Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 1 ANSWERS−AIEEE 2010 A B C D A B C D A B C D 1 3 1 1 2 31 4 4 4 2 61 4 2 3 1 2 2 2 1 3 32 2 4 4 2 62 2 3 4 2 3 1 3 1 1 33 4 3 1 1 63 3 2 1 4 4 3 1 4 4 34 4 4 2 4 64 4 3 4 3 5 4 1 1 4 35 1 2 4 2 65 4 4 4 4 6 1 3 2 3 36 3 4 3 4 66 2 1 2 4 7 2 2 2 2 37 3 1 2 1 67 1 4 1 2 8 3 1 4 2 38 2 4 4 2 68 2 3 3 4 9 2 2 4 3 39 2 1 2 2 69 2 1 4 3 10 3 1 4 3 40 1 1 1 3 70 4 1 1 3 11 4 3 3 4 41 4 4 2 2 71 1 2 3 3 12 3 3 4 2 42 3 2 1 2 72 1 1 2 2 13 2 2 3 1 43 2 2 2 3 73 1 3 2 4 14 2 2 3 2 44 3 2 3 1 74 1 1 4 2 15 2 3 2 1 45 1 2 4 2 75 3 3 2 3 16 3 1 4 1 46 4 3 3 1 76 4 4 4 4 17 3 1 2 1 47 1 2 4 1 77 3 1 1 2 18 1 4 1 3 48 4 1 3 4 78 3 1 1 2 19 3 2 3 4 49 1 3 3 2 79 2 2 1 1 20 1 3 1 4 50 1 4 4 3 80 2 2 4 2 21 3 4 1 4 51 4 1 4 3 81 2 3 2 3 22 3 2 2 2 52 3 1 1 4 82 2 2 3 1 23 1 4 3 3 53 3 3 2 4 83 3 1 2 4 24 1 4 1 2 54 1 2 1 4 84 4 4 1 1 25 4 4 3 2 55 3 1 2 3 85 4 1 4 1 26 4 1 4 3 56 2 3 4 4 86 3 3 2 3 27 4 3 2 1 57 2 3 1 2 87 3 1 3 3 28 1 1 3 1 58 4 4 3 1 88 1 4 3 1 29 4 4 4 4 59 2 2 2 3 89 1 3 3 1 30 1 4 1 2 60 3 4 4 4 90 4 4 3 1 Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 2 Solution AIEEE 2010 [SET D] PART A –CHEMISTRY 1. The standard enthalpy of formation of NH3 is −46.0 kJ mol−1. If the enthalpy of formation of H2 from its atoms is 1 436 kJmol− − and that of N2 is 1 712 kJmol− − the average bond enthalpy of N−H bond in NH3 is (1) 1 964 kJmol− − (2) 1 352 kJmol− + (3) 1 1056 kJmol− + (4) 1 1102 kJmol− − Sol. -2 2 3 1 3 N H NH 2 2 + ® ( )N H 3 1 46 436 712 3 BE 2 2 − − = × + × − × \ (B–E)N-H = = 352 kJ/mole 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 1 L− if its is a zero order reaction? (1) 4 th (2) 0.5 h (3) 0.25 h (4) 1 h Sol.-0 1/2 A 2 t k 1 2k 2 = = = .5 2 1 t t 1.5h = − × = .25 2 1 t t 1.75 = − × = \ .25 h 3. A solution containing 2.675 g of CoCl3. 6 NH3 (molar mass = 267.5 g mol 1 − ) is passed through a cation exchanger. The chloride ions obtaine in solutin 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) ( ) 3 3 6 Co NH Cl (2) ( ) 2 3 4 CoCl NH Cl (3) ( ) 3 3 3 CoCl NH (4) ( ) 3 2 5 CoCl NH Cl Sol.-Agcl 4.78 n .03 143.5 = = complex 2.675 n .01 267.5 = = hence ( ) 3 3 6 Co NH Cl 4. Consider the reaction : ( ) ( ) ( ) ( ) 2 2 Cl aq H S aq S s 2H aq + + ® + + ( ) 2 Cl aq − The rate equation for this reaction is rate = k [Cl2] [H2S] Which of these mechanism is /are consistent with this rate equation? A. ( ) 2 2 Cl H S H Cl Cl HS slow + − + − + ® + + ( ) Cl HS H Cl S fast + − + − + ® + + B. 2 H H HS + − Û + (fast equilibrium) ( ) 2 Cl HS 2Cl H S slow − − + + ® + + (1) B only (2) Both A and B (3) Neither A nor B (4) A only Sol.-Since [ ][ ] 2 2 r K Cl H S = hence Cl2 & H2S will participate is lowest step. 5. If 4 10− 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 1 mol − − ) (1) 3 5.56 10 mol − × (2) 2 1.53 10 mol − × (3) 2 4.46 10 mol − × (4) 3 1.27 10 mol − × Sol.-PV = nRT 3 3170 10 n 8.314 300 − × = × × \ 3 n 1.27 .10− = × 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) propene (2) 1-butene (3) 2 -butene (4) ethane Sol.-3 O 3 3 3 CH CH CH CH 2CH CHO − = − ¾¾® M = 44 Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 3 7. If sodium sulphate is considered to be completely dissociated in to cations and anions in aqueous solution, the change in freezing point of water ( ) f T D , when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is ( 1 f K 1.86kgmol− = ) (1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K Sol.-f f T i k m D = × × 2 2 4 4 Na SO 2Na SO + − ® = \ i = 3 3 1.86 .01 = × × .0558 = 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 Sol.