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IIT JEE Exam 2021 : Pace JEE MAINS AITS 2 (SOLVED)

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AITS 16 (Main) Physics PART (A) : PHYSICS SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 1. 1. 2. 2. A circular flexible loop of wire of radius r carrying a current I is placed in a uniform magnetic field B. If B is doubled, then tension in the loop (A) remains unchanged (B) is doubled (C) is halved (D) becomes 4 times (B) The position of a point P is r a cos i b sin j , where a and b are constants and angle between r and x-axis. If the rate of increasing of is . Find the equation of path of particle. (A) Circle (B) Parabola (C) Ellipse (D) Straight line (C) r xi yj x a cos But, r a cos i b sin j y b sin x 2 y2 2 cos 2 sin 2 1 2 a b x 2 y2 1 a 2 b2 This is an equation of ellipse 3. Two positively charged particles each having charge Q and are d distance apart. A third charge is introduced in midway on the line joining the two charges. Find nature and magnitude of third charge, so that the system is in equilibrium. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1 AITS 16 (Main) (A) q 3. Physics Q 4 (B) q Q 4 (C) q 3Q 4 (D) q 3Q 4 (A) Condition for equilibrium kQ 2 kqQ kQ 0 2 Q 4q 0 2 2 d d d / 2 a 4. 4. Q 4 Two wires A and B are of same length, radius and same material are in unison. If tension in A is increased by 4%, 4 beats are heard, then the frequency of the note produced when they were in unison, will be nearly (A) 50 Hz (B) 100 Hz (C) 150 Hz (D) 200 Hz (D) 2 104 n 4 or 100 n 5. Shell is fired from a cannon with a velocity v at an angle with the horizontal direction. At the highest point in its path, it explodes into two pieces, one of the particle follows its path back to the cannon then the speed of the other pieces immediately after the explosion is (A) 3v cos 5. 104 n 4 n 200Hz 100 n (B) 2v cos 3 (C) v cos 2 (D) 3 v cos 2 (A) As we know that at the highest point, the shell has only the horizontal component of velocity which is v cos . If u be the velocity of second exploded piece, then applying conservation of linear momentum along X-axis. 2mv cos mv cos mu Or u 3v cos 6. 6. 7. 7. 8. A vertical cylinder divided into two parts by a frictionless piston in the ratio of 5 : 4. The piston is free to slide along the length of the vessel and length of the vessel is 90 cm. Each of the two parts of the vessel contains 0.1 mole of an ideal gas and the temperature of gas is 300 K. Calculate the mass of the piston. g 9.8m / s 2 (A) 14 kg (B) (B) 12.7 kg (C) 16 kg (D) 15 kg A neckless weighing 50 g in air, but it weights 46 g in water. Assume copper is mixed with gold to prepare the neckless. Find how much copper is present in it. (Specific gravity of gold is 20 and that of copper is 10) (A) m = 25g (B) m = 30g (C) m = 35g (D) m = 20g (B) When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then (A) kinetic energy of body at any instant after releasing is inversely proportional to their masses CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 2 AITS 16 (Main) Physics (B) kinetic energy of body at any instant may or may not be inversely proportional to their masses KE of A mass of B (C) , When spring is massless KE of B mass of A 8. 9. 