AIEEE 2010 1 AIEEE -2010 Question paper with Solutions Disclaimer • The solutions have been created by the teachers of BASE with an intention to help students. • While care has been taken to ensure accuracy, BASE will not be responsible for any inadvertent error that may exist in the solution. However, BASE will welcome feedback on any errors noticed. The same may be sent via email to info@base-edu.in • BASE does not stake claim to the fact that the solutions it has provided are the only ways in which questions could be answered. • The solutions provided here cannot be used as a support for any legal disputes by any individual or organization. AIEEE 2010 2 AIEEE – 2010 Question paper (Code – A) Duration: 3 hours with Solutions Max. Marks: 432 Marks Pattern Part A – PHYSICS (144 marks) –Questions No. 1 to 20 and 23 to 26 consist of FOUR (4) marks each and Questions No. 21 to 22 and 27 to 30 consist of EIGHT (8) marks each for each correct response. Part B – CHEMISTRY (144 marks) – Questions No. 31 to 39 and 43 to 57 consist of FOUR (4) marks each and Questions No. 40 to 42 and 58 to 60 consist of EIGHT (8) marks each for each correct response. Part C – MATHEMATICS (144 marks) – Questions No. 61 to 66, 70 to 83 and 87 to 90 consist of FOUR (4) marks each and Questions No. 67 to 69 and 84 to 86 consist of EIGHT (8) marks each for each correct response. Candidates will be awarded marks as stated above 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. PART – A : PHYSICS Directions : Questions number 1-3 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index µ(I) = µ0 + µ2I, 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. 1. The initial shape of the wavefront of the beam is (1) Planar (2) Convex (3) Concave (4) Convex near the axis and concave near the periphery Ans (1) Wavefront of a parallel cylindrical beam is planar. 2. The speed of light in the medium is (1) Maximum on the axis of the beam (2) Minimum on the axis of the beam (3) The same everywhere in the beam (4) Directly proportional to the intensity I Ans (2) 3. As the beam enters the medium, it will (1) travel as a cylindrical beam (2) diverge (3) converge (4) diverge near the axis and converge near the periphery Ans (3) AIEEE 2010 3 Directions : Questions number 4-5 are based on the following paragraph. A nucleus of mass M + m is at rest and decays into two daughter nuclei of equal mass 2M each. Speed of light is c. 4. The speed of daughter nuclei is (1) m M m c D + D (2) m M m c D + D (3) Mm 2 c D (4) Mm c D Ans (3) Linear momentum is conserved. v v v 0 v 2M v 2M 2 1 2 1 = = = − say. Energy released is in the form of kinetic energy of daughter nuclei. Mm 2 c v v 2M 21 2 mc 2 2 D = = D 5. The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then (1) E1 = 2 E2 (2) E2 = 2 E1 (3) E1 > E2 (4) E2 > E1 Ans (4) Stability of the daughter nucleus is greater than stability of the parent nucleus. Hence, the answer. Directions : Questions number 6-7 contain Statement-I and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. 6. 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 false (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1 (3) Statement-1 is true, Statement-2 is true; Statement-2 is the not the correct explanation of Statement-1 (4) Statement-1 is false, Statement-2 is true Ans (1) Frequency and hence the energy of a X-rays is greater than ultraviolet light. Therefore, V0 and Kmax both increase. Though the energy gained by each electron is the same, the energy lost during collision with lattice points is different for different electrons. It thus, results in a range for speed. \ Statement II is false. 7. 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 false (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1 (3) Statement-1 is true, Statement-2 is true; Statement-2 is the not the correct explanation of Statement-1 (4) Statement-1 is false, Statement-2 is true Ans (2) AIEEE 2010 4 8. 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) 0.2 Ns (2) 0.4 Ns (3) 0.8 Ns (4) 1.6 Ns Ans (3) Impulse J = m |(v − u)| 0 2 0 2 2 4 2 0 4 . 0 −− − −− = = 0.4 |−2| = 0.8 Ns 9. 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 XXis given by (1) (2) (3) (4) Ans (2) ) j ˆ ( ) x d 2 ( 2 i j ˆ d 2 i B 0 0 bet − − p µ − p µ = for x = d, Bbet = 0 x < d, j ˆ Bbet x > d, ) j ˆ ( Bbet − 10. A ball is made of a material of density r where roil < r < rwater with roil and rwater 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 positions? (1) (2) (3) (4) Ans (3) AIEEE 2010 5 High density liquid (water) will be at the bottom. As density of ball is in between oil and water, it floats in between. 11. 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 ˆ r 2 q 2 0 2 e p (2) j ˆ r 4 q 2 0 2e p (3) j ˆ r 4 q 2 0 2e p − (4) j ˆ r 2 q 2 0 2e p − Ans (4) r pe = l − q = ) j ˆ ( sin dE E − q q p = ) j ˆ ( sin d r qr rK 2 2 j ˆ r 2 q 2 0 2e p − = 12. 