AIEEE 2010 Solution Resonance

Date : 25-04-2010 Duration : 3 Hours Max. Marks : 432

QUESTIONS & SOLUTIONS OF AIEEE 2010

IMPORTANT INSTRUCTIONS1. Immediately fill the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen.

Use of pencil is strictly prohibited.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and

fill in the particulars carefully.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are 432.

5. There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response.

Part A � CHEMISTRY (144 marks)

Question No. 4 to 9 and 13 to 30 consist of FOUR (4) marks each and Question No. 1 to 3 and 10 to 12 consist of EIGHT (8) marks

each for each correct response.

Part B � PHYSICS (144 marks)

Question No.33 to 49 and 54 to 60 consist of FOUR (4) marks each and Question No. 31 to 32 and 50 to 53 consist of EIGHT (8)

marks each for each correct response.

Part C � MATHEMATICS (144 marks)

Question No. 61 to 69, 73 to 81 and 85 to 90 consist FOUR (4) marks each and Question No. 70 to 72 and 82 to 84 consist of EIGHT

(8) marks each for each correct response.

6. Candidates will be awarded marks as stated above in Instructions No. 5 for correct response of each question. ¼ (one fourth) marks

will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is

indicated for an item in the Answer sheet.

7. Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil

is strictly prohibited.

8. No candidate is allowed to carry any textual material, printed or written, bits of papers, paper, mobile phone, any electronic device, etc.,

except the Admit Card inside the examination hall/room.

9. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each

page.

10. On completion of the test, the candiate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the

candidates are allowed to take away this Test Booklet with them.

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should immediately report the matter to the Invigilator fo replacement of both the Test Booklet and the Answer Sheet.

12. Do not fold or make any stray marks on the Answer Sheet.

Name of the Candiate (in Capital letters) : ____________________________________________________________

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Name of Examination Centre (in Capital letters) : ________________________________________

Candidate's Signature : ______________________________ Invigilator's Signature : ___________________________________

Page # 2

PART - A (CHEMISTRY)1. The standard enthalpy of formation of NH

3 is � 46.0 kJ mol�1. If the enthalpy of formation of H

2 from its atoms

is � 436 kJ mol�1 and that of N2 is � 712 kJ mol�1, the average bond enthalpy of N � H bond in NH

3 is

(1) � 964 kJ mol�1 (2) + 352 kJ mol�1 (3) + 1056 kJ mol�1 (4) � 1102 kJ mol�1

Ans. (2)

Sol. N2(g) +

23

H2(g) NH

3 (g) ; H

fº = � 46.0 kJ mol�1

2H(g) H2(g) ; H

fº = � 436 kJ mol�1

2N(g) N2(g) ; H

fº = � 712 kJ mol�1

NH3(g)

21

N2(g) +

23

H2 (g) ; H

= + 46

23

H2 3 H ; H

= + 436 ×

23

21

N2 N ; H

= + 712 ×

21

��

NH3(g) N

(g) + 3H (g) ; H

= + 1056 kJ mol�1

Average bond energy of N�H bond = 3

1056 = + 352 kJ mol�1

2. The time for half life period of a certain reaction A Products is 1 hour. When the initial concentration

of the reactant �A�, is 2.0 mol L�1 , how much time does it take for its concentration to come from 0.50 to 0.25

mol L�1. If it is a zero order reaction?

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

Ans. (3)

Sol. A product

For zero order reaction

t1/2

1na

1 a = initial concentration of reactant

t1/2 a

2

1

22/1

12/1

aa

)t()t(

50.02

)t(1

22/1

t1/2

= 25.0

= 0.25 h.

Page # 3

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

3 (molar mass = 267.5 g mol�1) is passed through a cation

exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl

(molar mass = 143.5 g mol�1). The formula of the complex is (At. mass of Ag = 108 u)

(1) [Co(NH3)

6 ] Cl

3(2) [CoCl

2 (NH

3)

4] Cl (3) [CoCl

3(NH

3)3] (4) [CoCl(NH

3)

5] Cl

2

Ans. (1)

Sol. Mole of CoCl3 . 6NH

3 =

5.267675.2

= 0.01

AgNO3 (aq) + Cl� (aq) AgCl (white)

Mole of AgCl = 5.143

78.4 = 0.03

0.01 mole of CoCl3 . 6NH

3 gives 0.03 mole of AgCl

1 mole of CoCl3 . 6NH

3 ionises to give 3 moles of Cl�.

Hence the formula of compound is [Co(NH3)

6] Cl

3

4. Consider the reaction

Cl2 (aq) + H

2S(aq) S(s) + 2H+ (aq) + 2Cl� (aq)

The rate equation for this reaction is

rate = k [Cl2][H

2S]

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

A. Cl2 + H

2S H+ + Cl� + Cl+ + HS� (slow)

Cl+ + HS� H+ + Cl� + S (fast)

B. H2S H+ + HS� (fast equilibrium)

Cl2 + HS� 2Cl� + H+ + S (slow)

(1) B only (2) Both A and B (3) Neither A nor B (4) A only

Ans. (4)

Sol. Mechanism (1) rate = K [Cl2] [H

2S]

Mechanism (2) rate = K1 [Cl

2] [HS�]

Keq

= ]SH[

]HS][H[

2

[HS�] = ]H[

]SH[K 2eq

= K1K

eq

]H[

]SH][Cl[ 22

Mechanism (1) is consistent with this rate equation.

5. If 10�4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour

phase when equilibrium is established ?

(Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R= 8.314 J K�1 mol�1)

(1) 5.56 × 10�3 mol (2) 1.53 × 10�2 mol (3) 4.46 × 10�2 mol (4) 1.27 × 10�3 mol

Ans. (4)

Page # 4

Sol. PV = nRT

V = 1 dm3 = 10�3 m3

P = 3170 Pa

R = 8.314 J K�1 mol�1

T = 300 K

3170 × 10�3 = n × 8.314 × 300

n = 300314.8

103170 3

= 1.27 × 10�3 mol.

6. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass

of 44 u. The alkene is :

(1) propane (2) 1-butene (3) 2-butene (4) ethene

Ans. (3)

Sol. CnH

2nO = 44

CnH

2n = 44 � 16

CnH

2n = 28

n = 2

CH3�CH=CH�CH

3

Zn/O3 CH3�CH=O

7. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,

the change in freezing point of water (Tf), when 0.01 mole of sodium sulphate is dissolved in 1 kg of water,

is (Kf = 1.86 K kg mol�1)

(1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K

Ans. (2)

Sol. Na2SO

4 (s)

OH2 2Na+ (aq.)+ SO4

2� (aq.)

Tf

= i Kfm

= 3 1.86 0.01

= 0.0558 K

8. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2,

is

(1) 2-Butanol (2) 2-Methylpropan-2-ol (3) 2-Methylpropanol (4) 1-Butanol

Ans. (2)

Sol. Reaction of alcohol with HCl and anhydrous ZnCl2 is an SN reaction.

3° alcohol react faster with HCl and anhydrous ZnCl2 since it forms more stable carbocation intermediate.

9. In the chemical reactions :

NH2

K278,HCl

NaNO2 A 4HBF B

the compounds �A� and �B� respectively are

(1) nitrobenzene and fluorobenzene (2) phenol and benzene

(3) benzene diazonium chloride and fluorobenzene (4) nitrobenzene and chlorobenzene

Ans. (3)

Page # 5

Sol. K278

HClNaNO2 4HBF

10. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl�s method and the

evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1

M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is :

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

Ans. (3)

Sol. Weight of organic compound = 29.5 mg

NH3 + HCl NH

4Cl

HCl (remaining) + NaOH NaCl + H2O

(1.5 m mole)

Total milimole of HCl = 2

mili mole of used by NH3 = 2 � 1.5 = 0.5

mili mole of NH3 = 0.5

weight of NH3

= 0.5 17 mg

= 8.5 mg

weight of nitrogen = 1714

8.5 mg

= 7 mg

% Nitrogen = 5.29

7 100 = 23.7 %

11. The energy required to break one mole of Cl � Cl bonds in Cl2 is 242 kJ mol�1 . The longest wavelength of light

capable of breaking a single Cl � Cl bond is

(c = 3 × 108 ms�1 and NA = 6.02 × 1023 mol�1)

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

Ans. (4)

Sol. Cl�Cl(g) 2Cl(g) H = 242 KJ mol

= 23

3

1002.6

10242

J molecule�1

E =

hc

02.61010242 323

=

834 103106.6

= 323

834

1010242

103106.6

= 242

02.636.6 10�6 = 0.494 10�6 = 494 10�9 m = 494 nm

Page # 6

12. Ionisation energy of He+ is

19.6 × 10�18 J atom�1 . The energy of the first stationary state (n = 1) of Li2+ 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

Ans. (2)

Sol. I.E. of He+ = 19.6 10�18 J atom�1

I.E. = � E1

E1 for He+ is = � 19.6 × 10�18 J atom�1

3Li1

He1

)E(

)E( = 2

Li

2He

)Z(

)Z(

2

2LI1

18

)E(106.19

= 94

E1(Li2+) =

41096.19 18

= � 44.1 10�18 = � 4.41 10�17 J atom�1

13. Consider the following bromides :

The correct, order of SN1 reactivity is

(1) B > C > A (2) B > A > C (3) C > B > A (4) A > B > C

Ans. (1)

Sol. Rate of SN1 reaction stability of carbocation

14. Which one of the following has an optical isomer ?

(1) [Zn(en)(NH3)2]2+ (2) [Co(en)

3]3+ (3) [Co(H

2O)

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

2]2+

(en = ethylenediamine)

Ans. (2)

Sol. Complex [Co(en)3]3+ lacks plane of symmertry and thus is optically active having following to enantiomeric

forms.

Page # 7

15. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid

components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution

obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane - 100 g 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

Ans. (1)

Sol. PT = X

Heptane

tanHepP + XOctane

etanOcP

= 557.025.0

105 + 557.0307.0

45

47.127 + 24.80 = 71.92 72 kPa

16. The main product of the following reaction is :

C6H

5CH

2CH(OH)CH(CH

3)2 42SOH.conc

(1) (2)

(3) (4)

Ans. (1)

Sol. 42SOH.Conc

shiftH

/H

+

17. Three reactions involving H2PO

4� are given below :

(i) H3PO

4 + H

2O H

3O+ + H

2PO

4�

(ii) H2PO

4� + H

2O HPO

42� + H

3O+

(iii) H2PO

4� + OH� 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

Ans. (1)

Sol. In IInd equation H2PO

4� give H+ ion to the H

2O therefore in the IInd equation it act as an acid.

18. In aqueous solution the ionization constants for carbonic acid are

K1 = 4.2 × 10�7 and K

2 = 4.8 × 10�11

Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

(1) The concentration of CO3

2� is 0.034 M.

(2) The concentration of CO3

2� is greater than that of HCO3�.

(3) The concentration of H+ and HCO3� are approximately equal.

(4) The concentration of H+ is double that of CO3

2�.

