CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com AIEEE EXAMINATION PAPER 2010 Code-A PHYSICS, CHEMISTRY , MATHEMATICS Time : -3 Hours Max. Marks:-432 Date : 25/04/10 Important Instructions: 1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited. 2 . The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 432. 5. There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. Part A – 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 6. Candidates will be awarded marks as stated above in instructions No. 5 for correct response of each question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. 7. Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited. 8. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, 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 and in 2 pages (Pages 38 – 39) at the end of the booklet. 10. On completion of the test, the candidate 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. The CODE for this Booklet is A. 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 candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet 12. Do not fold or make any stray marks on the Answer Sheet. CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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] 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. [1] 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 + (2) M m m c + (3) M2 m c (4) Mm c Sol. 2 2 v 2M 21 2 mc = × × CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com M2 mc v 2 2 = , M2 m v c = Ans. (3) 5. The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then – (1) E1 = 2E2 (2) E2 = 2E1 (3) E1 > E2 (4) E2 > E1 Ans. [4] Directions: Questions number 6 – 7 contain Statement-1 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 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 not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true. Ans. [1] 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 not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true. Sol. Both are true but not explain the Ist. Ans. (3) 8. The figure shows the position – time (x – t) graph of one-dimensional motion of the 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 CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Sol. From graph, v1 = 1 ms–1 , v2 = –1 ms–1 ∴ J = ∫Fdt = ∫dP = mV = 0.4 × 2 = 0.8 N.s. Ans. (3) 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 XX’ is given by -(1) (2) (3) (4) Ans. [2] 10. A ball is made of a material of density ρ where ρoil < ρ < ρwater with ρoil and ρwater 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. Q ρoil < ρ < ρwater, so ball will not sink in water but sink in oil. Ans. (3) 11. 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 – CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com (1) j ˆ 2 r q 2 0 π2ε (2) j ˆ 4 r q 2 0 π2ε (3) j ˆ 4 r q 2 0 π2ε − (4) j ˆ 2 r q 2 0 π2ε − Sol. dE ∫ ∫ π π −π θ λ θ = θ = /2 0 2 /2/2 cos Rk Rd E dE cos 2 [ ] [ 90 sin 0]( ˆ) 2 sin 4 2 2 0 2 /2 2 0 0 Sin j R q R R qR E = = − − π ε θ πε π π r ( ˆ) 2 2 0 2 j R q E = − π ε r Ans. (4) 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 Sol. η = − 12 TT 1 Q 1 2 2 1 1 1 T V T V γ− = γ− ∴ 1 1 21 12 32 1 VV TT γ− γ− = = Putting γ = 7 /5 0.75 43 41 1 = = η = − Ans. (3) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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 Sol. 23.023 significant fig. 5 0.0003 significant fig. 1 2.1 × 10–3 significant fig. 2 Ans. (2) 14. The combination of gates shown below yields – (1) NAND gate (2) OR gate (3) NOT gate (4) XOR gate Sol. X = A⋅B = A + B i.e. OR gate Ans. (2) 15. If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called – (1) γ-rays (2) X-rays (3) ultraviolet rays (4) microwaves Sol. λ = hc P n Ans. (2) 16. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be – (1) Z 2 A Z 4 − − − (2) Z 4 A Z 8 − − − (3) Z 8 A Z 4 − − − (4) Z 4A Z 12 −− − Sol. X X X A 12 Z 8 A 3 4Z 3 2 2 1 A (3 2 positron) Z −− − ×− × − × α+ → = ∴ Z 8 (A 12) (Z 8) No. of Protons No. of Neutrons − − − − = Z 8 A Z 4 − − − = Ans. (3) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 17. Let there be a spherically symmetric charge distribution with charge density varying as ρ = ρ − Rr 45 (r) 0 upto r = R, and ρ(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) − ε ρ Rr 45 3 r0 0 (2) − ε πρ Rr 35 3 4 r 00 (3) − ε ρ Rr 35 4 r0 0 (4) − ε ρ Rr 45 3 4 r 0 0 Sol. r < R ∫ ∫ ε ρ ⋅ = 0vdv E dS ∫ π − ερ ⋅ π = r0 2 02 0 4 r dr Rr 45 E 4 r − ε ρ π ⋅ π = ∫ ∫ r0 r0 3 2 0 2 0 dr Rr r dr 44 5 E 4 r − επρ ⋅ π = 4R r 3 r 44 5 E 4 r 3 4 0 2 0 − ερ = 4R r 3r 45 E 2 00 − ε ρ = Rr 35 4 r E 0 0 Ans. (3) 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 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) 242 W (2) 305 W (3) 210 W (4) Zero W Sol. XL = XC 242 W R V P 2 = = Ans. (1) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 19. In the circuit shown below, the key K is closed at t = 0. the current through the battery is – (1) 1 2 1 2 R R V(R + R ) at t = 0 and 2 RV at t = ∞ (2) 22 21 1 2 R R VR R + at t = 0 and 2 RV at t = ∞ (3) 2 RV at t = 0 and 1 2 1 2 R R V(R + R ) at t = ∞ (4) 2 RV at t = 0 and 22 21 1 2 R R VR R + at t = ∞ Sol. at t = 0 V R2 2 RV I = at t = ∞ V R1 R2 1 2 1 2 R R V(R R ) I + = Ans. (3) 20. A particle is moving with velocity ) j ˆ x i ˆ v = K(y + 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 Sol. j ˆ kx i ˆ v = ky + r ⇒ kx dt dy ky, dt dx = = ∴ yx dx dy = ⇒ ∫ ydy = ∫ x dx y x cons tant 2 = 2 + Ans. (1) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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 t1 /t2 will be – (1) 2 (2) 1 (3) 21 (4) 41 Sol. 2C Q e 2C Q U RC2t 20 2 − = = t /RC 0Q Q e= − 2 U U = 0 2C Q e 2 2C Q RC2t 20 20 − 1 = × RC t 0 0 2 Q e 4 Q − = RC2t1 e 21 − = RC t log 4 2 e = 2 log 2 1 e RC t = t RClog 4 2 e = 41 tt21 = Ans. (4) 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) I1 = I2 3R B , I 6R Blv lv = = (2) I1 = –I2 R2B , I R Blv lv = = (3) I1 = I2 3R 2B , I 3R Blv lv = = (4) I1 = I2 R B I lv = = CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Sol. R R BVl R R R BVl R I I1 I2 3R 2 I BVl = 3R I1 I2 BVl = = Ans. (3) 23. The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y = 0.02 (m) sin π − 0.50(m) x 0.04(s) t 2 . The tension in the string is – (1) 6.25 N (2) 4.0 N (3) 12.5 N (4) 0.5 N Sol. Putting 0.5 2 ,k .04 2 π = π ω = in equation T = μv2 = 2 k ω μ = 6.25 N Ans. (1) 24. Two fixed frictionless inclined planes making an angle 30° and 60° 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 vertical direction (2) 4.9 ms–2 in horizontal direction (3) 9.8 ms–2 in vertical direction (4) Zero Sol. 43g A g sin 60 2 vertical a = °= 4g a B g sin 30 2 vertical = = ° So, 4.9ms vertical 2g a 2 AB = = − CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Ans. (1) 25. For a particle in uniform circular motion, the acceleration ar at a point P(R, θ) on the circle of radius R is (Here θ is measured from the x-axis) (1) j ˆ R v i ˆ R v 2 2 + (2) sin ˆj R v i ˆ cos R v 2 2 − θ + θ (3) cos ˆj R v i ˆ sin R v 2 2 − θ + θ (4) sin ˆj R v i ˆ cos R v 2 2 − θ − θ Sol. Y ac X o ) j ˆ ( sin a ) i ˆ a a cos ( c c = θ − + θ − r sin ˆj R V i ˆ cos R V a 2 2 = − θ − θ r Ans. (4) 26. A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v0 in the x-y plane as shown in the figure. At a time gv sin t 0 θ < , the angular momentum of the particle is – (1) i ˆ mg v t cos 21 2 0 θ (2) mg v t cos ˆj 2 0 − θ (3) k ˆ mg v t cos 0 θ (4) k ˆ mg v t cos 21 2 0 − θ where k ˆ and j ˆ , i ˆ are unit vectors along x, y and z-axis respectively. Sol. at any time t gt ˆj 21 t ) sin v ( i ˆ r (v cos ) t 2 0 0 = θ + θ − r j ˆ ) gt sin v ( i ˆ v v cos 0 0 = θ + θ − r so, ( ) k ˆ mgv t cos 21 L m r v 2 0 = × = − θ r r r Ans. (4) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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 Sol. 2 1.6 .8 1 1 1 1 = − = σρ − Ans. (4) 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 = 2s is nearly. (1) 14 m/s2 (2) 13 m/s2 (3) 12 m/s2 (4) 7.2 m/s2 Sol. s t 5 = 3 + 2 3t dt ds v = = 6t dt dv = a (a a ) 2t 2r = + 2 2 2 2 14m/s dt dv R v = + = at t = 2s Ans. (1) 29. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by 12 6 xb xa U(x) = − , 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) 6a b2 (2) 2a b2 (3) 12a b2 (4) 4a b2 Sol. 12 6 xb xa U = − CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com so U 0 x =∞ = at equilibrium 12 ax 6bx 0 dx dU F 0 = = − = − −13 + −7 = ⇒ 2a b x16 = So 61 b2a x = 4a b U 2 eq = − 4a b 4a b D 0 2 2 = = − − Ans. (4) 30. Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are α1 and α2. the respective temperature coefficients of their series and parallel combinations are nearly – (1) 2 , 2 1 2 1 2 α + α α + α (2) 1 2 1 2 , 2 α + α α + α (3) , 2 1 2 1 2 α + α α + α (4) , 1 2 1 2 1 2 α + α α α α + α In series, R1 +R2 = Rs R(1 +α1T)+ R(1+ α2T) = 2R(1 + αs T) 2R +RT (α1 + α2) = 2R + 2R α sT 2 1 2 s α + α α = In parallel 1 2 R1 R1 Rp 1 = + (1 ) 1 (1 ) 1 (1 ) 2 1 1 2R T R T T R p α α +α + + = + 2(1 – αpT) = 1 – α1 T + 1 – α2T 2 1 2 α α α + = p CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com CHEMISTRY 31. In aqueous solution the ionization constants for carbonic acid are 7 1 K 4.2 x10= − and 11 K2 4.8 x10= − Selection the correct statement for a saturated 0.034 M solution of the carbonic acid. (1) The concentration of + H is double that of 3 2 CO − (2) The concentration of 3 2 CO − is 0.034 M. (3) The concentration of 3 2 CO − is greater than that of H 3 CO− (4) The concentration of + H and H 3 CO− are approximately equal. Sol. 2 3 H CO + + −3 H HCO 7 1 k 4.2 10= × − 0.034 M x x −3 HCO + + H 2 3 CO− 11 1 k 4.8 10= × − x – y x + 5 Sol. As 2 1 k << k , disso in second is negligible ∴ x + y ≅ x and hence (H ) + ≅ −3 [HCO ] Ans. (4) 32. Solution product of silver bromide is 5.0 x 13 10− . 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 x 10 −8 g (2) 1.2 x 10 −10 g (3) 1.2 x 9 10− g (4) 6.2 x 10 −5 g Sol. 13 sp k [Ag ][Br ] 5 10= + − = × − [Ag ] = 0.05M + 10 M 0.05 5 10 [Br ] 11 13 − − − = × = [Br ] = [kBr] − ∴ Mass added in gms = 10 120 g −11 × = 1.2 10 g × −9 Ans. (3) 33. The correct sequence which shows decreasing order of the ionic radii of the elements is (1) − − + > + 2 > > 3 O F Na Al (2) + + − − > > > > + 3 2 2 Al Mg Na F O (3) + > + > + > − > F− Na Mg Al O2 3 2 (4) + − + − + > > 2 2 3 Na F Mg O Al 33. Sol. Bioelectric series ↓ → ↑ ez r CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com ∴ 2 2 3 O F Na Mg Al− > − > + > + > + Ans. (1) 34. In the chemical reactions. The compounds 'A' and 'B' respectively are (1) nitrobenzene and chlorobenzene (2) nitrobenzene and flurobenzene (3) phenol and benzene (4) benzene diazonium chloride and flurobenzene Sol. NaNO + HCl 2 0 -5°C NH2 N Cl 2 F HBF4 – N2 – HCl Ans. (4) 35. If 4 10− dm −3 of water is introduced into a 1.0 dm −3 flask at 300 K, how many moles of water are in in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H O at 300 2 is 3170 pa; R = 8.314 J 1 K− mol −1 ) (1) 1.27 x 10 mol −3 (2) 5.56 x 10 mol −3 (3) 1.53 x10 mol −2 (4) 4346 x 10 mol −2 Sol. PV = nRT 10 m n 8.314 300 mN 3170 3 3 2 × − = × × 1.287 10 m 24.93 10 31.7 10 24.93 10 3170 10 n 1 2 1 2 3 − − − − − = × × × = ×× = Ans. (1) 36. From amongst the following alcohols the one that would react fastest with conc, HCI and anhydrous ZNCI , 2 is (1) 1-Butanol (2) 2-Butanol (3) 2-Methylpropan -2-ol (4) 2-Methylpropanol Sol. Reactivity of LuCaS reagent for alcohol = 3° > 2 ° > 1° > 3 CH CH3 CH3 C OH CH3 2-methyl. 