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) 0.25 (2) 0.5 (3) 0.75 (4) 0.99 Ans (3) For adiabatic process TVg−1 = constant 1 57 1 1 57 ) V 32 ( T TV − − = 4T T T 4 T = ¢ ¢ = 75 . 0 TT 1 = ¢ − = h 13. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are (1) 4, 4, 2 (2) 5, 1, 2 (3) 5, 1, 5 (4) 5, 5, 2 Ans (2) 14. The combination of gates shown below yields (1) NAND gate (2) OR gate (3) NOT gate (4) XOR gate Ans (2) dq dE dq q AIEEE 2010 6 B A B A B A + = + = × 15. If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called (1) -rays (2) X-rays (3) Ultraviolet rays (4) Microwaves Ans (2) p = nhg 34 20 3 10 023 . 6 10 10 4 − × × × = g = 6 × 1016 Hz This frequency belongs to X-rays. 16. 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) 2 Z 4 Z A − − − (2) 4 Z 8 Z A − − − (3) 8 Z 4 Z A − − − (4) 4 Z 12 Z A −− − Ans (3) Emission of a-particle results in decrease of mass number by 4 and decrease of atomic number by 2. Emission of position results in decrease of atomic (proton) number by 1 and increase neutron number by 1. \ np = Z − 3 × 2 − 2 × 1 = Z − 8 nh = A − Z − 3 × 2 + 2 × 1 = A − A − 4 \ Ratio 8 Z 4 Z A − − − = 17. Let there be a spherically symmetric charge distribution with charge density varying as − r = r Rr 45 ) 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 < R) from the origin is given by (1) − e r Rr 45 3 r0 0 (2) − e pr Rr 53 3 r 4 00 (3) − e r Rr 35 4 r0 0 (4) − e r Rr 45 3 r 4 0 0 Ans (3) − r p = Rr 45 dr r 4 dq 0 2 − pr = \ r0 3 2 0 dr Rr dr r 45 4 q Using e = 0 q EdA − e r = Rr 35 4 r E 0 0 AIEEE 2010 7 18. 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) 242 W (2) 305 W (3) 210 W (4) 0 W Ans (1) C L X X = W 242 R V P 2 = = \ 19. In the circuit shown below, the key K is closed at t = 0. The current through the battery is (1) ¥ = = + t at RV and 0 t at R R ) R R ( V 2 2 1 2 1 (2) ¥ = = + t at RV and 0 t at R R R VR 2 22 21 2 1 (3) ¥ = + = t at R R ) R R ( V and 0 t at RV 2 1 2 1 2 (4) ¥ = + = t at R R R VR and 0 t at RV 22 21 2 1 2 Ans (3) At t = 0, 2 RV i = 2 1 2 1 eq R R ) R R ( V RV i , t At + = = ¥ = 20. A particles is moving with velocity ) j ˆ x i ˆy ( K v + = r , where K is a constant. The general equation for its path is (1) y2 = x2 + constant (2) y = x2 + constant (3) y2 = x + constant (4) xy = constant Ans (1) ) j ˆ x i ˆy ( K v + = r ky dt dx vx = = Kx dt dy vy = = or yx dx dy = or ydy = x dx or C 2 x 2 y 2 2 + = or y2 = x2 + constant. AIEEE 2010 8 21. 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 21 tt will be (1) 2 (2) 1 (3) 21 (4) 41 Ans (4) t − t − = = = t 2 20 2 t 0 2 e C 2q e q C 21 C 2q U t − = t 2 i e U U 2 n 2T t e U 2 U 1 t 2 i 1 1 l = = t − t − = t 0 e q q 41 tt e q 4 q 21 T 2t 0 0 = = − 22. A rectangular loop has a sliding connector PQ of length l 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 I1, I2 and I are (1) Rv B I , Rv B I I 2 1 l l = = − = (2) R 3 v B 2 I , R 3 v B I I 2 1 l l = = − = (3) R 3 v B 2 I , I I 2 1 l = = (4) R 3 v B I , R 6 v B I I I 2 1 l l = = = = Ans (2) Induced emf, e = Blv. i = i1 + i2 KVL: i1R + iR − Blv = 0 and i2R + iR − Blv = 0 R 3Blv 2 i = R 3Blv i i 2 1 = = . 23. The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by − p = ) m ( 50 . 0 x ) s ( 04 . 0 t 2 sin ) m ( 02 . 0 y . The tension in the string is (1) 6.25 N (2) 4.0 N (3) 12.5 N (4) 0.5 N Ans (1) 1 ms 5 . 12 k v − = w = R R R i2 i2 AIEEE 2010 9 2 v T T v µ = µ = 22 kw µ = = 6.25 N 24. 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 m s–2 in vertical direction (2) 4.9 m s–2 in horizontal direction (3) 9.8 m s–2 in vertical direction (4) Zero Ans (1) aAy = g sin2 60° 4 3 8 . 9 a Ay × = aBy = g sin2 30° 48 . 9 a By = 2 By Ay ms 9 . 4 ) 1 3 ( 48 . 9 a a − = − = − 25. 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) j ˆ Rv i ˆRv 2 2 + (2) j ˆ sin Rv i ˆcos Rv 2 2 q + q − (3) j ˆ cos Rv i ˆsin Rv 2 2 q + q − (4) j ˆ sin Rv i ˆcos Rv 2 2 q − q − Ans (4) Refer to the diagram, 26. 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 gsin v t 0 q < , the angular momentum of the particle is (1) i ˆcos t mgv 21 2 0 q (2) j ˆ cos t mgv 2 0 q − (3) k ˆcos t mgv0 q (4) k ˆcos t mgv 21 2 0 q − where k ˆand j ˆ , i ˆ are unit vectors along x, y and z-axis respectively. Ans (4) v r ar q q q − cos Rv2 ar q − sin Rv2 AIEEE 2010 10 t = r r dt L d or dt L d t = r r or { }dt k ˆt cos v mg ( L d 0 q − = r or { }q − = = dt t k ˆcos v mg L d L 0 r r or k ˆcos t v mg 21 L 2 0 q − = r 27. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° 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) 1 (2) 4 (3) 3 (4) 2 Ans (4) q = tan mg F q = r − tan d 1 mg ) K /F ( and 2 d 1 1 K = r − = 28. 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) 14 m /s2 (2) 13 m /s2 (3) 12 m /s2 (4) 7.2 m /s2 Ans (1) s = t3 = 5 2 t 3 dt ds v = = at t = 2s, v = 12 ms−1. So 2 2 r ms 20 144 r v a − = = t 6 dt dv a t = = at t = 2s, at = 12 ms−2 Thus, 2t 2r a a a + = 2 2 12 2 . 