-3° alcohol reacts fastest with conc. HCl and ZnCl2 \ 3 3 3 OH | CH C CH | CH − − 2 methylproane 2 ol − − − 9. In the chemical reactions, 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 Sol.-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 percentageof nitrogen in the compound is (1) 59.0 (2) 47.4 (3) 23.7 (4) 29.5 Sol.-Meq. of NH3 + Meq. of NaOH = Meq. of HCl Meq. of NH3 + 15 .1 20 .1 × = × \ Meq. of NH3 = .5 = m mole of NH3 \ WN = .5×15 = 7 mg \ % of N = 7 100 23.7 29.5 × = 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 s (c = 3×108 ms 1 − and NA = 6.02 ×1023 mol 1 − ) (1) 594 nm (2) 640 nm (3) 700 nm (4) 494 nm Sol.-nhc E = l \ 3 23 34 8 1 242 10 6.02 10 6.6 10 3 10 − × = l × × × × × \ l = 494 nm 12. Ionisatin energy of He+ is 18 19.6 10− × J atom 1 − . The energy of the first stationary state ( ) n 1 = Li2+ is (1) 16 1 4.41 10 J atom − − × (2) 17 1 4.41 10 J atom − − − × (3) 15 1 2.2 10 J atom − − − × (4) 17 1 8.82 10 J atom − − × Sol.-( ) ( ) ( ) 2 18 2 2 IE He 19.6 10 J /atom k 1 + − = × = − × \ 18 K 4.9 10− = × \ So ( ) ( ) ( ) 2 2 17 2 Li 3 E K 4.41 10 J /atom 1 + − = − × = − × 13. Consider the following bromide : The correct order of SN1 reactivity is (1) B > C > A (2) B > A > C (3) C > B > A (4) A > B > C Sol.-N S 1 reacts with carbocation mechanism. hence B > C > A 14. Which one of the following has an optical isomer ? (1) ( )( ) 2 3 2 Zn en NH + (2) ( ) 3 3 Co en + (3) ( ) ( ) 3 2 4 Co H O en + (4) ( ) 2 2 Zn en + Sol.-( ) 3 3 Co en + Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 4 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 Sol.-Heptane = oc tan e .25 25 5 6 ,X .25 .3 55 11 11 = = + \ 5 6 P 105 45 11 11 = × + × = 72 kPa 16. The main product of the following reaction is ( ) ( ) 2 4 conc.H SO 6 5 2 3 2 C H CH CH OH CH CH ¾¾¾¾®? (1) (2) (3) (4) Sol.-17. Three reactions involving H2PH 4− are given below: (i) 3 4 2 3 2 4 H PO H O H O H PO + − + ® + (ii) 2 2 4 2 4 3 H PO H O HPO H O − − + + ® + (iii) 2 2 4 3 4 H PO OH H PO O − − − + ® + In which of the does 2 4 H PO− act as an acid ? (1) (ii) only (2) (i) and (ii) (3) (iii) only (4) (i) only Sol.-In (ii) 2 4 H PO− is donating H+ 18. In aqueous solution the ionization constants for carbonic acid are 7 1 K 4.2 10− = × and 11 2 K 4.8 10− = × Select the correct statement for a saturated 0.034 M solution of the carbonic acid. (1) The concentration of 23 CO − is 0.034 M. (2) The concentration of 23 CO − is greater (3) The concentrations of H+ and 3 HCO− are approximately equal. (4) The concentration of H+ is double that of 23 CO − Sol.-7 2 3 3 1 H CO H HCO k 4.2 10 + − − + = × ı ıı ı ı ı 2 11 2 3 3 2 H CO H HO k 4.8 10 + − − + = × ı ıı ı ı ı due to common ion effect total H+ will produce by first step only, so H+ and 3 HCO− should be same 19. The edge length of a face centered 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 Sol.-For fcc crystal, ( ) c a 2 r r a + = \ a c a r r 2 = − 508 110 2 = − = 114 pm 20. The correct order of increasing basicity of the given conjugate bases (R=CH3) is (1) 2 RCOO HC C R NH < º < < (2) 2 R HC C RCOO NH < º < < (3) 2 RCOO NH HC C R < < º < (4) 2 RCOO HC C NH R < º < < Sol.-2 RCCO HC C NH R − − − < º < < ı ı ı Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 5 21. The correct sequence which shows decreasing order of the ionic radii of the elements is (1) 3 2 2 Al Mg Na F O + + + − − > > > > (2) 2 3 2 Na Mg Al O F + + + − − > > > > (3) 2 2 3 Na F Mg O Al + − + − + > > > > (4) 2 2 3 O F Na Mg Al − − + + + > > > > Sol.