AIEEE 2010 Solution

8/9/2019 AIEEE 2010 Solution

1/37

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

QUESTIONS & SOLUTIONS OF AIEEE 2010

IMPORTANT INSTRUCTIONS

1. 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 of90 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 ofFOUR (4) marks each and Question No. 1 to 3 and 10 to 12 consist ofEIGHT (8) marks

each for each correct response.

Part B PHYSICS (144 marks)

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

marks each for each correct response.

Part CMATHEMATICS (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 ofEIGHT

(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. (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.

11. Make sure that the CODE printed on Side-2 of the Answer Sheet is the same as that on this booklet. In case of discrepancy, the condidate

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) : ____________________________________________________________

Roll Number : in figures : in words : _______________________________________________

Examination Centre Number :

Name of Examination Centre (in Capital letters) : ________________________________________

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


2/37

Page # 2

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

3is 46.0 kJ mol1. If the enthalpy of formation of H

2from its atoms

is 436 kJ mol1 and that of N2is 712 kJ mol1, the average bond enthalpy of N H bond in NH

3is

(1) 964 kJ mol1 (2) + 352 kJ mol1 (3) + 1056 kJ mol1 (4) 1102 kJ mol1

Ans. (2)

Sol. N2(g) + 2

3H2(g) NH3 (g) ; Hf = 46.0 kJ mol

1

2H(g) H2(g) ; H

f = 436 kJ mol1

2N(g) N2(g) ; H

f = 712 kJ mol1

NH3(g)

2

1N

2(g) +

2

3H

2(g) ; H

= + 46

2

3H

2 3 H ; H

= + 436

2

3

2

1N

2 N ; H

= + 712

2

1

NH3(g) N

(g) + 3H (g) ; H

= + 1056 kJ mol1

Average bond energy of NH bond =3

1056= + 352 kJ mol1

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 L1 , how much time does it take for its concentration to come from 0.50 to 0.25

mol L1. 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

a

a

)t(

)t(

50.0

2

)t(

1

22/1

t1/2

=

2

5.0= 0.25 h.


3/37

Page # 3

3. A solution containing 2.675 g of CoCl3

. 6 NH3

(molar mass = 267.5 g mol1) is passed through a cation

exchanger. The chloride ions obtained in solution were treated with excess of AgNO3to give 4.78 g of AgCl

(molar mass = 143.5 g mol1). 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

. 6NH3

=

5.267

675.2= 0.01

AgNO3(aq) + Cl (aq) AgCl (white)

Mole of AgCl =5.143

78.4= 0.03

0.01 mole of CoCl3. 6NH

3gives 0.03 mole of AgCl

1 mole of CoCl3.6NH

3ionises to give 3 moles of Cl.

Hence the formula of compound is [Co(NH3)6] Cl

3

4. Consider the reactionCl

2(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

+ H2S 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

[Cl2] [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 104 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 K1 mol1)

(1) 5.56 103

mol (2) 1.53 102

mol (3) 4.46 102

mol (4) 1.27 103

molAns. (4)


4/37

Page # 4

Sol. PV = nRT

V = 1 dm3 = 103 m3

P = 3170 Pa

R = 8.314 J K1 mol1

T = 300 K

3170 103 = n 8.314 300

n =300314.8

103170 3

= 1.27 103 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

CH3CH=CHCH

3

Zn/O3 CH3CH=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 mol1)

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

Ans. (2)

Sol. Na2SO4 (s) OH2

2Na+ (aq.)+ SO42

(aq.)T

f= i K

fm

= 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 ZnCl2is 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)


5/37

Page # 5

Sol.

K278

HClNaNO2 4HBF

10. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahls 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 NH4Cl

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 =17

14 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 Cl2is 242 kJ mol1 . The longest wavelength of light

capable of breaking a single Cl Cl bond is

(c = 3 108 ms1 and NA

= 6.02 1023 mol1)

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

Ans. (4)

Sol. ClCl(g) 2Cl(g) H = 242 KJ mol

= 23

3

1002.6

10242

J molecule1

E =

hc

02.6

1010242 323 =

834 103106.6

= 323834

1010242

103106.6

=242

02.636.6 106 = 0.494 106 = 494 109 m = 494 nm


6/37

Page # 6

12. Ionisation energy of He+ is

19.6 1018 J atom1 . The energy of the first stationary state (n = 1) of Li2+ is :

(1) 4.41 1016 J atom1 (2) 4.41 1017 J atom1

(3) 2.2 1015 J atom1 (4) 8.82 1017 J atom1

Ans. (2)

Sol. I.E. of He+ = 19.6 1018 J atom1

I.E. = E1

E1

for He+ is = 19.6 1018 J atom1

3Li1

He1

)E(

)E(= 2

Li

2He

)Z(

)Z(

2

2LI1

18

)E(

106.19=

9

4

E1(Li2+) =

4

1096.19 18

= 44.1 1018 = 4.41 1017 J atom1

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.