Ans. (3)

Page # 8

Sol. H2CO

3 H+ + HCO

3� K

1 = 4.2 × 10�7

HCO3� H+ + CO

32� K

2 = 4.8 × 10�11

K1 >> K

2

[H+] = [HCO3�]

K2 =

]HCO[

]CO][H[

3

23

but [H+] = [HCO3�]

[CO3

2�] = K2 = 4.8 × 10�11

19. The edge length of a face centred cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110

pm, the radius of the anion is

(1) 288 pm (2) 398 pm (3) 618 pm (4) 144 pm

Ans. (4)

Sol.

2 × 110 + 2 × r� = 508

2r� = 288

r� = 144 pm

20. The correct order of increasing basicity of the given conjugate bases (R = CH3) is :

(1) 2HNRCHCORCO (2) 2HNORCOCHCR

(3) RCHC2HNORCO (4) R2HNCHCORCO

Ans. (4)

Sol. Basicity ativityElectroneg1

(In period)

If lone pair of electron takes part in conjugation then avialibality of lone pair of electron decrease and basic

strength decrease.

21. The correct sequence which shows decreasing order of the ionic radii of the elements 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+

Ans. (4)

Sol. O2�, F�, Na+, Mg and Al3+ have same number of electrons (i.e. 10 electrons) but different nuclear charges and,

therefore, they are isoelectronic species.

For isoelectronic species ionic radii eargchnuclear1

.

So, correct order is 8O2� >

9F� >

11Na+ >

12Mg2+ >

13Al3+

22. Solubility product of silver bromide is 5.0 × 10�13 . This quantity of potassium bromide (molar mass taken as

120 g mol�1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :

(1) 1.2 × 10�10 g (2) 1.2 × 10�9 g (3) 6.2 × 10�5 g (4) 5.0 × 10�8 g

Ans. (2)

Page # 9

Sol. Ksp

= [Ag+] [Br�] = 5.0 × 10�13

[Ag+] = 0.05 M

[0.05] [Br�] = 5.0 × 10�13

[Br�] = 05.0100.5 13

= 1 × 10�111 M

moles of KBr = M × V

= 1 × 10�11 × 1

= 1 × 10�11

weight of KBr = 1 × 10�11 × 120

= 1.2 × 10�9 g

23. The Gibbs energy for the decomposition of Al2O

3 at 500ºC is as follows :

32

Al2O

3

34

Al + O2,

rG = + 966 mol�1 . The potential difference needed for electrolytic reduction of AlAl

2O

3

at 500ºC is at least :

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

Ans. (3)

Sol.32

Al2 O

3

34

Al + O2

rG = +966 kJ mol�1 = 966 × 103 J mol�1

G = � nFEcell

966 × 103 = � 4 × 96500 × Ecell

Ecell

= 2.5 V

24. At 25 °C, the solubility product of Mg(OH)2 is 1.0 10�11. At Which pH, will Mg2+ ions start precipitating in the

form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?

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

Ans. (2)

Sol. KSP

= 1.0 × 10�11 = (Mg+2) (OH�)2

1.0 × 10�11 = (0.001) (OH�)2

(OH�) = 10�4

POH = 4

PH = 14 � 4 = 10.

25. Percentages of free space in cubic close packed structure and in body centered packed structure are

respectively.

(1) 30% and 26% (2) 26% and 32% (3) 32% and 48% (4) 48% and 26%

Ans. (2)

Sol. Packing fraction of CCP = 23

= 0.74 74%

Percentage of free space in CCP = 100 � 74 = 26%

Packing fraction of BCC = 8

3 = 0.68 68%

Percentage of free space in BCC = 100 � 68 = 32%.

Page # 10

26. Out of the following, the alkene that exhibits optical isomerism is.

(1) 3-methyl-2pentene (2) 4-methyl-1-pentene (3) 3-methyl-1-pentene (4) 2-methyl-2-pentene

Ans. (3)

Sol.

It is optical active since it has chiral carbon atom.

27. Biuret test is not given by

(1) carbohydrates (2) polypeptides (3) urea (4) proteins

Ans. (1)

Sol. Biuret test is characteristic of compound containing �CONH� functional group.

28. The correct order of M/M2E 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

Ans. (1)

Sol. Generally across the first transition series, the negative values for standard electrode potential decrease

(exception Mn- due to stable d5 configuration)

Standard electrode potential �

Mn Cr Fe Co

M2+/M �1.18 �0.90 �0.44 �0.28 E/V

So, correct order is Mn > Cr > Fe > Co.

29. The polymer containing strong intermolecular forces e.g. hydrogen bonding is

(1) teflon (2) nylon 6,6 (3) polystyrene (4) natural rubber

Ans. (2)

Sol. Nylon 6,6 has group which forms intermolecular H-bonding.

30. For a particular reversible reaction at temperature T, H and S were found to be both +ve. If Te is the

temperature at equilibrium, the reaction would be spontaneous when.

(1) Te > T (2) T > T

e(3) T

e is 5 times T (4) T = T

e

Ans. (2)

Sol. G = H � TS

For spontaneous reaction G must be negative

At equilibrium temperature

G = 0

to maintain the negative value of G

T should be greater than Te

Page # 11

PART - B (PHYSICS)31. A rectangular loop has a sliding connector PQ of length and resistance R and it is moving with a speed

v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three

currents 1, 2 and are :

(1) 1 = � 2 = R

vB, =

RvB2

(2) 1 = 2 = R3vB

, = R3

vB2

(3) 1 = 2 = R

vB(4) 1 = 2 =

R6vB

, = R3vB

Ans. (2)

Sol. Current = R2/R

vB

= R3

vB2

1 = 2 = R3

vB

32. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the

energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to

reduce to one-fourth its initial value. Then the ratio t1/t2 will be

(1) 1 (2) 21

(3) 41

(4) 2

Ans. (3)

Sol. U0 = C2

q20 U =

C2

eq /t220

1 ô

= 2

U0 = C4

q20

21

e /t2 1 ô

t1 = 2ô

n2 ....(1)

and q = q0ô/t2e ;

4

q0 = q0 ô/t2e ,41

e /t2 ô

t2 = 2 ln 2 ....(2)

41

tt

2

1

Page # 12

Direction : Question number 33-34 contain statement-1 and statement-2. Of the four choices given after the state-

ments, choice the one that best describes the two statements.

33. Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely

inelastic collision.

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

(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

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

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

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

Ans. (1)

Sol. If initial momentum of particles is zero, then they loss all their energy in inelastic collision but here initial

momentum is not zero.

Principle of conservation of momentum holds good for all collision.

34. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum

kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by X-rays, both V0 and

Kmax increase.

Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of

the range of frequencies present in the incident light.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

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

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


Ans. (4)

Sol. Energy of X-rays-photon is greater then ultraviolet photon.

So, V0 and Kmax increases.

Electrons have speed ranging from 0 to maximum, because before emitting a large number of collisions take

place and energy is lost in collision.

35. A ball is made of a material of density where oil < < water with oil and water representing the densities of

oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of

this oil and water, which of the following pictures represents its equilibrium position?

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

Ans. (2)

Sol. For equilibrium, weight should be balanced by buoyant force.

density of oil < density of water

and ball should be in between oil and water.

Page # 13

36. A particle is moving with velocity )j�xi�y(Kv

, where K is a constant. The general equation for its path is:

(1) y = x2 + constant (2) y2 = x + constant

(3) xy = constant (4) y2 = x2 + constant

Ans. (4)

Sol. )j�xi�y(Kdtrd

dtdx

= y ,dtdy

= x

So, yx

dxdy

dxxdyy

C2x

2y 22

y2 = x2 + constant

37. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane

of the paper as shown. The variation of magnetic field B along the line XX´ is given by

(1) (2)

(3) (4)

Ans. (1)

Sol.

Towards left of both wires direction of B is downward and at mid point between two wires, magnetic field is

zero

Page # 14

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

(1) 22

21

21

RR

RVR

at t = 0 and

2RV

at t = (2) 2R

V at t = 0 and

21

21

RR)RR(V

at t =

(3) 2R

V at t = 0 and

22

21

21

RR

RVR

at t = (4)

21

21

RR)RR(V

at t = 0 and 2R

V at t =

Ans. (2)

Sol. At t = 0, current does not flow through inductor.

i = 2R

V

At t = Req = 21

21

RRRR

i = 21

21

RR)RR(V

39. The figure shows the position - time (x � t) graph of one-dimensional motion of a body of mass 0.4 kg. The

magnitude of each impulse is

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

Ans. (2)

Sol. V1 = 122

dtdx

1

V2 = 1dtdx

2

Impulse = |P| = |m(V2 � V1)| = |0.4 (�1 �1)| = 0.8 Ns

Page # 15

Directions : Question number 40 � 41 are based on the following paragraph.

The nucleus of mass M + m is at rest and decays into two daughter nuclei of equal mass 2M

each.. Speed

of light is c.

40. This binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then :

(1) E1 = 2E1 (2) E1 > E2 (3) E2 > E1 (4) E1 = 2E2

Ans. (3)

Sol. Energy is released

(B.E.)product > (B.E.)Reactant

41. The speed of daughter nuclei is

(1) c mM

m

(2) c

Mm2

(3) c Mm

(4) cmM

m

Ans. (2)

Sol. Q = m c2 = 21

×

2M

v2 + 21

×

2M

v2

mc2 = 21

× Mv2

v = cM

m2

42. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 -particles and 2 positrons. The

ratio of number of neutrons to that of protons in the final nucleus will be

(1) 4Z

8ZA

(2)

8Z4ZA

(3)

4Z12ZA

(4)

2Z4ZA

Ans. (2)

Sol.

2e2YHe3X 01

12A8Z

42

AZ

number of proton = Z � 8

number of neutron = (A � 12) � (Z � 8) = A � Z � 4

ratio is 8Z

4ZA

Page # 16

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

at the

centre O is :

(1) j�r4

q2

02åð

(2) � j�r4

q2

02åð

(3) � j�r2

q2

02åð

(4) j�r2

q2

02åð

Ans. (3)

Sol. )j�(r

k2E

ë

)j�(r2

E0

ðå

ë

rqð

ë )j�(r2

qE

20

2

åð

44. The combination of gates shown below yields

(1) OR gate (2) Not gate (3) XOR gate (4) NAND gate

Ans. (1)

Sol.

Truth Table

A B A B BA X = BA 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 1 1 0 0 0 1

A B X 0 0 0 0 1 1 1 0 1 1 1 1

The second table is of OR Gate.

Page # 17

45. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion

part of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is :

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

Ans. (2)

Sol. TV � 1 = constant

T1 1

57

V

= T2

157

)V32(

1

2

TT

= 5/2)32(

1 =

41

= 1 � 1

2

TT

= 1 � 43

41

46. If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of the spectrum

called :

(1) X-rays (2) ultraviolet rays (3) microwaves (4) -rays

Ans. (1)

Sol. Energy of each photon = 2010

4000 = 4 × 1017

= 17

19

104

106.112400

AA0 = 49.6 Å

It is in X-ray spectrum.

47. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10�3 are

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

Ans. (1)

Sol. Rule : I. We know all non zero digits are significant.