9. (D) Both (B) and (C) are correct (D) When spring is massless. Then according to momentum conservation principle, Pi Pf or O m1v1 m 2 v 2 m1v1 m 2 v 2 m1v1 m 2 v 2 or p1 p 2 K1 K1 m 2 K 2 m1 p12 p2 K2 2 2m1 2m 2 p1 p2 A charged particle of mass m and charge q, enters onto a plane horizontally uniform magnetic field B acting into the plane. The plane is frictional having coefficient of friction . The speed of charged particle just before entering into the region is v0. The radius of curvature of the path after the time v0 is 2 g mv0 mv0 mv0 (A) (B) (C) (D) None of these qB 2qB 4qB (B) Here, centripetal acceleration is provided by magnetic force but tangential acceleration is provided by force of friction mv 2 qvB r mv r qB Tangential acceleration a mg m v v v g 0 0 2 g 2 mv mv0 r qB 2qB 10. 10. A galvanometer of 50 resistance has 25 divisions. A current of 4 10 4 A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of (A) 2500 as a shunt (B) 2950 as a shunt (C) 2550 in series (D) 2450 in series (D) Here, Ig 25 4 10 4 A 10 2 A CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 3 AITS 16 (Main) Physics To convert the galvanometer into a voltmeter, we must join a series resistance of V 25 R G 2 50 2500 50 2450 Ig 10 11. Choose the correct relation between the rms speed v rms of the gas molecules and the velocity of sound in that gas vs in identical situations of pressure and temperature. (A) Vrms VS 11. 3 (B) Vrms vs (C) Vrms vs 3 (D) Vrms 3VS (B) 3RT RT Vrms and VS m m 12. 12. 3 Vrms VS The instantaneous velocity of point B of the given rods of length 0.5 is 3m/s in the represented direction. The angular velocity of the rod for minimum velocity of end A is (A) 1.5rad / s (B) 5.2 rad / s (C) 2.5rad / s (D) 3.6 rad / s (B) If rod is rotated about end A, then vertical component of velocity, V of end A will be zero. 13. 13. V sin 60 3V 3 3 5.2rad / s l 2l 2 0.5 A capacitor of capacitance 10 F is charged by connecting through a resistance of 20 and battery of 20 V. What is the energy supplied by the battery? (A) Less than 2 mJ (C) More than 2 mJ (C) (B) Equal to 2 mJ (D) Cannot be predicted CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 4 AITS 16 (Main) 14. 14. 15. Physics In the ideal double slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength ). The intensity at the position where the central maximum occurred previously remain unchanged. The minimum thickness of the glass plate is 2 (A) 2 (B) (C) (D) 3 3 (A) 1 tD = fringe width , D We can write, shift d d Thickness, t 2 A metal rod AB of length l is rotated with a constant angular velocity about an axis passing through O and normal to its length. Potential difference between ends of rod in absence of external magnetic field (where, e= electric charge) (A) zero 15. (B) m 2l 2 4e (C) m 2l 2 2e (D) m 2l 2 8e (B) V eV m 2 x m 2 x e 3/ 4 dV 16. 16. m 2 m 2 m 2l 2 dx V x dx e e 1/4 4e The displacement of a particle is given by x t 2 where, x is in metres and t in seconds. The distance covered by the particle in first 4 s is (A) 4 m (B) 8 m (C) 12 m (D) 16 m (B) 2 Here, x t 2 2 dx 2 t 2 m / s dt dv Acceleration , a 2ms 2 (i.e uniform) dt Velocity v When t 0, v 4m / s t 2s, v 0 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 5 AITS 16 (Main) Physics t 4s, v 4m / s Velocity (v)-time (t) graph of this motion is as shown in figure Distance travelled = area AOB + area BCD 4 2 4 2 8m 2 2 17. A rectangular box is sliding on a smooth inclined plane of inclination . At t 0 , the box starts to move on the inclined plane. A bolt starts to fall from point A. Find the time after which bolt strikes the bottom surface of the box. (A) 17. 2l g cos (B) 2l g sin (C) 2l g (D) l g (A) The actual acceleration of bolt a g sin i g cos j 1 a 2 = The actual acceleration of box g sin i s rel l j.u rel 0 1 Srel u rel t a rel t 2 2 1 Or l j 0 a1 a 2 t 2 2 gt 2 cos l 2 2l t g cos B2 . Here, B = magnetic field strength 2 0 0 magnetic permeability of vacuum. The name of physical quantity u is 18. A physical quantity u us given by the expression u 18. (A) Energy (B) (B) energy density (C) pressure (D) None of these CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 6 AITS 16 (Main) u Physics B2 2 N / Am u 2 N 2A 2 N Nm J 2 3 3 2 2 2 N/A NA m m m m energy per unit volume = energy density Unit of 19. From the figure describing photoelectric effect, we may infer correctly that 19. (A) (B) (C) (D) (B) 20. 