7 + = 144 50 + @ 194 @ 2 ms 14 − @ y x v r v0 q g mr t ) cos v ( 0 q AIEEE 2010 11 29. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by , xb xa ) x ( U 6 12 − = where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U(x = ) – Uat equilibrium], D is (1) a 6b2 (2) a 2b2 (3) a 12 b2 (4) a 4b2 Ans (4) 6 12 xb xa ) x ( U − = [ ] 7 13 bx 6 ax 12 dx dU F − − + − − = − = At equilibrium, F = 0 or − 12 ax−13 + 6bx−7 = 0 Or a 2b x1 a 2b x 6 6 = = − Thus, − = 6 2 6 eg x1 b x1 a v a 4b U a 2b a 4b a 2 eq 2 2 2 − = − = Thus, D = U (¥) − U (eq) = 0 − − a 4b2 Or a 4b D 2 = 30. Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are a1 and a2. The respective temperature coefficients of their series and parallel combinations are nearly (1) 2 , 2 2 1 2 1 a + a a + a (2) 2 1 2 1 , 2 a + a a + a (3) 2 , 2 1 2 1 a + a a + a (4) 2 1 2 1 2 1 , a + a a a a + a Ans (1) R1 = R0 (1 + a1t1) R2 = R0 (1 + a2t) Rs = R1 + R2 2R0 (1 + ast) = 2R0 + R0 (a1t + a2t) 2 2 1 s a + a = a ) t 1 ( R ) t 1 ( R ) t 1 ( R ) t 1 ( R R R R R R 2 0 1 0 2 0 1 0 2 1 2 1 p a + + a + a + a + = + = ] t t 2 [ R ] t t t 1 [ R ) t 1 ( 2 R 2 1 0 2 2 1 2 1 20 p 0 a + a + a a + a + a + = a + On simplifying 2 2 1 p a + a = a AIEEE 2010 12 NaNO2 HCl, 278 K NH2 A B HBF4 NH2 N NCl HBF4 F A B PART – B : CHEMISTRY 31. In aqueous solution the ionisation constants for carbonic acid are K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11 Select the correct statement for a saturated 0.034 M solution of the carbonic acid. (1) The concentration of H+ is double that of − 23 CO (2) The concentration of − 23 CO is 0.034 M (3) The concentration of − 23 CO is greater than that of −3 HCO (4) The concentrations of H+ and −3 HCO are approximately equal Ans (4) 32. 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) 5.0 × 10–8 g (2) 1.2 × 10–10 g (3) 1.2 × 10–9 g (4) 6.2 × 10–5 g Ans (3) Ksp (AgBr) = [Ag+][Br−] 5.0 × 10−13 = (0.05) [Br−] M 10 1 05 . 0 10 0 . 5 ] Br [ 11 13 − − − × = × = \ Weight of KBr required = Molarity × GMM × Vol (in L) = 1 × 10−11 × 120 × 1 = 1.2 × 10−9 g. 33. The correct sequence which shows decreasing order of the ionic radii of the elements is (1) O2– > F– > Na+ > Mg2+ > Al3+ (2) Al3+ > Mg2+ > Na+ > F– > O2– (3) Na+ > Mg2+ > Al3+ > O2– > F– (4) Na+ > F– > Mg2+ > O2– > Al3+ Ans (1) 34. In the chemical reactions, the compounds 'A' and 'B' respectively are (1) Nitrobenzene and chlorobenzene (2) Nitrobenzene and fluorobenzene (3) Phenol and benzene (4) Benzene diazonium chloride and fluorobenzene Ans (4) AIEEE 2010 13 H5C6CH2CH2 C CH2 H3C H5C6 C C H HCH(CH3)2 C C H CH3 CH3 C6H5CH2 C C H CH(CH3)2 H C6H5 35. 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) 1.27 × 10–3 mol (2) 5.56 × 10–3 mol (3) 1.53 × 10–2 mol (4) 4.46 × 10–2 mol Ans (1) mol 10 27 . 1 300 314 . 8 10 3170 RT PV n 3 3 − − × = × × = = 36. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is (1) 1-Butanol (2) 2-Butanol (3) 2-Methylpropan-2-ol (4) 2-Methylpropanol Ans (3) Order of reactivity of alcohols towards Lucas reagent (Zn + conc. HCl) is 3° > 2° > 1°. 37. 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 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1) (1) 0.0186 K (2) 0.0372 K (3) 0.0558 K (4) 0.0744 K Ans (1) DT = Kf × molality = 1.86 × 0.01 = 0.0186 K. 38. Three reactions involving −4 2PO H are given below (i) H3PO4 +H2O ® H3O+ + −4 2PO H (ii) + − − + ® + O H HPO O H PO H 3 24 2 4 2 (iii) − − − + ® + 2 4 3 4 2 O PO H OH PO H In which of the above does −4 2PO H act as an acid? (1) (i) only (2) (ii) only (3) (i) and (ii) (4) (iii) only Ans (2) Only in equation (II), −4 2PO H acts as proton donar. 39. The main product of the following reaction is ¾ ¾ ¾ ¾ ® ¾ 4 2SO H . conc 2 3 2 5 6 ) CH ( CH ) OH ( CH CH H C (1) (2) (3) (4) Ans (2) Acid catalysed dehydration involves formation of stable carbocation and is anti-elimination. 40. 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 m s–1 and NA = 6.02 × 1023 mol–1) (1) 494 nm (2) 594 nm (3) 640 nm (4) 700 nm Ans (1) AIEEE 2010 14 Energy = Nhn per mole l = c Nh l × × × × × = − 8 34 23 10 3 10 62 . 6 10 02 . 6 242000 242000 10 3 10 62 . 6 10 02 . 6 8 34 23 × × × × × = l \ − = 0.000494 × 10−3 m = 494 nm. 41. 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 neutralisation. The percentage of nitrogen in the compound is (1) 29.5 (2) 59.0 (3) 47.4 (4) 23.7 Ans (4) Mass of organic compound (m) = 29.5 mg = 0.0295 g Normality of HCl = 0.1 N Volume of HCl used up = (20 − 15) = 5.0 cm3 Percentage of nitrogen = mNV 4 . 1 × % 7 . 23 0295 . 0 0 . 5 1 . 0 4 . 1 = × × = 42. Ionisation energy of He+ is 19.6 × 10–18 atom–1. The energy of the first stationary state (n = 1) of Li2+ is (1) 8.82 × 10–17 J atom–1 (2) 4.41 × 10–16 J atom–1 (3) –4.41 × 10–17 J atom–1 (4) –2.2 × 10–15 J atom–1 Ans (3) According to Bohr’s atomic model, 22 22 2 21 21 1 22 Zn E Zn E and nZ E = µ For n1 = n2 = 1, we can write 22 2 18 222 211 3E 210 6 . 