Rule : II. If zero is between two non-zero digits this is also significant.

Rule : III. If zero left to the non-zero digit they are non-significant.

Significant figures for number 23.023 is 5. Using I & II.

Significant figures for number 0.0003 is 1. Using I, II & III.

Significant figures for number 2.1 × 10�3 is 2. Using I.

48. In a series LCR circuit R = 200 and the voltage and the frequency of the main supply is 220 V and 50 Hz

respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On

taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR

circuit is

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

Ans. (4)

Page # 18

Sol. tan 30º = RXL

XL = 3

R =

3

200

tan 30º = R

XC Xc =

3

200

Z = 2CL )XX(R = 200

irms = 200220

= 1.1

P = (irms )2 × R = (1.1)2 × 200

P = 242 W

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

Rr

45

)r( 0ññ

upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r <

R) from the origin is given by

(1)

R

r

3

5

3

r4

0

0

å

ðñ(2)

R

r

3

5

4

r

0

0

å

ñ(3)

R

r

4

5

3

r4

0

0

å

ñ(4)

R

r

4

5

3

r

0

0

å

ñ

Ans. (2)

Sol. Consider a spherical shell of radius x and thickness dx.

Charge on it dq

dq = × 4x2 . dx

dq = 0

R

x

4

5 x 4x2 dx

q = 40

r

0

32

dxRx

4x5

q = 40

R4r

43r5 43

E = 2r

kq = 2r4

1

× 40

R4r

43r5 43

E = 0

0

4

r

R

r

3

5

Page # 19

50. The potential energy function for the force between two atoms in a diatomic molecule is approximately given

by U(x) = 612 x

b

x

a , where a and b are constants and x is the distance between the atoms. If the dissocia-

tion energy of the molecule is D = [U (x = ) � Uat equilibrium], D is

(1) a2

b2

(2) a12

b2

(3) a4

b2

(4) a6

b2

Ans. (3)

Sol. U = 12X

a � 12X

a

At equilibrium dXdU

= 13x

a12 + 7x

b6 = 0

x =

6/1

ba2

Uat equilibrium = a

2

a2b

� b

a2b

= a2

b

b

2b

Uat r = a4b2

Uat = 0

D = a4

b2

51. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º

with each other. When suspended in a liquid of density 0.8 g cm�3, the angle remains the same. If density of

the material of the sphere is 1.6 g cm�3, the dielectric constant of the liquid is

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

Ans. (3)

Sol. At equilibrium

tan /2 = mg

Fe = 2

2

0 )]2/sin([

q4

1

èð . mg

1

When suspended in liquid

Page # 20

tan 2

= 2

2

0 )]2/sin([

qK4

1

èð )F�mg(1

B

= 2

2

0 )]2/sin([

qK4

1

èð .

)g8.06.1

m�mg(

1

on comparing the two equation we get

K

6.18.0

�1 = 1 K = 2.

52. Two conductors have the same resistance at 0ºC but their temperature coefficients of resistance are 1 and

2. The respective temperature coefficients of their series and parallel combinations are nearly

(1) 2

21 áá , 1 + 2 (2) 1 + 2 , 2

21 áá (3) 1 + 2 ,

21

21

áá

áá

(4)

221 áá

, 2

21 áá

Ans. (4)

Sol. Let R be their individual resistance at 0ºC. Their resistance at any other temperature t is

R1 = R (1 + 1 t) and R2 = R (1 + 2 t).

In series

Rseries = R1 + R2 = R [2 +(1 + 2) .t]

= 2R

t

21 21

.

Series = 2

21

In Parallel

RParallel = 21

21

RRRR

= )t)2(R

)t1(R)t1(R

21

21

)t2

1(R2

)t)1(R

21

212

2R

t

21 21

Parallel = 2

21 .

Page # 21

53. A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of �P�

is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path

is 20 m. The acceleration of �P� when t = 2 s is nearly.

(1) 13 m/s2 (2) 12 m/s2 (3) 7.2 m/s2 (4) 14 m/s2

Ans. (4)

Sol. S = t3 + 5

Linear speed of the particle

= dtdS

= 3 t2

at t = 2 s

v = (3 × 22) m/s

= 12 m/s

Linear acceleration

a1 = dtd

= 6 t

at t = 2 s,

a1 = 12 m/s2

The centripetal acceleration

a2 = R

2

= 20

122 m/s2

= 7.2 m/s2

anet = 22

21 aa = 22 2.712 = 14 m/s2

Page # 22

54. Two fixed frictionless inclined planes making an angle 30º and 60º with the vertical are shown in the figure.

Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect

to B?

(1) 4.9 ms�2 in horizontal direction (2) 9.8 ms�2 in vertical direction

(3) Zero (4) 4.9 ms�2 in vertical direction

Ans. (4)

Sol. Vertical component of acceleration of A

a1 = (g sin ). sin

= g sin 60º . sin 60º = g . 43

That for B

a2 = g sin 30º . sin 30º = g 41

(aAB)= 4g3

� 4g

= 2g

= 4.9 m/s2

55. For a particle in uniform circular motion, the acceleration a

at a point P (R, ) on the circle of radius R is

(Here is measured from the x-axis)

(1) � Rv2

cos i� + Rv2

sin j� (2) � Rv2

sin i� + Rv2

cos j�

(3) � Rv2

cos i� � Rv2

sin j� (4) Rv2

i� + Rv2

j�

Ans. (3)

Page # 23

Sol.

aC = �RV2

cos i� �RV2

sin j�

Direction : Questions number 56 � 58 are based on the following paragraph.