20. Na and Al both have the same threshold frequency Maximum kinetic energy for both the metals depend linearly The stopping potentials are same for Na and Al Al is a better photosensitive material than Na For the given combination of gates, if the logic states of inputs A B, and C are as follows. A B C 0 and A B 1, C 0 , then the logic states of output D are (A) 0, 0 (D) (B) 0, 1 The output D for the given combination (C) 1, 0 (D) 1, 1 D A B .C A B C If, A B C D , then D 0 0 0 0 0 1 1 1 If A B 1, C 0 , then D 1 1 0 1 0 0 1 1 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 7 AITS 16 (Main) Physics NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 21. 21. 22. 22. 23. 23. Find the wavelength of the emitted radiation, if electron in hydrogen atom jumps from the third orbit 36 to second orbit. If answer is find the value of n. nR (5) An EMV is travelling along z-axis is B 5 10 8 jT.c 3 108 m / s and frequency of wave is 25 Hz, volt then electric field in . m (15) 5 g of water at 30 C and 5 g of ice at 20 are mixed together in a calorimeter. The water equivalent of calorimeter is negligible and specific heat and latent heat of ice are 0.5 cal / g and 80 cal/g respectively. The final temperature of the mixture is (ans. In C ) (0) Total heat required to convert the temperature of ice from 20 C to 0 C water is H mL ms 5 80 5 0.5 0 20 400 25 20 400 50 450cal The total heat released to loss the temperature of water from 30 C to water at 0 C is H ' 5 1 30 0 150 Since, heat released in lesser than heat required. Hence total ice is not converted into water. Hence, final temperature of mixture is 0 C 24. 24. Suppose the gravitational force varies inversely as the nth power of the distance. Then, the time period of a planet in circular orbit of radius R around the sun will be proportional to R x n 1 where x (0.5) 2 R T v 1 GMm 2GM E mv2 n 1 or v n 1 2 R R 1/2 2 R 2 n 1 /2 R 2GM 2GM n 1 R R T R n 1 / 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 8 AITS 16 (Main) 25. 25. 26. 26. 27. 27. 28. 28. 29. 29. 30. 30. Physics A 5 C rise in temperature is observed in a conductor by passing a current. When the current is doubled, then rise in temperature will be approximately (Express ans. In C ) (20) The diameter of a gas bubble formed at the bottoms of a pond is d = 4cm. When the bubble rises to the surface, its diameter tension of water = T = 0.07 N m 1 . (5) A substance breaks down under a stress of 105 Pa. If the density of the wire is 2 103 kg / m3 , find the minimum length of the wire which will break under its own weight ( g 10ms 12 ) . (5) In the spectrum of singly ionized helium, the wavelength of a line observed is almost the same as the first line of Balmer series of hydrogen. It is due to transition of electron from n1 6 to n2 * . What is the value of * . (4) The half-life of a radioactive nuclide is 20 h. It is found that the fraction (1/x) of original activity remains after 40 hours? What is the value of x? (4) U 238 changes to (an integer). (7) 92 85 U 210 by a series of and decays. Find the number of decays undergone CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 9 AITS 16 (Main) Chemistry PART (B) : CHEMISTRY SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 31. PV Compressibility factor Z is plotted against pressure nRT What is the correct order of liquefiability of the gases shown in the above graph? (A) H 2 N 2 CH 4 CO 2 (B) CO 2 CH 4 N 2 H 2 31. 