2-propanol (3° -alcohol) highest reactivity CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Ans. (3) 37. If sodium sulphate is considered to be completely dissociated into cations and anions in equeous solution, the change in freezing point of water ( f T ), When 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is ( 1 f K 1.86K Kg mol= − ) (1) 0.0186 K (2) 0.0372 K (3) 0.0558 K (4) 0.0744 K Sol. 2 4 Na SO 2 4 2 2Na SO+ − i = 3 f T iK .m. f = = 3 × 1.86 × 0.0 = 0.0558 K Ans. (3) 38. Three reactions involving − 2 4 H Po are given below: (1) 4 H Po H O H O H O H Po 3 4 2 3 3 2 − + → + → + (2) + + − + → − H O 4 2 H O HPO 4 H Po 2 2 3 (3) + − → + − − 2 2 3 4 OH H PO O 4 H Po In which of the above does 4 H Po 2 − act as an acid? (1) (i) Only (2) (ii) Only (3) (iii) and (ii) (4) (iii) only 38. Sol. (ii) only − + → − + O+ H PO H O HPO H3 2 2 4 2 4 acid base Ans. (2) 39. The main product of the following reaction is C H CH CH(OH)CH(CH ) 2 4 ? conc.H SO 6 5 2 3 2 → CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Sol. → ⊕ H Benzyl Carbon cation stable by Resonance Removal of ⊕ H Trans is more stable than its Akene CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Ans. (2) 40. The energy required to break one mole of CI–Cl bonds in 2 Cl is 242 kJ 1 mol− . The longest wavelength of light capable of breaking a single Cl–Cl bond is (C = 3 x 10 ms and N 6.02 x10 mol ) 1 A 23 8 −1 = − (1) 494 nm (2) 594 (3) 640 nm (4) 700 nm Sol. λ = hc E = λ × × × = ×× + −34 8 23 3 6.62 10 3 10 6.02 10 242 10 λ = 242 6.62 10 3 6.02 10× −26 × × × 20 = 242 6.62 18.06 10 × × −6 = 0.494 × 6 10− = 4.94 × 10 m −7 = 494 nm Ans. (1) 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 neutralization. The percentage of nitrogen in the compound is (1) 29.5 (2) 59.0 (3) 47.4 (4) 23.7 41. Sol. Equation of 3 NH = (0.1× 20) − (0.1×15) = 0.5 wt. of 3 NH = 0.5×17 = 8.5mg wt. of 'N' = 8.5mg 7mg 17 14 × = % of 'N' = 100 23.7 29.5 7 × = Ans. (4) 42. Ionisation energy of + He is 19.6 x 18 10− J atom −1 . The energy of the first stationary state (n = 1) of + 2 Li is (1) 8.82 x 17 1 10 J atom− − (2) 4.41 x 16 1 10 J atom− − (3) 17 1 4.41x10 J atom− − − CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com (4) -2.2 x 15 1 10 J atom− − Sol. 2221 21 ZZ I.EI.E = = 94 x19.6 10 18 = × − x = 19.6 10 44.1 10 J /atm. 49 × × −18 = × −18 = 4.41 10 J /atm × −17 Ans. (3) 43. 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 1 mol− and of octane =114 g 1 mol− ) (1) 144.5 kPa (2) 72.0 kPa (3) 36.1 kPa (4) 96.2 kPa Sol. PT = P0x0 + Phepxhep = 0.55 0.25 105 25 0.55 0.3 45× + × = 45 × 0.545 + 105 × 0.454 = 72.25 kPa. Ans. (2) 44. Which one of the following has an optical isomer ? (1) 2+ 2 [Zn(en) ] (2) 2+ 3 2 [Zn(en)(NH ) ] (3) 3+ 3 [CO(en) ] (4) 3+ 2 4 [Co(H O) (en)] Sol. [M(AA) ] 3 type of compound ∴ optically active Ans. (3) 45. Consider the following bromides The correct order of S 1 N reactivity is (1) A > B > C (2) B > C > A (3) B > A > C (4) C > B > A CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Sol. Reactivity for SN'∝ Stability of corbocation 1° carbocation stable by 2° Carbocation resonance (A) (B) (C) B > C > A Ans. (2) 46. 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) ethene (2) propene (3) 1-butene (4) 2-butene Sol. Molecular weight = 44 [CH CHO] 3 ∴ − 3 3 3 3 CH −CH = O+O = CH −CH →CH −CH = CH −CH 2-butene Ans. (4) 47. Consider the reaction : Cl (aq) H S(aq) S(s) 2H (aq) 2 Cl (aq) 2 2 + → + + + − The rate equation for this reaction is Rate = k [Cl 2 ][H 2S] Which of these mechanisms is/are consistent with this rate equation ? (A) Cl HS H Cl S (fast) Cl H S H Cl Cl HS (slow) 2 2 + + → + + + → + + + − + − + − + − (B) Cl HS 2Cl H S (slow)H S H HS (fast equilibrium) 2 2 + → + + ⇔ + − − + + − (1) A only (2) B only (3) Both A and B (4) Neither A nor B Sol. r k[Cl ][H S] 2 2 = ∴ According to A [ ][ ] 2 2 →r = k H S Cl ∴ According to B r k[Cl ][HS] 2 → = or [H S] [H ][HS] K 2 eq + = + = H [H S] [HS] K 