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AIEEE 2006 Exam Papers

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Tags : AIEEE,2003,Paper,Exam,Chemistry,Maths,Physics,IIT,JEE,IIT JEE,Joint Entrance Examination, iit jee solved question paper with answers and detailed solutions, online question paper free,important questions,aieee exam paper, jee mains, iit jee papers. FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 CHEMISTRY PART – C 96. HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give (1) CH3CHO and CH3Br (2) BrCH2CHO and CH3OH (3) BrCH2 – CH2 – OCH3 (4) H3C – CHBr – OCH3 Ans. (4) Sol. Electrophilic addition reaction more favourable. H2C CH OCH3 HBr H2C CH H OCH3 Br H C 3 CH Br OCH3 97. The IUPAC name of the compound shown below is Cl Br (1) 2-bromo-6-chlorocyclohex-1-ene (2) 6-bromo-2-chlorocyclohexene (3) 3-bromo-1-chlorocyclohexene (4) 1-bromo-3-chlorocyclohexene Ans. (3) 98. The increasing order of the rate of HCN addition to compounds A – D is (A) HCHO (B) CH3COCH3 (C) PhCOCH3 (D) PhCOPh (1) A < B < C < D (2) D < B < C < A (3) D < C < B < A (4) C < D < B < A Ans. (3) 99. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? (1) 0.02 (2) 3.125 × 10–2 (3) 1.25 × 10–2 (4) 2.5 × 10–2 Ans. (2) Sol. Mg3(PO4)2 ‘n’ moles 8n = 0.25 n 0.25 8 = 25 3.125 10 2 8 100 = = × − × 100. According to Bohr’s theory, the angular momentum of an electron in 5th orbit is (1) 25 h π (2) 1.0 h π (3) 10 h π (4) 2.5 h π Ans. (4) Sol. mvr nh 2 = π 5h 2.5 h 2 = = π π FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 101. Which of the following molecules/ions does not contain unpaired electrons? (1) 22 O − (2) B2 (3) N2+ (4) O2 Ans. (1) 102. Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius) (1) 20 r3 3 π (2) 24 r3 3 π (3) 12 r3 3 π (4) 16 r3 3 π Ans. (4) Sol. V n 4 r3 3 = ×  π      4 4 r3 3 = ×  π      16 r3 3 = π 103. A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will (1) remain unchanged (2) triple (3) increase by a factor of 4 (4) double Ans. (3) Sol. R ∝ [W]2 R′ ∝ [2CO]2 R ∝ 4[W]2 R ∝ 4M 104. Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4? (1) 2HI +H2SO4 →I2 + SO2 + 2H2O (2) Ca(OH)2 +H2SO4 →CaSO4 + 2H2O (3) NaCl +H2SO4 →NaHSO4 +HCl (4) 2PCl5 +H2SO4 →2POCl3 + 2HCl + SO2Cl2 Ans. (1) 105. The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is (1) nitrito-N-pentaamminecobalt (III) chloride (2) nitrito-N-pentaamminecobalt (II) chloride (3) pentaammine nitrito-N-cobalt (II) chloride (4) pentaammine nitrito-N-cobalt (III) chloride Ans. (4) 106. The term anomers of glucose refers to (1) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4) (2) a mixture of (D)-glucose and (L)-glucose (3) enantiomers of glucose (4) isomers of glucose that differ in configuration at carbon one (C-1) Ans. (4) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 107. In the transformation of 238 234 92 U to 92 U, if one emission is an α-particle, what should be the other emission(s)? (1) Two β– (2) Two β– and one β+ (3) One β– and one γ (4) One β+ and one β– Ans. (1) Sol. 238 234 4 0 92 92 2 1 U U He2e− → + + 108. Phenyl magnesium bromide reacts with methanol to give (1) a mixture of anisole and Mg(OH)Br (2) a mixture of benzene and Mg(OMe)Br (3) a mixture of toluene and Mg(OH)Br (4) a mixture of phenol and Mg(Me)Br Ans. (2) 109. CH3Br +Nu− →CH3 −Nu + Br− The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is [Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O–] (1) D > C > A > B (2) D > C > B > A (3) A > B > C > D (4) B > D > C > A Ans. (1) 110. The pyrimidine bases present in DNA are (1) cytosine and adenine (2) cytosine and guanine (3) cytosine and thymine (4) cytosine and uracil Ans. (3) 111. Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is (1) CH3CH2CH(OH)CH2CH3 (2) C6H5CH2CH2OH (3) H3C CH3 OH (4) PhCHOHCH3 Ans. (4) 112. The increasing order of stability of the following free radicals is (1) (CH3 )2 CH (CH3 )3 C (C6H5 )2 CH (C6H5 )3 C • • • • < < < (2) (C6H5 )3 C (C6H5 )2 CH (CH3 )3 C (CH3 )2 CH • • • • < < < (3) (C6H5 )2 CH (C6H5 )3 C (CH3 )3 C (CH3 )2 CH • • • • < < < (4) (CH3 )2 CH (CH3 )3 C (C6H5 )3 C (C6H5 )2 CH • • • • < < < Ans. (1) 113. Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1, accurate upto 0.001%, will be (1) 19.2 × 10–2 m (2) 5.76 × 10–2 m (3) 1.92 × 10–2 m (4) 3.84 × 10–2 m (h = 6.63 × 10–34 Js) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 Ans. (3) Sol. x. V h 4 m Δ Δ ≥ π 34 31 x h 6.63 10 4 m V 4 3.14 9.1 10 300 0.001 100 − − × Δ ≥ = π Δ × × × × × 34 31 3 6.63 10 4 3.14 9.1 3 10 10 − − − × = × × × × × = 0.01933 = 1.93 × 10–2 114. Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl5(g)􀁙􀁚􀁚􀁚􀁚􀁘􀁚􀁚PCl3(g) + Cl2(g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be (1) x P x 1    +    (2) 2x P 1 x    −    (3) x P x 1    −    (4) x P 1 x    −    Ans. (1) Sol. PCl5(g) PCl3(g) Cl2(g) (1 x) x x + − 􀁙􀁚􀁚􀁚􀁚􀁘􀁚􀁚 PCl3 P x P 1 x =   ×  +    115. The standard enthalpy of formation o (ΔfH ) at 298 K for methane, CH4(g), is –74.8 kJ mol–1. The additional information required to determine the average energy for C – H bond formation would be (1) the dissociation energy of H2 and enthalpy of sublimation of carbon (2) latent heat of vapourization of methane (3) the first four ionization energies of carbon and electron gain enthalpy of hydrogen (4) the dissociation energy of hydrogen molecule, H2 Ans. (1) 116. Among the following mixtures, dipole-dipole as the major interaction, is present in (1) benzene and ethanol (2) acetonitrile and acetone (3) KCl and water (4) benzene and carbon tetrachloride Ans. (2) 117. Fluorobenzene (C6H5F) can be synthesized in the laboratory (1) by heating phenol with HF and KF (2) from aniline by diazotisation followed by heating the diazonium salt with HBF4 (3) by direct fluorination of benzene with F2 gas (4) by reacting bromobenzene with NaF solution Ans. (2) 118. A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct? (1) MCl2 is more volatile than MCl4 (2) MCl2 is more soluble in anhydrous ethanol than MCl4 (3) MCl2 is more ionic than MCl4 (4) MCl2 is more easily hydrolysed than MCl4 Ans. (3) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 119. Which of the following statements is true? (1) H3PO3 is a stronger acid than H2SO3 (2) In aqueous medium HF is a stronger acid than HCl (3) HClO4 is a weaker acid than HClO3 (4) HNO3 is a stronger acid than HNO2 Ans. (4) 120. The molar conductivities o o ∧NaOAc and ∧HCl at infinite dilution in water at 25oC are 91.0 and 426.2 S cm2/mol respectively. To calculate o ∧HOAc , the additional value required is (1) o H2O ∧ (2) o ∧KCl (3) o ∧NaOH (4) o ∧NaCl Ans. (4) Sol. o o o CH3COONa CH COO Na 3 − + λ =λ +λ ……….. (1) o o o HCl H+ Cl− λ = λ +λ ………………….. (2) o o o NaCl Na Cl− λ = λ +λ …………………. (3) o CH3COOH λ = (1) + (2) − (3) 121. Which one of the following sets of ions represents a collection of isoelectronic species? (1) K+ , Cl–, Ca2+, Sc3+ (2) Ba2+, Sr2+, K+, S2– (3) N3–, O2–, F–, S2– (4) Li+, Na+, Mg2+, Ca2+ Ans. (1) 122. The correct order of increasing acid strength of the compounds (a) CH3CO2H (b) MeOCH2CO2H (c) CF3CO2H (d) CO2H Me Me is (1) b < d < a < c (2) d < a < c < b (3) d < a < b < c (4) a < d < c < b Ans. (3) 123. In which of the following molecules/ions are all the bonds not equal? (1) SF4 (2) SiF4 (3) XeF4 (4) BF4− Ans. (1) 124. What products are expected from the disproportionation reaction of hypochlorous acid? (1) HClO3 and Cl2O (2) HClO2 and HClO4 (3) HCl and Cl2O (4) HCl and HClO3 Ans. (4) 125. Nickel (Z = 28) combines with a uninegative monodentate ligand X– to form a paramagnetic complex 2 [NiX4 ] − . The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively (1) one, tetrahedral (2) two, tetrahedral (3) one, square planar (4) two, square planar FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 Ans. (2) Sol. 28Ni: ………. 3s2, 3p6, 3d8, 4s2 Ni2+: 3s2, 3p6, 3d8 3d 4s 4p sp3 Tetrahedral geometry 126. In Fe(CO)5, the Fe – C bond possesses (1) π-character only (2) both σ and π characters (3) ionic character (4) σ-character only Ans. (2) 127. The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is (1) F < S < P < B (2) P < S < B < F (3) B < P < S < F (4) B < S < P < F Ans. (4) 128. An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct? (1) (Tf)irrev > (Tf)rev (2) Tf > Ti for reversible process but Tf = Ti for irreversible process (3) (Tf)rev = (Tf)irrev (4) Tf = Ti for both reversible and irreversible processes Ans. (1) 129. In Langmuir’s model of adsorption of a gas on a solid surface (1) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered (2) the adsorption at a single site on the surface may involve multiple molecules at the same time (3) the mass of gas striking a given area of surface is proportional to the pressure of the gas (4) the mass of gas striking a given area of surface is independent of the pressure of the gas Ans. (3) 130. Rate of a reaction can be expressed by Arrhenius equation as: k = A e−E /RT In this equation, E represents (1) the energy above which all the colliding molecules will react (2) the energy below which colliding molecules will not react (3) the total energy of the reacting molecules at a temperature, T (4) the fraction of molecules with energy greater than the activation energy of the reaction Ans. (2) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 131. The structure of the major product formed in the following reaction I Cl NaCN DMF → is (1) CN CN (2) I Cl NC (3) CN Cl (4) I CN Ans. (4) 132. Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces (1) 4-phenylcyclopentene (2) 2-phenylcyclopentene (3) 1-phenylcyclopentene (4) 3-phenylcyclopentene Ans. (4) Sol. According to E2 mechanism. 133. Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is (1) Eclipse, Gauche, Anti (2) Gauche, Eclipse, Anti (3) Eclipse, Anti, Gauche (4) Anti, Gauche, Eclipse Ans. (3) 134. The structure of the compound that gives a tribromo derivative on treatment with bromine water is (1) CH3 OH (2) CH2OH (3) CH3 OH (4) CH3 OH Ans. (1) 135. The decreasing values of bond angles from NH3 (106o) to SbH3 (101o) down group-15 of the periodic table is due to (1) increasing bp-bp repulsion (2) increasing p-orbital character in sp3 (3) decreasing lp-bp repulsion (4) decreasing electronegativity Ans. (4) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 136. Me N Et Me n-Bu Δ→ OH The alkene formed as a major product in the above elimination reaction is (1) Me (2) CH2 = CH2 (3) Me (4) Me Ans. (2) 137. The “spin-only” magnetic moment [in units of Bohr magneton, (μB )] of Ni2+ in aqueous solution would be (Atomic number of Ni = 28) (1) 2.84 (2) 4.90 (3) 0 (4) 1.73 Ans. (1) 138. The equilibrium constant for the reaction 3 2 2 SO (g) SO (g) 1O (g) 2 􀁙􀁚􀁚􀁚􀁚􀁘􀁚􀁚 + is Kc = 4.9 × 10–2. The value of Kc for the reaction 2SO2(g) + O2(g)􀁙􀁚􀁚􀁚􀁚􀁚􀁘􀁚2SO3(g) will be (1) 416 (2) 2.40 × 10–3 (3) 9.8 × 10–2 (4) 4.9 × 10–2 Ans. (1) Sol. 2 c 2 K 1 4.9 10− ′ =    ×    104 100 100 4.9 4.9 24.01 × = = × = 4.1649 × 100 = 416.49 139. Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture? (1) The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group (2) In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group (3) Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens (4) In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group Ans. (4) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 140. Given the data at 25oC, Ag + I− →AgI + e− ; Eo = 0.152 V Ag→Ag+ + e−; Eo = −0.800 V What is the value of log Ksp for AgI? 2.303RT 0.059 V F  =      (1) –8.12 (2) +8.612 (3) –37.83 (4) –16.13 Ans. (4) Sol. AgI(s) + e− 􀁙􀁚􀁚􀁚􀁚􀁘􀁚􀁚 Ag(s) + I− ; Eo = −0.152 Ag(s) Ag e ; Eo 0.8 → + + − = − AgI(s) Ag I ; Eo 0.952 → + + − = − o cell E 0.059 logK n = sp 0.952 0.059 log K 1 − = sp log K 0.952 16.135 0.059 = − = − 141. The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr: NO(g) + Br2(g)􀁙􀁚􀁚􀁚􀁚􀁚􀁘􀁚NOBr2(g) NOBr2(g) +NO(g)→2NOBr(g) If the second step is the rate determining step, the order of the reaction with respect to NO(g) is (1) 1 (2) 0 (3) 3 (4) 2 Ans. (4) Sol. NO(g) + Br2(g)􀁙􀁚􀁚􀁚􀁚􀁚􀁘􀁚NOBr2(g) NOBr2(g) +NO(g)→2NOBr(g) 2 R = K[NOBr ] [NO] 2 c 2 c 2 K.K [NO] [Br ][NO], where K [NOBr ] [NO] [Br ] = = 2 2 = K′[NO] [Br ] 142. Lanthanoid contraction is caused due to (1) the appreciable shielding on outer electrons by 4f electrons from the nuclear charge (2) the appreciable shielding on outer electrons by 5d electrons from the nuclear charge (3) the same effective nuclear charge from Ce to Lu (4) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge Ans. (4) 143. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω. The molar conductivity of 0.02 M solution of the electrolyte will be (1) 124 × 10–4 S m2 mol–1 (2) 1240 × 10–4 S m2 mol–1 (3) 1.24 × 10–4 S m2 mol–1 (4) 12.4 × 10–4 S m2 mol–1 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 Ans. (4) Sol. There is one mistake in Question paper. Assuming concentration of solution is 0.2 M instead of 0.02 M. Since resistance of 0.2 M is 520 Ω. R = 100 Ω K 1 R a =       􀁁 1.29 1 100 a =       􀁁 129 m 1 a   = −   􀁁 R = 520Ω , C = 0.2 M K 1 1 (129) 1m 1 R a 520 =   = Ω− −   􀁁 in cm3 μ = K × V 1 129 1000 10 6 m3 520 0.2 = × × × − 129 1000 10 6 520 0.2 = × × − = 1.24 ×10−3 = 12.4 ×10−4 144. The ionic mobility of alkali metal ions in aqueous solution is maximum for (1) K+ (2) Rb+ (3) Li+ (4) Na+ Ans. (2) 145. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is (1) 1.14 mol kg–1 (2) 3.28 mol kg–1 (3) 2.28 mol kg–1 (4) 0.44 mol kg–1 Ans. (3) 146. The enthalpy changes for the following processes are listed below: Cl2(g) = 2Cl(g), 242.3 kJ mol–1 I2(g) = 2I(g), 151.0 kJ mol–1 ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1 I2(s) = I2(g), 62.76 kJ mol–1 Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is (1) –14.6 kJ mol–1 (2) –16.8 kJ mol–1 (3) +16.8 kJ mol–1 (4) +244.8 kJ mol–1 Ans. (3) Sol. 2 2 1I (s) 1Cl ICl(g) 2 2 + → [ ] I2(s) I2(g) I I Cl Cl I Cl H 1 H 1 1 2 2 2 → − − − Δ =  Δ + μ + μ  − μ   1 62.76 1 151.0 1 242.3 (211.3) 2 2 2 =  × + × + ×  −     = 228.03 – 211.3 ΔH = 16.73 147. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca2+ ion? (1) Six (2) Three (3) One (4) Two Ans. (3) FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. FIITJEE Solutions to AIEEE−2006 148. OH + CHCl3 + NaOH → O Na CHO The electrophile involved in the above reaction is (1) dichloromethyl cation (CHCl2 ) (2) dichlorocarbene( :CCl2 ) (3) trichloromethyl anion (CCl3 ) (4) formyl cation (CHO) Ans. (2) 149. 18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100oC is (1) 759.00 Torr (2) 7.60 Torr (3) 76.00 Torr (4) 752.40 Torr Ans. (4) Sol. o s s P P n P N − = s s 18 1 760 P 180 10 0.1 P 178.2 9.9 9.9 18 − = = = s s 760 P 1 P 99 − = 760 × 99 – Ps × 99 = Ps 760 × 99 = 100 Ps s P 760 99 752.4 100 × = = 150. (ΔH− ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 J K–1 mol–1) (1) –1238.78 J mol–1 (2) 1238.78 J mol–1 (3) –2477.57 J mol–1 (4) 2477.57 J mol–1 Ans. (1) Sol. g ΔH− ΔU = Δn RT 1 8.314 298 2 = − × × = –1238.78 FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. MATHEMATICS PART − A 1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC 􀁊􀁊􀁊􀁇 􀁊􀁊􀁊􀁇 with magnitudes 1 AB and 1 AC respectively is the force along AD 􀁊􀁊􀁊􀁇 , where D is the foot of the perpendicular from A onto BC. The magnitude of the resultant is (1) 2 2 2 2 AB AC (AB) (AC) + (2) (AB)(AC) AB + AC (3) 1 1 AB AC + (4) 1 AD Ans. (4) Sol: Magnitude of resultant = 1 2 1 2 AB2 AC2 AB AC AB AC     +   +   =     ⋅ BC BC 1 AB AC AD BC AD = = = ⋅ ⋅ A B C D 2. Suppose a population A has 100 observations 101, 102, … , 200, and another population B has 100 observations 151, 152, … , 250. If VA and VB represent the variances of the two populations, respectively, then A B V V is (1) 1 (2) 9/4 (3) 4/9 (4) 2/3 Ans. (1) Sol: 2 i 2x d n σ = Σ . (Here deviations are taken from the mean) Since A and B both has 100 consecutive integers, therefore both have same standard deviation and hence the variance. ∴ A B V 1 V = ( 2 ) As Σdi is same in both the cases . 