32. 32. (C) H 2 CH 4 N 2 CO 2 (D) CH 4 H 2 N 2 CO 2 (A) Lesser is the value of Z at low pressure, easy will be liquefaction of the gas. Z< 1, shows the domination of attractive forces. If the radius of first orbit of H atom is a 0 , the de-Broglie wavelength of an electron in the third orbit is (A) 2 a 0 (B) 4 a 0 (C) 6 a 0 (D) 8 a 0 (C) n 2 r n 3 r 9a0 3 2 9a0 6 a0 33. 33. 34. 34. 35. 35. Concentration of 10 vol H 2 O 2 is (A) 1.786 N (B) 1.786 M (A) (C) 0. 893 N The set representing the correct order of first ionization potential is (A) K Na Li (B) Be Mg Ca (C) B C N (B) (D) 0.0893 M (D) Ge Si C Fixed volume of 0.1 M benzoic acid pK a 4.2 solution is added into 0.2 M sodium benzoate solution and formed a 300 mL, resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid: (A) 100 mL (B) 150 mL (C) 200 mL (D) None of these (B) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1 AITS 16 (Main) Chemistry C6 H 5 COO 2 C6 H5COOH Let volume of acid is VmL 0.2 300 V 2 V 150mL 0.1 V 36. 36. 37. 37. The IUPAC name of Ni NH 3 4 NiCl4 is (A) Tetrachloronickel (II)tetraamminenickel (II) (B) Tetraamminenickel (II)tetrachloronickel (II) (C) Tetraamminenickel (II)tetrachloronickelate (II) (D) Tetrachloronickel (II)tetraamminenickelate (0) (C) A mixture of A and B, which are two miscible liquids, is distilled under equilibrium conditions at atmospheric pressure. The mole fraction of A in solution is 0.3 and in vapour phase is 0.6. If the solution behaves ideally the ratio of p 0A to p 0B is: (A) 4.0 (B) 2.5 (C) 3.5 (D) 1.5 (C) In solution X A 0.3 X B 0.7 In vapours YA 0.6 YA 38. pA pT YB 0.4 i.e. YA p 0A X A p 0A X A p 0B X B YB p 0B X B p 0A X A p 0B X B YA p0A X A p 0A YA X B 0.6 0.7 3.5 YB p 0B X B p 0B YB X A 0.4 0.3 A certain reaction proceeds in a sequence of three elementary steps with the rate constant k1 , k 2 and 1/2 k k 3 . If the observed rate constant of the reaction expressed as k (obs) 1 k 3 , the observed k2 energy of activation of the reaction is 38. 1 E (A) 1 E 3 2 E2 (D) K K obs K 3 1 K2 A 3 A1 A2 1/2 1/ 2 E (C) E 3 1 E2 E E1 (B) 3 2 A e E1 /RT A 3e E3 / RT 1 E 2 / RT A 2e (D) E 3 1 E1 E 2 2 1/2 E E E2 e 3 1 RT R1 K obs Ae E a obs / RT E E E obs E 3 1 2 2 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 2 AITS 16 (Main) 39. 39. 40. Chemistry Which is not correctly matched? (1) Basic strength of oxides: Cs 2O Rb 2O K 2 O Na 2O Li 2O (2) (3) (4) (A) (D) Mobility of hydrated ions: Be 2 Li Na K Thermal stability: NaHCO3 KHCO 3 RbHCO3 CsHCO3 Melting point: NaF NaCl NaBr NaI 1 and 3 (B) 1 and 2 (C) 2 and 3 (D) 1 and 4 40. Zinc reacts with caustic soda to form: (A) Na 2 ZnO 2 (B) Zn OH 2 (A) 41. Select the best indicator for titration of 20ml of 0.02 M CH 3COOH with 0.02 NaOH ; Given (C) ZnO (D) ZnH 2 a PKa CH 3COOH 4.74 41. Indicator pH-range (I) Bromothymol Blue 6.0-7.6 (II) Thymopthalein 9.3-10.5 (III) Malachite green 11.4-13 (IV) P- cresol purple 7.4-9 (A) I (B) II (C) III (D) IV (D) At equivalence point 1 PH PKw PKa log c 2 1 14 4.74 2 8.37 2 For best indicator, PH at equivalence point should be between color transition range of indicator 42. Which of the following compounds is most rapidly hydrolysed by SN1 mechanism? 42. (A) CH 3CH CHCl (B) CH 2 CHCH 2Cl (C) (D) (D) C6H5 3 CCl 43. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 3 AITS 16 (Main) Chemistry 43. (A) (C) (B) 44. Which of the following complexes is most stable? (A) CoF6 3 (B) Co NH 3 6 (C) 3 (C) Co CN 6 (D) 3 (D) Co en 3 3 44. (D) 45. The C N bond length in the molecules: 45. is most correctly in the order (A) II III I IV (B) II III I IV (C) III IV I II (D) IV III IV I (A) As the s-character of the hybridized orbital of C-atom increases, the bond length decreases. A double bond is shorter than a single bond. Participation of lone pair on N atom in resonance induces partial double-bond character between C and N. 46. 46. 47. Gas released during Bhopal tragedy was (A) Potassium isothiocyanate (C) Methyl isocyanate (C) (B) Phosgene (D) Ammonia 47. Distillation under reduced pressure is used to purify liquids which (A) Are highly volatile (B) Have high boiling points (C) Are explosives (D) Decomposes below their boiling points (D) 48. Compound (A), C5 H10 O 4 is oxidized by Br2 H 2O to the acid, C5 H10 O 5 which readily forms a lactone. (A) forms a triacetate with Ac 2 O and a hydrazone with PhNHNH 2 . A is oxidized by HIO 4 , only one molecule of which is consumed. The structure of A is CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 4 AITS 16 (Main) Chemistry (A) (B) (C) (D) (D) (A) (B) (C) (D) (D) 49. 50. A hydrogen electrode placed in a buffer solution of CH 3COONa and acetic acid in the ratio s x:y 50. and y : x has electrode potential values E1 volts and E 2 volts respectively at 25 C . The pK a value of acetic acid is ( E1 and E 2 are oxidation potential): E E2 E E1 E E2 E E2 (A) 1 (B) 2 (C) 1 (D) 1 0.118 0.118 0.118 0.118 (A) 48. 49. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 5 AITS 16 (Main) Chemistry NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 51. A list of species having the formula XZ 4 is given below: 51. XeF4 , SF4 , SiF4 , BF4 , BrF4 ,[Cu ( NH 3 ) 4 ]2 ,[ FeCl 4 ]2 ,[CoCl4 ]2 and [ PtCl4 ]2 . Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is: (4) 52. 52. Number of moles of KMnO4 required to oxidise 10 moles of Fe C 2O 4 in acidic medium is(6) 53. H in the fusion of ice 1.5 k cal mol 1 & Sm for the phase change ice to water is 5.3 cal/K. The 53. water free energy change involved in ice (0) 54. 54. How many moles of acetyl chloride are used per mole of sucrose for esterification? (8) 55. The number of compounds which give white precipitate of AgCl with aqueous AgNO3 is 55. (4) 56. When BrO3 ion reacts with Br ion in acidic medium, Br2 is liberated. Calculate the ratio of 56. molecular mass and equivalent mass of KBrO3 . (5) 57. 57. A certain dye absorbs lights of 400 nm and then fluorescence light of wavelength 500 nm. Assuming that 40% of the absorbed energy is re-emitted as fluorescence, calculate ratio of quanta absorbed to number of quanta emitted out? (2) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 6 AITS 16 (Main) Chemistry 40 = E emitted 100 hc 40 hc n absorbed n emitted 400 100 500 n absorbed 400 100 2 n emitted 500 40 E absorbed 58. 58. Total no. of optical isomers in Ma3b3 is/are : (0) 59. 59. Total no. of lone pairs in I3 is/are (9.00) 60. No. of products obtained by ozonolysis of 2-methyl hepta-2,5-diene (3) 60. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 7 AITS 16 (Main) Mathematics PART (C) : MATHEMATICS SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 61. Let f :[0, 4] R, be a differentiable function. Then there exists real numbers a, b (0,4) such that ( f (4))2 ( f (0)) 2 Kf 1 (a)f(b) Where K, is (A) 61. 1 4 (B) 8 (B) By LMVT, a 0, 4 f 4 f 0 2 a 0, 4 (C) f 4 f 0 4 0 1 12 (D) 4 f 1 a f 4 f 0 4 f 1 a lies between f 0 and f 4 , by Intermediate value theorem f 4 f 0 2 f b Hence, f 4 f 0 62. ( p q) (~ p q) q 62. (A) a tautology (A) 2 2 8f 1 a f b is logically equivalent to (B) a contradiction (C) (~ pvp) q (D) ( p p) q p q ~ p q q ~ p q p q q ~ p q q c q q q q ~ q q t 63. 