2 eq r = [H ][H S] k[Cl ] K 2 2 eq + = [H ] [Cl ][H S] K' 2 2 + ∴ (A) Only Ans. (1) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 48. The Gibbs energy for the decomposition of 2 3 Al O at 500 °C is as follows: 1 2 3 Al O2 , rG 966kJ mol 34 Al O 32 → + = + − The potential difference needed for electrolytic reduction of 2 3 Al O at 500 °C is at least (1) 5.0 V (2) 4.5 V (3) 3.0 V (4) 2.5 V Sol. G = −nFE n = 4 966 10 4 96500 E × 3 = − × × = 2.5 V Ans. (4) 49. The correct order of increasing basicity of the given conjugate bases (R CH ) 3 = is (1) RCOO HC C NH R 2 < ≡ < < (2) 2 RCOO < HC ≡ C < R < NH (3) 2 R < HC ≡ C < RCOO < NH (4) RCOO NH HC C R 2 < < ≡ < Sol. Confugated acid RCOOH CH ≡ CH 3 NH R −H Order to A.S. RCOOH CH CH NH R H 3 ⇒ > = > > − Order to B.S. ⇒ RCOO− < CH ≡ C− < NH < R− 2 Ans. (1) 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 (1) 288 pm (3) 398 pm (4) 618 pm Sol. •− •⊕+ ( ) r r = 2q 110 + •(−) r = 2 508 ( ) r − = 254 – 110 = 144 nm Ans. (1) 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-pentene (4) 3-methyl-1-pentene Sol. CH3 CH2 CH= CH2 CH CH3 Due to presence of chiral carbon atom is 4 is show optical isomerism. Ans. (4) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 52. For a particular reversible reaction at temperature T, H and S were found to be both +ve. If e T is the temperature at equilibrium, the reaction would be spontaneous when (1) T = e T (2) e T > T (3) T > e T (4) e T is 5 times T Sol. G = H − TS +ve +ve T > e T for G = −ve Ans. (3) 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% Sol. ccp : p – f = 74%; 100 – 74 = 26% bcc : p – f = 68%; 100 – 68 = 32% Ans. (3) 54. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is (1) natural rubber (2) teflon (3) nylon 6,6 (4) polystyrene Sol. Fact Ans. (3) 55. At 25°C, the solubility product of 2 Mg(OH) is 1.0 × 11 10− . At which pH, will 2+ Mg ions start precipitating in the form of 2 Mg(OH) from a solution of 0.001 M 2+ Mg ions ? (1) 8 (2) 9 (3) 10 (4) 11 Sol. Ksp = 2 2 [ ][ ] + − Mg OH 11 2 1 10 [0.001][ ] × − = OH− [oh ] − = 4 10− P on = 4; pn = 10 Ans. (3) 56. The correct order of ° + M /M E 2 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 Sol. 0 M 2/M E + 0.76 Zn o.34 Cu 0.24 Ni 0.28 Co 0.44 Fe 1.18 Mn 0.91 Cr 1.18 V 1.67 Ti − − − − − − − + − Ans. (2) 57. Biuret test is not given by (1) proteins (2) carbohydrates CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com (3) polypeptides (4) urea Sol. Carbohydrate does not give biuret test. Due to absence of amide group. Ans. (2) 58. 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 1 L− , how much time does it take for its concentration come from 0.50 to 0.25 mol 1 L− if it is a zero order reaction ? (1) 1h (2) 4 h (3) 0.5 h (4) 0.25 h Ans.[4] 59. A solution containing 2.675 g of 3 3 CoCl ⋅ 6NH (molar mass = 267.5 g 1 mol− ) is passed through a cation exchanger . The chloride ions obtained in solution were treated with excess of 3 AgNO to give 4.78 g of AgCl (molar mass =143.5 g 1 mol− ). The formula of the complex is (At. mass of Ag = 108 u) (1) 3 5 2 [CoCl(NH ) ]Cl (2) 3 6 3 [Co(NH ) ]Cl (3) [CoCl(NH ) ]Cl 3 4 (4) [CoCl (NH ) ] 3 3 3 Sol. COCl .6NH AgCl 3 3 → 4.75 g or 0.03 m/s 143.5 4.78 = Ans. (2) 60. The standard enthalpy of formation of 3 NH is 1 46.0 kJ mol− − . If the enthalpy of formation of 2 H from its atoms is 1 436 kJ mol− − and that of 2 N is 1 712 kJ mol− − , the average bond enthalpy of N − H bond in 3 NH ] (1) 1 1102 kJ mol− − (2) 1 964 kJ mol− − (3) 1 352 kJ mol+ − (4) 1 1056 kJ mol+ − Sol. 2 H(g) NH3 23 N (g) 21 + → f N N BEH H 3.B.EN H 23 B E 21 H = − − + − − − –46 = ( 436) 3 x 23 ( 712) 21 × − + × − − × x = 3 1056 = 352 kJ/ml. Ans. (3) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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 ≠ 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 