-More is the (+)ve charge on cation smaller will be size. So 2 2 3 O F Na Mg Al − − + + + > > > > 22. Solubility product of silver bromide is 5.0 ×10−13. The 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) 10 1.2 10 g − × (2) 9 1.2 10 g − × (3) 5 6.2 10 g − × (4) 8 5.0 10 g − × Sol. -Al Br kSp + − = [ ] 13 11 .05 Br 5 10 Br 10 − − − × = × = = \ 11 9 kBr W 120 10 1.2 10 gm − − = × = × 23. The Gibbs energy for the decomposition of Al2O3 at 500 °C is as follows: 1 2 3 2 r 2 4 Al O Al O , G 966kJ mol 3 3 − ® + D = + The potential difference needed for electrolytic reduction of Al2O3 at 500ºC is at least (1) 4.5 V (2) 3.0 V (3) 2.5 V (4) 5.0 V Sol.-G nFE D = − 3 10 966 4 96500 E × = − × × \ E = 2.5 V 24. At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10−11. At with pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M 2 Mg + ions? (1) 9 (2) 10 (3) 11 (4) 8 Sol.-2 2 Mg OH Ksp + − = 2 3 11 4 10 OH 10 OH 10 POH 4 − − − − − × = \ = \ = \ pH= 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% Sol.-CCP = fcc so packing afficienty = 74% \ free space = 26% for BCC, packing efficieny = 68% \ free space = 32% 26. Out of the following, the alkene that exhibits optical isomerism is (1) 3-methyl-2-pentene (2) 4-methyl-1-pentene (3) 3-methyl-1-pentene (4) 2-methyl-2-pentene Sol.-3 3 2 3 3 methyl 2 pentene CH | CH CH C CH CH − − − − = − − 3 3 2 3 3 Methyl 1 pentene CH | CH CH CH CH CH − − − = − − − 3 2 2 3 3 Methyl 1 pentene CH | CH CH CH CH CH − − − = − − − 3 2 2 3 2 Methyl 2 pentene CH | CH C CH CH CH − − − − = − − \ Chiral carbon is present in (3) 27. Biuret test is not given by (1) carbohydrates (2) polypeptides (3) urea (4) proteins Sol.-Carbohydrates do not give Biuret test 28. The correct order of 2 0M /M E + 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) electrochemical series shows the same order. 29. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is (1) Teflon (2) nylon 6,6 (3) polystyrene (4) natural rubber Sol.-Natural rubber 30. For a particular reversible reaction at temperature at temperature T H D and S D were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when (1) Te > T (2) T > Te (3) Te is 5 times (4) T = Te Sol.-DG = DH −TDS at Eq. DG = 0 So e H T S D = D For speontaneity G 0 H T S 0 D < \D − D < H T S D \ > D e T T \ > Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 6 PART B – PHYSICS 31. A rectangular loop has a sliding connector PQ of length l and resistance R W 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 I1, I2 and I are (1) 1 2 B v 2B v I I , I R R = − = = l l (2) 1 2 B v 2B v I I , I 3R 3R = = = l l (3) 1 2 B v I I I R = = = l (4) 1 2 B v B v I I , I 6R 3R = = = l l Sol. AS e = B l n As I = 2B3Rn l pq V e IR = − = 2B B v 3 n − l l V PQ = B3n l pq 1 V B I R 3Rn = = l pq 2 V B I R 3Rn = = l 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) 12 (3) 14 (4) 2 Sol. Instantaneous energy in capacitor is – 2t. /RC O U U e− = 0 e U t RC log U = 0 e 0 e 12 0 e e 0 U log U /2 log 2 t log 2 1 t 2log 2 2 U log 4 log U /4 = = = = Answer :– 2 Directions : Questions number 33 – 34 contain Statement – 1 and Statemetn -2 . Of the four choice s given after the statements, choose 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 ture 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. Sol. (1) As momentum is conserved, final momentum will be non zero hence, kE is non zero, hence total energy can not lost in 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 increage. (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 ture, Statement – 2 is false. Sol. (4) Photoelectrons are having different speeds because of different frequency of photons and also because of collision, so statement 2 is incorrect. Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 7 35. A ball is made of a materil of density p where poil < p < pwater with poil and pwater 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) Sol. As rwater > roil \ water will settled down & as roil < r < rwater \ ball will sink in oil but floate on water. 36. A particle is moving with velocity ( ) ˆ ˆ v K yi xj , = + r where K is a constant. The general equation for its path is (1) y = x2 + constant (2) y2 = x + constant (3) xy = constnat (4) y2 = x2 + constant Sol. ( ) V k yi x j = + ur $ $ As x V ky = y V kx = dx ky dt = dy kx dt = dy x dx y = ydy = xdx 2 2 y x 2 2 = + constant 2 2 y x = + 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 the magnetic field B along the line XX’ is given by (1) (2) (3) (4) Sol. As in middle region field nearer to A will be upwards and nearer to B will be downwards and at middle point net field will be zero. Towards right of B net field will be in upward direction. Towards left of A net field be in downward direction. 38. In the circuit shown below, the key K is closed at t = 0. The current through the battery is (1) 1 2 2 2 1 2 VR R R R + at t = 0 and 2 V R at t = ¥ (2) ( ) 1 2 2 1 2 V R R V at t 0and at t R R R+ = = ¥ (3) 1 2 2 2 2 1 2 VR R V at t 0and at t R R R = = ¥ + (4) ( ) 1 2 1 2 2 V R R V at t 0 and at t R R R = = ¥ Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 8 Sol. As ( ) t /T 0 I I 1 e− = − where T = time constant 1 2 L R R = + and ( ) 1 2 0 1 2 V R R I R R+ = So at t = ¥ ( ) 1 2 0 1 2 V R R I I R R+ = = and at t = 0 current will flow only in R2 resistance So, 2 V I R = 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 Na Sol. Because of impulse velocity changes from V = 1 m/s to V = – 1m/s or vice versa, So impuse = ( ) f i m v v 0.8Ns − = Directions : Question number 40-41 are based on the following paragraph. A nucleus of mass M + Dm is at rest and decays into two daughter nuclei of equal mass M2 each. Speed of light is c. 40. The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then (1) E2 = 2E1 (2) E1 > E2 (3) E2 > E1 (4) E1 = 2E2 41. The speed of daughter nuclei is : (1) m cM m D+ D (2) 2 m c MD (3) m c M D (4) m c M m D+ D Sol. mass defect Dm converted into 2 2 1 m mC v 2 2 2 D = × 2 2 m mc v 2 D 2 2 2 mc v m D = 2 m v c mD = 42. A radioactive nucleus (initial mass number A and atomic mass number Z) emits 3 a−particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be (1) 8 4 A Z Z− − − (2) 4 8 A Z Z− − − (3) 12 4 A Z Z − − − (4) 4 2 A Z Z− − − Sol. 4 0 2 1 3 He 2 e A A 12 A 12 z z 6 Z 8 X Y D + − − − − ¾¾¾® ¾¾¾® no. of neutrons = (A – 12) – (Z – 8) = A – 12 – Z + 8 no. of protons = A – Z – 4 Ratio A Z 4 Z 8 − − = − 43. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field Er at the centre O is (1) 2 2 0 ˆ 4 q j r p e (2) 2 2 0 ˆ 4 q j r p e − (3) 2 2 0 ˆ 2 q j r p e − (4) 2 2 0 ˆ 2 q j r p e Sol. ( ) 2 sin /2 kQ E R 2a = a ( ) 0 sin /2 1 q 4 r 2p = × × p pe 2 0 2q E 4 r = p e 2 0 q ˆ E j 2 r − = p e ur Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 9 44. The combination of gates shown below yields (1) OR gate (2) NOT gate (3) XOR gate (4) NAND gate Sol, Final expression x A.B = A B = + x = A + B 45. A diatomic ideal gas is used in a Carno 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 Sol. for carnot cycle, 21 T 1 T h = − In adiabatic expansion, T1V1y-1 = T2V2y-1 V2 = 32V1 y 1 2 1 2 2 T V 1 T V 4 − = = 3 0.75% 4 h = = 46. If a source of power 4 kW produces 1020 photon/second, the radiation belongs to a part of the spectrum called (1) X-rays (2) ultraviolet rays (3) microwaves (4) g-rays Sol. Power = kJ J 4 4000 sec. sec. = Energy of a photons 17 20 4000 4 10 10 − = = × 17 3 19 4 10 1 ev 10 eV 4 1.6 10−− × = = × × ( ) ( ) ( ) o 3 ev nm 1242 1 1242 E 10 49.6A x rays 4 = × = l = − l l 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 Sol. 5, 1, 2 48. In a series LCR circuit r = 200 W 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 Sol. So, VL = Vc XL = Xc So For P = Vv IV cos 0º 2 v v V V V V 220 220 Z R 200 × = = = = 242 W 49. Let there be a spherically symmetric charge distribution with charge density varying as ( ) 0 54 r r R r r = − upto r = R, and r (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) 0 0 4 5 3 3 r rR pre − (2) 0 0 5 4 3 r rR re − (3) 00 4 5 3 4 r rR re − (4) 00 5 3 4 r rR re − Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 10 Sol. Charge inside sphere of radius ‘r’ is r 3 4 0 0 5 r r q dv 4 4 3 4R = r = pr − 0 2 0 0 r q 5 r E 4 3 R 4 r r = = − e pe 50. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U(x) = 12 6 a b x x − , where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is : ( ) at equilibrium D U x U ,D = = ¥ − is (1) 2 b2a (2) 2 b 12a (3) 2 b4a (4) 2 b6a Sol. ( ) 12 6 a b U x x x = − 13 7 dy 12a 6b F dx x x = − = − F = 0, at equilibrium 6 2a x b = then Uat equilibrium = 2 b4a − Uas = 0 2 b D 4a = 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 Sol. F F/k tan mg mg upthlus q = = − k s = s − r 1.6 2 1.6 0.8 = = − 52. Two conductors have the same resistance at 0ºC but their temperature coefficients of resistance are a1 and a2. The respective temperature coefficients of their series and parallel combinations are nearly : (1) 1 2 1 2 , 2 a + a a + a (2) 1 2 1 2 , 2 a + a a + a (3) 1 2 1 2 1 2 , a a a + a a + a (4) 1 2 1 2 , 2 2 a + a a + a Sol. 1 1 2 2 series 1 2 R R x R R a + a = + 1 2 1 2 parallel 1 2 R R x 1 1 R R a a + = + 1 2 series x 2 a + a = 1 2 parallel x 2 a + a = 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 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 Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 11 Sol. So, arelative = aqsin 60 – a2 sin 30 = 2 3 1 g g 4.9m/s 4 4 2 = − = = 55. For a particle in uniform circular motion, the acceleration ar at a point P (R, q) on the circle of radius R is (Here q is measure from the x-axis) (1) 2 2 v v ˆ ˆ cos i sin j R R − q + q (2) 2 2 v v ˆ ˆ sin i cos j R R − q + q (3) 2 2 v v ˆ ˆ cos i sin j R R − q − q (4) 2 2 v v ˆ ˆ i j R R + Directions : Questions number 56-58 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index ( ) 0 2 I I, µ = µ + µ where µ0 and µ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. 56. An 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 Sol. I decreases with radios and µ increase with I so, µ is higher at axis and cower at periphery Rays would continue to move along straight line with different speeds. 57. The initial shape of the wave front of the beam is : (1) convex (2) concave (3) convex near the axis and concave near the periphery (4) planar Sol. Rays at periphery move faster and at axis leaves behind so shape of wave front is concave. 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 I (4) maximum on the axis of the beam Sol. (1) 59. A small particle of mass m is projected at an angle q with the x-axis with an initial velocity v0 in the x – y plane as shown in the figure. At a time 0 v sin t , g q < the angular momentum of the particle is : (1) – mg v0 t2 cos ˆj q (2) 0 ˆ mg v t cos k q (3) 2 0 1 ˆ mg v t cos k 2 − q (4) 2 0 1 ˆ mg v t cos i 2 q where ˆ ˆ i, j and ˆk are unit vectors along x, y and z-axis respectively. Sol. ( ) ( ) 2 x 0 0 1 ˆ L mv cos v sin t gt k 2 = q q − − [ ] ( ) y 0 0 ˆ L m v sin gt cos t k = q − ×n q + uur 2 2 0 0 mv cos gt L mv cos sin t 2 q = q q − ur 20 0 mv sin cos t mv cos gt − q q + q 2 0 1 ˆ L mv cos gt k 2 = − q ur 60. The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by ( ) ( ) ( ) t x y 0.02 m sin 2 . 