An initially parallel cylindrical beam travels in a medium of refractive index () = 0 + 2, where 0 and 2 are

positive constants and is the intensity of the light beam. The intensity of the beam is decreasing with

increasing radius.

56. As the beam enters the medium, it will

(1) diverge

(2) converge

(3) diverge near the axis and converge near the periphery

(4) travel as a cylindrical beam

Ans (4)

Sol.medium

Beam is incident normally so does not diverge or converge so it travel as a cylindrical beam.

57. The initial shape of the wavelength of the beam is :

(1) convex

(2) concave

(3) convex near the axis and concave near the periphery

(4) planar

Ans. (4)

Sol. Beam does not converge or diverge so shape of wave front remain planar.

58. The speed of light in the medium is

(1) minimum on the axis of the beam (2) the same everywhere in the beam

(3) directly proportional to the intensity (4) maximum on the axis of the beam

Ans. (1)

Page # 24

Sol. = 0 + 2 ()

r

as it is given that intensity of beam is decreasing with increasing radius, and as decreases also de-

creases.

Now by V =

C

speed of light is minimum at axis of cylinder.

59. A small particle of mass m is projected at an angle with the x-axis with an initial velocity v0 in the x-y plane

as shown in the figure. At a time t < g

sinv0 è, the angular momentum of the particle is

(1) �mg v0 t2 cos j� (2) mg v0 t cos k�

(3) � 21

mg v0 t2 cos k� (4)

21

mg v0 t2 cos i�

where j�,i� and k� are unit vectors along x, y and z-axis respectively..

Ans. (3)

Page # 25

Sol.

V0

Angular momentum = m

2

0000 gt21

tsinV)cosV()tcosV)(gtsinv(

= � 21

mg V0 t2 cos 0 k�

60. The equation of a wave on a string of linear mass density 0.04 kg m�1 is given by y = 0.02 (m) sin

)m(50.0x

)s(04.0t

2ð . The tension in the string is :

(1) 4.0 N (2) 12.5 N (3) 0.5 N (4) 6.25 N

Ans. (4)

Sol. By equation

f = 04.01

and = 0.5

V = 04.01

× 0.5 = 2

25

by V =

T

2

225

=

04.0

T = 4

625 × 0.04

T = 6.25 N

Page # 26

PART - C (MATHEMATICS)

61. Let cos( + ) = 54

and let sin( � ) = 135

, where 0 , 4

. Then tan 2 =

(1) 3356

(2) 1219

(3) 720

(4) 1625

Ans. (1)

Sol. tan 2 = tan (( + ) + ( � )) = )tan()tan(1)tan()tan(

=

125

.43

1

125

43

= 1548

4)59(

=

33414

= 3356

Hence correct option is (1)

62. Let S be a non-empty subset of R. Consider the following statement :

P : There is a rational number x S such that x > 0.

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

(1) There is no rational number x S such that x 0.

(2) Every rational number x S satisfies x 0.

(3) x S and x 0 x is not rational.

(4) There is a rational number x S such that x 0.

Ans. (2)

Sol. P : At least one rational number x S such that x > 0

Negation : all rational numbers x S are x 0

63. Let k��j�a

and k��j��i�c

. Then the vector b

satisfying 0cba

and 3b.a

is

(1) k�2j��i�2 (2) k�2�j��i� (3) k�2�j�i� (4) k�2�j�i��

Ans. (4)

Sol. Since 0cba

0ca)ba(a

0cab)a.a(a)b.a(

Since ca

= k�j�i�2

0k�j�i�2b2)k�j�(3

k�2j�i�b


Page # 27

64. The equation of the tangent to the curve y = x + 2x

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

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

Ans. (3)

Sol. y = x + 2x

4

y = 1 � 3x

8 = 0 x3 = 8 x = 2 y = 2 + 22

4 = 3

(2, 3) is point of contact . Thus y = 3 is tangent . Hence correct option is (3)

65. Solution of the differential equation cosx dy = y(sinx � y) dx, 0 < x < 2

is

(1) y sec x = tan x + c (2) y tan x = sec x + c (3) tanx = (sec x + c)y (4) secx = (tanx + c) y

Ans. (4)

Sol. cos x dy � y sin x dx = � y2 dx

cos x dy + y d(cos x) = � y2 dx

xcosy

)xcosy(d22 = �

xcos

dx2

� xcosy1

= � tan x + c

� sec x = y (� tan x + c)

sec x = y(tan x + k)


66. The area bounded by the curves y = cos x and y = sinx between the ordinates x = 0 and x = 2

3is

(1) 224 (2) 1�24 (3) 124 (4) 2�24

Ans. (4)

Sol. Required area =

4/

0

dx)xsinx(cos +

4/5

4/

dx)xcosx(sin +

2/3

4/5

dx)xsinx(cos

= 2 4/0xcosxsin

+ 4/54/xsin�xcos

= 2�24


Page # 28

67. If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is

(1) 2x + 1 = 0 (2) x = � 1 (3) 2x � 1 = 0 (4) x = 1

Ans. (2)

Sol. Locus of P will be directrix of parabola which is x = � 1


68. If the vectors k�2j��i�a

, k�j�4i�2b

and k�µj�i�c

are mutually orthogonal, then ( , µ) =

(1) (2, � 3) (2) (�2, 3) (3) (3, � 2) (4) (�3, 2)

Ans. (4)