19 or ZE ZE = × = − 9 410 6 . 19 E 18 2 × × = \ − 1 17 atom J 10 41 . 4 − − × = \ The energy of the first stationary state of Li2+ = − 4.41 × 10−17 J atom−1. 43. On mixing, heptane and octane from 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) 144.5 kPa (2) 72.0 kPa (3) 36.1 kPa (4) 96.2 kPa Ans (2) We know that heptane and octane form an ideal mixture. meq NH3 = 0.5 mass of N in mg = 0.5 × 14 % % 7 . 23 100 14 5 . 295 . 0 N = × × = AIEEE 2010 15 O H ozonolysis 2 Co Co enen en en en en 3+ 3+ o B B o A A total P X P X P + = = 0.45 × 105 + 0.55 × 45 = 47.25 + 24.75 = 72.0 k Pa Where XA is the mole fraction of heptane 45 . 0 55 . 0 25 . 0 3 . 0 25 . 0 25 . 0 114 35 100 25100 25 n n n X B A A A = = + = + = + = XB is the molefraction of octane. 55 . 0 55 . 0 3 . 0 3 . 0 25 . 0 3 . 0 114 35 100 25114 35 n n n X B A B B = = + = + = + = 44. Which one of the following has an optical isomer? (1) [Zn(en)2]2+ (2) [Zn(en)(NH3)2]2+ (3) [Co(en)3]3+ (4) [Co(H2O)4(en)]3+ Ans (3) [Co(en)3]3+ has an optical isomer 45. Consider the following bromides The correct order of SN1 reactivity is (1) A > B > C (2) B > C > A (3) B > A > C (4) C > B > A Ans (1) Higher the stability of carbocation greater the reactivity towards 1 N S reactions. 46. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene (1) Ethane (2) Propene (3) 1-butene (4) 2-butene Ans (4) AIEEE 2010 16 47. Consider the reaction Cl2 (aq) + H2S (aq) ® S(s) + 2H+(aq) + 2Cl–(aq). The rate equation for this reaction is rate = k [Cl2] [H2S]. 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) A only (2) B only (3) Both (A) & (B) (4) Neither (A) nor (B) Ans (1) 48. The Gibbs energy for the decomposition Al2O3 at 500°C is as follows 1 r 2 3 2 mol kJ 966 G , O Al 34 O Al 32 − + = D + ® The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least (1) 5.0 V (2) 4.5 V (3) 3.0 V (4) 2.5 V Ans (4) 2 3 2 O Al 34 O Al 32 + ® DrG = +966 kJ mol−1 2Al2O3 ® 4Al + 3O2 Al2O3 ® 2Al + 2 O 23 mol /kJ 1449 2 3 966 G r = × + = D We know that DG° = −nFE° − DG° = nFE° 1449000 = 6 × 96500 × E° V 5 . 2 96500 61449000 E = × = ° \ 49. The correct order of increasing basicity of the given conjugate bases (R = CH3) is (1) R H N C HC O RCO 2 < < º < (2) 2 H N R C HC O RCO < < º < (3) 2 H N O RCO C HC R < < º < (4) R C HC H N O RCO 2 < º < < Ans (1) Lesser the stability of conjugate base greater the basicity and weaker be its conjugate acid. 50. 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) 144 pm (2) 288 pm (3) 398 pm (4) 618 pm Ans (1) Edge length = 508 pm = 2(r+ + r−) ; pm 144 110 2 508 r = − = − Radius of cation = 110 pm AIEEE 2010 17 r r r r * cation anion cation \ Diameter of anion = 508 − 2 × 110 = 288 pm \ Radius of anion = pm 144 2 288 = 51. Out of the following, the alkene that exhibits optical isomerism is (1) 2-methyl-2-pentene (2) 3-methyl-2-pentene (3) 4-methyl-1-pentene (4) 3-methyl-1-pentene Ans (4) (3-methyl-1-pentene) has one chiral carbon, hence, exhibits optical activity. 52. 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) T = Te (2) Te > T (3) T > Te (4) Te is 5 times T Ans (3) ) ve ( S T ) ve ( H G + D − + D = D At equilibrium, DG = 0. DG becomes −ve (i.e., the reaction becomes spontaneous) only if the temperature T is greater than the temperature at equilibrium (Te). 53. Percentages of free space in cubic close packed structure and in body centered packed structure are respectively (1) 48% and 26% (2) 30% and 26% (3) 26% and 32% (4) 32% and 48% Ans (3) In CCP lattice, packing efficiency is 74 % \ Free space = 100 − 74 = 26 % In bcc lattice, packing efficiency is 68 % \ Free space = 100 − 68 = 32 % 54. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is (1) Natural rubber (2) Teflon (3) Nylon 6, 6 (4) Polystyrene Ans (3) 55. 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) 8 (2) 9 (3) 10 (4) 11 Ans (3) Ksp = [Mg2+][OH−]2 1.0 × 10−11 = (0.001) (OH−)2 or 8 11 2 10 1 001 . 0 10 0 . 1 ] OH [ − − − × × = M 10 1 10 1 ] OH [ or 4 8 1 − − − × = × = \ pOH = 4 or pH = 10 AIEEE 2010 18 56. The correct order of 0M M2 E + values with negative sign for the four successive elements Cr, Mn, Fe and Co is (1) Cr > Mn > Fe > Co (2) Mn > Cr > Fe > Co (3) Cr > Fe > Mn > Co (4) Fe > Mn > Cr > Co Ans (2) 57. Biuret test is not given by (1) Proteins (2) Carbohydrates (3) Polypeptides (4) Urea Ans (1) Biuret test is not answered by carbohydrates 58. The time for half life period of a certain reaction A ® Products is 1 h. 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) 1 h (2) 4 h (3) 0.5 h (4) 0.25 h Ans (4) h 25 . 0 1 2 5 . 