0.04 s 0.50 m = p − The tension in the string is : (1) 4.0 N (2) 12.5 N (3) 0.5 N (4) 6.25 N Sol. m = 4 × 10–2 T f m l = 2 2 T mf = l = 6.25 × 4 × 10-2 = 2500 × 10–2 × 25 × 10-2 = 6.25 Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 12 PART C – MATHEMATICS 61. Let cos (a + b) = 45 and let sin (a − b) = 5 13 , where 0 £ a, b 4p £ . Then tan 2a = (1) 56 33 (2) 19 12 (3) 20 7 (4) 25 16 Sol. 0 £ a + b £ 2p = ( ) 3 tan 4 a b + = 4 4 p p a b − £ − £ ( ) 5 tan 12 a b = − = tan 2a = tan (a + b) + (a − b) 3 5 56 4 12 3 5 33 1 4 12 + = = − × 62. Let S be a non-empty subset of R. Consider the following statement : P : There is a rational number x Î S such that x > 0. Which of the following statements is the negation of the statement P ? (1) There is no rational number x Î S such that x £ 0. (2) Every rational number x Î S satisfies x £ 0. (3) x Î S and x £ 0 x is not rational. (4) There is a rational number x Î S such that x £ 0. Sol. ~P = every rational number x Î s Satisfies x £ 0 63. Let ˆ ˆ a j k = − r and ˆ ˆ ˆ c i j k = − − r . Then the vector br satisfying 0 a b c × + = r r r r and . 3 a b = r r is (1) ˆ ˆ ˆ 2 2 i j k − + (2) ˆ ˆ ˆ 2 i j k − − (3) ˆ ˆ ˆ 2 i j k + − (4) ˆ ˆ ˆ 2 i j k − + − Sol. Only fourth option satisfy . 3 a b = r r and a b c × = − r r r . ( ) ( ) ( ) ˆ ˆ 0 1 1 2 1 1 1 1 1 2 i j k i j k − = − + − − + − − ˆ ˆ ˆ i j k = − + + c = −r 64. The equation of the tangent to the curve 2 4 y x x = + , that is parallel to the x-axis (1) y = 1 (2) y = 2 (3) y = 3 (4) y = 0 Sol. Y = 2 3 4 8 ' 1 x y x x + = − ' 0 2 y x = = at x = 2 2 4 2 2 y = + y = 3 65. Solution of the differential equation cos x dy = y (sin x −y) dx, 0 < x < 2p is (1) y sec x = tan x + c (2) y tan x = sec x + c (3) tan x = (sec x + c) y (4) sec x = (tan x + c) y Sol. cos x dy = y (sin x −y) dx 2 tan sec dy y x y x dx = − 2 tan sec dy y x y x dx − = − [Q Bernoulli’s Form] 2 1 tan sec dy x x dx y y − + = Now let 2 1 1 dt dy t y dx dx y− = = = ( ) tan sec dt x t x dx + = [Q linear differential equation[ P = tan x Q = sec x Integrating factor I. F. = tan ln sec x dx x e e = = sec x Solution of the equation ( )( ) .sec sec sec t x x x dx = ( ) sec tan x y x C = + 66. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = 32p is (1) 4 2 2 + (2) 4 2 1 − (3) 4 2 1 + (4) 4 2 2 − Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 13 Sol. Bounded area between x = 0 and x = 32p ( ) 4 0 cos sin A x x dx p = − ( ) ( ) 3 4 /4 2 /4 4 sin cos cos sin x x x x dx p p p p 5 + − + − [ ] /4 0 2 sin cos x x p = + + [ ]5 /4 /4 cos sin x x p p + − − ( ) 1 1 2 0 1 2 2 = + − + 1 1 1 1 2 2 2 2 − − + − − + + 2 4 2 1 2 2 = − + 8 2 2 − 4 2 2 = − 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 Sol. Let P (h, k) Equation of tangent Y = mx + 1m k = mh + 1m mk = m2h + 1 m2h − mk + 1 = 0 m1m2 = 1h −1 = 1h h = −1 Locus of P, x = −1 68. If the vectors ˆ ˆ ˆ 2 a i j k = − + r , ˆ ˆ ˆ 2 4 b i j k = + + r and ˆ ˆ ˆ c i j k l µ = + + r are mutually orthogonal, then (l, µ) = (1) (2, −3) (2) (−2, 3) (3) (3, −2) (4) (−3, 2) Sol. ˆ ˆ ˆ 1 1 2 9 3 6 2 4 1 i j k a b i j k × = − = − + + r r 9 3 6 1 l µ − = = l = −3 µ = 2 69. Consider the following relations : R = {(x, y)| x, y are real numbers and x = wy for some rational number w} ; S = {( , m p n q ) | m, n, p and q are integers such that n, q ¹ 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 Sol. Both R and S satisfies three relations, reflexive, symmetric and transitive. 