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AKASH MULTIMEDIA 1

AIEEE-2010 CHEMISTRY, PHYSICS & MATHS

PART-A : CHEMISTRY1. The standard enthalpy of formation of NH

3 is

–46.0kJmol–1. If the enthalpy of formation ofH

2 from its atoms is – 436 kJmol–1 and that of

N2 is – 712 kJmol–1, the average bond enthalpy

of N–H bond in NH3 is

1) –964 kJ mol–1 2) +352 kJ mol–1

3) +1056 kJmol-1 4) –1102 kJ mol–1

2. The time for half life period of a certain reactionAm products is 1 hour. When the initialconcentration of the reactant ‘A’, is 2.0 molL–1,how much time does it take for its concentrationto come from 0.50 to 0.25 mol L–1 if it is zeroorder reaction?

1) 4h 2) 0.5h 3) 0.25h 4) 1h

3. A solution containing 2.675 g of CoCl3. 6NH

3

(molar mass = 267.5g mol–1) is passed througha cation exchanger. The chloride ions obtainedin solution were treated with excess of AgNO

3

to give 4.78g of AgCl (molar mass =143.5 g mol–1). The formula of the complex is(At. Mass of Ag = 108u)

1) [Co(NH3)

6]Cl

32) [CoCl

2(NH

3)

4]Cl

3) [CoCl3(NH

3)

3] 4) [CoCl(NH

3)

5]Cl

2

4. Consider the reaction :Cl

2(aq)+ H

2S

(aq) m S

(s) + 2H+

(aq) + 2Cl–

(aq)

The rate equation for this reaction is,

rate = k[Cl2][H

2S]

Which of these mechanisms is /are consistentwith this rate equation ?

(A) Cl2+ H

2 m H+ + Cl– + Cl+ + HS– (slow)

Cl+ + HS– m H+ + Cl– + S (fast)

(B) H2S � H+ + HS– (fast equilibrium)

Cl2 + HS– m 2Cl– + H+ + S (slow)

1) B only 2) Both A and B3) Neither A nor B 4) A only

5. If 10–4 dm3 of water is introduced into a1.0 dm3 flask to 300k, how many moles of waterare in the vapour phase when equilibrium isestablished ? (Given : Vapour pressure of H

2O

at 300 K is 3170 Pa; R = 8.314 JK–1 mol–1)1) 5.56 × 10–3 mol 2) 1.53 × 10–2 mol3) 4.46 × 10–2 mol 4) 1.27 × 10–3 mol

6. One mole of a symmetrical alkene on ozonolysisgives two moles of an aldehyde having amolecular mass of 44u. The alkene is1) propene 2) 1-butene3) 2-butene 4) ethene

7. If sodium sulphate is considered to be completelydissociated into cations and anions in aqueoussolution, the change in freezing point of water tT% , when 0.01 mol of sodium sulphate isdissolved in 1 kg of water, is(K

f = 1.86 K kg mol–1)

1) 0.0372 K 2) 0.0558 K

3) 0.0744 K 4) 0.0186 K

8. From amongst the following alcohols the onethat would react fastest with conc.HCl andanhydrous ZnCl

2, is

1) 2 – Butanol

2) 2- Methylpropan-2-ol

3) 2-Methylpropanaol 4) 1 -Butanol

9. In the chemical reactions,NH2

2 4NaNO HBFHCl,278K A B}}}}m }}}m

the compound ‘A’ and ‘B’ respectively are

1) nitrobenzene and fluorobenzene

2) phenol and benzene

3) benzene diazonium chloride andfluorobenzene

4) nitrobenzene and chlorobenzene

10. 29.5 mg of an organic compound containingnitrogen was digested according to Kjeldahl’smethod and the evolved ammonia was absorbedin 20 mL of 0.1 M HCl solution. The excess ofthe acid required 15mL of 0.1 M NaOH solutionfor complete neutralization. The percentage ofnitrogen in the compound is

1) 59.0 2) 47.4 3) 23.7 4) 29.5

11. The energy required to break one mole ofCl – Cl bonds in Cl

2 is 242 kJ mol–1. The longest

wavelength of light capable of breaking asingle Cl – Cl bond is (c = 3 × 108 ms–1 andN

A = 6.02 × 1023 mol–1)

1) 594 nm 2) 640 nm 3) 700 nm 4) 494 nm

AIEEE-2010 QUESTION PAPER

AKASH MULTIMEDIA2

AIEEE-2010CHEMISTRY, PHYSICS & MATHS

12. Ionisation energy of He+ is 19.6×10–18J atom–1.

The energy of the first stationary state (n = 1) of

Li 2+ is

1) 4.41 × 10–16 J atom–1

2) – 4.41 × 10–17 J atom–1

3) – 2.2 × 10–15 J atom–1

4) 8.82 × 10–17 J atom–1

13. Consider the following bromides :

Me Br

(A)

Me

(B) Br

Me

(C)Br

Me

The correct order of SN

1 reactivity is

1) B > C > A 2) B > A > C

3) C > B > A 4) A > B > C

14. Which one of the following has an opticalisomer?

1) [Zn(en) (NH3)

2]2+ 2) [Co(en)

3]3+

2) [Co(H2O)

4(en)]3+ 4) [Zn(en)

2]2+

(en = ethylenediamine)

15. On mixing, heptane and octane form an idealsolution. At 373 K, the vaopur pressure of the twoliquid components (heptane and octane) are105 kPa and 45 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing25.0g of heptane and 35g of octane will be (molarmass of heptane = 100g mol–1 and of octane =114 g mol–1).

1) 72.0 kPa 2) 36.1 kPa

3) 96.2 kPa 4) 144.5 kPa

16. The main product of the following reaction is

2 4conc.H SO6 5 2 3 2

C H CH CH(OH)CH CH ?}}}}m

1)

5 6

3 2

H C H

C = C

H CH CH2)

C6H5CH2 CH3

CH3H

C = C

3)

CH(CH3)2

H

C = C

H

C6H5

4) C = CH2

H5C6CH2CH2

H3C

17. Three reactions involving H2PO

4– are given

below:i) H

3PO

4 + H

2O m H

3O+ + H

2PO

4–

ii) H2PO

4– + H

2O m 2

4HPO � + H3O+

iii) H2PO

4– + OH– m H

3PO

4 + O2–

In which of the above does H2PO

4– act as an

acid?1) ii only 2) i and ii 3) iii only 4) i only

18. In aqueous solution the ionization constants forcarbonic acid are

K1 = 4.2 × 10–7 and K

2 = 4.8 × 10–11

select the correct statement for a saturated 0.034M solution of the carbonic acid.1) The concentration of 2

3CO � is 0.034 M.2) The concentration of 2

3CO � is greater thanthat of 3HCO� .

3) The concentration of H+ and 3HCO� areapproximately equal.

