

dE dr100 2 100

E r

dE dr100 4% 100 2%

E r

56. An electric bulb is suspended at a vertical height 2 m from the centre of a square table

of side 2 m. If the luminous intensity of bulb is 60 cd (candela), then the illumination at

one corner of the table is :

(a) 8.16 cd/m2 (b) 6.24 cd/m2 (c) 9.25 cd/m2 (d) 8.72 cd/m2

Sol. (a)

The diagonal

BD 8 m

SB2 = SO2 + BO2

=

2

2 82

2

= 8

4 64



24

2cos BSO

6

Now, illumination at a corner B

02 2

I 60 2cos

r 66

= 120 20

2.456 6

= 8.16 cd/m2

57. A thin glass (refractive index 1.5) lens has optical power of –5D in air. Its optical

power in a liquid medium with refractive index 1.6 will be :

(a) 1D (b) –1D (c) 25 D (d) –25D

Sol : (a)

0 1 2

1 1 11

f R R

= 1 2

1 11.5 1

R R ..(i)

and g m

m m 1 2

1 1 1

f R R

m 1 2

1 1.5 1 11

f 1.6 R R …(ii)



25

Thus, m

a

1.5 1f8

1.5f1

1.6

m af 8 f

= 1

85

a

1 1f m

p 5

= 1.6 cm

mm

1.6P 1 D

f 1.6

58. A clock fixed on a wall shows time 04 : 25 : 37. What time will its image in a plane

mirror show ?

(a) 07 : 43 : 32 (b) 07 : 33 : 32 (c) 07 : 35 : 23 (d) 43 : 27 : 36

Sol : (c)

Due to planar mirror, clockwise watch is converted into anticlockwise watch (shown

in figure). Here, (c) is correct.

59. A point object P is situated infront of plane mirror shown in figure. The width of

mirror AB is d. The visual region on a line passing through point P and parallel to the

mirror is:

(a) d (b) 2d (c) 3d (d) none of these

Sol : (b)



26

The length CF gives visual region

From the figure AP’ = d cos 450 & FD = d cos 450

FP’ = 2d cos 450

Since triangle AP’B and FP’C are similar

FP' FC

AP' AB

0

0

2d cos 45 FC

d cos 45 d

2d = FC

60. A flat mirror revolves at a constant angular velocity making 2 revolutions/s. With

what

velocity will a light spot move along a spherical screen with a radius of 10 m, if the

mirror is at a centre of curvature of the screen ?

(a) 251.2 m/s (b) 261.2 m/s (c) 271.2 m/s (d) 241.2 m/s

Sol : (a)

Angular velocity of the mirror

= 2 rev/s

= 4 rad/s

Angular velocity of the reflected ray

= 2 × 4

= 8 rad/s

Velocity of light spot over the screen v = r

v = 10 × 8

= 80 = 251.2 m/s



27

Section – C {Chemistry}

61. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid

(H3PO3), the volume of 0.1 M aqueous KOH solution required is

(a) 10 mL (b) 20 mL (c) 40 mL (d) 60 mL

Sol: (c)

H3PO3 is a dibasic acid

N1V1(acid) = N2V2(base)

0.1 2 20 = 0.1 1 = V2

V2 = 0.1 2 20

40mL.0.1 1

62. The total vapour pressure of a solution of components A and B is 600 torr. The mole

fraction of component A in liquid and vapour phase are 0.70 and 0.35 respectively. The

vapour pressure of pure A and B are

(a) 300 torr, 130 torr (b) 1300 torr , 300 torr

(c) 300 torr, 1300 torr (d) 300 torr, 300 torr

Sol: (c)

XA =0.70 XB = 0.30;

YA = 0.35 XB = 0.65;