7/37

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 mol1 and of

octane = 114 g mol1)

(1) 72.0 kPa (2) 36.1 kPa (3) 96.2 kPa (4) 144.5 kPa

Ans. (1)

Sol. PT = XHeptane

tanHepP + XOctane

etanOcP

=557.0

25.0 105 +

557.0

307.0 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 H2PO4

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

= 4.2 107 and K2

= 4.8 1011

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

(1) The concentration of CO32 is 0.034 M.

(2) The concentration of CO32 is greater than that of HCO

3.

(3) The concentration of H+

and HCO3

are approximately equal.(4) The concentration of H+ is double that of CO

32.

Ans. (3)


8/37

Page # 8

Sol. H2CO

3H+ + HCO

3 K

1= 4.2 107

HCO3 H+ + CO

32 K

2= 4.8 1011

K1

>> K2

[H+] = [HCO3]

K2

=]HCO[

]CO][H[

3

23

but [H+] = [HCO3]

[CO3

2] = K2

= 4.8 1011

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 radiieargchnuclear

1 .

So, correct order is8O2 >

9F >

11Na+ >

12Mg2+ >

13Al3+

22. Solubility product of silver bromide is 5.0 1013 . This quantity of potassium bromide (molar mass taken as

120 g mol1) 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

gAns. (2)


9/37

Page # 9

Sol. Ksp

= [Ag+] [Br] = 5.0 1013

[Ag+] = 0.05 M

[0.05] [Br] = 5.0 1013

[Br] =05.0

100.5 13= 1 10111 M

moles of KBr = M V

= 1 1011 1

= 1 1011

weight of KBr = 1 1011 120

= 1.2 109 g

23. The Gibbs energy for the decomposition of Al2O

3at 500C is as follows :

3

2Al

2O

3

3

4Al + O

2,

rG = + 966 mol1 . The potential difference needed for electrolytic reduction of AlAl

2O

3

at 500C is at least :

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

Ans. (3)

Sol.3

2Al

2O

3

3

4Al + O

2

rG = +966 kJ mol1 = 966 103 J mol1

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 1011. At Which pH, will Mg2+ ions start precipitating in theform of Mg(OH)

2from a solution of 0.001 M Mg2+ ions ?

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

Ans. (2)

Sol. KSP

= 1.0 1011 = (Mg+2) (OH)2

1.0 1011 = (0.001) (OH)2

(OH) = 104

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%.


10/37

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

eis 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


11/37

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 = RvB

, =R

vB2 (2) 1 = 2 = R3

vB, =

R3

vB2

(3) 1 = 2 = RvB (4) 1 = 2 = R6

vB , =R3vB

Ans. (2)

Sol. Current =R2/R

vB

=R3

vB2

1 = 2 = R3vB

32. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for theenergy 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)2

1(3)

4

1(4) 2

Ans. (3)

Sol. U0 = C2

q20U =

C2

eq /t2201

=2

U0=

C4

q20

2

1e

/t2 1

t1 = 2

n2 ....(1)

and q = q0/t2e ; 4

q0= q0

/t2e , 4

1e

/t2

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

4

1

t

t

2

1


12/37

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 andKmax 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.

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

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 waterand ball should be in between oil and water.


13/37

Page # 13

36. A particle is moving with velocity )jxiy(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. )jxiy(Kdtrd

dt

dx= y ,

dt

dy= x

So, y

x

dx

dy

dxxdyy

C2

x

2

y 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


14/37

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

2R

Vat t = (2)

2R

Vat t = 0 and

21

21

RR

)RR(V at t =

(3)2R

Vat t = 0 and

22

21

21

RR

RVR

at t = (4)

21

21

RR

)RR(V at t = 0 and

2R

Vat t =

Ans. (2)Sol. At t = 0, current does not flow through inductor.

i =2R

V

At t = Req =21

21

RR

RR

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 =1

2

2

dt

dx

1

V2 = 1dt

dx

2

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


15/37

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 mass2

Meach.. 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 = 2E2Ans. (3)

Sol. Energy is released

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

41. The speed of daughter nuclei is

(1) cmM

m

(2) cM

m2(3) c

M

m(4) c

mM

m

Ans. (2)

Sol. Q = m c2 =2

1

2

Mv2 +

2

1

2

Mv2

mc2 =2

1 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)8Z

4ZA

(3)4Z

12ZA

(4)2Z

4ZA

Ans. (2)

Sol.