3. If the roots of the quadratic equation x2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q − p is (3) 2 (2) 3 (3) 0 (4) 1 Ans. (2) Sol: x2 + px + q = 0 tan 30° + tan 15° = − p tan 30° ⋅ tan 15° = q FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. tan 45° = tan30 tan15 p 1 1 tan30 tan15 1 q ° + ° − = = − ° ° − ⇒ − p = 1 − q ⇒ q − p = 1 ∴ 2 + q − p = 3. 4. The value of the integral, 6 3 x dx 9 − x + x ∫ is (1) 1/2 (2) 3/2 (3) 2 (4) 1 Ans. (2) Sol: I = 6 3 x dx 9 − x + x ∫ I = 6 3 9 x dx 9 x x − − + ∫ 2I = 6 3 ∫dx = 3 ⇒ I = 3 2 . 5. The number of values of x in the interval [0, 3π] satisfying the equation 2sin2x + 5sinx − 3 = 0 is (1) 4 (2) 6 (3) 1 (4) 2 Ans. (1) Sol: 2 sin2 x + 5 sin x − 3 = 0 ⇒ (sin x + 3) (2 sin x − 1) = 0 ⇒ sin x = 1 2 ∴ In (0, 3π), x has 4 values 6. If (a × b)× c = a × (b × c), where a, b and c are any three vectors such that a ⋅ b ≠ 0 , b ⋅ c ≠ 0 , then a and c are (1) inclined at an angle of π/3 between them (2) inclined at an angle of π/6 between them (3) perpendicular (4) parallel Ans. (4) Sol: (a × b) × c = a × (b × c), a ⋅ b ≠ 0, b ⋅ c ≠ 0 ⇒ (a ⋅ c) b − (b ⋅ c)a = (a ⋅ c) b − (a ⋅ b)c (a ⋅ b)c = (b ⋅ c)a a 􀀦 c 7. Let W denote the words in the English dictionary. Define the relation R by : FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is (1) not reflexive, symmetric and transitive (2) reflexive, symmetric and not transitive (3) reflexive, symmetric and transitive (4) reflexive, not symmetric and transitive Ans. (2) Sol: Clearly (x, x) ∈ R ∀ x ∈ W. So, R is reflexive. Let (x, y) ∈ R, then (y, x) ∈ R as x and y have at least one letter in common. So, R is symmetric. But R is not transitive for example Let x = DELHI, y = DWARKA and z = PARK then (x, y) ∈ R and (y, z) ∈ R but (x, z) ∉ R. 8. If A and B are square matrices of size n × n such that A2 − B2 = (A − B) (A + B), then which of the following will be always true ? (1) A = B (2) AB = BA (3) either of A or B is a zero matrix (4) either of A or B is an identity matrix Ans. (2) Sol: A2 − B2 = (A − B) (A + B) A2 − B2 = A2 + AB − BA − B2 ⇒ AB = BA. 9. The value of 10 k 1 sin 2k icos 2k = 11 11  π π   +    Σ is (1) i (2) 1 (3) −1 (4) −i Ans. (4) Sol: 10 10 10 k 1 k 1 k 1 sin 2k icos 2k sin 2k i cos 2k 11 11 11 11 = = =  π π  π π  +  = +   Σ Σ Σ = 0 + i (− 1) = − i. 10. All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval (1) −2 < m < 0 (2) m > 3 (3) −1 < m < 3 (4) 1 < m < 4 Ans. (3) Sol: Equation x2 − 2mx + m2 − 1 = 0 (x − m)2 − 1 = 0 (x − m + 1) (x − m − 1) = 0 x = m − 1, m + 1 − 2 < m − 1 and m + 1 < 4 FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. m > − 1 and m < 3 − 1 < m < 3. 11. A particle has two velocities of equal magnitude inclined to each other at an angle θ. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then θ is (1) 90° (2) 120° (3) 45° (4) 60° Ans. (2) Sol: u sin tan 2 4 u u cos 2 θ θ = + θ ⇒ sin 1 sin cos 1 sin cos 4 2 4 2 4 θ θ θ + θ= θ ∴ 2sin sin 3 3sin 4sin3 4 4 4 4 θ θ θ θ = = − ∴ sin2 1 4 4 θ = ⇒ 30 4 θ = ° or θ = 120°. u u θ/2 θ/4 θ/4 R2 R1 u/2 12. At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is (1) e 6 5 (2) 5 6 (3) 6 55 (4) 5 6 e Ans. (4) Sol: P (X = r) = e mmr r ! − P (X ≤ 1) = P (X = 0) + P (X = 1) = e−5 + 5 × e−5 = 5 6 e . 13. A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10 m/s2, then the height above the point P from where the body began to fall is (1) 720 m (2) 900 m (3) 320 m (4) 680 m Ans. (1) FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Sol: We have h 1 gt2 2 = and h + 400 = 1 g(t 4)2 2 + . Subtracting we get 400 = 8g + 4gt ⇒ t = 8 sec ∴ h 1 10 64 320m 2 = × × = ∴ Desired height = 320 + 400 = 720 m. h 400m Q(t) P(t+4) 14. 0 xf(sin x)dx π∫ is equal to (1) 0 f(cos x)dx π π∫ (2) 0 f(sinx)dx π π∫ (3) / 2 0 f(sinx)dx 2 π π∫ (4) / 2 0 f(cos x)dx π π ∫ Ans. (4) Sol: I = 0 0 xf(sin x)dx ( x) f(sin x)dx π π ∫ = ∫ π − = 0 f(sin x)dx I π π∫ − 2I = 0 f(sin x)dx π π∫ I = / 2 0 0 f(sin x)dx f(sin x)dx 2 π π π ∫ = π ∫ = π / 2 0 f(cos x)dx π ∫ . 15. A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is (1) x + y = 7 (2) 3x − 4y + 7 = 0 (3) 4x + 3y = 24 (4) 3x + 4y = 25 Ans. (3) Sol: The equation of axes is xy = 0 ⇒ the equation of the line is x 4 y 3 12 2 ⋅ + ⋅ = ⇒ 4x + 3y = 24. 16. The two lines x = ay + b, z = cy + d; and x = a′y + b′, z = c′y + d′ are perpendicular to each other if (1) aa′ + cc′ = −1 (2) aa′ + cc′ = 1 (3) a c 1 a c + = − ′ ′ (4) a c 1 a c + = ′ ′ FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Ans. (1) Sol: Equation of lines x b y z d a c − − = = x b y z d a c − ′ − ′ = = ′ ′ Lines are perpendicular ⇒ aa′ + 1 + cc′ = 0. 17. The locus of the vertices of the family of parabolas a3x2 a2x y 2a 3 2 = + − is (!) xy 105 64 = (2) xy 3 4 = (3) xy 35 16 = (4) xy 64 105 = Ans. (1) Sol: Parabola: y = a3x2 a2x 2a 3 2 + − Vertex: (α, β) α = 2 3 a /2 3 2a / 3 4a − = − , β = 4 3 4 3 3 a 4 a 2a 1 8 4 3 a 4 3 a 4 a 4 3 3     −  + ⋅ ⋅  −  +    = −   = 35 a 3 35 a 12 4 16 − × =− αβ = − 3 35 a 105 4a 16 64  −  =     . 