4) The concentration of H+ is double that of 23CO �

19. The edge length of a face entered cubic cell ofan ionic substance is 508 pm. If the radius ofthe cation is 110 pm, the radius of the anion is1) 288 pm 2) 398 pm3) 618 pm 4) 144 pm

20. The correct order of increasing basicity of thegiven conjugate bases (R= CH

3) is

1) 2RCOO HC C R NH� � � �

2) 2R HC C RCOO NH� y � �

3) 2RCOO NH HC C R� � y �

4) 2RCOO HC C NH R� y � �

21. The correct order of increasing basicity of thegiven conjugate bases (R = CH

3) is

1) Al3+ > Mg2+ > Na+ > F– > O2–

2) Na+ > Mg2+ > Al3+ > O2– > F–

3) Na+ > F– > Mg2+ > O2– > Al3+

4) O2– > F– > Na+ > Mg2+ > Al3+

22. Solubility product of silver bromide is5.0 × 10–13. The quantity of potassium bromide(molar mass taken as 120g of mol–1) to be addedto 1 litre of 0.05 M solution of silver nitrate tostart the precipitation of AgBr is

1) 1.2 × 10–10g 2) 1.2 × 10–9g

3) 6.2 × 10–5g 4) 5.0 × 10–8 g

AKASH MULTIMEDIA 3


23. The Gibbs energy for the decomposition ofAl

2O

3 at 5000C is as follows :

12 3 2 r

2 4Al O Al O , G 966kJmol

3 3�

m � % � �

The potential difference needed for electrolyticreduction of Al

2O

3 at 5000C is at least

1) 4.5 V 2) 3.0 V 3) 2.5 V 4) 5.0 V

24. At 250C, the solubility product of Mg(OH)2 is

1.0 10–11. At which pH, will Mg2+ ions startprecipitating in the form of Mg(OH)

2 from a

solution of 0.001 M Mg2+ ions ?

1) 9 2) 10 3) 11 4) 8

25. Percentage of free space in cubic close packedstructure and in body centered packed structureare respectively

1) 30% and 26% 2) 26% and 32%

3) 32% and 48% 4) 48% and 26%

26. Out of the following, the alkene that exhibitsoptical isomerism is

1) 3 – methyl– 2 – pentane

2) 4 – methyl – 1– pentane

3) 3 – methyl – 1 – pentane

4) 2 – methyl – 2 – pentane

27. Biuret test is not given by

1) carbohydrates 2) polypeptides

3) urea 4) proteins

28. The correct order of 20

M / ME

� values with negative

sign for the four successive elements Cr, Mn,Fe and Co is

1) Mn > Cr > Fe > Co 2) Cr > Fe > Mn > Co

3) Fe > Mn > Cr > Co 4) Cr > Mn > Fe > Co

29. The polymer containing strong intermolecularforces e.g. hydrogen bonding, is1) teflon 2) nylon 6, 63) polystyrene 4) natural rubber

30. For a particular reversible reaction attemperature T, H% and S% were found to beboth +ve. If T

e is the temperature at equilibrium,

the reaction would be spontaneous when

1) Te > T 2) T > T

e

3) Te is 5 times T 4) T = T

e

PART-B : PHYSICS

31. A rectangular loop has a sliding connector PQof length l and resistance R8 and it is movingwith a speed v as shown . The set-up is placedin a uniform magnetic field going into the planeof the paper. The three currents I

1,I

2 and I are

VR8R8 R8

I

I1

I2

P l

1) 1 2

Bl 2BlI I , I

R R

V V

� � � �

2) 1 2

Bl 2BlI I ,I

3R 3R

V V

� � �

3) 1 2

BlI I I

R

V

� � �

4) 1 2

Bl BlI I ,I

6R 3R

V V

� � �

32. Let C be the capacitance of a capacitordischarging through a resistor R. Suppose t

1 is

the time taken for the energy stored in thecapacitor to reduce to half its initial value and t

2

is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t

1/t

2 will

be

1) 1 2) 1

23)

1

43) 2

Directions : Questions number 33 – 34 containStatement -1 and Statement-2. Of the fourchoices given after the statements, choose theone that best describes the two statements.

33. Statement -1 : Two particles moving in the samedirection do not lose all their energy in acompletely inelastic collision.

Statement-2 : Principle of conservation ofmomentum holds true for all kinds of collisions.

1) Statement -1 is true, Statement -2 is true,Statement-2 is the correct explanation ofStatement1

2) Statement -1 is true, Statement-2 is true;Statement-2 is not the correct explanation ofStatement -1.

3) Statement -1 is false, Statement -2 is true.

4) Statement-1 is true, Statement -2 is false

AKASH MULTIMEDIA4


34. Statement -1 : When ultraviolet light is incidenton a photocell, its stopping potential is V

0 and

the maximum kinetic energy of the photo-electrons is K

max. When the ultraviolet light is

replaced by X-rays , both V0 and K

max increase

Statement-2 : Photoelectrons are emitted withspeeds ranging from zero to a maximum valuebecause of the range of frequencies present inthe incident light

1) Statement -1 is true, Statement -2 is true,Statement-2 is the correct explanation ofStatement1

2) Statement -1 is true, Statement-2 is true;Statement-2 is not the correct explanation ofStatement -1.

3) Statement -1 is false, Statement -2 is true.

4) Statement-1 is true, Statement -2 is false

35. A ball is made of a material of density Swhere

oil water oilwithS � S � S S and waterS representingthe densities of oil and water, respectively. Theoil and water are immiscible . If the above ballis in equilibrium in a mixture of this oil andwater, which of the following pictures representsits equilibrium position ?

1)

water

oil2)

water

oil

3)

water

oil4)

water

oil

36. A particle is moving with velocity

ˆ ˆK yi xj ,V � �

JG

where K is a constant . Thegeneral equation for its path is

1) 2 2y x constant� � 2) 2y x constant� �

3) 2y x constant� � 4) xy = constant

37. Two long parallel wires are at a distance 2d apart.They carry steady equal currents flowing out ofthe plane of the paper as shown. The variationof the magnetic field B along the line XX’ is givenby

1)

B

d d

x x’

2)

B

d d

x x’

3)

B

d d

x x’

4)

B

d d

x x’

38. In the circuit shown below, the key K is closedat t = 0. The current through the battery is

v k

R1L

R2

1) 1 2

1 2 2

V R R Vat t 0 and at t

R R R

�

� � d

2) 1 2

2 221 2

VR R Vat t 0 and at t

RR R� � d

3) 1 2

2 1 2

V R RVat t 0 and at t

R R R

�

� � d

4) 1 2

2 22 1 2

VR RVat t 0 and at t

R R R� � d

39. The figure shows the position -time (x -t) graphof one-dimensional motion of a body of mass0.4kg. The magnitude of each impulse is

0 2 4 8 106 12 14 16

t (s)

x(m)2

1) 0.4 Ns 2) 0.8 Ns 3) 1.6 Ns 4) 0.2Ns

AKASH MULTIMEDIA 5


Directions : Questions number 40 -41 are basedon the following paragraph.

A nucleus of mass M + m% is at rest and decays

into two daughter nuclei of equal mass M

2 each,

Speed of light is c.

40. The bindign energy per nucleon for the parentnucleus is E

1 and that for the daughter nuclei is E

2.