Sol. Probable part of R is {(0, 1), (0, 2)} But (1, 0) ∉ R as 1 = (w) 0 So not symmetric ie. not equivalence Relation qm pn qp S nm → = Reflexive mm mn nm S nm → = hence function reflexive . Let qm pn qp S nm → = Then pn mq nm S qp → = hence function symmetric mq pn (1) qp S nm → = ps qr (2) sr S qp → = eqn. (1)/(2) sr S nm sr nm = → hence transitive So S is equivalence relation Ans. (3) 62. The number of complex numbers z such that |z –1 | = | z + 1| = | z – i| equals -(1) 0 (2) 1 (3) 2 (4) ∞ CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Sol. |z – 1| = |z + 1| = |z – i| The point z is equidistance from (–1, 0), (1, 0) and (0, 1) is only (0, 0) hence z is only point (0, 0) Ans. (2) 63. If α and β are the roots of the equation x2 – x + 1 = 0, then α2009 + β2009 = (1) – 2 (2) – 1 (3) 1 (4) 2 Sol. Here roots of x2 – x + 1 = 0 are – ω and – ω2. (– ω)2009 + (– ω2)2009 = (– ω)2007 × (– ω)2 + (– ω2)2007 × (– ω2)2 –( ω2 + ω) = 1. Ans. (3) 64. Consider the system of linear equations : 3 5 2 1 2 3 3 2 3 1 2 3 1 2 3 1 2 3 + + = + + = + + = x x x x x x x x x The system has (1) Infinite number of solutions (2) Exactly 3 solutions (3) a unique solution (4) No solution Sol. Here 1 3 5 22 3 11 2 1 = = (1) – 2(1) + 1 (1) = 0 3 1 5 23 3 13 2 1 1 = = (1) – 2 (5) + 1 (12) = 5 1 ≠ 0 When = 0 and if 1, 2, 3, are not zero then no solution Ans. (4) 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 Sol. By 3C2 way we can select 2 balls from A and By 9C2 ways we can select 2 balls from B Total no. of ways 108 2 9 2 3C × C = Ans. (4) CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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 Sol. g'(x) = 2 [f(2f(x) + 2)]. f '(2 f(x) + 2) .2f ' (x) g'(0) = 2 [f(2.f(0)+2)] . f ' (2. f(0) + 2). 2f '(0) = 2[f(0)]. f '(0) .2 = 2(–1) .(1).2 = – 4 Ans. (2) 67. Let f : R → R be a positive increasing function with 1. ( )(3 ) lim = →∞ f xf x x Then = →∞ ( )(2 ) lim f xf x x (1) 1 (2) 32 (3) 23 (4) 3 Sol. Function (↑) f(x) < f(2x) < f(3x) ( )(3 ) ( )(2 ) 1 f xf x f xf x ≤ ≤ given that 1 ( )(3 ) = f xf x hence 1 ( )(2 ) 1 ≤ ≤ f xf x hence 1 ( )(2 ) lim = →∞ f xf x x (by sandwich theorem) Ans. (1) 68. 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 ∫ 1 0 p(x)dx equals -(1) 41 (2) 21 (3) 41 (4) 42 Sol. P'(x) = P' (1 – x) integrate P(x) = – P (1 – x) + k -------(1) put x = 1 P(1) = – P(0) + k 41 = – 1 + k CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com K = 42 Put in (1) P(x) = –P(1 – x) + 42 ------(2) Now = ∫10 I P(x)dx also = ∫ − 10 I P(1 x)dx = ∫ + − 10 2I (P(x) P(1 x))dx using (2) I = ∫ dx 10 2 42 10 = 42(x) I = 21 Ans. (2) 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 Sol. a1 = a2 = a3 ..... a9 = 150 a1 + a2 + a3 + .... + a9 = 1350 a10 + a11 + ......... + an = 4500 – 1350 = 3150 [2 150 ( 1)( 2)] 3150 2 × + n − − = n 150 n – n2 + n = 3150 n2 – 151n + 3150 = 0 n = 25 min hence total time = 25 + 9 = 34 min Ans. (2) 70. The equation of the tangent to the curve 2 4 x y = x + , that is parallel to the x – axis, is -(1) y = 0 (2) y = 1 (3) y = 2 (4) y = 3 Sol. 2 4 x y = x + 0 8 1 3 = − = dx x dy 3 8 1 x = CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com x3 = 8 x = 2 at x = 2, 2 4 x y = x + 3 44 = 2 + = tangent y – 3 = 0 (x – 2) y = 3 Ans. (4) 71. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and 2 3π x = is -(1) 4 2 − 2 (2) 4 2 + 2 (3) 4 2 −1 (4) 4 2 +1 Sol. π/4 5π/4 3π/2 y = 0 x = 0 cos x sin x Area = ∫ − + ∫ − + ∫ − 5 2 34 5 44 40 (cos sin ) (sin cos ) (cos sin ) ππ π π π x x dx x x dx x x dx 2 34 5 4 54 4 0 [sin cos ] [ cos sin ] [sin cos ] ππ π π π = x + x + − x − x + x + x − − − + − + − − − − − = + − + 2 1 2 1 1 0 2 1 2 1 2 1 2 1 (0 1) 2 1 2 1 1 2 2 4 = 2 −1+ − + = 4 2 − 2 Ans. (1) 72. Solution of the differential equation cos x dy = y (sin x – y) dx, 2 0 π < x < is -(1) sec x = (tan x + c) y (2) y sec x = tan x + c (3) y tan x = sec x + c (4) tan x = (sec x + c) y Sol. 2 cos y sin x y dx dy x = − 2 cos sin x.y y dx dy x − = − CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com x x dx y dy y tan sec 1 1 2 − = z y − = 1 dx dz dx dy y = 2 1 x z x dx dz + tan . = sec I.F. = ∫ x dx e tan Solution of above differential equation is z. sec x = ∫ x dx 2 sec x c y x = tan + sec sec x = y (tan x + c) Ans. (1) 73. Let a = ˆj − kˆ and c = iˆ − ˆj − kˆ. r r Then the vector br satisfying 0r r r r a ×b + c = and a.b = 3 r r is -(1) − iˆ + ˆj − 2kˆ (2) 2iˆ − ˆj + 2kˆ (3) iˆ − ˆj − 2kˆ (4) iˆ + ˆj − 2kˆ Sol. Let b b iˆ b ˆj b kˆ 1 2 3 = + + given a . b = 3 r r b2 – b3 = 3 (1) and a × b + c = 0 r r r a b cr r r × = − i j k b b bi j k 0 1 1 ˆ ˆ ˆ 1 2 3 − = − + + b3 + b2 = – 1 (2) – b1 =1 -----(3) – b1 = 1 ------(4) b1 = – 1 from (1) and (2) b2 = 1 b3 = – 2 b = −iˆ + ˆj − 2kˆ r Ans. (1) 74. If the vectors a = iˆ − ˆj + 2kˆ r , b = 2iˆ + 4 ˆj + kˆ r and c =λiˆ + ˆj +μkˆ r are mutually orthogonal, then (λ, μ) = CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com (1) (–3, 2) (2) (2, –3) (3) (–2, 3) (4) (3, –2) Sol. . 0 . 0 ⊥ ∴ = ⊥ ∴ = b c b c a b a b r r r r r r r r 2λ + 4 + μ = 0 .....(1) a ⊥ c∴a.c = 0 r r r r λ – 1 + 2μ = 0 .....(2) solving (1) and (2), we get λ = – 3 μ = 2 Ans. (1) 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 Sol. y2 = 4x comparing with y2 = 4ax a = 1 Locus of point P will be directrix of given parabola as tangents drawn from P are at right angles, therefore required locus is x = – a x = – 1 Ans. (3) 76. The line L given by 1 5 + = bx y passes through the point (13, 32). The line K is parallel to L and has the equation 1 3 + = y cx . Then the distance between L and K is -(1) 15 23 (2) 17 (3) 15 17 (4) 17 23 Sol. 1 (1) 5 + = bx y Passess through (13, 32) 1 32 5 13 + = b ⇒ 13b + 160 = 5b ⇒ b = – 20 so line is –20x + 5y = – 100 (1) second line 1 3 + = y cx 3x + cy = 3c (2) (1) and (2) are parallel CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 20 5 3 c = − 4−3 c = Line 49 43 3x − y = − 12x –3y = – 9 − + = − × − 35 20x 5y 9 –20x + 5y = 15 ......(2) Distance between (1) and (2) 400 25| 100 15 | +− − = 17 23 5 17 115 425 115 = = = Ans. (4) 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 θ with the positive z – axis, then θ equals -(1) 30° (2) 45° (3) 60° (4) 75° Sol. cos2 α + cos2 β + cos2 γ = 1 cos2 45° + cos2 120° + cos2 γ = 1 = °= ± = + + = 60 21 cos 41 cos cos 1 41 21 2 2 γ γγ γ Ans. (3) 78. Let S be a non-empty subset of R. Consider the following statement : P : There is a rational number x ∈ S such that x > 0 Which of the following statements is the negation of the statement P ? (1) There is a rational number x ∈ S such that x < 0. (2) There is no rational number x ∈ S such that x < 0. (3) Every rational number x ∈ S satisfies x < 0. (4) x ∈ S and x < 0 ⇒ x is not rational Ans. [3] CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 79. Let 54 cos(α +β ) = and let , 13 5 sin(α −β ) = where 4 0 , π ≤α β ≤ .Then tan 2α = (1) 16 25 (2) 33 56 (3) 12 19 (4) 7 20 Sol. [( ) ( )] 1 tan( ) tan( ) tan( ) tan( ) tan 2 tan α β α β α β α β α α β α β − + − + + − = + + − = as cos(α + β) = 4/5, sin (α – β) = 5/13 33 56 16 16 5 129 5 12 5 . 43 1 12 5 43 tan 2 = − + = − + α = Ans. (2) 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 Sol. x2 + y2 –4x –8y –5 = 0 centre = (2, 4) and radius = 5 P < r for if line is intersecting the circle at two points |10 | 25| 10 | 25 5 3 43(2) 4(4)2 2 + < − − < < + − − = mm m P –35 < m < 15 Ans. (2) 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 Sol. n1 = 5 n2 = 5 4 2 1 σ = 5 22 σ = 2 1 x = 4 2 x = sum of data = 10 sum of data = 20 CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com ( ) 4 51 4 = sum of squares − ( ) 16 51 5 = sum of squares − (as variance 2 2 = − Σ Σnx nx i i ) sum of squares = 40 sum of squares = 105 3 1010 20 = + x = new variance = 2 11 (145) 