Sol. c,b,a

are mutually orthogonal

0c.a

� 1 + 2 = 0 ........(i)

and

0c.b

+ 4 + = 0 ........(ii)

solving (i) and (ii), we get = � 3 and = 2


69. Consider the following relations :

R : {(x, y)| x ,y are real numbers and x = wy for some rational number w}

S =

qp

,nm

{ | m, n, p and q are integers such that n, q 0 and qm = pn}

Then

(1) neither R nor S is an equivalence relation

(2) S is an equivalence relation but R is not an equivalence relation

(3) R and S both are equivalence relations

(4) R is an equivalence relation but S is not an equivalence relation

Ans. (2)

Sol. (x, x) R for w = 1

R is reflexive

If x 0, then (0, x) R for w = 0 but (x, 0) R for any w

R is not symmetric

R is not equivalence relation

n

m,

n

m S as mn = mn

S is reflexive

q

p,

n

m S qm = pn np = mq

n

m,

q

p S

S is symmetric

q

p,

n

m S and

b

a,

q

p S

Page # 29

qm = pn and bp = aq

nm

= qp

and qp

= ba

nm

= ba

ba

,nm

S

S is transitive

S is equivalence


70. Let f : R R be defined by

f(x) =

1�x

1�x

if

if

,3x2

,x2�k

If f has a local minimum at x = � 1, then a possible value of k is

(1) 0 (2) 21

� (3) �1 (4) 1

Ans. (3)

Sol. 1xlim f(x) = 1

f(�1) = k + 2

)1(xlim f(x) = k + 2

f has a local minimum at x = � 1

f(�1+) f(�1) f(�1�)

1 k + 2 k + 2

k � 1

possible value of k is � 1


71. The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is

(1) 5 (2) 6 (3) at least 7 (4) less than 4

Ans. (3)

Sol. Let A =

321

321

321

ccc

bbb

aaa

det (A) = a1 (b

2c

3 � c

2b

3) � a

2 (b

1c

3 � c

1b

3) + a

3 (b

1c

2 � c

1b

2)

= a1b

2c

3 � a

1c

2b

3 + a

2c

1b

3 � a

2 b

1c

3 + a

3b

1c

2 � a

3 c

1b

2

if any of the terms is non-zero, then det (A) will be non-zero and all the element of that term will be unity

Now there are 6 elements remaining out of which any one can be unity.

Hence number of non-singular matrices =

tripletoneanygsinchoo

16 C

×

elementoneanygsinchoo

16 C


Page # 30

Directions : Question number 72 to 76 are Assertion-Reason type questions. Each of these questions

contains two statements.

Statement - 1 : (Assertion) and

Statement - 2 : (Reason).

Each of these questions also has four alternative choices, only one of which is the correct answer. You have

to select the correct choice.

72. Four numbers are chosen at random (without replacement) from the set {1,2,3,.....,20}.

Statement -1 : The probability that the chosen numbers when arranged in some order will form an AP is

851

.

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

difference is {±1, ±2, ±3, ±4, ±5}

(1) Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement -1.


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

(4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.

Ans. (2)

Sol. Statement-1 Total ways = 20C4

number of AP's of common difference 1 is = 17






_______

total = 57

probability = 4

20 C

57 =

851

Statement-2 common difference can be ± 6 , so statement -2 is false


73. Statement -1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x � y + z = 5.

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





Ans. (1)

Sol. Let image be ()

11

= 13

=

14

= � 2

35431

1

1 =

13

=

14

= 2

Page # 31

= 3, = 1, = 6

A(3, 1, 6) statement 1 is true

Now midpoint of A(3, 1, 6) and B(1, 3, 4) is (2, 2, 5)

equation of plane is x � y + z = 5

coordinates of midpoint lies on the plane so plane bisects the line segment AB. But it is not correct explana-

tion of statement-1


74. Let S1 =

10

1j

)1�j(j 10Cj , S

2 =

10

1j

j 10Cj and S

3 =

10

1j

2j 10Cj.

Statement -1 : S3 = 55 × 29 .

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





Ans. (2)

Sol. S1 =

10

1j)1j(j

)110(10.)1j(j 8C

j�2

S1 = 9 × 10

10

2j2j

8C

S1 = 90 . 28

S2 = j

910

1j

Cj

10.j

�1

= 10.29

S3 =

10

1jj

10C)j)1j(j( =

10

1j

10

1jj

10j

10 CjC)1j(j = 90

10

1j1j

910

2j2j

8 C10C

= 90 × 28 + 10 × 29 = (45 + 10) . 29 = (45 + 10) . 29 = 55.29

so statement-1 is true and statement 2 is false.


75. Let A be a 2 × 2 matrix with non-zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define

Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A.

Statement -1 : Tr(A) = 0

Statement -2 : |A| = 1





Ans. (2)

Page # 32

Sol. Let A =

dc

ba

A2 =

dc

ba

dc

ba =

2

2

dbc)da(c

)da(bbca =

10

01

a + d = 0 and a2 + bc = 1

Tr(A) = 0

Statement-1 is true

Statement-2 |A| = ad � bc = � a2 � bc = � 1

Statement-1 is true statement-2 is false.


76. Let f : R R be a continuous function defined by

f(x) = x�x e2e

1

.

Statement -1 : f(c) = 31

, for some c R.

Statement -2 : 0 < f(x) 22

1, for all x R.





Ans. (4)

Sol. ex + 2e�x 2 2 (AM GM)

xx e2e

1

22

1

22

1 f(x) > 0 so statement- 2 is correct

so f(c) = 31

for some c. because f(x) is continuous [since f(0) = 31

]


77. For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement

among the following is

(1) There is a regular polygon with 2

1Rr . (2) There is a regular polygon with

32

Rr .

(3) There is a regular polygon with 23

Rr . (4) There is a regular polygon with

21

Rr .