Then

1) E2 = 2E

12) E

1 > E

2

3) E2 > E

14) E

1 = 2E

2

41. The speed of daughter nuclei is

1) m

cM m

%

� %

2) m

cM m

%

� %

3) 2 m

cM

%4)

mc

M

%

42. A radioactive nucleus (initial mass number Aand atomic number Z) emits 3B � particles and2 positrons. The ratio of number of neutrons tothat of protons in the final nucleus will be

1) A Z 4

Z 2

� �

�

2) A Z 8

Z 4

� �

�

3) A Z 4

Z 8

� �

�

4) A Z 12

Z 4

� �

�

43. A thin semi-circular ring of radius r has apositive charge q distributed uniformly over it.The net field E

G

at the centre O is

1) 2 20

qj

2 rQ F2) 2 2

0

qj

4 rQ F

3) 2 20

qj

4 r�

Q F

O i�

j�

4) 2 20

qj

2 r�

Q F

44. The combination of gates shown below yields

1) OR gate

2) NOT gate XA

B3) XOR gate

4) NAND gate

45. A diatiomic ideal gas is used in a Car engine asthe working substance. If during the adiabaticexpansion part of the cycle, volume of the gasincreases from V to 32V the efficiency of theengine is

1) 0.5 2) 0.75 3) 0.99 4) 0.25

46. If a source of power 4 kW produces1020 photons/second, the radiation belong to apart of the spectrum called1) X-rays 2) ultraviolet rays3) microwaves 4) raysH �

47. The respective number of significant figures forthe numbers 23.023, 0.0003 and 2.1s 10–3 are

1) 5, 1, 2 2) 5, 1, 5 3) 5, 5, 2 4) 4, 4, 2

48. In a series LCR circuit R = 2008 and thevoltage and the frequency of the main supplyis 220V and 50Hz respectively. On taking outthe capacitance from the circuit the current lagsbehind the voltage by 300. On taking out theinductor from the circuit the current leads thevoltage by 300. The power dissipated in the LCRcircuit is

1) 305W 2) 210W 3) Zero W 4) 242W

49. Let there be a spherically symmetric chargedistribution with charge density varying as

0

5 rr

4 R¥ ´

S � S �¦ µ§ ¶

upto r=R, and r 0S � for

r > R, where r is the distance from the origin.The electric field at a distance r(r < R) from theorigin is given by

1) 0

0

4 r 5 r

3 3 R

QS ¥ ´

�¦ µ

§ ¶F

2) 0

0

r 5 r

4 3 R

S ¥ ´

�¦ µ

§ ¶F

3) 0

0

4 r 5 r

3 4 R

S ¥ ´

�¦ µ

§ ¶F

4) 0

0

r 5 r

3 4 R

S ¥ ´

�¦ µ

§ ¶F

50. The pontential energy function for the forcebetween two atoms in a diatomic moleculeis approximately given by

U (X) = 12 6

a bU X

X X� � , where a and b are

constants and X is the distance between theatoms. If the dissociation energy of the moleculeis at equilibriumD [U(X ) U ],D is� � d �

1) 2b

2a2)

2b

12a3)

2b

4a4)

2b

6a

AKASH MULTIMEDIA6


51. Two identical charged spheres are suspendedby strings of equal lengths. The strings makean angle of 300 with each other. Whensuspended in a liquid of density 0.8 g cm-3, theangle remans the same. If density of the materialof the sphere is 1.6 g cm-3, the dielectric constantof the liquid is

1) 4 2) 3 3) 2 4) 1

52. Two conductors have the same resistance at 00Cbut their temperature coefficients of resistance

are 1 2 and B B . The respective temperature

coefficients of their series and parallelcombinations are nearly

1) 1 21 2,

2

B �B

B �B 2) 1 21 2 ,

2

B �B

B �B

3) 1 2

1 21 2

,B B

B �B

B �B 4) 1 2 1 2,

2 2

B �B B � B

53. A point P moves in counter-clockwise directionon a circular path as shown in the figure. Themovement of ‘P’ is such that it sweeps out alength s = t3 + 5, where s is in metres and t is inseconds. The radius of hte path is 20 m. Theacceleration of ‘P’ when t = 2 s is nearly.

1) 13 m/s2

2) 12 m/s2

3) 7.2 m/s2

O A

B

20 m

y

x

P(x,y)

4) 14 m/s2

54. Two fixed frictionless inclined planes makingan angle 300 and 600 with the vertical are shownin the figure. Two blocks A and B are placed onthe two planes. what is the relative verticalacceleration of A with respect to B?

A

600 300

B

1) 4.9 ms-2 in vertical direction

2) 9.8 ms-2 in vertical direction

3) Zero

4) 4.9 ms-2 in horizontal direction

55. For a particle in uniform circle motion, theacceleration a

G

at a point P R,R on the circle ofradius R is (HereR is measured from theX-axis)

1) 2 2v vˆ ˆcos i sin j

R R� R � R 2)

2 2v vˆ ˆsin i cos jR R

� R � R

3) 2 2v vˆ ˆcos i sin j

R R� R � R 4)

2 2v vˆ ˆi jR R

�

Directions : Questions number 56-58 are basedon the following paragraph

An initially parallel cylindrical beam travels ina medium of refractive index 0 2(I) I,N � N � N

where 0 2andN N are positive constants and I isthe intensity of the light beam. The intensity ofthe beam is decreasing with increasing radius.

56. As the beam enters the medium , it will1) diverge 2) converge3) diverge near the axis and converge near the

periphery4) travel as a cylindrical beam

57. The initial shape of the wave front of the beamis1) convex 2) concave3) convex near the axis and concave near the

periphery4) planar

58. The speed of light in the medium is1) minimum of the axis of the beam2) the same everywhere in the beam3) directly proportional to the intensity4) maximum on the axis of hte beam

59. A small particle of mass m is projected at anangle Rwith the X-axis with an intial velocity

0O in the x-y plane as shown in the figure. At a

time 0v sin

tg

R� ,the angular momentum of the

particle is

1) 20

ˆmgv t cos j� R 2) 0ˆmg v t cos kR

3) 2

01 ˆmgv t cos k2

� R V0

y

xR4)

20

1 ˆmgv t cos i2

R

where ˆ ˆi, j and k are unit vectors along x, y andz– axis respectively.

AKASH MULTIMEDIA 7


60. The equation of a wave on astring of linear massdensity 0.04 kg m–1 is given by

t xy 0.02(m)sin 2

0.04(s) 0.50(m)

¨ ·¥ ´

� Q �© ¸¦ µ

§ ¶ª ¹

The tension in the string is

1) 4.0N 2) 12.5N

3) 0.5 N 4) 6.25 N

PART-C : MATHEMATICS

61. Let 4

cos( )5

B �C � and let 5

sin( ) ,13

B �C �

where 0 , ,4

Q

b B C b then tan 2B �

1) 56

332)

19

123)

20

74)

25

16

62. Let S be a non-empty subset of R. Consider thefollowing statement :P : There is a rational number x S� such thatx > 0.

Which of the following statements is thenegation of the statement P ?

1) There is no rational number x S� such that0x b

2) Every rational number x S� satisfies 0rx

3) x S� and 0x xb � is not rational

4) There is a rational number x S� such that

0x b

63. Let ˆˆa j k� �

G and ˆˆ ˆ .c i j k� � �

G Then vectorbG

satisfying 0a b cs � �

G GG G and . 3a b �

GG is

1) ˆˆ ˆ2 2i j k� � 2) ˆˆ ˆ 2i j k� �

3) ˆˆ ˆ 2i j k� �

4) ˆˆ ˆ 2i j k� � �

64. The equation of the tangent to the curve

2

4,y x

x� � that is parallel to the x-axis, is

1) y = 1 2) y = 2 3) y = 3 4) y = 0

65. Solution of the differential equation

cos x dy = y(sinx – y)dx, 02

xQ

� � is

1) y sec x = tan x + c 2) y tan x = sec x + c3) tanx = (secx + c)y 4) sec x = (tanx + c)y

66. The area bounded by the curves y = cosx and

y = sinx between the ordinates x = 0 and 3

2x

Q

�

is

1) 4 2 2� 2) 4 2 1�

3) 4 2 1� 4) 4 2 2�

67. If two tangents drawn from a point P to theparabola y2 = 4x are at right angles, then thelocus of P is1) 2x + 1 = 0 2) x = –13) 2x – 1 = 0 4) x = 1

68. If the vectors ˆ ˆˆ ˆ ˆ ˆ2 , 2 4a i j k b i j k� � � � � �

G

G andˆˆ ˆc i j k� M � � N

G are mutually orthogonal, then

( , )M N �

1) (2, –3) 2) (–2, 3) 3) (3, –2) 4) (–3, 2)

69. Consider the following relations :R = {(x, y) |x, y are real numbers an x = wy forsome rational number w};

, . . and are integers m p

S m n p qn q

«¥ ´®

� ¬¦ µ§ ¶®

^such that , 0 and n q qm pnx � .Then1) neither R nor S is an equivalence relation2) S is an equivalence relation but R is not an

equivalence relation3) R and S both are equivalence relations4) R is an equivalence relation but S is not an

equivalence relation

70. Let :f R Rm be defined by

2 , if 1( ) .