PA = 600 0.35 0A

600 0.35P

0.70

0Ap 300 torr

PB =600 0.65 0B

600 0.35P

0.70

0BP 1300torr

63. CsBr crystal has bcc structure. It has an edge length of 4.3o

A . The shortest interionic

distance beween Cs+ and Br– ions is

(a) 1.86 o

A (b) 2.86o

A (c) 3.72 o

A (d) 4.72o

A



28

Sol: (c)

Closest approach in bcc lattice

1 1of body diaginal 3a

2 2

o34.3 3.72A

2

64. The lattice points of a crystal of hydrogen iodide are occupied by

(a) HI molecules (b) H atoms and I atoms

(c) H+ cations and I– anions (d) H2 molecules and I2 molecules

Sol: (a)

Since, HI is a covalent molecule, and crystallizes in face centre cubic structure, HI

molecule are present at the lattice points of the crystal.

65. KCl crystallises in the same type of lattice as does NaCl. Given that Na Cl

r / r 0.55 and

K Clr / r 0.74 Calculate the ratio of the side of the unit cell for KCl to that of NaCl.

(a) 1.123 (b) 1.414 (c) 1.732 (d) 0.0891

Sol: (a)

Given

Na Cl

K

Cl

r / r 0.55

r0.74

r

Na

Cl

r1 0.55 1 ........ 2

r

K

Cl

r1 0.74 1 ......... 1

r

From eq. (i)/ (ii)

K Cl

Na Cl

r r 1.741.123

r r 1.55

66. The limiting molar conductivities o for NaCl, KBr and KCl are 126, 152 and 150 S cm2

mol–1 respectively. The o for NaBr is



29

(a) 128 S cm2 mol–1 (b) 248 S cm2 mol–1 (c) 328 S cm2 mol–1 (d) 348 S cm2

mol–1

Sol: (a)

(126 Scm2 mol–1) NaCl Na Cl …………..(i)

(152Scm2 mol–1) KBr K Br ………..(ii)

(150 Scm2 mol–1) KCl K Cl …………….(iii)

By eq. (i) + (ii) – (iii)

NaBr Na Br

= 126 +152 –150 =128 Scm2 mol–1

67. How many faradays of electricity are required to electrolyse 1 mole CuCl2 to copper

metal and chlorine gas ?

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

(d) 4 F

Sol: (b)

The cathode and anode reactions respectively are

Cu2+ + 2e– Cu

2 Cl– Cl2 + 2e–

The two moles of electrons have been transferred from anode to cathode to produce

Cu and Cl2 in a mole ratio of 1 : 1. Thus, 2F electricity is required.

68. The one which decreases with dilution is

(a) molar conductance (b) conductance

(c) specific conductance (d) equivalent conductance

Sol: (c)

The number of ions per cc decreases with the dilution and therefore, specific

conductance decreases with dilution .

69. For the reaction 2 2 3N 3H 2NH the rate of change of concentration for hydrogen is –

0.3 10–4 Ms–1. The rate of change of concentration of ammonia is



30

(a) –0.2 10–4 (b) 0.2 10–4 (c) 0.1 10–4 (d) 0.3 10–4

Sol: (b)

2 2 3N 3H 2NH

2 4 1d H

0.3 10 Msdt

2 3

3 2

4

4 1

d H d NH1 1Rate

3 dt 2 dt

d NH d H2

dt 3 dt

20.3 10

3

0.2 10 Ms

70. The given reaction 3 2 2 42FeCl SnCl 2FeCl SnCl is an example of

(a) third order reaction (b) second order reaction

(c) first order reaction (d) None of these

Sol: (a)

3 2 2 42FeCl SnCl 2FeCl SnCl

It is a third order reaction. As the concentration of both FeCl3 and SnCl2 affect the rate

of reaction.