18. The values of a, for which the points A, B, C with position vectors 2ˆi − ˆj + kˆ, ˆi − 3ˆj − 5kˆ and aˆi − 3ˆj + kˆ respectively are the vertices of a right-angled triangle with C 2 π = are (1) 2 and 1 (2) −2 and −1 (3) −2 and 1 (4) 2 and −1 Ans. (1) Sol: ⇒ BA = ˆi − 2ˆj + 6kˆ 􀁊􀁊􀁊􀁇 CA = (2 − a)ˆi + 2ˆj 􀁊􀁊􀁊􀁇 CB = (1− a)ˆi − 6kˆ 􀁊􀁊􀁊􀁇 CA ⋅CB 􀁊􀁊􀁊􀁇 􀁊􀁊􀁊􀁇 = 0 ⇒ (2 − a) (1 − a) = 0 ⇒ a = 2, 1. FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 19. ( ) ( ) / 2 3 2 3 /2 x cos x 3 dx −π − π  + π + + π  ∫   is equal to (1) 4 32 π (2) 4 32 2 π π + (3) 2 π (4) 1 4 π − Ans. (3) Sol: I = / 2 3 2 3 /2 (x ) cos (x 3 ) dx −π − π  + π + + π  ∫   Put x + π = t I = / 2 / 2 3 2 2 / 2 0 t cos t dt 2 cos tdt π π −π  +  = ∫   ∫ = / 2 0 (1 cos 2t)dt 0 2 π π ∫ + = + . 20. If x is real, the maximum value of 2 2 3x 9x 17 3x 9x 7 + + + + is (1) 1/4 (2) 41 (3) 1 (4) 17/7 Ans. (2) Sol: 2 2 y 3x 9x 17 3x 9x 7 + + = + + 3x2(y − 1) + 9x(y − 1) + 7y − 17 = 0 D ≥ 0 ∵ x is real 81(y −1)2 − 4x3(y −1)(7y −17) ≥ 0 ⇒ (y − 1) (y − 41) ≤ 0 ⇒ 1 ≤ y ≤ 41. 21. In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is (1) 3 5 (B) 1 2 (C) 4 5 (D) 1 5 Ans. (1) Sol: 2ae = 6 ⇒ ae = 3 2b = 8 ⇒ b = 4 b2 = a2(1 − e2) 16 = a2 − a2e2 a2 = 16 + 9 = 25 a = 5 e 3 3 a 5 ∴ = = FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 22. Let A = 1 2 3 4       and B = a 0 0 b       , a , b ∈ N. Then (1) there cannot exist any B such that AB = BA (2) there exist more than one but finite number of B’s such that AB = BA (3) there exists exactly one B such that AB = BA (4) there exist infinitely many B’s such that AB = BA Ans. (4) Sol: 1 2 a 0 A B 3 4 0 b     =   =       a 2b AB 3a 4b   =    a 0 1 2 a 2a BA 0 b 3 4 3b 4b      =     =        AB = BA only when a = b 23. The function f(x) = x 2 2 x + has a local minimum at (1) x = 2 (2) x = −2 (3) x = 0 (4) x = 1 Ans. (1) Sol: x 2 2 x + is of the form x 1 2 x + ≥ & equality holds for x = 1 24. Angle between the tangents to the curve y = x2 − 5x + 6 at the points (2, 0) and (3, 0) is (1) 2 π (2) 2 π (3) 6 π (4) 4 π Ans. (2) Sol: dy 2x 5 dx = − ∴ m1 = (2x − 5)(2, 0) = −1, m2 = (2x − 5)(3, 0) = 1 ⇒ m1m2 = −1 25. Let a1, a2, a3, … be terms of an A.P. If 2 1 2 p 2 1 2 q a a a p , p q a a a q + + ⋅ ⋅ ⋅ = ≠ + +⋅⋅⋅+ , then 6 21 a a equals (1) 41 11 (2) 7 2 (3) 2 7 (4) 11 41 Ans. (4) FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Sol: ( ) ( ) ( ) ( ) 1 2 1 2 1 1 p 2a p 1 d 2a p 1 d 2 p p q 2a q 1 d q 2a q 1 d q 2  + −  + − = ⇒ =  + −  + −   1 1 a p 1 d 2 p a q 1 d q 2  −  +    =  −  +    For 6 6 21 21 a a 11 , p 11, q 41 a a 41 = = → = 26. The set of points where f(x) x 1 |x| = + is differentiable is (1) (−∞, 0) ∪ (0, ∞) (2) (−∞, −1) ∪ (−1, ∞) (3) (−∞, ∞) (4) (0, ∞) Ans. (3) Sol: ( ) ( ) 2 2 x 1 , x 0 , x 0 (1 x) f x 1 x f (x) x 1 , x 0 , x 0 1 x 1 x   <  <  − =  − ⇒ ′ =     ≥  ≥  +  + ∴ f′(x) exist at everywhere. 27. A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is (1) 3 x2 2 (2) x3 8 (3) 1 x2 2 (4) πx2 Ans. (3) Sol: Area = 1 x2 sin 2 θ 2 max A 1 x at sin 1, 2 2  π  =  θ = θ =    θ x x 28. At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is (1) 5040 (2) 6210 (3) 385 (4) 1110 Ans. (3) Sol: 10C1 + 10C2 + 10C3 + 10C4 = 10 + 45 + 120 + 210 = 385 FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 29. If the expansion in powers of x of the function 1 (1− ax)(1− bx) is a0 + a1x + a2x2 + a3x3 + … , then an is (1) bn an b a − − (2) an bn b a − − (3) an 1 bn 1 b a + − + − (4) bn 1 an 1 b a + − + − Ans. (4) Sol: (1 ax) 1 (1 bx) 1 (1 ax a2x2 ......)(1 bx b2x2 ....) − − − − = + + + + + + ∴ coefficient of xn = bn + abn−1 + a2bn−2 + .... + an−1b + an = bn 1 an 1 b a + − + − ∴ n 1 n 1 n a b a b a + − + = − 30. For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is (1) (20, 45) (2) (35, 20) (3) (45, 35) (4) (35, 45) Ans. (4) Sol: ( )m ( )n m m 2 n n 2 1− y 1+ y = 1− C1y + C2y − .... 1+ C1y + C2y + ... = ( ) ( ) ( ) m m 1 n n 1 2 1 n m mn y ..... 2 2  − −  + − +  + −  +   2 2 1 2 a n m 10 and a m n m n 2mn 10 2 + − − − ∴ = − = = = So, n − m = 10 and (m − n)2 − (m + n) = 20 ⇒ m + n = 80 ∴ m = 35, n = 45 31. The value of a 1 ∫[x] f′(x)dx , a > 1, where [x] denotes the greatest integer not exceeding x is (1) af(a) − {f(1) + f(2) + … + f([a])} (2) [a] f(a) − {f(1) + f(2) + … + f([a])} (3) [a] f([a]) − {f(1) + f(2) + … + f(a)} (4) af([a]) − {f(1) + f(2) + … + f(a)} Ans. (2) Sol: Let a = k + h, where [a] = k and 0 ≤ h < 1 [ ] ( ) ( ) ( ) ( ) ( ) a 2 3 k k h 1 1 2 k 1 k x f ' x dx 1f ' x dx 2f ' x dx ........ k 1 dx kf ' x dx + − ∴ ∫ =∫ + ∫ + ∫ − + ∫ {f(2) − f(1)} + 2{f(3) − f(2)} + 3{f(4) − f(3)}+…….+ (k−1) – {f(k) − f(k − 1)} + k{f(k + h) − f(k)} = − f(1) − f(2) − f(3)……. − f(k) + k f(k + h) = [a] f(a) − {f(1) + f(2) + f(3) + …. + f([a])} FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 32. If the lines 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 are two diameters of a circle of area 49π square units, the equation of the circle is (1) x2 + y2 + 2x − 2y − 47 = 0 (2) x2 + y2 + 2x − 2y − 62 = 0 (3) x2 + y2 − 2x + 2y − 62 = 0 (4) x2 + y2 − 2x + 2y − 47 = 0 Ans. (4) Sol: Point of intersection of 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 is (1 , − 1), which is the centre of the circle and radius = 7. ∴ Equation is (x − 1)2 + (y + 1)2 = 49 ⇒ x2 + y2 − 2x + 2y − 47 = 0. 33. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants is of (1) second order and second degree (2) first order and second degree (3) first order and first degree (4) second order and first degree Ans. (4) Sol: Ax2 +By2 = 1 … (1) Ax By dy 0 dx + = … (2) 2 2 2 A By d y B dy 0 dx dx + +   =     … (3) From (2) and (3) 2 2 2 x By d y B dy By dy 0 dx dx dx     − −    + =     ⇒ 2 2 2 xy d y x dy y dy 0 dx dx dx +   − =     34. Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of 2 3 π at its centre is (1) x2 y2 3 2 + = (B) x2 + y2 = 1 (3) x2 y2 27 4 + = (D) x2 y2 9 4 + = Ans. (4) Sol: 2 2 cos h k h2 k2 9 3 3 4 π + = ⇒ + = 35. If (a, a2) falls inside the angle made by the lines y x , x 0 2 = > and y = 3x, x > 0, then a belongs to (1) 0, 1 2       (2) (3, ∞) (3) 1, 3 2       (4) 3, 1 2  − −      FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Ans. (3) Sol: a2 − 3a < 0 and a2 a 0 1 a 3 2 2 − > ⇒ < < 36. The image of the point (−1, 3, 4) in the plane x − 2y = 0 is (1) 17 , 19 , 4 3 3  − −      (2) (15, 11, 4) (3) 17 , 19 ,1 3 3  − −      (4) (8, 4, 4) Sol: If (α, β, γ) be the image then 1 2 3 0 2 2 α −  β +  −   =   ∴ α − 1 − 2β − 6 ⇒ α − 2β = 7 … (1) and 1 3 4 1 2 0 α + β − γ − = = − … (2) From (1) and (2) 9, 13, 4 5 5 α = β = − γ = No option matches. 37. If z2 + z + 1 = 0, where z is a complex number, then the value of 2 2 2 2 2 3 6 2 3 6 z 1 z 1 z 1 z 1 z z z z  +  +  +  +  +  + ⋅ ⋅ ⋅ +  +                  is (1) 18 (2) 54 (3) 6 (4) 12 Ans. (4) Sol: z2 + z + 1 = 0 ⇒ z = ω or ω2 so, 2 2 2 3 3 3 2 3 z 1 1, z 1 1, z 1 2 z z z + = ω+ω = − + = ω +ω = − + = ω +ω = 4 5 6 4 5 6 z 1 1, z 1 1 and z 1 2 z z z + = − + = − + = ∴ The given sum = 1 + 1 + 4 + 1 + 1 + 4 = 12 38. If 0 < x < π and cosx + sinx = 1 2 , then tanx is (1) (1 7) 4 − (B) (4 7) 3 − (3) (4 7) 3 + − (4) (1 7) 4 + Ans. (3) Sol: cos x sinx 1 1 sin2x 1 sin2x 3 , 2 4 4 + = ⇒ + = ⇒ =− so x is obtuse and 2 2 2tanx 3 3tan x 8tanx 3 0 1 tan x 4 = − ⇒ + + = + FIITJEE Solutions to AIEEE−2006 FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. ∴ tanx 8 64 36 4 7 6 3 − ± − − ± = = tanx 0 tanx 4 7 3 − − ∵ < ∴ = 39. If a1, a2, … , an are in H.P., then the expression a1a2 + a2a3 + … + an−1an is equal to (1) n(a1 − an) (2) (n − 1) (a1 − an) (3) na1an (4) (n − 1)a1an Ans. (4) Sol: 2 1 3 2 n n1 1 1 1 1 ..... 1 1 d a a a a a a− − = − = = − = (say) Then 1 2 2 3 n 1 n 1 2 2 3 n 1 n a a a a a a a a , a a ,......., a a d d d − − − − − = = = ∴ 1 n 1 2 2 3 n 1 n a a a a a a ....... a a d − − + + + = Also, ( ) n 1 1 1 n 1 d a a = + − ⇒ 1 n ( ) 1 n a a n 1aa d − = − 40. If xm ⋅ yn = (x + y)m+n , then dy dx is (1) y x (2) x y xy + (3) xy (4) x y Ans. (1) Sol: xm.yn (x y)m n mlnx nlny (m n)ln(x y) + = + ⇒ + = + + ∴ m n dy m n 1 dy m m n m n n dy x y dx x y dx x x y x y y dx +    +   +  + =  +  ⇒  −  =  −  +    +   +  ⇒ ( ) ( ) my nx my nx dy dy y x x y y x y dx dx x −  −  =   ⇒ = +  +  FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 41. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity ‘v’ that varies as v = α x . The displacement of the particle varies with time as (1) t3 (2) t2 (3) t (4) t1/2 Ans: (2) Sol. dx dx dt dt x = α ∫ = ∫α ⇒ x α t2 42. A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is (1) Mg( 2 −1) (2) Mg( 2 +1) (3) Mg 2 (4) Mg 2 Ans: (1) Sol. F 􀁁 sin 45 = Mg (􀁁 − 􀁁 cos 45) F = Mg (√2 − 1) 43. A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms−1. The kinetic energy of the other mass is (1) 96 J (2) 144 J (3) 288 J (4) 192 J Ans: (3) Sol. m1v1 = m2v2 2 2 2 KE 1m v 1 4 144 288J 2 2 = = × × = 44. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by the force of gravity during the time the particle goes up is (1) 0.5 J (2) −0.5 J (3) −1.25 J (4) 1.25 J Ans: (3) Sol. − mgh = − mg v2 1.25J 2g     = −   . 45. A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms−1. The velocity of sound in air is 300 ms−1. If the person can hear frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear the whistle is (1) 30 ms−1 (2) 15 2 ms−1 (3) 15 / 2 ms−1 (4) 15 ms−1 Ans: (4) Sol. ( ) app f 300 f v 15m/ s 300 v = ⇒ = − 46. A electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. (1) a torque only (2) a translational force only in the direction of the field (3) a translational force only in a direction normal to the direction of the field (4) a torque as well as a translational force Ans: (4) Sol. A torque as well as a translational force 47. A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio 􀁁A /􀁁B of their respective lengths must be (1) 2 (2) 1 (3) 1 2 (4) 1 4 Ans: (1) Sol. A A 1 2 A R R ρ = π 􀁁 B B 2 2B R R ρ = π 􀁁 2 2 A B A A A 2 2 B A B A A R 2 R R 4R ρ ρ = = ρ ρ ⋅ 􀁁 􀁁 ⇒ B A 􀁁 = 2 􀁁 48. The Kirchhoff’s first law (Σi = 0) and second law (ΣiR =ΣE) , where the symbols have their usual meanings, are respectively based on (1) conservation of charge, conservation of energy (2) conservation of charge, conservation of momentum (3) conservation of energy, conservation of charge (4) conservation of momentum, conservation of charge Ans: (1) Sol. Conservation of charge, conservation of energy 49. In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a (1) circle (2) helix (3) straight line (4) ellipse Ans: (3) Sol. Straight line 50. Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will (1) attract all three of them (2) attract N1 and N2 strongly but repel N3 (3) attract N1 strongly, N2 weakly and repel N3 weakly (4) attract N1 strongly, but repel N2 and N3 weakly Ans: (3) Sol. attracts N1 strongly, N2 weakly and Repel N3 weakly 51. Which of the following units denotes the dimensions ML2/Q2, where Q denotes the electric charge? FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. (1) Weber (Wb) (2) Wb/m2 (3) Henry (H) (4) H/m2 Ans: (3) Sol. Henry (H) 52. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to (1) 300 N (2) 150 N (3) 3 N (4) 30 N Ans: (4) Sol. (mv − 0)⇒0.15× 20 F 3 30N 0.1 = = 53. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2 (1) 22 N (2) 4 N (3) 16 N (4) 20 N Ans: (4) Sol. mgh = Fs F = 20 N 54. Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position? (1) d (2) 2 1 m m d (3) 1 1 2 m m +m d (4) 1 2 m d m Ans: (4) Sol. m1 d + m2 x = 0 1 2 m d x m = 55. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? (1) 1 s 12 (2) 1 s 6 (3) 1 s 4 (4) 1 s 3 Ans: (2) Sol. 2 2 max 1mv 3 1mv 2 42 =       A2 2 cos2 t 3 A2 2 4 ω ω ⇒ ω FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. cos t 3 2 ω = t t 1 sec 6 6 π ω = ⇒ = 56. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is (1) 100 s (2) 0.01 s (3) 10 s (4) 0.1 s Ans: (2) Sol. Aω = vmax max T 2 2 A 0.01sec v π π = = = ω 57. A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is (1) 10.5 Hz (2) 105 Hz (3) 1.05 Hz (4) 1050 Hz Ans: (2) Sol. ( ) ( ) n n 1 v 315, v 420 2 2 + = = 􀁁 􀁁 Solving v 105 2 = 􀁁 58. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on Earth, at a distance r from the Sun. (1) 2 4 2 R T r σ (2) 2 2 4 0 2 4 r R T r π σ (3) 2 2 4 0 2 r R T r π σ (4) 2 2 4 0 2 r R T 4 r σ π where r0 is the radius of the Earth and σ is Stefan’s constant. Ans: (3) Sol. 2 0 4 2 2 r ( T 4 R ) 4 r π σ ⋅ π π 4 2 2 0 2 T R r r σπ = 59. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be an of minimum deviation for red and blue light respectively in a prism of this glass. Then (1) D1 > D2 (2) D1 < D2 (3) D1 = D2 (4) D1 can be less than or greater than depending upon the angle of prism Ans: (2) Sol. D = (μ − 1)A D2 > D1 FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 60. In a Wheatstone’s bridge, there resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for bridge to be balanced will be (1) 1 2 P R Q S S = + (2) 1 2 P 2R Q S S = + (3) 1 2 1 2 P R(S S ) Q SS + = (4) 1 2 1 2 P R(S S ) Q 2SS + = Ans: (3) Sol. 1 2 1 2 P R(S S ) Q SS + = 61. The current I drawn from the 5 volt source will be (1) 0.17 A (2) 0.33 A (3) 0.5 A (4) 0.67 A Ans: (3) Sol. i 5 0.5 10 = = 5Ω 10Ω 5Ω 5 Volt 10Ω I 10Ω + − 62. In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2μF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is (1) 4 × 10−3 V (2) 2.5 × 10−2 V (3) 40 V (4) 250 V Ans: (4) Sol. i 100 0.1A 1000 = = L C 6 V V 0.1 250 200 2 10− = = = × × V 63 Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 −V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10−19 C, me = 9.11 × 10−31 kg) (1) 32 × 10−19 m/s (2) 2.65 × 106 m/s (3) 7.02 × 1012 m/s (4) 1.87 × 106 m/s 0.1 m Y X 1 2 Ans: (2) Sol. 1mv2 eV 2 = v 2eV m = = 2.65 × 106 m/s 64. The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 Ω at a temperature of (1) 200°C (2) 300°C (3) 400°C (4) 500°C Ans: (2) FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Sol. 200 = 100[1 +(0.005 × Δt)] T− 100 = 200 T = 300° C 65. In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is ‘ (1) N.A.B.ω (2) N.A.B.R.ω (3) N.A.B (4) N.A.B.R Ans: (1) Sol. NBAω 66. The flux linked with a coil at any instant ‘t’ is given by φ = 10t2 − 50t + 250 The induced emf at t = 3 s is (1) 190 V (2) −190 V (3) −10 V (4) 10 V Ans: (3) Sol. e d (20 t 50) 10 dt φ = − = − − = − volt 67. A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will (1) flow from Antimony to Bismuth at the cold junction (2) flow from Antimony to Bismuth at the hot junction (3) flow from Bismuth to Antimony at the cold junction (4) not flow through the thermocouple Ans: (1) Sol. Flow from Antimony to Bismuth at cold junction 68. The time by a photoelectron to come out after the photon strikes is approximately (1) 10−1 s (2) 10−4 s (3) 10−10 s (4) 10−16 s Ans: (3) Sol. 10−10 sec. 69. An alpha nucleus of energy 1mv2 2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to (1) 1 Ze (2) v2 (3) 1 m (4) 4 1 v Ans: (3) FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Sol. 1 m 70. The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface 5 V. The incident radiation lies in (1) X-ray region (2) ultra-violet region (3) infra-red region (4) visible region Ans: (2) Sol. 1242eVnm 1100 11.2 λ = ≈ Å Ultraviolet region 71. The energy spectrum of β-particles [number N(E) as a function of β-energy E] emitted from a radioactive source is (1) N(E) E0 E (2) N(E) E0 E (3) N(E) E0 E (4) N(E) E0 E Ans: (4) Sol. N(E) E0 E 72. When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8, the emitted particles will be (1) neutrons (2) alpha particles (3) beta particles (4) gamma photons Ans: (4) Sol. Gamma-photon 73. A solid which is transparent to visible light and whose conductivity increases with temperature is formed by (1) Metallic binding (2) Ionic binding (3) Covalent binding (4) Van der Waals binding Ans: (3) Sol. Covalent binding FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 74. If the ratio of the concentration of electrons that of holes in a semiconductor is 7 5 and the ratio of currents is 7 , 4 then what is the ratio of their drift velocities? (1) 4 7 (2) 5 8 (3) 4 5 (4) 5 4 Ans: (4) Sol. e n n 7 n 5 = e n I 7 I 4 = d e d n (V ) (V ) ⇒ e n n e I n 5 I n 4 × = 75. In a common base mode of a transistor, t collector current is 5.488 mA for an emit current of 5.60 mA. The value of the base current amplification factor (β) will be (1) 48 (2) 49 (3) 50 (4) 51 Ans: (2) Sol. Ib = Ie − Ic β = c b I 49 I = 76. The potential energy of a 1 kg particle free move along the x-axis is given by x4 x2 V(x) J 4 2   =  −    The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is (1) 2 (2) 3 / 2 (3) 2 (4) 1/ 2 Ans: (2) Sol. k Emax = ET − Umin Umin (±1) = −1/4 J KEmax = 9/4 J ⇒ U = 3 2 J 77. A force of −Fkˆ acts on O, the origin of the coordinate system. The torque about the point (1, −1) is (1) −F(ˆi − ˆj) (2) F(ˆi − ˆj) (3) −F(ˆi + ˆj) (4) F(ˆi + ˆj) Ans: (3) Sol. 