2 3, if 1

k x xf x

x x

� b �«

� ¬

� � �

If f has a local

minimum at x = –1, then a possible value of k is

1) 0 2) 1

2� 3) –1 4) 1

71. The number of 3 × 3 non-singular matrices, withfour entries as 1 and all other entries as 0, is1) 5 2) 63) at least 7 4) less than 4

72. Four numbers are chosen at random (withoutreplacement) from the set {1,2,3....,20}.Statement-1 : The probability that the chosennumbers when arranged in some order will form

an AP is 1

.85

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Statement-2 : If the four chosen numbers froman AP, then the set of all possible values of

common difference is { 1, 2, 3, 4, 5}.p p p p p

1) Statement-1 is true, Statement-2 is true;Statement-2 is a correct explanation forStatement-1

2) Statement-1 is true, Statement-2 is true;Statement-2 is not correct explanation forStatement-1

3) Statement-1 is true, Statement-2 is false4) Statement-1 is false, Statement-2 is true

73. Statement-1 : The point A(3, 1, 6) is the mirrorimage of the point B(1,3,4) in the planex–y+z = 5.

Statement-2 : The plane x – y + z = 5 bisects theline segment joining A(3, 1, 6) and B(1, 3, 4).



3) Statement-1 is true, Statement-2 is false

4) Statement-1 is false, Statement-2 is true

74. Let

10 1010 10

1 21 1

( 1) , j jj j

S j j C S j C� �

� � �¤ ¤ and

102 10

31

.jj

S j C�

�¤

Statement-1 : S3 = 55×29

Statement-2 : S1 = 90 × 28 and S2 = 10×28.





75. Let A be a 2 × 2 matrix with non-zero entriesand let A2 = I, where I is 2 × 2 identity matrix.Define Tr(A) = sum of diagonal elements ofA and |A| = determinant of matrix A.

Statement-1 : Tr(A) = 0

Statement-2 : |A| = 1





76. Let mR R be a continuous funciton defined

by 1

( ) .2 �

�

�x x

f xe e

Statement-1 : 1

( ) ,3

�f c for some .�c R

Statement-2 : 1

0 ( ) ,2 2

� bf x for all �x R

1) Statement-1 is true, Statement-2 is true;Statement-2 is not the correct explanation forStatement-1



4) Statement-1 is true, Statement-2 is true;Statement-2 is the correct explanation forStatement-1

77. For a regular polygon, let r and R be the radii ofthe inscribed and the circumscribed circles. Afalse statement among the following is

1) There is a regular polygon with 1

2

r

R�


3

r

R�


2

r

R�


2

r

R�

78. If B and C are the roots of the equationx2–x+1=0, then 2009 2009

B �C �

1) –1 2) 1 3) 2 4) –2

79. The number of complex number z such that|z–1| = |z + 1| = |z–i| equals

1) 1 2) 2 3) d 4) 0

AKASH MULTIMEDIA 9


80. A line AB in three-dimensional space makesangles 450 and 1200 with the positive x-axis andthe positive y-axis respectively. If AB makes anacute angle R with the positive z-axis, then Requals1) 450 2) 600 3) 750 4) 300

81. The line L given by 15

x y

b� � passes through

the point (13, 32). The line K is parallel to L

and has the equation 1.3

x y

c� � Then the

distance between L and K is

1) 17 2) 17

153)

23

174)

23

15

82. A person is to count 4500 currency notes. Letan denote the number of notes he counts in thenth minute. If a1 = a2 = .....=a10 = 150 and a10,a11,....are in A.P. with common difference –2,then the time taken by him to count all notes is1) 34 minutes 2) 125 minutes3) 135 minutes 4) 24 minutes

83. Let :f R Rm be a positive increasing function

with (3 )

lim 1.( )x

f x

f xmd

� Then (2 )

lim( )x

f x

f xmd

�

1) 2

32)

3

23) 3 4) 1

84. Let p(x) be a function defined on R such that

p'(x) = p'(1–x), for all [0,1], (0) 1x p� � and

p(1) = 41. Then

1

0

( )p x dx° equals

1) 21 2) 41 3) 42 4) 41

85. Let f : (–1, 1) m R be a differentiable functionwith f(0) = –1 and f'(0) = 1.

Let g(x) = [f(2f(x) + 2)]2. Then g'(0) =

1) –4 2) 0 3) –2 4) 4

86. There are two urns. Urn A has 3 distinct redballs and urn B has 9 distinct blue balls. Fromeach urn two balls are taken out at random andthen transferred to the other. The number ofways in which this can be done is1) 36 2) 66 3) 108 4) 3

87. Consider the system of linear equations :x1 + 2x2 + x3 = 32x1 + 3x2 +x3 = 33x1 + 5x2 + 2x3 = 1

The system has

1) exactly 3 solutions

2) a unique solution

3) no solution

4) infinite number of solutions

88. An urn contains nine balls of which three arered, four are blue and two are green. Three ballsare drawn at random without replacement fromthe urn. The probability that the three balls havedifferent colour is

1) 2

72)

1

21

3) 2

234)

1

3

89. For two data sets, each of size 5, the variancesare given to be 4 and 5 and the correspondingmeans are given to be 2 and 4, respectively.The variance of the combined data set is

1) 11

22) 6 3)

13

24)

5

2

90. The circle x2 + y2= 4x + 8y + 5 intersects the line3x – 4y = m at two distinct points if

1) –35 < m < 15 2) 15 < m < 65

3) 35 < m < 85 4) –85 < m < –35

PART-A : CHEMISTRY

1) 2 2) 3 3) 1 4) 4 5) 4

6) 3 7) 2 8) 2 9) 3 10) 3

11) 4 12) 2 13) 1 14) 2 15) 1

16) 1 17) 1 18) 3 19) 4 20) 4

21) 4 22) 2 23) 3 24) 2 25) 2

26) 3 27) 1 28) 1 29) 2 30) 2

123456789012345678901234123456789012345678901234123456789012345678901234AIEEE 2010 ANSWERS

AKASH MULTIMEDIA10


PART-B : PHYSICS

31) 2 32) 3 33) 1 34) 4 35) 2

36) 1 37) 1 38) 3 39) 2 40) 3

41) 3 42) 2 43) 3 44) 3 45) 1

46) 2 47) 1 48) 1 49) 4 50) 2

51) 3 52) 3 53) 4 54) 4 55) 4

56) 3 57) 2 58) 4 59) 1 60) 3


61) 1 62) 4 63) 4 64) 3 65) 4

66) 4 67) 2 68) 4 69) 2 70) 3

71) 3 72) 3 73) 2 74) 3 75) 3

76) 4 77) 2 78) 2 79) 1 80) 2

81) 3 82) 1 83) 4 84) 1 85) 1

86) 3 87) 3 88) 1 89) 1 90) 1

PART-A : CHEMISTRY

1. (2) Enthalpy of formation of NH3= –46 kJ/mole

2 2 3N 3H 2NH= � m fH 2 46kJmol% � � s

Bond breaking is endothermic and Bondformation is exothermic

Assuming ‘x’ is the bond energy of N–H bond(kJ mol–1)