70. Let f : R ® R be defined by ( ) 2 , 1 2 3, 1 k x if x f x x if x − £ − = + > − If f has a local minimum at x = −1, then a possible value of k is (1) 0 (2) 12 − (3) −1 (4) 1 Sol. ( ) 2 1 2 3 1 k x if x F x x if x − £ − = + > − Minima exists . if ( ) ( ) ( ) ( ) ( ) 1 1 lim 1 lim x x F x F F x − + ®− ®− = − £ ( ) ( ) ( ) 1 1 lim 2 2 1 lim 2 3 x x k x k x ®− ®− − = − − £ + k + 2 £ −2 + 3 k £ −1 k Î (−¥, −1] Only k = −1 satisfy given condition. Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 14 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 Sol. 10 1 0 0 1 A − − = − has 3 space vacant and | A | = 1 (non singular) Also 1 0 0 1 01 A = − − − Atleast 7 matrix are there. Directions : Questions 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 1 85 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. (2) Statement -1 is true, Statement-2 is false. (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. Sol. {1, 2, 3,…….20} Total outcomes = 20C4 4 terms of A.P. with common difference ± 1, ±2, ±3, ±4, ±5, ±6 are {(1, 2, 3, 4), (2, 3, 4, 5),….(17, 18, 19, 20)}, {(1,3,5,7), (2,4,6,8),……..(14,16,18,20)} {(1,4,7,10), (2,5,8,11),……..(11,14,17,20)} {(1,5,9,13), (2,6,10,14),……..(8,12,16,20)} {(1,6,11,16), (2,7,12,17),……..(5,10,15,20)} {(1,7,13,19), (2,8,14,20)} respectively \ Required probability 20 4 17 14 11 8 5 2 1 85 C + + + + + = = 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). (1) Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement -1. (2) Statement -1 is true, Statement-2 is false. (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. Sol. The midpoint of AB lies in the plane. Both statement are correct and reason is correct. 74. Let ( ) 10 10 10 10 1 2 1 1 1 , j j j j S j j C S j C = = = − = and 10 2 10 3 1 j j S j C = =Statement-1 : S3 = 55 × 29 Statement-2 : S1 = 90 × 28 and S2 = 10 × 2 (1) Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement -1. (2) Statement -1 is true, Statement-2 is false. (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. Sol. ( )10 10 1 1 1 n j j S j j C = = − 10 10 2 10 10 1 1 . j j j j j C j C = = = − ( ) 10 10 9 10 1 1 1 10 1 1 . j j j j j C j C − = = = − + − 10 10 8 9 2 1 2 1 90 10 ....... j j j j C C − − = = = + − 8 9 9 90.2 10.2 10.2 = + − S1 = 90 . 28 Similarly, S2 = 10 10 9 1 . 10.2 j j j C = = S3 = 90. 23 + 10. 29 = 29 . 55 St. 1 is true, St. 2 is false. Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 15 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 (1) Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement -1. (2) Statement -1 is true, Statement-2 is false. (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. Sol. 1 0 1 0 0 1 0 1 A or A − = = − tr (A) ¹ 0 | A | = 1 76. . Let f : R ® R be a continuous function define by ( ) 12 x x f x e e− = + . Statement-1 : f(c) 13 = , for some c Î R. Statement-2 : 0 < f(x) 1 2 2 £ , for all x Î R. (1) Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement -1. (2) Statement -1 is true, Statement-2 is false. (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. Sol. ( ) 12 x x F x e e− = + 2 .2. 2 x x x x e e e e − − + ³ 2 ³ 2 2 2 x x e e− + ³ 1 1 0 2 2 2 x x e e− < £ + ( ) 1 0 2 2 F x < £ So statement 2 is correct Now 1 1 (0, ] 3 2 2 Î Hence Statement 1 is correct. 