0 t , L mol 1 k k 2A t 21 1 0 21 = × = = = − 59. A solution containing 2.675 g of CoCl3.6NH3 (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) [CoCl(NH3)5]Cl2 (2) [Co(NH3)6]Cl3 (3) [CoCl2(NH3)4]Cl (4) [CoCl3(NH3)3] Ans (2) Number of moles of mol 01 . 0 gmol 5 . 267 g 675 . 2 NH 6 . COCl 1 3 3 = = − Number of moles of AgCl precipitated mol 033 . 0 gmol 5 . 143 g 78 . 4 1 = = − \ 0.01 mole of the complex precipitates 0.033 mole of Cl− ions in the form of AgCl. As, such the formula of the complex is [Co(NH3)6]Cl3 60. The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 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 is NH3 is (1) –1102 kJ mol–1 (2) –964 kJ mol–1 (3) +352 kJ mol–1 (4) +1056 kJ mol–1 Ans (3) 3 2 2 NH H 23 N 21 ® + DHr × n = BE(reactants) − BE(products) ) H N ( BE 3 ) 436 ( 23 ) 712 ( 21 0 . 46 − × − − + − = − −46.0 = −356 − 654 − 3 × BE (N − H) \ 3 × BE(N − H) = −964 1 mol kJ 3 . 321 3 964 ) H N ( BE − − = − = − \ Average bond dissociation energy of N−H bond in NH3 is » 352 kJ mol−1 AIEEE 2010 19 PART – C : MATHEMATICS 61. Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; = qp , nm S m, n, p and q are integers such that n, q ¹ 0 and qm = pn}. Then (1) R is an equivalence relation but S is not an equivalence relation (2) Neither R nor S is an equivalence relation (3) S is an equivalence relation but R is not an equivalence relation (4) R and S both are equivalence relations Ans (3) { } w w = = number rational some for y x and numbers real are y , x /) y , x ( R = ¹ = pn qm and 0 q , n that such egers int are q , p , n , m /qp , nm S Let us consider the relation R (i) R is reflexive: Q for any x Î R, x = 1.x where 1 is rational (x, x) Î R "x (ii) R is not symmetric: Clearly ( ) R 2 , 0 Î since 2 . 0 0 = where 0 is rational But ( ) R 0 , 2 Ï since 0 . 2 w = is not true for any w rational. R is not an equivalence relation Let us now consider the relation S (i) S is reflexive: nm for Q such that m, n Î Z and n 0 We have mn = nm S nm , nm Î " such nm (ii) S is symmetric: we have nm and qp such that n, q 0 and m, n, p, q Î Z qm = pn iff np = mq Û S nm , qp iff S qp , nm Î Î (iii) S is transitive: S sr , qp and S qp , nm Î Î Clearly m, n, p, q, r, s Î Z and n, q, s 0 And qm = pn … (1) sp = qr … (2) Take product of equation (1) and (2), we get pqms = pqrn ms = rn S sr , nm Î S is an equivalence relation AIEEE 2010 20 62. The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals (1) 0 (2) 1 (3) 2 (4) Ans (2) We need to find the number of complex numbers z obeying the equation |z − 1| = |z + 1| = |z − i| Clearly the above equation says that z is equidistant from the points 1, −1, i. Since 1, −1, i are non-collinear points, it forms a unique triangle for which there is a unique circumcentre. Hence, there exist only one such z. 63. If and are the roots of the equation x2 – x + 1 = 0, then 2009 + 2009 = (1) –2 (2) –1 (3) 1 (4) 2 Ans (3) Given a and b are the roots of the equation x2 − x + 1 = 0. Let us multiply the given equation by x + 1 both sides, we get x3 + 1 = 0 x = 31 ) 1 (− x = − 1, −w, −w2 where w is the complex cube root of unity. Q x + 1 is a factor we introduced, we do not consider its corresponding root x = − 1. \ x = −w or −w2 So, 2009 2 2009 2009 2009 ) ( ) ( w − + w − = b + a 2009 2 2009 ( ) = − w + w 2 2 2 ( ) = − w + w (Q 2009 = 3k + 2 for some k Î Z) 2 1 = − w + w = 64. Consider the system of linear equations: x1 + 2x2 + x3 = 3 2x1 + 3x2 + x3 = 3 3x1 + 5x2 + 2x3 = 1 The system has (1) Infinite number of solutions (2) Exactly 3 solutions (3) A unique solutions (4) No solution Ans (4) x1+ 2x2 + x3 = 3 . . . (1) 2x1 + 3x2 + x3 = 3 . . . (2) 3x1 + 5x2 + 2x3 = 1 . . . (3) Adding equations (1) and (2), we get 3x1 + 5x2 + 2x3= 6. But according to (3), 3x1 + 5x2 + 2x3 = 1. Hence, no solution. 65. 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) 3 (2) 36 (3) 66 (4) 108 Ans (4) 3C2 × 9C2 = 3 × 36 = 108 AIEEE 2010 21 66. 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) –4 (3) 0 (4) –2 Ans (2) Given f : (−1, 1) ® R is a differentiable function with f(0) = −1 And f¢(0) = 1 and g(x) = [f(2f(x) +2)]2. We need to find g¢ (0) )] x ( f 2 )][ 2 ) x ( f 2 ( f )][ 2 ) x ( f 2 ( f [ 2 ) x ( g ¢ + ¢ + = ¢ Put x = 0, g¢ (0) = 2[f(2f(0)+2)][f¢(2f(0) + 2)] [2f¢(0)] = −4 67. Let f : R R be a positive increasing function with ) x ( f ) x 2 ( f lim Then . 