712 3 436 6x 46 2= � s � � � s

= x = 352 kJ/mol

2. (3) For a zero order reaction x

kt

� ...... (1)

Where x = amount decomposedk = zero order rate constant for a zero orderreaction

< >0

1

2

Ak

2t�

................. (2)

Since [A0] = 2M, t

1/2 = 1 hr; k = 1

= from equation (1) 0.25

t 0.25hr1

� �

3. (1) CoCl3.6NH

3 m xCl– 3AgNO

}}}m x AgCln

n(AgCl) = xn (CoCl3.6NH

3)

4.78 2.675

143.5 267.5� s x 3= �

= The complex is [Co(NH3)

6]Cl

3

4. (4) Rate equation is to be derived with respectto slow step, from mechanism (A)

Rate = k[Cl2][H

2S]

5. (4) 5PVn 128 10

RT�

� � s moles

5

1 1

3170 10 atm 1L

0.0821L atm k mol 300K

�

� �

s s

�

s

= 1.27 × 10–3 mol

6. (3) 2-butene is symmetrical alkene3

2

O3 3 3Zn / H OCH CH H CH 2CH CHO� � � }}}}m

Molar mass of CH3CHO is 44u.

7. (2) Vant Hoff’s factor (i) for Na2SO

4 = 3

0.013 1.80 0.0558K

1� s s �

8. (2) 30 alcohols react fastest with ZnCl2/conc. HCl

due to formation of 30 carbocation and

=2– methyl propan – 2 – ol is the only30 alcohol

9. (3)

NH2

2NaNOHCl,278K}}}}m

4HBF}}}m

(A) (B)

+ N2 + BF3 + HCl

N2 Cl F

10. (3) Moles of HCl reacting with ammonia

= (moles of HCl absorbed) – (moles of NaOHsolution required)

= (20 × 0.1 × 10–3) – (15 × 0.1 × 10–3)

= moles of NH3 evolved

= moles of nitrogen in organic compound

= wt. of nitrogen in org.comp = 0.5 × 10–3 × 14

= 7 × 10–3g

% wt = 3

3

7 1023.7%

29.5 10

�

�

s

�

s

11. (4) Energy required for 1 Cl2 molecule =

3

A

242 10

N

s

joules.

This energy is contained in photon ofwavelength ‘M ’.

.. ..AIEEE 2010 HINTS AND SOLUTIONS

AKASH MULTIMEDIA 11


34 8 3

23

hc 6.626 10 3 10 242 10E

6.022 10

�

s s s s

� � �

M M s

0

4947 A 494nmM � �

12. (2) 22 2He He He

1 1IE 13.6Z where Z 2

1� � �

¨ ·

� � �© ¸

dª ¹

Hence 13.6 x 2

HeZ

� = 19.6 × 10–18 J atom–1.

2

2 2

22 2 Li

1 2 2Li Li HeHe

Z1E 13.6Z 13.6Z

1 Z

�

� � �

�

¨ ·

� � s � � s © ¸

© ¸ª ¹

= – 19.6 × 10–18 × 9

4 = - 4.41 × 10–17 J / atom

13. (1) SN1 proceeds via carbocation intermediate,

the most stable one forming the product faster.Hence reactivity order for A, B, C depends onstability of carbocation created.

Me

>Me

Me> Me

14. (2) Only option (2) is having non-superimposable mirror image & hence one opticalisomer

1)

Zn+2

NH3NH3

en

No optical isomer. It is Tetrahedral with a plane of symmetry

2) Co+3en

en

en

Co+3 en

en

en

Optical isomer

3)

H2O

H2O

Co+3

H2O

H2O

en

Horizontal plane is plane of symmetry

4)

Zn+2

en

No optical isomer. It is Tetrahedral with a plane of symmetry

15. (1) Mole fraction of Heptane =25 /100 0.25

0.4525 35 0.557

100 114

� �

�

XHeptane

= 0.45

=Mole fraction of octane = 0.55 = Xoctane

Total pressure = 0i iX P¤

= (105 × 0.45) + (45 × 0.55) kPa

= 72.0 kPa

16. (1)

conc. H2SO4

CH2-CH-CH-CH3

OH CH3

loss of proton

CH2-C-CH-CH3

CH3

CH=CH-HCCH3

CH3

(conjugated system)

Trans isomers is more stable & main product hereCH(CH3)2

C = C

H

H

(trans isomer)

17. (1) i) 3 4 2 3 2 4conjugate baseacid

H PO H H O H PO� �

� m �

ii) 2

2 4 2 4 3acid conjugate base

H PO H O HPO H O� � �

� m �

iii) 2

2 4 3 4acid conjugate base

H PO OH H PO O� � �

� m �

Only in reaction (ii) 2 4H PO� acts as ‘acid’

18. (3) 2 3 3A H CO H HCO� �

m �U K1 = 4.2 × 10–7

23 3B HCO H CO� � �

m �U

As K2 << K

1

All major [H+]total

= [H+]A and from I equilibrium,

[H+]A z 3 total

HCO H� �

¨ · ¨ ·zª ¹ ª ¹

23CO�¨ ·ª ¹ is negligible compared to 3 total

HCO or H� �¨ · ¨ ·ª ¹ ª ¹

19. (4) For an ionic substance in FCC arrangement,

2 r r edge length� �

� �

2(110 + r–) = 508r– = 144 pm

20. (4) Correct order of increasing basic strength isR – COO(–) < CH y C(–) <

2NH � < R(–)

21. (4) For isoelectronic species higher the Z/e ratio,smaller the ionic radius. Z/e values for,

AKASH MULTIMEDIA12


2 8O 0.8;

10�

� � 9

F 0.910

�

� �

11Na 1.1;

10�

� � Mg2+ = 12

1.210

�

Al 3+ = 13

1.310

�

22. (2) Ag Br AgBr� �

� ZZXYZZ

Precipitation starts when ionic product justexceeds solubility productK

sp = [Ag+][Br –]

[Br–] =

13sp 11

K 5 1010

0.05Ag

�

�

�

s

� �

¨ ·ª ¹

Precipitation just starts when 10–11 moles of KBris added to 1L of AgNO

3 solution.

Number of moles of KBr to be added = 10–11

Weight of KBr = 10–11 × 120

= 1.2 × 10–9g23. (3) g nFE% � �

WG 960000E 2.5V

nF 4 96500

� �

� � � �

s

24. (2) 2

2Mg 2OH Mg OH� �

� U

22spK Mg OH� �

¨ · ¨ ·�ª ¹ ª ¹

sp 4

2

KOH 10

Mg� �

�

¨ · � �ª ¹

¨ ·ª ¹

=pOH = 4 and pH = 10.