Ans. (2)

Page # 33

Sol.Rr

= cos

n

Let cos n

= 32

for some n 3, n N

As 21

32

2

1 cos

3

cos n

cos 4

3

n

4

3 n < 4, which is not possibleso option (2) is the false statementso it will be the right choice

78. If and are the roots of the equation x2 � x + 1 = 0, then 2009 + 2009 =

(1) � 1 (2) 1 (3) 2 (4) �2

Ans. (2)

x2 � x + 1 = 0

x = � , �2

2009 + 2009 = �2009 � 4018 = � 2 � = 1


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

(1) 1 (2) 2 (3) (4) 0

Ans. (1)

Sol. |z � 1|2 = |z + 1|2 x = 0

|z � 1|2 = |z � i|2

(x � 1)2 + y2 = x2 + (y � 1)2 1 + y2 = (y � 1)2 ( x = 0)

y = 0 (0, 0) satisfies


80. A line AB in three-dimensional space makes angles 45º and 120º with the positive x-axis and the positive y-

axis respectively. If AB makes an acute angle with the positive z-axis, then equal

(1) 45º (2) 60º (3) 75º (4) 30º

Ans. (2)

Sol. = 2

1, m = �

21

2 + m2 + n2 = 1

n2 = 41

n = ± 21

cos = 21

, = 60º


Page # 34

81. The line L given by by

5x = 1 passes through the point (13, 32). The line K is parallel to L and has the

equation 3y

cx = 1. Then the distance between L and K is

(1) 17 (2) 15

17(3)

17

23(4)

15

23

Ans. (3)

Sol.5x

+ by

= 1

513

+ b

32 = 1

b32

= � 58

b = � 20

5x

� 20y

= 1 4x � y = 20

Line K has same slope � c3

= 4

c = � 43

4x � y = � 3

distance = 17

23


82. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If

a1 = a

2 = .....= a

10 = 150 and a

10, a

11,....are in an AP 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

Ans. (1)

Sol. a1 + a

2 + .... + a

n = 4500 notes

As a1 + a

2 + .... + a

10 = 150 × 10 = 1500 notes

a11

+ a12

+ ..... + an = 3000

148 + 146 ...... = 3000

2

)10n( [2 × 148 + (n � 10 � 1) (�2)] = 3000

n = 34, 135

a34

= 148 + (34 � 1) (�2) = 148 � 66 = 82

a135

= 148 + (135 � 1) (�2) = 148 � 268 = � 120 < 0

so answer in 34 minutes is taken


Page # 35

83. Let f : R R be a positive increasing function with )x(f)x3(f

limx

= 1. Then )x(f)x2(f

limx

.

(1) 32

(2) 23

(3) 3 (4) 1

Ans. (4)

Sol. 0xlim

)x(f)x3(f

= 1

f(x) < f(2x) < f(3x) Divide by f(x)

)x(f)x3(f

)x(f)x2(f

1

using sandwitch theorem x

lim )x(f)x2(f

= 1


84. Let p(x) be a function defined on R such that p(x) = p(1 � x), for all x [0, 1], p(0) = 1 and p(1) = 41. Then

dx)x(p1

0 equals

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

Ans. (1)

Sol. p(x) = p(1 � x)

p(x) = �p(1 � x) + c

put x = 0

p(0) = � p(1) + c

c = 42

= 1

0

dx)x(p ; =

1

0

dx)x1(p

2 =

1

0

dx))x1(p)x(p( = 1

0

1

0

dx42dxc

2 = 42 = 21


85. Let f : (�1, 1) R be a differentiable function with 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

Ans. (1)

Sol. g(x) = 2f(2f(x) + 2) . f (2f(x) + 2) . 2f(x)

g(0) = 2f (2f(0) + 2) f (2f(0) + 2) . 2f(0)

= 2f(0) f(0) 2f(0) = (2) (�1) (1) (2) (1) = � 4


Page # 36

86. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls

are taken out at random and then transferred to the other. The number of ways in which this can be done is

(1) 36 (2) 66 (3) 108 (4) 3

Ans. (3)

Sol. 3C2 × 9C

2 = 3 ×

1289

= 12 × 9 = 108


87. Consider the system of linear equations :

x1 + 2x

2 + x

3 = 3

2x1 + 3x

2 + x

3 = 3

3x1 + 5x

2 + 2x

3 = 1

The system has

(1) exactly 3 solutions (2) a unique solution (3) no solution (4) infinite number of solutions

Ans. (3)

Sol. Equation (2) � equation (1) x1 + x

2 = 0

(3) � 2(1) x1 + x

2 = � 5

No solution


88. An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at

random without replacement from the urn. The probability that the three balls have different colours is

(1) 72

(2) 211

(3) 232

(4) 31

Ans. (1)

Sol. = 3

91

21

41

3

C

CCC =

1.2.37.8.92.4.3

= 72


89. For two data sets, each of size 5, the variance are given to be 4 and 5 and the corresponding means are given

to be 2 and 4, respectively. The variance of the combined data set is

(1) 211

(2) 6 (3) 2

13(4)

25

Ans. (1)

Page # 37

Sol. x2 = 4

nx2

i �

2

i

n

x

= 4

5x 2

i � (2)2 = 4 x

i2 = 40

similarly yi2 = 105

2 = 10

yx 2i

2i

�

2ii

10

yx

=

10145

�

2

102010

= 5.5


90. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x � 4y = m at two distinct points if

(1) � 35 < m < 15 (2) 15 < m < 65 (3) 35 < m < 85 (4) � 85 < m < � 35

Ans. (1)

Sol. r = 5164 = 5

5

m166 < 5

� 25 < m + 10 < 25

� 35 < m < 15


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AIEEE 2010 Solution Resonance

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