1 ) x ( f ) x 3 ( f lim x x ¥ ® ¥ ® = (1) 1 (2) 32 (3) 23 (4) 3 Ans (1) Given that f : R ® R is a positive increasing function. and 1 ) x ( f ) x 3 ( f lim x = ¥ ® ) x ( f Q is an increasing function f(x) < f(2x) < f(3x) 0 x > " Let us divide the above inequality by f(x) throughout We get ) x ( f ) x 3 ( f ) x ( f ) x 2 ( f 1 < < 0 x > " (note that the inequality is preserved only because f(x) > 0 ) R x Î " Now we use sandwich theorem to the above inequality and conclude that 1 ) x ( f ) x 2 ( f lim x = ¥ ® 68. Let p(x) be a function defined on R such that ), x 1 ( p ) x ( p , 1 ) x ( f ) x 3 ( f lim x − ¢ = ¢ = ¥ ® for all x Î [0, 1], p(0) = 1 and p(1) = 41. Then 1 0 dx ) x ( p equals (1) 41 (2) 21 (3) 41 (4) 42 Ans (2) Given, p(x) is function defined on 1R such that ) x 1 ( p ) x ( p − ¢ = ¢ ] 1 , 0 [ xÎ " and p(0) = 1 p(1) = 41 We need to find 1 0 dx ) x ( p Since, we have ) x 1 ( p ) x ( p − ¢ = ¢ ] 1 , 0 [ xÎ " Let us take definite integral over the interval [0, y] where ] 1 , 0 [ yÎ On both sides, we get − ¢ = ¢ y0 y0 dx ) x 1 ( p ) x ( p y0 y0 )] x 1 ( p [ )] x ( p [ − − = [by fundamental theorem of calculus] p(y) − p(0) = −p(1 − y) + p(1) p(y) + p(1 − y) = p(0) + p(1) = 42 " y Î [0, 1] AIEEE 2010 22 Again take definite integral over the interval [0, 1], we get = − + 10 10 10 dy 42 dy ) y 1 ( p dy ) y ( p = 10 42 dy ) y ( p 2 − + = ba ba dx ) x b a ( f dx ) x ( f Q = 10 21 dy ) y ( p 69. 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) 24 minutes (2) 34 minutes (3) 125 minutes (4) 135 minutes Ans (2) A person is to count 4500 current notes. an denotes the number of notes he counts in the nth minute if ai = 150 " i = 1, 2, . . . 10 and a10, a11 . . . are in A.P. with common difference −2 Now, we need to find the time taken by him to count all the notes. Let it take ‘p’ minutes for him to count all the notes. Clearly p >10. Otherwise if p £ 10 he can count maximum of 1500 notes only ] 2 ) 10 i ( 150 [ 150 a a a P11 i 101 i i P11 i i 101 i i P1 i − − S + S = S + S = S \ = = = = = ) 10 p ( 170 1500 ) i 2 170 ( 150 P11 i 101 i − + = − S + S = = = 10 ) 1 p ( p + + − = 90 p 170 ) 1 p ( p a 4500 i P1 i − + + − = S = = 0 4590 p 170 ) 1 p ( p = + − + 0 4590 p 169 p2 = + − 0 ) 34 p )( 135 p ( = − − Clearly, we can seen that 170 − 2i < 0 of i > 85 and 170 − 2i represents the number of notes counted on ith minute when i > 10. So, 170 − 2i cannot be negative. p = 34 minutes. 70. The equation of the tangent to the curve 2 x4 x y + = , that is parallel to the x-axis, is (1) y = 0 (2) y = 1 (3) y = 2 (4) y = 3 Ans (4) The given curve is 2 x4 x y + = and it is understood that x ¹ 0. A tangent of the above curve will be parallel to x-axis iff its derivative is zero as the given function is a differentiable function. 3 y 2 x 8 x 0 x8 1 y 3 3 = = = = − = ¢ \ The point of contact of the tangent with zero slope is (2, 3) Equation of the tangent is y = 3. AIEEE 2010 23 71. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and 2 3 x p = is (1) 2 2 4 − (2) 2 2 4 + (3) 1 2 4 − (4) 1 2 4 + Ans (1) We need to find the area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and 2 3 x p = Let us draw a rough diagram of the graph as per the given data. clearly, cosx ³ sinx " x Î p4 , 0 cosx £ sinx " x Î p p 4 5 , 4 cosx ³ sinx " x Î p p 2 3 , 4 5 Required area = dx x sin x cos 2 3 0 p − dx x sin x cos dx x sin x cos dx x sin x cos 2 34 5 4 5 4 4 0 p p p p p − + − + − = ( ) ( ) ( )dx x sin x cos dx x cos x sin dx x sin x cos 2 34 5 4 5 4 4 0 p p p p p − + − + − = [ ] [ ] [ ] 2 34 5 4 54 40 x cos x sin x sin x cos x cos x sin pp p p p + + − − + + = ( ) ( ) 2 1 2 2 1 2 + − + + − = 2 2 4 − = 72. Solution of the differential equation 2 x 0 , dx ) y x (sin y xdy cos p < < − = is (1) secx = (tanx + c)y (2) ysecx = tanx + c (3) ytanx = secx + c (4) tanx = (secx + c)y Ans (1) Given differential equation is cos x dy = y (sin x − y)dx cos x dy = y sinx dx − y2 dx cosx dy − ysinx dx = −y2 dx dividing throughout by −y2 cos2 x, we get xdx sec xdx tan x sec y1 xdy sec y1 2 2 = + − O y x 4p 2p 2 3p 4 5p AIEEE 2010 24 ) x (tan d y x sec d = y ) c x (tan x sec c x tan y x sec + = + = 73. Let k ˆj ˆ i ˆc and k ˆj ˆ a − − = − = r r . Then the vector br satisfying 0 c b a r r r r = + × and 3 b a = × r r is (1) k ˆ2 j ˆ i ˆ− + − (2) k ˆ2 j ˆ i ˆ2 + − (3) k ˆ2 j ˆ i ˆ− − (4) k ˆ2 j ˆ i ˆ− + Ans (1) Given k ˆj ˆ i ˆc and k ˆj ˆ a − − = − = r r we need to find br such that 3 b . a and 0 c ) b a ( = = + × r r r r r r Let k ˆz j ˆ y i ˆx b + + = r Now c k ˆj ˆ i ˆ z y x 1 1 0 k ˆj ˆ i ˆ b a r r r − = + + − = − = × k ˆj ˆ i ˆk ˆx j ˆ x i ˆ) z y ( + + − = − − + x = −1, y + z = − 1 … (1) Now, 3 b a = × r r 3 ) k ˆz j ˆ y i ˆx ( ). k ˆj ˆ ( = + + − y − z = 3 … (2) From (1) and (2), we get y = 1, z = −2 k ˆ2 j ˆ i ˆb − + − = \ r 74. If the vectors k ˆj ˆ i ˆc and k ˆj ˆ 4 i ˆ2 b , k ˆ2 j ˆ i ˆa µ + + l = + + = + − = r r r are mutually orthogonal, then (l, µ) = (1) (−3, 2) (2) (2, −3) (3) (−2, 3) (4) (3, −2) Ans (1) Let k ˆ2 j ˆ i ˆa + − = r , k ˆj ˆ 4 i ˆ2 b + + = r and k ˆj ˆ i ˆc µ + + l = r and it is given that c , b , a r r r are mutually orthogonal. c k b a r r r = × for some scalar ( ) k ˆj ˆ i ˆk 1 4 2 2 1 1 k ˆj ˆ i ˆ µ + + l = − ( ) k ˆj ˆ i ˆk k ˆ6 j ˆ 3 i ˆ9 µ + + l = + + − By comparing the x, y, z components, we get k = 3, l = −3, µ = 2 (l, µ) = (−3, 2) 75. If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is (1) x = 1 (2) 2x + 1 = 0 (3) x = –1 (4) 2x – 1 = 0 Ans (3) Given P