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 12 rR = (2) There is a regular polygon with 23 rR = (3) There is a regular polygon with 3 2 rR = (4) There is a regular polygon with 12 rR = Sol. cos rR np = If n = 4 12 rR = If n = 3 12 rR = If n = 6 3 2 rR = Ans. 2 is false 78. If a and b are the roots of the equation x2−x + 1 = 0, then a2009 + b2009 = (1) −1 (2) 1 (3) 2 (4) −2 Sol. Roots of the x = −w, x = −w2 Then = (−w)2009 + (−w2)2009 ( )2 2007 2 2007 2 . . w w w w = − + 2 4 w w = − + = −[−1]= 1 79. The number of complex numbers z such that |z −1| = |z + 1| = |z − i| equals (1) 1 (2) 2 (3) ¥ (4) 0 Sol. Let z = x + iy 1 1 z z z i − = + = − ( ) ( ) ( ) 1 1 1 x yi x iy x i y − + = + + = + − ( ) ( ) ( ) 2 2 2 2 2 2 1 1 1 x y x y x y = − + = + + = + − 2 2 2 2 2 2 1 2 1 2 1 2 x y x x y x x y y + + − = + + + = + + − −2x – 2x = −2y x = 0, y = x = 0 Only solution (0, 0) Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 16 80. A line AB in three-dimensional space makes angles 45º and 120º with the positive x-axis and the positive yaxxi respectively. If AB makes an acute angle q with the positive z-axis, then q equals (1) 45º (2) 60º (3) 75º (4) 30º Sol. a = 45º, b = 120º, x = q cos2a + cos2b + cos2x = 1, 2 1 1 cos 1 2 4 q + + = 2 3 cos 1 4 q= − 2 1 cos 4 q= 1 cos 2 q= ± q = 60º 81. The line L given by 1 5x yb + = 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 15 (3) 23 17 (4) 23 15 Sol. 13 32 1 5 b + = gives b = −20 Now slope of first line = slope of II line 3 5b c − − = 4 = 3 c − 3 4 c − = Now distance 23 17 = 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 Sol. In 10 minutes, person will count 1500 currency. Now, he has to count remaining 3000 currency. A = 148, d = −2 Sn = 3000 ( ) 2 1 3000 2n a n d + − = n2 − 149 n + 3000 = 0 \n = 24 \ time taken by him = 34 minutes 83. Let f : R ® R be a positive increasing function with ( ) ( ) 3 lim 1 x f x f x ®¥ = . Then ( ) ( ) 2 lim x f x f x ®¥ = (1) 23 (2) 32 (3) 3 (4) 1 Sol. ( ) ( ) 3 lim 1 x F x F x ®¥ = ( ) ( ) ( ) ( ) lim 3 lim x x F x F x l ®¥ ®¥ = = Now ( ) ( ) 2 lim x F x F x ll ®¥ = = 1 84. Let p(x) be a function defined on R such that p’(x) = p’(1 −x), for all x Î [0, 1], p(0) =1 and p(1) = 41. Then ( ) 10 p x dx equals (1) 21 (2) 41 (3) 42 (4) 41 Sol. Using L.M.V.T. ( ) ( ) ( ) 1 0 ' 41 1 40 1 P P P x − = = − = ( ) ( ) ( ) 1 1 1 0 0 0 .1 . ' . P x dx P x x P x x dx = − ( ) 10 1 40. P x dx = − ( ) 1 2 0 1 40 2 x P = − = P(1) − 40 2 = 41 − 20 = 21 Solution SET -D Indore : 3-A, Residency Area, Manoramaganj, : 0731 -2495865 Web : www.guptatutorials.com Mhow : 18, Vikram Nagar, Kishanganj, A.B. Road : 07324 -324944 e-mail : Info@guptatutorials.com AIEEE 2010 A4.doc 17 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 Sol. ( ) ( ) ( ) ( ) ( ) ( ) ' 2 2 2 ' 2 2 .2 ' g x f f x f f x f x = + + ( ) ( ) ( ) ( ) ( ) ( ) ' 2 2 2 ' 2 2 .2 ' g x f f x f f x f x = + + ( ) ( ) ( ) 2 0 . ' 0 .2 ' 0 f f f = ( ) ( ) ( ) ( ) 2 4 0 . ' 0 4. 1 4 f f = = − = − 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 Sol. No. of ways = 3C2 × 9C2 = 3 × 36 = 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 Sol. 1 2 1 2 3 1 0 3 5 2 D = = 3 2 1 3 3 1 5 0 1 5 2 x D = = ¹ \ By cramer’s Rule, system of equations has 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) 27 (2) 1 21 (3) 2 23 (4) 13 Sol. Required prob. 3 4 2 1 1 1 9 3 27 C C C C × × = 89. For two data sets, each of size 5, the variances 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) 11 2 (2) 6 (3) 13 2 (4) 52 Sol. Variance of combined series ( ) ( ) 2 2 2 1 1 2 2 1 2 1 2 2 1 2 1 2 n n n n m m n n n n s s + = + − + + ( ) ( ) 2 2 5 5 2 4 5 4 5 5 5 5 5 5 × − × + × = + + + 45 25 4 9 11 1 10 100 2 2 × = + = + = = 11 2 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 Sol. Centre : (2, + 4) Radius : 4 16 5 5 + + = CP < R 6 18 5 5 m − − < 10 25 m − − < −25 < 10 + m < 25 −35 < m <15