25. (2) packing fraction of cubic close packing andbody centered packing are 0.74 and 0.68respectively.

26.(3) H2C=HC C2H5

H

CH3

Only 3 – methyl – 1 – pentene has a chiralcarbon

27. (1) It is a test characteristic of amide linkage.Urea also has amide linkage like proteins.

28. (1) Mn2+ is more stable.

29. (2) nylon 6, 6 is a polymer of adipic acid andhexamethylene diamine

2 24 6 n

O O || ||

C CH C NH CH NH� � � � � � �

30. (2) G H T S% � % � %

at equilibrium, G% = 0

for a reaction to be spontaneous G% should benegative.

Therefore T > Te

PART-B : PHYSICS

31. (2) A moving conductor is equivalent to abattery of emf = v B l (motion emf)

Equivalent circuit l = l1 + l

2

applying kirchoff’s law

l1R + lR – vBl = 0 .................. (1)

l2R + lR – vBl = 0 .................. (2)

RR

I1 I2

adding (1) & (2)

2|R + |R = 2vBl

2vB|

3R�

l; 1 2

vB| |

3R� �

l

32. (3) 22

t / T 2 2t / T00

q1 q 1U (q e ) e

2 C 2C 2C� �

� � �

(where U =CR)

U = Ui e–2t/U

2t1/i i

1U U e

2� U

�

l2t /1

1 Te t | n2

2 2� U

� � �

Now q = 0q

4 = q

0 e–t/2T

t2 = T|n4 = 2T|n2

1

2

t 1

t 4= �

AKASH MULTIMEDIA 13


33. (1)

m1

v1

m2

v2

If it is a completely inelastic collision then

m1v

1 + m

2v

2 = m

1v + m

2v

1 1 2 2

1 2

m v m vv

m m

��

�

2 21 2

1 2

p pK.E

2m 2m� �

as 1 2p and pG G

both simultaneously cannot be zero

therefore total KE cannot be lost.

34. (4) Since the frequency of ultraviolet light isless than the frequency of X–rays, the energyof each incident photon will be more for X–rays

K.Ephotoelectron

= hO �K

Stopping potential is to stop the fastestphotoelection

0

hV

e e

O K

� �

so, K.Emax

and V0 both increases.

But K.E ranges from zero to K.Emax

because foloss of energy due to subsequent collisionsbefore getting ejected and not due to range offrrequencies in the incident light.

35. (2) oil waterS � S � S

Oil the least dense of them so it should settle atthe top with water at the base. Now the ball isdenser than oil but less denser than water. So, itwill sink through oil but will not sink in water.So it will stay at the oil-water interface.

36. (1) ˆ ˆv Kyi Kxj� �

G

dx dyKy, Kx

dt dt� � ;

dy dy dt Kx

dx dt dx Ky� s �

y dy = x dx

y2 = x2 + c

37. (1) The magnetic field in between because ofeach will be in opposite direction

Bin between

= 0 0i iˆ ˆj ( j)

2 x 2 (2d x)

N N� �

Q Q �

0i 1 1 ˆ( j)2 x 2d x

N ¨ ·

� �© ¸

Q �ª ¹

at x = d, Bin between

= 0

for x < d, Bin between

= ˆ( j)

for x > d, Bin between

= ˆ( j)�

towards x net magnetic field will add up anddirection will be ˆ( j)�

towards x’ net magnetic field will add up anddirection will be ˆ( j)

38. (3) At t = 0, inductor behaves like an infiniteresistance

So at t = 0, i = 2

V

R and at t = d , inductor behaves

like a conducting wire

1 2

eq 1 2

V(R R )Vi

R R R

�

� �

39. (2) From the graph, it is a straight line so, uniformmotion. Because of impulse direction of velocitychanges as can be seen from the slope of thegraph.

Initial velocity = 2

2= 1 m/s

Final velocity = 2

1m /s2

� � �

iP 0.4N s� �

G

1Pj 0.4N s� � �

JJG

f iJ P P 0.4 0.4� � � � �

G G G

0.8N s(J impulse)� � � �

G

J 0.8N s� �

G

40. (3) After decay, the daughter nuclei will bemore stabel hence binding energy per nucleonwill be more than that of their parent nucleus.

41. (3) Conserving the momentum

1 2

M M0 V V

2 2� �

V1 = V

2............ (1)

2 2 21 2

1 M 1 Mmc . V . .V

2 2 2 2% � � ................ (2)

2 21

Mmc V

2% � ;

22

1

2 mcV

M

%

� ; 1

2 mV c

M

%�

42. (3) In positive beta decay a proton istransformed into a neutron and a positron isemitted.

0p n e� �}}m �

no. of neutrons initially was A – Z

AKASH MULTIMEDIA14


no. of neutrons after decay (A – Z) – 3 × 2(due to alpha particles) + 2 x 1 (due to positivebeta decay)

The no. of proton will reduce by 8. [as 3 × 2(due to alpha particles) +2 (due to positive betadecay)]

Hence atomic number reduces by 8.

43. (3) Linear charge density q

r¥ ´

M �¦ µ§ ¶Q

y

R

dR

xR

2

K.dqˆ ˆE dEsin ( j) sin ( j)r

� R � � R �° °

2

K qr ˆE d sin ( j)rr

� R R �

Q

°

= 20

K q ˆsin ( j)r

Q

R �

Q

° 2 20

q ˆ( j)2 r

�

Q F

44. (1)

Truth table for given combination is

A B X

0 0 0

0 1 1

1 0 1

1 1 1

This comes out to be truth of OR gate

45. (2) The efficiency of cycle is

2

1

T1

TI � �

for adiabatic process1TV H � = constant

For diatomic gas 7

5H �

111 1 2 2T V T V H �

H �

�

1

21 2

1

VT T

V

H �

¥ ´�

¦ µ§ ¶

71

51 2T T (32)

�

�

5 2 / 52T (2 )� = 2T 4s

1 2T 4T�

1 31 0.75

4 4¥ ´

I � � � �¦ µ§ ¶

46. (1) 3 204 10 10 hfs � s

3

20 34

4 10f

10 6.023 10�

s

�

s s

f = 6.03 s 1016 Hz

The obtained frequency lies in the band ofX-rays.

47. (1)

48. (4) The given circuit is under resonance asX

L = X

C

Hence power dissipated in the circuit is

2VP 242W

R� �

49. (2) Apply shell theorem the total charge uptodistance r cab be calculated as followeddq = 24 r .dr.Q S

20

5 r4 r .dr.