is a point such that two tangents drawn from P to the parabola y2 = 4x are at angles or in other words, P is a point of intersection a pair of perpendicular tangents of a parabola y2 = 4x. AIEEE 2010 25 We know that the locus of points of intersection of perpendicular tangents of a parabola is a line and that line is the directrix. P is a point on the directrix of y2 = 4x locus of P is x = −1 76. The line L given by 1 by 5x = + passes through the point (13, 32). The line K is parallel to L and has the equation 1 3y cx = + . Then the distance between L and K is (1) 15 23 (2) 17 (3) 15 17 (4) 17 23 Ans (4) Equation of line L is 1 by 5x = + and it passes through (13, 32) 1 b 32 5 13 = + b = −20 Equation of line K is 1 3y cx = + and it is parallel to line L 43 c c3 4 c3 5b − = = − − = − Equation of line L can be written as 4x − y = 20 Equation of line K can be written as 4x − y = −3 \ Distance between lines L and K is 17 23 ) 1 ( 4 ) 3 ( 20 2 2 = − + − − 77. 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 equals (1) 30° (2) 45° (3) 60° (4) 75° Ans (3) Given that the direction angles of a line segment AB is 45°, 120°, q° where 0 < q < 90 We need to find q. DCs of the line segment AB is cos 45°, cos 120°, cos q° cos2 45° + cos2 120° + cos2 q = 1 21 cos 41 cos 1 cos 41 21 2 2 = q = q = q + + ° = p = q 60 3c q − ¹ q acute is 21 cos Q 78. 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 a rational number x Î S such that x £ 0 (2) There is no rational number x Î S such that x £ 0 (3) Every rational number x Î S satisfies x £ 0 (4) x Î S and x £ 0 ⇒ x is not rational Ans (3) Given that S is a non-empty subset of R. P : There is a rational number x Î S such that x > 0 AIEEE 2010 26 Now, we need to find the negation of P Clearly, P is equivalent to saying that “There is a positive rational number in S” So, its negation − P is “There exist no positive rational in S” − P: There exist no positive rational number in S Û −P: Every rational number x Î S satisfies x £ 0. 79. Let 13 5 ) sin( let and 54 ) cos( = b − a = b + a , where 4 , 0 p £ b a £ . Then tan 2a = (1) 16 25 (2) 33 56 (3) 12 19 (4) 7 20 Ans (2) Given that 54 ) cos( = b + a and 13 5 ) ( sin = b + a , 4 , 0 p £ b a £ we need to find tan 2a. First let us try to write cos (a + b) and sin (a − b) both in terms of tan. p Î b + a = b + a = b + a 2 , 0 also method triangle by 43 ) tan( 54 ) cos( Q > b − a p − Ï b − a p Î b − a = b − a = b − a 0 ) sin( 0 , 4 : Note 4 , 0 , also method triangle by 12 5 ) tan( 13 5 ) sin( QQ [ ] ) ( ) ( tan 2 tan b − a + b + a = a \ = 33 56 12 5 . 43 1 12 5 43 ) ( tan ) ( tan 1 ) tan( ) tan( = − + = b − a b + a − b − a + b + a 80. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if (1) –85 < m < –35 (2) –35 < m < 15 (3) 15 < m < 65 (4) 35 < m < 85 Ans (2) Given circle is x2+ y2 = 4x + 8y + 5 Which can be rewritten as (x − 2)2 + (y − 4)2 = 25 Now we need to find the possible values of m for which the line 3x − 4y = m intersects the circle at two distinct points. A straight line will intersect a given circle at two distinct points iff the distance of the line from the centre of the circle is strictly less than the radius of the circle. 5 ) 4 ( 3 m ) 4 ( 4 ) 2 ( 3 2 2 < − + − − 25 10 m 25 25 10 m < + < − < + −35 < m < 15 81. 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) 25 (2) 2 11 (3) 6 (4) 2 13 Ans (2) E(X2) − (E(X))2 = 4 \ E(X2) = 4 + 4 = 8 AIEEE 2010 27 40 X2i = E(Y2) − (E(Y))2 = 5 E(Y2) = 5 + 16 = 21 105 Y2 i = \ 20 Y , 10 X i i = = 30 ) Y X ( i i = + \ 145 ) Y X ( 2 i 2i = + . \ Variance (combined data) = 2 11 10 55 9 10 145 = = − 82. 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) 31 (2) 72 (3) 21 1 (4) 23 2 Ans (2) 72 7 8 9 6 24 C C C C obability Pr 3 9 1 2 1 4 1 3 = × × × = × × = 83. 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 21 Rr = (2) There is a regular polygon with 2 1 Rr = (3) There is a regular polygon with 32 Rr = (4) There is a regular polygon with 23 Rr = Ans (3) Consider a regular n-polygon, from its centre to each of the vertex draw a line segment, those n-line segments will divide the n-polygon into n number of similar triangles which will look as in the following diagram. OA1 = OA2 = R OD is a median Also OD = r And n 2 OA A 2 1 p = Ð n DOA2 p = Ð Q OD is also an angle bisector of top angle. So, from right triangle ODA2, we get Rr n cos = p If n = 4 then 2 1 4 cos Rr = p = If n = 6 then 23 6 cos Rr = p = If n = 3 then 21 3 cos Rr = p = 2 1 32 21 < < Q 4 cos cos 3 cos p < q < p where 32 cos 1 − = q O A2 A1 D AIEEE 2010 28 4 3 p > q > p Q cos x is decreasing over p q 2 , But $ no. nÎZ such np = q because if $ such a n, then 4 n 3 p > p > p 3 < n < 4 and n Î Z which is a contradiction. 84. The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is (1) Less than 4 (2) 5 (3) 6 (4) At least 7 Ans (4) We need to find the number of 3 × 3 non-singular matrices with four entries as 1 and all other entries as 0. 