Rate =k[FeCl3]2 [SnCl2]

Hence, order of reaction = 2 + 1 =3

71. The rate constant for a zero order reaction is

(a) 0 tc ck ln

2t (b) 0 tc c

kt

(c) 0

t

ck

c (d) 0c

k2t

Sol: (b)



31

For a zero order of reaction

t

0

0 t

0 t

A B

d Ak

dt

C t 0or d A k dt

C t 0

C C kt

Or k C C / t

72. The graph plotted between log k versus 1

T for calculating activation energy is shown

by

Sol: (b)

A graph plotted b/w log k versus 1

Tfor calculating activation energy is shown as



32

From Arrhenius equation

log K = log A – aE

2.303RT

73. Given the hypothetical reaction mechanism

I II III IVA B C D E and the data as,

Specifies formed Rate of its formation

B

C

D

E

0.002 mol/h, per mole of A

0.030 mol/h, per mole of B

0.011 mol/h, per mole of C

0.042 mol/h, per mole of D

The rate determining step is

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

Sol: (a)

The slowest step is the rate determining step. Formation of B (i.e., step 1) is the

slowest step, therefore, step 1 is rate determining step.

74. Which is more powerful to coagulate the negative colloid ?

(a) ZnSO4 (b) Na3PO3 (c) AlCl3 (d) K4[Fe(CN)6]

Sol: (c)

Negative colloid is coagulated by positive ion or vice versa. Greater the valency of

coagulating ion, greater will be coagulating power.

Since, in AlCl3, the valency of positive ion is highest , it is most powerful coagulating

agent among the given to coagulate negative colloid .

75. Which one is correct statement ?

(a) Basicity of H3PO4 and H3PO3 is 3 and 3 respectively

(b) Acidity of H3PO4 and H3PO3 is 3 and 3 respectively

(c) Acidity of H3PO4 and H3PO3 is 3 and 2 respectively

(d) Basicity of H3PO4 and H3PO3 is 3 and 2 respectively

Sol: (d)



33

Orthophosphoric acid (H3PO4) is a tribasic acid because it has 3 replaceable hydrogen

atom .

Hence, the basicity of H3PO4 is 3.

Its structure is as :

While phosphorus acid (H3PO3) is a dibasic acid because it has two replaceable

hydrogen atoms.

Hence, basicity of H3PO3 is 2. Its structure is as

76. In pyrophosphoric acid, H4P2O7, number of and d p bonds are respectively

(a) 8 and 2 (b) 6 and 2 (c) 12 and zero (d) 12 and 2

Sol: (d)

Draw bond structure and then count bonds.

12 2d p bonds

77. Nitrogen dioxide cannot be prepared by heating

(a) KNO3 (b) Pb(NO3)2 (c) Cu(NO3)2 (d) AgNO3

Sol: (a)



34

Nitrate of alkali metals on heating evolve oxygen gas (eg, KNO3) while nitrates of p and

d-block elements [eg,Pb (NO3)2Cu(NO3)2 and AgNO3] gives out nitrogen dioxide on

heating.

2KNO3 2KNO2 + O2

Nitrogen dioxide cannot be prepared from KNO2

78. S-S bond is present in

(a) 3 n(SO ) (b) 2 9 n(S O ) (c) 2 2 3H S O (d) 2 2 8H S O

Sol: (c)

H2S2O3,

79. The correct order of pseudohalide, polyhalide and interhalogen are

(a) 2 5BrI ,OCN ,IF (b) 5 2IF ,BrI , OCN (c) 5 2OCN ,IF ,BrI (d) 2 5OCN ,BrI ,IF

Sol: (d)

Pseudohalide They are of combination more than one electronegative atoms which

among themselves are called eg, OCN–, CN–.

Polyhalide ions The complex ions which are formed by reaction of halogens among

themselves are called polyhalide ions eg, 3 2I ,BrI .

Interhalogens They are the compounds which are formed when halogens react

among themselves. One of the halogens behave as cation and other acts as anion, eg.

7 5 3IF ,ICl ,BrF .