4 R¨ ·

� Q S �© ¸

ª ¹

32

0

5 r4 r dr dr

4 R

¨ ·

� QS �© ¸

ª ¹

r 32

0

0

5 rdq q 4 r dr dr

4 R

¥ ´� � QS �

¦ µ§ ¶

° °

3 4

0

5 r 1 r4

4 3 R 4

¨ ·

� QS �© ¸

ª ¹

2

kqE

r�

3 4

020

1 1 5 r r.4

4 4 3 4Rr

¨ ·¥ ´

� QS �© ¸¦ µ

QF § ¶ª ¹

0

0

r 5 rE

4 3 R

S ¨ ·

� �© ¸

F ª ¹

50. (3) 12 6

a bU(x)

x x� �

U(x ) 0�d �

as, 13 7

dU 12a 6bF

dx x x¨ ·

� � � �© ¸

ª ¹

at equilibrum, F = 0 6 2a

xb

= �

2

at equilibrium 2

a b bU

2a 4a2abb

�

= � � �

¥ ´¥ ´¦ µ¦ µ § ¶§ ¶

2

at equilibrium

bD U(x ) U

4a¨ ·= � � d � �ª ¹

AKASH MULTIMEDIA 15


51. (3) From F.B.D fo sphere, using Lami’s theorem

Ftan

mg� R ............... (i)

T

F

mg

R

when suspended in liquid, asR remains same,

F’tan

mg 1d

= � R

S¥ ´�¦ µ

§ ¶

............. (ii)

using (i) and (ii)F F ’

mgmg 1

d

�

S¥ ´�¦ µ

§ ¶

where, F

F’K

�

F F

mgmg K 1

d

= �

S¥ ´�¦ µ

§ ¶

or 1

K 21

d

� �

S

�

52. (4) Let R0 be the initial resistance of both

conductors

= At temperature R their resistance will be,

1 0 1R R (1 )� � B R and 2 0 2R R (1 )� � B R

for , series combination, Rs = R1 + R

2

s0 s 0 1 0 2R (1 ) R (1 ) R (1 )� B R � � B R � � B R

where so 0 0 0R R R 2R� � �

0 s 0 0 1 22R (1 ) 2R R ( )= � B R � � R B � B

or 1 2s 2

B �B

B �

for parallel combination, 1 2p

1 2

R RR

R R�

�

0 1 0 2p0 p

0 1 0 2

R (1 )R (1 )R (1 )

R (1 ) R (1 )

�B R �B R

� B R �

�B R � � B R

where, 0 0 0p0

0 0

R R RR

R R 2� �

�

20 0 1 2 1 2

p0 1 2

R R (1 )(1 )

2 R (2 )

� B R � B R � B B R

= � B R �

� B R � B R

as 1B and 2B are small quantities

1 2= B B is negligible

or 1 2 1 2p 1 2

1 2

[1 ( ) ]2 ( ) 2

B �B B �B

B � � � B �B R

� B �B R

as 21 2( )B �B is negligible

1 2p 2

B �B

=B �

53. (4) S = t3 + 5 2ds

speed,v 3tdt

= � �

and rate of change of speed = dv

6tdt

�

=tangential acceleration at t = 2s,

at = 6s 2 = 12 m/s2

at t = 2s, v=3(2)2 = 12 m/s

= centripetal acceleration,2

2c

v 144a m / s

R 20� �

= net acceleration 2 2t ia a� �

214m / sz

54. (4) mg sinR = ma a gsin= � R

where a is along the inclined plane

= vertical component of acceleration is g sin2R

= relative vertical acceleration of A with respectto B is

g[sin2 60 –sin2 30] 2g

4.9m / s2

� �

in vertical direction.

55. (3) For a particle in uniform circular motion,2v

aR

�

G

towards centre of circle

2v ˆ ˆa ( cos i sin j)R

= � � R � R

G

ac

P(R, )R

ac

y

x

56. (2) As intensity is maximum at axis,

= µ will be maximum an speed will beminimum on the axis of the beam

= beam will converge.

AKASH MULTIMEDIA16


57. (4) For a parallel cylinderical beam, wavefrontwill be planar.

58. (1)

59. (3) L m(r v)� s

G

G G

2o 0

1ˆ ˆL m v cos t i (v sin t gt ) j2

¨ ·

� R � R � s© ¸

ª ¹

G

0 0ˆ ˆv cos i (v sin gt) j¨ ·R � R �

ª ¹

0

1 ˆmv cos t gt k2

¨ ·

� R �© ¸

ª ¹

2

0

1 ˆmgv t cos k2

� � R

60. (4) 2

22

T vk

X� N � N =0.04

2

2

(2 / 0.004)6.25N

(2 / 0.50)

Q

�

Q


61. (1) 4

cos( )5

B �C �

3tan( )

4� B�C �

5sin( )

13B �C �

5tan( )

12� B�C �

3 5564 12tan 2 tan( )

3 5 3314 12

�

B � B � C � B �C � �

�

62. (4) P : there is a rational number x S� suchthat x > 0

:P� Every rational number x S� satisfies

0x b

63. (4) c b a� s

GG G . 0b c� �

G G

1 2 3ˆ ˆˆ ˆ ˆ ˆ. 0b i b j b k i j k� � � � � �

b1 – b2 – b3 = 0

and . 3a b �

GG

� b2 – b3 = 3

b1 = b2 + b3 = 3 + 2b3 take b3 = –2

3 3 3ˆˆ ˆ(3 2 ) (3 ) .b b i b j b k� � � � �

G

64. (3) Parallel to x-axis 0dy

dx� � 3

81 0

x� � �

2x� � 3y� �

Equation of tangent is y – 3 = 0(x–2) 3 0y� � �

65. (4) cos x dy = y(sinx – y)dx

2tan secdy

y x y xdx

� �

2

1 1tan sec

dyx x

dx yy� � �

Let 1

ty�

2

1 dy dt

dx dxy� �

tan secdy

t x xdx

� � � � (tan ) secdt

x t xdx

� � �

tan. . secxdxI F e x°� �

Solution is ( . ) ( . )sect I F I F xdx� °

1sec tanx x c

y� �

66. (4)

5

4 4

04

(cos sin ) (sin cos )x x dx x x dx

Q Q

Q

� � � �° °

3

2

54

(cos sin ) 4 2 2x x

Q

Q

� � �°

cos x sin x

0

4

Q

5

4

Q 3

2

Q

Q

2Q

67. (2) The locus of perpendicular tangents isdirectrixi.e., x = –a; x = –1

68. (4) . 0,a b �

G

G

. 0,b c �G

G

. 0c a �

G G

2 4 0� M � � N � 1 2 0M � � N �

solving we get 3; 2M � � N �

69. (2) xRy need not implies yRx

:m p

S s qm pnn q

� �

m ms

n n reflexive

AKASH MULTIMEDIA 17


�m p p m

s sn q q n symmetric

p r

, s , transitive.q s

� � � � �

m ps qm pn ps rq ms rn

n q

S is an equivalence relation.

70. (3) f(x) = k–2x if 1x b � = 2x +3 if x > –1

f(x) = k-2x if x < -1 = 2x+3 if x > -1

k-2x1

-1

2x+3

Thi is truewhere k = -1

11

xLt f xm �

b �

71. (3) First row with exactly one zero; total numberof cases = 6First row 2 zeros we get more casesTotal we get more than 7.Directions : Questions Number 72 to 76 areAssertion-Reason type questions. Each of thesequestions contains two statements.Statement-1:(Assertion) and Statement-2:(Reason)Each of these questions also has four alternativechoices, only one of which is the correct answer.You have to select the correct choice.

72. (3) N(S) = 20C4

Statement-1:common difference is 1; total number of cases = 17common difference is 2; total number of cases = 14common difference is 3; total number of cases = 11common difference is 4; total number of cases = 8common difference is 5; total number of cases = 5common difference is 6; total number of cases = 2

Prob. = 204

17 14 11 8 5 2 1.

85C

� � � � �

�

73. (2) A=(3, 1, 6); B = (1, 3, 4)

Mid-point of AB = (2, 2, 5) lies on the plane.

and d.r's of AB = (2, –2, 2)

d.r's of normal to plane = (1, –1, 1)

AB is perpendicular bisector

= A is image of BStatement-2 is correct but it is not correctexplanation.