1 0 0 0 1 0 0 0 1 Here we can fill any one of those 6 boxes with 1 and rest all with leading diagonal elements being 1. Hence these are the matrices with four 1 and five 0 and it is non-singular. We see that there are six possible matrices in this particular format. And consider 0 0 1 0 1 0 1 0 1 , this is another matrix with four 1 and five 0 and it is non-singular. Hence, there are at least 7 such matrices. 85. Let f : R R be defined by − > + − £ − = 1 x if , 3 x 2 1 x if , x 2 k ) x ( f If f has a local minimum at x = −1, then a possible value of k is (1) 1 (2) 0 (3) 21 − (4) −1 Ans (4) Given f: R ® R is defined by − > + − £ − = 1 x if , 3 x 2 1 x if , x 2 k ) x ( f And f has a local minimum at x = −1 $ d > 0 such that f(−1) £ f(x) " x Î (−1 − d, −1 + d) k + 2 £ f(x) " x Î (−1−d, −1+d) Clearly k + 2 £ f(x) " x Î (−1−d, −1] Q f(x) = k − 2x And k + 2 £ f(x) " x Î (−1, −1 + d) k + 2 £ 2x + 3 " x Î (−1, −1 + d) 1 3 x 2 lim 2 k 1 x = + £ + + − ® k £ −1 Directions : Questions number 86 to 90 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. AIEEE 2010 29 86. 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 85 1 Statement-2 : If the four chosen numbers from 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 a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Ans (3) Let d be common difference. Let x be numbers of arrangements with common difference d. Let d = 1. {{1, 2, 3, 4}, …. , {17, 18, 19, 20}} \ x = 17 Let d = 2 {{1, 3, 5, 7}, …., {14, 16, 18, 20}} x = 14 Let d = 3 {{1, 4, 7, 10}, …. , {11, 14, 17, 20}} x = 11 Let d = 4 {{1, 5, 9, 13}, …. , {8, 12, 16, 20}} x = 8 Let d = 5 {{1, 6, 11, 16}, …. , {5, 10, 15, 20}} x = 5 Let d = 6 {{1, 7, 13, 19}, {2, 8, 14, 20}} x = 2 \ Probability = 85 1 17 18 19 20 24 57 C 574 20 = × × × × = Since d can be ±6 also, statement 2 is false. 87. Let j 101 j 2 3 j 101 j 10 2 j 101 j 10 1 C j S and C j S , C ) 1 j ( j S = = = = = − = Statement-1 : S3 = 55 × 29 Statement-2 : S1 = 90 × 28 and S2 = 10 × 28 (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Ans (3) = = = − − = − = = 101 j 101 j 2 101 j j 2 3 )! j 10 ( ! ) 1 j ( ! 9 j 10 )! j 10 ( ! j ! 10 j C 10 j S = − = + − − − = 101 j 1 j 101 j 3 C 9 )! j 10 ( ! ) 1 j ( ! 9 ) 1 j ( 10 S + − − = = − = 101 j 1 j 101 j 3 C 9 )! j 10 ( ! ) 2 j ( ! 8 9 10 S AIEEE 2010 30 + = = − = − 101 j 1 j 101 j 2 j 3 C 9 C 8 9 10 S ( ) [ ] 9 8 9 8 3 2 . 55 2 . 110 2 2 9 10 S = = + = 9 101 j 1 j 101 j 2 101 j j 2 2 . 10 C 9 10 )! j 10 ( ! j ! 10 j C 10 j S = = − = = = − = = 8 9 2 3 101 j j 1 2 . 90 2 . 45 S S C 10 ) 1 j ( j S = = − = − = = Clearly, statement 1 is true and statement 2 is false. 88. 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 a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Ans (2) 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. Let M be the midpoint AB M = (2, 2, 5) Clearly M is a point of the given plane as it satisfies the equation x − y + z = 5. D.Rs of the line segment joining AB is (2, −2, 2) and D.Rs of the normal of the given plane is (1, –1, 1) Q these two D.Rs are proportional and the plane bisects AB. We can conclude that A is the mirror image of B in the given plane. Statement 1 is true. Statement 2: The plane x − y + z = 5 bisects the plane segment joining A (3, 1, 6) and B (1, 3, 4) Statement 2 is also true. But, statement 2 by itself cannot imply statement 1. \ statement 1 is true, statement 2 is true; statement 2 is not correctly explanation for statement 1. 89. Let f : R R be a continuous function defined by x x e 2 e 1 ) x ( f − + = Statement-1 : 31 ) c ( f = , for some c Î R. Statement-2 : 2 2 1 ) x ( f 0 £ < , for all x Î R. (1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Ans (1) Let f : R ® R be a continuous function defined as follows x x e 2 e 1 ) x ( f − + = We know that if a, b ³ 0 then ab 2 b a ³ + Put a = ex, b = 2e−x AIEEE 2010 31 Clearly a, b > 0 2 2 g 2 . e 2 e 2 e x x x x = ³ + − − " xÎR 2 21 e 2 e 1 0 x x £ + < − 2 21 ) x ( f 0 £ < " xÎR \ Statement 2 is true and Î 2 21 , 0 31 Q $ cÎR such that 31 ) c ( f = \ so statement 1 is also true \ Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1. 90. 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 a correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is false, Statement-2 is true Ans (3) Given A is a 2×2 matrix with non-zero entries and A2 = I Let = d c b a A a, b, c, d 0 And A2 = I = 1 0 0 1 d c b a d c b a = + + + + 1 0 0 1 d bc ) d a ( c ) d a ( b bc a 2 2 a + d = 0 0 c , b ¹ Q Tr(A) = 0 Now, A2 = I |A2| = |I| |A|2 = 1 All that we can conclude is |A| = ± 1 We cannot say |A| = ± 1. In fact $ a counter example to this statement 2. Let − = 2 1 2 1 2 1 2 1 A Clearly, A2 = I and A has only non-zero entries. But |A| = − 1 \ statement 1 is true, statement 2 is false. * * *