9 10 1 − = Ans. (2) 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 Sol. Total balls = 3 red balls + 4 blue balls + 2 green balls = 9 balls required probability = 72 3 9 1 2 1 4 1 3 = × × C C C C Ans. (2) 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 Sol. r x rx n 2 2 tan = = π = n x r π cot 2 and R x n 2 sin = π n ec x R π cos 2 = CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com = = n n ecn Rr π π π cos cos cot (1) .5 21 = 3, = = Rr n (2) .707 2 1 = 4, = = Rr n (3) .6 32 = 5, = = Rr n (4) 23 = 6, = Rr n (3) is not possible because .6 comes between n = 3 and n = 4 but no integer between n = 3 and n = 4 Ans. (3) 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 Sol. 0 1 01 0 01 0 1 ; 0 0 10 1 11 0 0 ; 0 0 10 1 01 0 1 ; 0 0 10 1 01 1 0 ; 0 1 10 1 01 0 0 ; 0 0 11 1 01 0 0 ; 1 0 10 1 01 0 0 A = So at least 7 non singular matrices are there Ans. (4) 85. Let f : R → R be defined by + > −− ≤ − = 2 3, 12 , 1 ( ) x if xk x if x f x If f has a local minimum at x = –1, then a possible value of k is (1) 1 (2) 0 (3) 21 − (4) –1 Sol. f:R→R f(x) = + > −− ≤ −2x 3 x 1k 2x x 1 f ' (x) = >− < −2 12 1 x x o 1 2 1 = − − = + k k x CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com Ans. (4) 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. 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 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 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. Sol. S : 1 required no of groups (1,2,3,4) ……………………………….. (17,18,19,20) = 17 ways (1,3,5,7) ……………………………….. (14,16,18,20) = 14 ways (1,4,7,10) ……………………………….. (11,14,17,20) = 11 ways (1,5,9,13) ……………………………….. (8,12,16,20) = 8 ways (1,6,11,16) ……………………………….. (5,10,15,20) = 5 ways (1,7,13,19) ………………………………….. (2,8,14,20) = 2 ways required arability = 4! (17 14 11 8 5 2)4! 4 20C + + + + + = 20.19.18.17 57 4! = 20.18.17. 3.4.3.2.1 = 85 1 S : 1 is true. S : 2 possible cases of common difference are [±1,± 2,±3,± 4,±5,±6] S:2 is false Ans. (3) 87. Let Σ Σ = = = − = 101 101 10 2 10 1 ( 1) , j j j j C j S C j j S and Σ= = 101 2 10 3 . j j S j C Statement – 1 : S3 = 55 × 29. CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 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 ture. Sol. 8 102 101 1 90 2 ( 2)!(8 ( 2))! 8! 90 ( 1)( 2)!(10 )! 10! ( 1) = × − − − = − − − =Σ − Σj= j= j j j j j j S j j 9 101 101 2 10 2 ( 1)!(9 ( 1))! 9! 10 ( 1)!(9 ( 1))! 10! = × − − − = − − − =Σ Σj= j= j j j j j S j 8 9 8 9 101 10 101 10 101 3 ( 1) ( ) 90.2 10.2 110.2 55.2 (10 )! 10! [ ( 1) ] = − = = + = = − =Σ − + Σ Σ= = j= j j j j j C j C j j S j j j Hence statement 1 is true, statement 2 is false Ans. (3) 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 ture. Sol. Mid point of A(3, 1, 6) and B(1, 3, 4) should lie in the plane mid point : (2, 2, 5) it satisfies the plane x – y + z = 5. Also AB ⊥ to plane. Hence Dr's of AB are < 1, – 1, 1 > statement 1 and 2 are true Ans. (1) 89. Let f : R → R be a continuous function defined by x x e e f x + − = 2 1 ( ) Statement – 1 : 31 f (c) = , for some c ∈ R. Statement – 2 : , 2 2 1 0 < f (x) ≤ for all x ∈ 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 ture. Sol. AM > GM ≥ + x x x x e e e e 2 ( ) 2 2 CODE -A AIEEE 2010 EXAMINATION CAREER POINT CORPORATE OFFICE, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-2434159 Website : www.careerpointgroup.com, Email: info@careerpointgroup.com 2 2 (1) 2 + ≥ x x e e 0 (2) 2 > 0⇒ + > x x x e Qe e 0 < x x e e 2 1+ < 2 2 1 also f(c) = 1/3 for c = 0 so statement 1 : is true statement 2 : is also true with correct explanation Ans. (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 ture. Sol. let c da b A2 =I = 0 11 0 c da b c da b = + ++ + 0 11 0 2 2 ac dc bc da bc ab bd ab + bd = 0 b (a + d) =0 b ≠ 0 so, a = – d = c da b A a + d = 0 T (A) = 0 r But |A| ≠ 1. So, statement I is true and statement 2 is false. Ans. (3)