􀁇τ = (−ˆi + ˆj)×(−Fkˆ) = – F( ˆi + ˆj) FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 78. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ω′ = (1) m (m 2M) ω + (2) (m 2M) m ω + (3) (m 2M) (m 2M) ω − + (4) m (m M) ω + Ans: (1) Sol. Li = Lf mR2ω = (mR2 + 2MR2)ω′ ω′ = m m 2M  ω   +    79. If the terminal speed of a sphere of gold (density = 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3) of the same size in the same liquid. (1) 0.2 m/s (2) 0.4 m/s (3) 0.133 m/s (4) 0.1 m/s Ans: (4) Sol. s s g g v ( ) v ( ) ρ − ρ = ρ −ρ 􀁁 􀁁 vs = 0.1 m/s 80. The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7° C. The gas is (R = 8.3 J mol−1 K−1) (1) monoatomic (2) diatomic (3) triatomic (4) a mixture of monoatomic and diatomic Ans: (2) Sol. 146 = CvΔT ⇒ Cv = 21 J/mol K 81. The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is (1) 3.3 × 10−3 J/m3 (2) 4.58 × 10−6 J/m3 (3) 6.37 × 10−9 J/m3 (4) 81.35 × 10−12 J/m3 Ans: (2) Sol. Uav = 2 6 0Erms 4.58 10ε = × − J/m3 2 6 3 rms E = 4.58 ×10− J/m 82. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time (1) at the highest position of the platform (2) at the mean position of the platform (3) for an amplitude of 2 g ω (4) for an amplitude of 2 2 g ω Ans: (3) Sol. Aω2 = g ⇒ A = g/ω2 FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 83. An electric bulb is rated 220 volt − 100 watt. The power consumed by it when operated on 110 volt will be (1) 50 watt (2) 75 watt (3) 40 watt (4) 25 watt Ans: (4) Sol. 2 2 1 2 1 2 V V Resistance P P = = ⇒ P2 = 25 W 84. The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows : (1) O λ I (2) O λ I (3) O λ I (4) O λ I Ans: (3) 85. The ‘rad’ is the correct unit used to report the measurement of (1) the rate of decay of radioactive source (2) the ability of a beam of gamma ray photons to produce ions in a target (3) the energy delivered by radiation to a target. (4) the biological effect of radiation Ans: (4) 86. If the binding energy per nucleon in 73 Li and 4 2He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction 7 4 p +3 Li→ 2 2He energy of proton must be (1) 39.2 MeV (2) 28.24 MeV (3) 17.28 MeV (4) 1.46 MeV Ans: (3) Sol. EP = (8 × 7.06 – 7 × 5.60) MeV = 17.28 MeV 87. If the lattice constant of this semiconductor is decreased, then which of the following is correct? EC EV Eg Condutton band width Band gap Valence badn width (1) All Ec, Eg, Ev decrease (2) All Ec, Eg, Ev increase (3) Ec, and Ev increase but Eg decreases (4) Ec, and Ev, decrease Eg increases Ans: (4) FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. 88. In the following, which one of the diodes is reverse biased? (1) R + 5 V (2) R + 10 V + 5 V (3) − 10 V R − 5 V (4) R −10 V Ans: (1) 89. The circuit has two oppositely connect ideal diodes in parallel. What is the current following in the circuit? 4 Ω 3 Ω 2 Ω D1 12 V D2 (1) 1.33 A (2) 1.71 A (3) 2.00 A (4) 2.31 A Ans: (3) Sol. D1 is reverse biased therefore it will act like an open circuit. i 12 2.00 A 6 = = 90. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10−2 Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre is (1) 1.05 × 10−4 Weber/m2 (2) 1.05 × 10−2 Weber/m2 (3) 1.05 × 10−5 Weber/m2 (4) 1.05 × 10−3 Weber/m2 Ans: (2) Sol. B2 = 2 1 2 2 2 1 1 B n i (6.28 10 )(100 i / 3) 1.05 10 n i 200(i) − × × − = = × W/m2 91. Four point masses, each of value m, are placed at the corners of a square ABCD of side 􀁁. The moment of inertia through A and parallel to BD is (1) m􀁁2 (2) 2 m􀁁2 (3) 3m􀁁2 (4) 3 m􀁁2 Ans: (4) Sol. I = 2m (􀁁/ 2 )2 = 3 m􀁁2 92. A wire elongates by 􀁁 mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm) (1) 􀁁/2 (2) 􀁁 (3) 2􀁁 (4) zero FIITJEE Solutions to AIEEE−2006- PHYSICS FIITJEE Ltd., ICES House, 29 – A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 26515949, 26569493, Fax: 011-26513942. Ans: (2) 93. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while Box B contains one mole of helium at temperature (7/3) T0. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, Tf, in terms of T0 is (1)f 0 T 5 T 2 = (2) f 0 T 3 T 7 = (3)f 0 T 7 T 3 = (4) f 0 T 3 T 2 = Ans: (4) Sol. ΔU = 0 ⇒ f 0 f 0 3R(T T ) 1 5R(T 7 T ) 0 2 2 3 − + × − = Tf = 0 3 T 2 94. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is (1) 1 : 4 (2) 4 : 1 (3) 1 : 2 (4) 2 : 1 Ans: (4) Sol. A B B A E r 2 E r 1 = = 95. An inductor (L = 100 mH), a resistor (R = 100 Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 mm after the circuit is E A B R L (1) 1 A (2) 1/e A (3) e A (4) 0.1 A Ans: (2) Sol. I = I0 e−Rt /L = 1 A e * * * * * *