74. (3) 10

11

10!( 1)

( 1)( 2)!(10 )!�

� �

� � �

¤j

S j jj j j j

108

2

8!90 90.2

( 2)! 8 ( 2) !�

� �

� � �

¤j j j

10

21

10!

( 1)! 9 ( 1) !�

�

� � �

¤j

S jj j j

109

1

9!10 10.2 .

( 1)! 9 ( 1) !�

� �

� � �

¤j j j

< >

10

31

10!( 1)

! 10 !�

� � �

�

¤j

S j j jj j

10 1010 10 8 9

1 1

( 1) 90.2 10.2� �

� � � � �¤ ¤j jj j

j j C j C

= 90.28 + 20.28 = 110.28 = 55.29.

75. (3) Let , 0a b

A abcdc d

¥ ´� x¦ µ§ ¶

2 .a b a b

Ac d c d

¥ ´ ¥ ´�¦ µ ¦ µ§ ¶ § ¶

22

2

a bc ab bdA

ac cd bc d

¥ ´� �� ¦ µ

� �§ ¶

2 21, 1a bc bc d� � � � �

ab + bd = ac + cd = 0

0c x and 0b x 0a d� � �

Trace A = a + d = 0|A| = ad – bc = –a2 – bc = –1.

76. (4) Let :f R Rm be a continuous function

defined by 2

1( )

2 2

x

x x x

ef x

e e e�

� �

� �

2 2

22 2

2 2 .’( )

x x x x

x

e e e ef x

e �

� �

�

f'(x) = 0 2 22 2x xe e� � �

e2x = 2 2xe� �

maximum 2 1

( )4 2 2

f x � �

AKASH MULTIMEDIA18


10 ( )

2 2f x x R� b � �

Since 1 1

0 for some3 2 2

� � � �c R

1( )

3f c �

77. (2) cot2

ar

n

Q

�

'a' is side of polygon

cosec2

aR

n

Q

�

cotcos

cosec

r nR n

n

Q

Q

� �

Q

2cos

3n

Q

� for some .n N�

78. (2) x2 – x + 1 = 0 1 1 4

2x

p �

� � ; 1 3

2

ix

p

�

1 3,

2 2

iiB � �

1 3

2 2

iC � �

cos sin ,3 3

iQ Q

B � � cos sin3 3

iQ Q

C � �

2009 2009 2cos20093

Q¥ ´B �C � ¦ µ

§ ¶

2 22cos 668 2cos

3 3

Q Q¨ · ¥ ´

� Q � Q � � Q �¦ µ© ¸§ ¶ª ¹

2 12cos 2 1

3 2

Q ¥ ´� � � � �¦ µ

§ ¶

79. (1) Let z = x + iy|z–1| = |z + 1| �Re z = 0 �x = 0|z – 1| = |z –i| � x = y|z + 1| = |z – i| � y = –xOnly (0,0) will satisfy all conditions.

� Number of complex number z = 1

80. (2) 0 1

cos452

� �A

0 1cos120

2m � � � ; cosn � R

where R is the angle which line makes withpositive z-axis.

Now 2 2 2 1m n� � �A

21 1cos 1

2 4� � � R �

2 1cos

4R �

1cos

2� R � (R Being acute) .

3

Q

�R �

81. The line L given by 15

x y

b� � passes through

the point (13, 32). The line K is parallel to L

and has the equation 1.3

x y

c� � Then the

distance between L and K is

1) 17 2) 17

153)

23

174)

23

15

Sol. (3) Slope of line 5

bL � �

Slope of line 3

Kc

� �

Line L is parallel to line k.

3

5

b

c� � 15bc� �

(13, 32) is a point on L.

13 321

5 b� � �

32 8

5b� � �

20b� � �

3

4c� � �

Equation of K : y – 4x = 3

Distance between L and 52 32 3 23

17 17K

� �

� �

82. A person is to count 4500 currency notes. Letan denote the number of notes he counts in thenth minute. If a1 = a2 = .....=a10 = 150 and a10,a11,....are in A.P. with common difference –2,then the time taken by him to count all notes is

1) 34 minutes 2) 125 minutes

3) 135 minutes 4) 24 minutes

Sol. (1) Till 10th minute number of

counted notes = 1500

3000 [2 148 ( 1)( 2)] [148 1]2

� s � � � � � �

nn n n

AKASH MULTIMEDIA 19


n2 – 149n + 3000 = 0n = 125, 24n = 125, 24n = 125 is not possible.Total time = 24 + 10 = 34 minutes

83. (4) f(x) is a positive increasing function

0 ( ) (2 ) (3 )f x f x f x� � � �

(2 ) (3 )0 1

( ) ( )

f x f x

f x f x� � � �

(2 ) (3 )lim 1 lim lim

( ) ( )x x x

f x f x

f x f xmd md md

� b b

By sandwich theorem

(2 )lim 1

( )x

f x

f xmd

� �

84. (1) p'(x) = p'(1–x)

( ) (1 )p x p x c� � � � �

at x = 0

p(0) = –p(1) + c 42 c� �

now p(x) = –p(1–x) + 42� p(x) + p(1–x) = 42

1 1

0 0

( ) (1 )I p x dx p x dx� � �° °1

0

2 (42)I dx� ° 21.I� �

85. (1) ’( ) 2( (2 ( ) 2)) (2 ( ) 2)d

g x f f x f f xdx

¥ ´� � � �¦ µ

§ ¶

2f(2f(x)+2)f'(2f(x) +2).(2f'(x))

’(0) 2 (2 (0) 2). ’(2 (0) 2).g f f f f� � � �

2( ’(0) 4 (0) ’(0)f f f�

= 4(–1) (1) = –4

86. (3) Total number of ways = 3C2 × 9C2

9 83 3 36 108

2

s

� s � s �

87. (3)

1 2 1

2 3 1 0

3 5 2

D � � ; 1

3 2 1

3 3 1 0

1 5 2

D � x

� Given system, does not have any solution.

� No solution.

88. An urn contains nine balls of which three arered, four are blue and two are green. Three ballsare drawn at random without replacement fromthe urn. The probability that the three balls havedifferent colour is

1) 2

72)

1

213)

2

234)

1

3

Sol. (1) n(S) = 9C3

n(E) = 3C1 × 4C1 × 2C1

Probability = 93

3 4 2 24 3! 24 6 26! .

9! 9 8 7 7C

s s s s

� s � �

s s

89. For two data sets, each of size 5, the variancesare given to be 4 and 5 and the correspondingmeans are given to be 2 and 4, respectively.The variance of the combined data set is

1) 11

22) 6 3)

13

24)

5

2

Sol. (1) 2 4xT � ; 2 5yT �

2x � ; 4y �

2,5

¤

�

ix10, 20,¤ � ¤ �xi yi

2 2 2 21 1( ) ( ) 16

2 5x i ix x y¥ ´

T � ¤ � � ¤ �¦ µ§ ¶

2 40ix¤ � ; 2 105iy¤ �

2

2 2 21

10 2z i ix y

x y�¥ ´

T � ¤ � ¤ � ¦ µ§ ¶

1 145 90 55 11(40 105) 9

10 10 10 2

�

� � � � � �

90. (1) Circle x2 + y2– 4x – 8y – 5 = 0

Centre = (2, 4), Radius = 4 16 5 5� � �

If circle is intersecting line 3x – 4y = mat two distinct points.� length of perpendicular from centre < radius

6 165

5

m� �

� � 10 25m� � �

35 15.m��

P

These Solutions Prepared by Srichaitanya educational Institutions, Vijayawada

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