80. End product of the hydrolysis of XeF6 is

(a) XeF4 (b) XeF2O2 (c) XeO3 (d) 3XeO

Sol: (c)

The end product of the hydrolysis of XeF6 is XeO3

2 2 2H O H O H O

6 4 2 2 32HF 2HF 2HFXeF XeOF XeO F XeO



35

81. The shape of XeF4 molecule and hybridization of xenon in it are

(a) tetrahedral and sp3 (b) square planar and dsp2

(c) square planar and sp3d2 (d) octahedral and sp3d2

Sol: (c)

In XeF4, the central atom, Xe, has eight electrons in it outer most shell. Out of these

four are used for forming for bonds with F and four remain as lone pair,

4XeF 4 bonds 2 lone pairs

6 hybridised orbitals , that is sp3d2 hybridization.

82. The IUPC name of [Pt(NH3)4(NO2)Cl]SO4 is

(a) chloronitro tetrammine platinum (IV) sulphate

(b) tetrammine chloronitro platinum (II) sulphate

(c) tetrammine chloronitro platinum (IV) sulphate

(d) chlorotetrammine nitroplatinum (IV) sulphate

Sol: (c)

83. Both CO3+ and Pt4+ have a coordination number of six. Which of the following pairs of

complexes will show approximately the same electrical conductance for their 0.001 M

aqueous solutions ?

(a) CoCl3. 4NH3 and PtCl4. 4NH3 (b) CoCl3. 3NH3 and PtCl4. 5NH3

(c) CoCl3. 6NH3 and PtCl4. 5NH3 (d) CoCl3. 6NH3 and PtCl4. 3NH3

Sol: (c)

CO3+ and Pt4+ = coordination no. CoCl3. 6NH3 and PtCl4. 5NH3 and PtCl4.5NH3

3In solution

3 3 36 6Co NH Cl Co NH 3Cl

3Insolution

3 3 35 5[PtCl NH Cl PtCl NH 3Cl

Number of ionic species are same in the solution of both complexes therefore their

equimolar solution will show same conductance.

84. What is the EAN of nickel in Ni(CO)4 ?

(a) 38 (b) 30 (c) 36 (d) 32



36

Sol: (c)

EAN = Atomic no. – O.S. + 2×C.N.

= 28 – 0 + 2×4 = 36

85. In the reaction, 2BrAlc. KOH KCN3 2CH CH I X Y Z is

(a) CH3CH=CN (b) CH2BrCH2CN (c) CNCH2CH2CN (d) BrCH=CHCN

Sol: (c)

C2H5I 2BrAlcKOH KCN2 2 2 2 2 2HI ZX Y

CH CH BrCH CH Br NCCH CH CN

86. Chlorobenzene can be obtained from benzene diazonium chloride by the

(a) Friedel-Crafts reaction (b) Sandmayer reaction

(c) Wurtz reaction (d) Fittig reaction

Sol: (b)

CuCl/HCl6 5 6 5 2C H N NCl C H Cl N

87. Chloroform on reduction with zinc dust and water gives

(a) methyl chloride (b) dichloro methane (c) chloro methane (d) methane

Sol: (d)

2

2

6 H ;Zn/H O

3 4HCl , ZnClmethaneChloroform

CHCl CH

88. The major product of the following reaction is



37

Sol: (a)

PhS– is a strong nucleophile and dimethylformamide (DMF) is a highly polar aprotic

solvent. 2SN takes place at 20 benzylic place. It involves inversion of configuration.

89. Ethyl orthoformate is formed by heating ………..with sodium ethoxide.

(a) CHCl3 (b) C2H5OH (c) HCOOH (d) CH3CHO

Sol: (a)

3 2 5 2 5 3ethylorthoformate

CHCl 3C H ONa CH OC H 3NaCl

90. In the following sequence of reactions, AgCN Reduction2 5C H Br X Y; is

(a) n-propyl amine (b) iso-propylamine (c) ethylamine (d) ethylmethyl

amine.

Sol: (d)

AgCN Reduction2 5 2 5 2 5 3

X Y

C H Br C H NC C H NH.CH

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