2e2YHe3X0

112A

8Z4

2A

Z

number of proton = Z 8

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

ratio is8Z

4ZA


16/37

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) jr4

q2

02

(2) jr4

q2

02

(3) jr2

q2

02

(4) jr2

q2

02

Ans. (3)

Sol. )j(

r

k2E

)j(r2

E0

r

q

)j(

r2

qE

20

2

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 11 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.


17/37

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

T11

5

7

V

= T21

5

7

)V32(

1

2

T

T= 5/2)32(

1=

4

1

= 1 1

2

T

T= 1

4

3

4

1

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

104106.112400

AA0 = 49.6

It is in X-ray spectrum.

47. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 103 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 2.1 103 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 30. On

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

circuit is

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


18/37

Page # 18

Sol. tan 30 =R

XL XL = 3

R=

3

200

tan 30 =R

XC Xc = 3

200

Z = 2CL )XX(R = 200

irms = 200

220= 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

R

r

4

5)r( 0

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 0

Negation : all rational numbers x S are x 0

63. Let kja

and kjic

. Then the vector b

satisfying 0cba

and 3b.a

is

(1) k2ji2 (2) k2ji (3) k2ji (4) k2ji

Ans. (4)

Sol. Since 0cba

0ca)ba(a

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

Since ca

= kji2

0kji2b2)kj(3

k2jib

Hence correct option is (4)


27/37

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 + 2

x

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 0 so statement- 2 is correct

so f(c) = 3

1

for some c. because f(x) is continuous [since f(0) = 3

1

]


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

among the following is

(1) There is a regular polygon with2

1

R

r . (2) There is a regular polygon with

3

2

R

r .

(3) There is a regular polygon with

2

3

R

r . (4) There is a regular polygon with

2

1

R

r .

Ans. (2)


33/37

Page # 33

Sol.R

r= cos

n

Let cosn

=

3

2for some n 3, n N

As2

1

3

2

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 = 20094018 = 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) satisfiesHence correct option is (1)

80. A line AB in three-dimensional space makes angles 45 and 120 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) 45 (2) 60 (3) 75 (4) 30

Ans. (2)

Sol. =2

1, m =

2

1

2 + m2 + n2 = 1

n2 =4

1 n =

2

1

cos =2

1, = 60



34/37

Page # 34

81. The line L given byb

y

5

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

equation3

y

c

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

(1) 17 (2) 15

17(3) 17

23(4) 15

23

Ans. (3)

Sol.5

x+

b

y= 1

5

13+

b

32= 1

b

32=

5

8 b = 20

5

x

20

y= 1 4x y = 20

Line K has same slope c

3= 4

c = 4

3 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

= a2

= .....= a10

= 150 and a10

, a11

,....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

+ a2

+ .... + an

= 4500 notes

As a1

+ a2

+ .... + a10

= 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



35/37

Page # 35

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

)x3(flimx

= 1. Then)x(f

)x2(flimx

.

(1)3

2(2)

2

3(3) 3 (4) 1

Ans. (4)

Sol.0x

lim )x(f

)x3(f= 1

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

)x(f

)x3(f

)x(f

)x2(f1

using sandwitch theorem xlim

)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(p

1

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 = 0p(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



36/37

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

12

89

= 12 9 = 108


87. Consider the system of linear equations :

x1

+ 2x2

+ x3

= 3

2x1

+ 3x2

+ x3

= 3

3x1

+ 5x2

+ 2x3

= 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 + x2 = 0(3) 2(1) x

1+ 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) 7

2

(2) 21

1

(3) 23

2

(4) 3

1

Ans. (1)

Sol. =3

9

12

14

13

C

CCC=

1.2.3

7.8.9

2.4.3

=7

2


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)2

11(2) 6 (3)

2

13(4)

2

5

Ans. (1)


37/37

Sol. x2 = 4

n

x2i

2

i

n

x

= 4

5

x2i (2)2 = 4 x

i2 = 40

similarly yi2 = 105

2 =10

yx2

i2

i

2

ii

10

yx

=

10

145

2

10

2010

= 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


Date post:	30-May-2018
Category:	Documents
Upload:	resonance-kota
View:	215 times
Download:	0 times

AIEEE 2010 Solution

Documents