63. A variable straight line of slope 4 intersects the hyperbola xy = 1 at two points. The locus of point which divides the line segment between these two points in the ratio 1 : 2 is (A) 16 x 2 10 xy y 2 0 (B) 16 x 2 10 xy y 2 2 (C) 16 x 2 10 xy y 2 4 (B) (D) 16 x 2 10 xy y 2 4 Let P h, k y k 4 x h Let it meets xy 1 ... 1 ... 2 At A x1 , y1 and B x2 , y2 4h k 1 , x1 x2 Also. 4 4 2 x1 x2 8h k 2h k h x1 .x2 3 4 2 x1 x2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1 AITS 16 (Main) Mathematics 16 x 2 10 xy y 2 2 64. The centre of a circle of radius 4 5 lies on the line y = x and satisfies the inequality 3x + 6y > 10. If the line x + 2y = 3 is a tangent to the circle, then the equation of the circle is 2 2 17 17 (B) x y 80 3 3 2 2 17 17 (D) x y 80 3 3 23 23 (A) x y 80 3 3 64. 23 23 (C) x y 80 3 3 (A) 2 2 2 2 c a, a radius 4 5 length of the from a, a to the line i.e., a 2 a 3 4 1 4 5 a 23 17 , 3 3 23 23 17 17 Centre , or , 3 3 5 3 3 x 6 y 10 23 23 C , 3 3 2 65. 2 23 23 x y 80 3 3 If the two adjacent sides of two rectangles are represented by the vectors p = 5a - 3b; q = a - 2b and r = 4a - b; s = -a b respectively, then the angle between he vectors 1 1 x = (p + r s ) and y = (r s) 3 5 19 (A) is cos 1 5 43 65. 19 (C) is cos 1 5 43 (B) 19 (B) is cos 1 5 43 (D) cannot be evaluated p.q =0 (5a 3b) . ( a 2b) 0 2 2 5a 10ab 3ab 6b 0 2 2 6b 6a 7a . b 0 Also r . s 0 (i) 2 2 ( 4a 6b)(a b) 0 or 4a 4a .b a . b b = 0 2 2 Or 4a b 3a .b 0 ..(ii) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 2 AITS 16 (Main) Mathematics 1 1 p r s (5 a 3b 4 a b a b ) b 3 3 1 1 y ( r s ) ( 5a ) a 5 5 Now x x. y a.b Angle between x and y i.e. cos .. (iii) | x || y | | a || b | 25 43 a . b and |b| = From (i) and (ii) | a | 19 19 25 43 | a | |b| a.b 19 a. b 19 19 cos 1 | a | |b| 5 43 5 43 66. a.b The coefficient of x 4 and x 2 in the expansion of x x2 1 x 6 6 x 2 1 is and , then - is equal to 66. (A) 48 (B) x (B) 132 x 2 1 x 6 x2 1 (C) 60 (D) 60 6 Expand then find and 96, 36 132 67. Solution of the differential equation 2 y sin x dy 2sin x cos x y 2 cos x satisfying y ( / 2) 1 is given dx by 67. (A) y 2 sin x (A) (B) y sin 2 x The given equation can be written as 2y 2 y sin x (C) y 2 cos x 1 (D) y 2 sin x 4cos2 x dy y 2 cos x sin 2 x dx d ( y 2 sin x ) sin 2 x y 2 sin x ( 1 / 2) cos 2 x C dx So y ( / 2) sin ( / 2) ( 1 / 2)cos(2 / 2) C C 1 / 2 2 Hence 68. 68. y2 sin x (1/ 2)(1 cos2x) sin2 x y2 sin x The reflection of the point P(1, 0, 0) in the line x 1 y 1 z 10 is 2 3 8 (C) (1, -1, -10) (D) (2, -3, 8) (A) (3, -4, -2) (B) (5, -8, -4) (B) Hint Coordination of any Point Q on the given line are (2r + 1, -3r 1, 8r 10) for some r R So the direction ratio of PQ are 2r, -3r 1, 8r 10 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 3 AITS 16 (Main) Mathematics Now PQ is perpendicular to the given line If 2(2r) 3 (-3r 1) + 8(8r 10) = 0 77r 77 0 And the coordinates of Q, the foot of the perpendicular from P on the line are (3, -4, 2) Let R (a, b, c) be the reflection of P in the given lines when Q is the mid-point of PR a 1 b c 3, 4, 2 2 2 2 a 5, b 8, c 4 And the coordinate of the required point are (5, -8, -4) 69. a 4 If lim 1 2 x x x 2x e3 then a is equal to (A) 2 (B) 2 3 (C) 3 69. (D) 1 form 70. Area bounded between the curves y 4 x 2 and y 2 = 3|x| is/are (A) 70. 1 (B) 3 2 1 3 3 (C) 3 2 (D) 2 3 3 3 (C) 1 Required area = 2 0 4 x 2 3 x dx x 4 3.2 x3/ 2 x 4 x 2 sin 1 = 2 2 2 2 2 = 71. 2 3 3 (D) 1 0 2 3 3 x2 1 The range of the function f(x) = tan 1 2 , x R is x 3 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 4 AITS 16 (Main) 71. Mathematics (A) , 6 2 (C) We have (B) , 6 3 (C) , 6 4 (D) , 3 2 x2 1 f(x) = tan-1 2 , x R x 3 Let y = x2 1 x2 3 yx2 3 y x2 1 3 y 1 (1 y ) x 2 3y 1 x2 0 1 y 3y 1 0 y 1 So y , 1 3 72. 72. 1 6 4 Range of f ( x) , If A and B are square matrices of the same order and A is non-singular, then for a positive integer n, A -1 BA is equal to n (A) A -n Bn A n (C) Hint (C) A 1Bn A (B) A n Bn A -n (D) n(A 1BA) (A-1 BA)2 = (A-1BA) (A-1BA) = A-1B(AA-1)BA = A-1BIBA =A-1B2A (A-1BA)3 = (A-1B2A)(A-1BA) = A-1B2 (AA-1)BA = A-1B2IBA = A-1B3A (A-1BA)n A-1BnA 73. 73. For 20 observations mean and variance is given as 10 and 4, later it was observed that by mistake 9 was taken in place of 11 then the correct variance is (A) 3.98 (B) 4.01 (C) 3.99 (D) 4.02 (C) Mean = 10 x 200 variance = 4 x i 2 i 2080 New mean = 200 9 11 10.1 20 2080 81 121 New variance = (10.1) 2 3.99 20 74. If the system of linear equations x + ay + z = 3, x + 2y + 2z = 6, x + 5y + 3z = b has no solution then CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 5 AITS 16 (Main) 74. Mathematics (A) a = 1, b = 9 (C) (B) a = -1, b = 9 (C) a = -1, b 9 (D) a -1, b = 9 0, 0 75. If n arithmetic means are inserted between two sets of numbers a, 2b and 2a, b, where a, b, R. Suppose that mth arithmetic mean between these two sets of numbers is same, then the ratio a:b equals 75. (A) n m +1; m (C) Let (B) n m + 1: m (C) m: n m +1 A1, A2, ....An be airthmetricments between a and b, then Again let (D) n: n m + 1 2b a Am a m n 1 B1, B2 , ....Bn be arithmetic means b 2a n 1 Between 2a and b then B m 2 a m 2b a 2a n 1 Now A m Bm a m b 2a b a a m m m a b n m 1 n 1 n 1 76. 76. A group of students decided to buy a Alarm Clock priced between Rs. 170 to Rs. 195. But at the last moment, two students backed out of the decision so that the remaining students had to pay 1 Rupee more than they had planned. If the students paid equal shares, the price of the Alarm Clock is (A) 190 (B) 196 (C) 180 (D) 171 (C) Let cost of clock = x Number of student = n Then x x n 2 2n 1 x n 2 n 2 n2 2n 170 195 2 77. A window is in the shape of a rectangle surmounted by a semi-circle. If the perimeter of window is of fixed length l then the maximum area of the window is (A) 77. l2 2 4 (C) l = 2x + 2r + (B) l2 8 (C) l2 2 8 (D) l2 8 4 r A = 2 rx 1 r 2 2 dA l 0 r dV 4 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 6 AITS 16 (Main) sin 78. If A = 1 78. Mathematics t dt dt , B 1 t2 cosec 1 1 A 1 A B dt then e t (1 t 2 ) 1 (B) cosec (A) sin (C) B= cos ec A2 B2 A2 B 2 (C) 0 B 1 = ? 1 (D) 1 1 t(1 + t 2 ) 1 Let u B t sin 1 u du 1 u2 Hint : A + B = 0 A = -B A2 A2 -A -1 A 2 + B2 -1 A e0 1 79. Normal at (2, 2) to curve x 2 2 xy 3 y 2 0 is L. The perpendicular distance from origin to line L is 79. (A) 4 2 (C) Given curve (B) 4 (C) 2 2 (D) 2 x2 2xy 3y2 0 (x 3 y) (x y) 0 (2, 2) lies on (x y) = 0 Equation of normal is x + y = Distance = 2 2 80. 80. AC C If A and B are two events such that P(A) = 0.3, P(B) = 0.25, P ( A B ) 0.2, then P C is B equal to 2 11 13 14 (A) (B) (C) (D) 15 15 15 15 (A) A C C AC P C 1 P C B B P (A C B C ) 1 P( BC ) 1 1 P( A B) 1 P ( B) 1 1 P( A) P( B) P( A B) 1 P( B) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 7 AITS 16 (Main) Mathematics NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 81.The number of different words that can be formed using all the letters of the word SHASHANK such that in any word the vowels are separated by atleast two consonants, is 81. (2700) The letters other than vowels are SHSHNK which can be arranged in 6! ways. 2!2! Now in its each case, let the first A be placed in the rthgap then the number of ways to place the 2nd A 6! 5 will be (7 r 1). So, the total number of ways = (6 r ) 2!2! r 1 82. 82. 6! (5 4 3 2 1) 2700 2!2! The coordinates of the point on the parabola y = x 2 7 x 2, which is nearest to the straight line y = 3x 3 is (-a, -b) then a + b = (10) Sol.: Hint : Any point on the parabola is x, x 2 7 x 2 Its distance from the line y =3x 3 is given by 3x ( x2 7 x 2) 3 x2 4x 5 x2 4x 5 P= as x2 4 x 5 0 x R 9 1 10 10 dp 0 x 2 then required point = (-2, -8) dx 83. A sequence is obtained by deleting all perfect square from set of natural numbers. The remainder when the 2003rd term of new sequence is divided by 2048 is 83. (0) Hint: Since 2046 2047 2048 2049 45 2003 term is 2003 + 45 = 2048 rd Hence remainder is 0 tan x tan y tan z 38 and x + y + z = , if tan 2 x tan 2 y tan 2 z then K = 2 3 5 K 84. If 84. (3) tan x = 2t, tan y = 3t tanz = 5t tan x (tan x ) t 2 1 3 tan2 x tan2 y tan2 z t2 (4 9 25) 38t 2, K 3 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 8 AITS 16 (Main) 85. 85. Mathematics One vertex of an equilateral triangle is at the origin and the other two vertices are, roots o 2 z 2 2 z k 0 then 3k is (2) 2z2 2z k 0 2 4 8k 4 Since z is a complex number 4- 8k will be negative z k 1/ 2 1 2k 1 1 1 (0,0), , , 2k 1 2 2 2 2 Since triangle is equilateral 1 1 (2 k 1) (2 k 1) 4 4 2 k 3 86. If the parabola y ax 2 bx c has vertex at (4, 2) and a [1, 3] such that difference between extreme values of abc is , then is 1728 86. (2.00) b 4 b 8a 2a 4ac b 2 b2 2 c 2 4a 4a c 2 16a f a abc 8a 2 2 16a f a 16a 2 128a3 d f a 0 when a (1, 3] da f 1 144 f 3 3600 Difference = 3456 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 9 AITS 16 (Main) Mathematics 87. Let f be a continuous function satisfying the equation x x 0 0 f t dt t f x t dt e x 1 , then value of f 0 is 87. (1.00) x 0 x f t dt x t f t e x 1 0 d.w.r. to x x x f x f t dt 1 x f x e x 0 d.w.r. to x f x f x e x I.f. e x f x x 1 e x f 0 1 sin 2 A 88. If A, B and C are the angles of a triangle and sin C sin B 88. sin C sin B sin 2 B sin A sin A sin B sin C , then is sin A sin 2C (0.00) 2ka cos A LHS kc kb 2a cos A k 3 c b kc kb 2kb cos B ka (From sine rule) ka 2kc cos C c b 2b cos B a a 2c cos C cos A a 0 a cos A 0 k cos B b 0 b cos B 0 0 cos C c 0 c cos C 0 3 0 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 10 AITS 16 (Main) Mathematics x2 y 2 1 and strike a 25 16 circle of radius 65 units and reflected from it. The reflected ray entering inside the ellipse through the focus ( 3, 0) and strike lower surface of ellipse and gets reflected from it. If the reflected ray returns to the point p 0, 4 , then ordinate of the centre of circle is ______ 89. A ray of light emerging from the point p 0, 4 along the tangent of the ellipse 89. (5.00) Using reflection property of ellipse equation of reflected ray SP is 4 x 3 y 12 ... 1 Solving equation (1) with ellipse, we get 75 32 A , 17 17 Equation of reflected ray BS is 16 x 63 y 48 0 ... 2 Equation of incident ray PB is y 4 ... 3 Equation of bisector of (2) and (#) with negative slope gives normal to circle at 3 which is 4 x 32 y 53 0 75 Point B is , 4 , using parametric equation of line ordinate of centre is 5. 4 90. If ax 2 2hxy by 2 2 gx 2 fy c 0 and ax 2 2hxy by 2 2 gx 2 fy c 0 each represent pair of straight line and area of parallelogram enclosed by them is c h 2 ab , then is 90. (2.00) Let lines for first equation is y m1 x c1 y m2 x c2 0 y 2 x m1 m2 xy m1m2 x 2 x m1c2 c1m2 y c1 c2 c1c2 0 By comparison m1m2 a 2h c and c1c2 , m1 m2 b b b CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 11 AITS 16 (Main) tan 2 h 2 ab a b Mathematics ; sin 2 h 2 ab a b 2 4h 2 Another pair has lines y m1 x c1 0 and y m2 x c2 0 4 c1c2 a b 4h 2 PP Area 1 2 sin 1 m 2 2 h 2 ab 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 12

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