ENGLISH Form Number : Paper Code : 1001CT103516012 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE : III, IV & V Test Type : MINOR Test Pattern : JEE-Main TEST DATE : 18 - 12 - 2016 PAPER – 2 Important Instructions Do not open this Test Booklet until you are asked to do so. 1. Immediately fill in the form number on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited. 2. The candidates should not write their Form Number anywhere else (except in the specified space) on the Test Booklet/Answer Sheet. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the question paper A,B,C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 6. One Fourth mark will be deducted for indicated 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 writting 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, mobile phone any electronic device etc, except the Identity 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. 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 candidate are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet. Your Target is to secure Good Rank in JEE (Main) 2017 Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PART A - PHYSICS 1. A smooth sphere A of mass m collides elastically with an identical sphere B at rest. The velocity of A before collision is 2M 8 m/s in a direction making 60° with the M line of centres at the time of impact. (i) The sphere A comes to rest after 3m 2M collision. (ii) The sphere B will move with a speed of 8 m/s after collision. (iii) The directions of motion A and B after (1) 0.5m (2) 0.6m collision are at right angles. (3) 1.5m (4) 2m (iv) The speed of B after collision is 4 m/s. 3. In a parallel plate air capacitor, a cathode The correct option is :- beam comprising n = 106 electrons is (1) (i), (ii) (2) (ii), (iii), (iv) (3) (iii), (iv) (4) (ii), (iii) emitted with a velocity v0 = 108 m/s into 2. A block of mass M is tied to one end of the space between the plates. The massless rope. The other end of rope is in potential difference between the plate is the hands of a man of mass 2M as show in = 400 V, the seperation between the figure. Initially the block and the man are plates is d = 2 cm and the area of each resting on a rough plank of mass 2M as plate is 2 = 100 cm2. The deflection of the shown in figure. The whole system is resting on a smooth horizontal surface. The electron beam, is :- man pulls the rope. Pulley is massless and (1) 1.6 mm frictionless. What is the magnitude of (2) 1.76 mm displacement of the plank when the block meets the pulley (Man does not leave his (3) 0 mm position on the plank during the pull). (4) 5 mm SPACE FOR ROUGH WORK 1001CT103516012 E-1/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 4. A parallel plate capacitor, partially filled (1) Q1 ' K 1 (2) Q2 ' (K 1) with a dielectric slab of dielectric constant Q1 K Q2 2 K, is connected with a cell of emf V volt, as shown in the figure. Separation between (3) Q2 ' K 1 (4) Q1 ' K the plates is D. Then Q2 2K Q1 2 V 6. A and C are concentric conducting D P P' spherical shells of radius a and c (1) electric field at point P is less than that respectively. A is surrounded by a at point P. concentric dielectric radius a, outer radius b and dielectric constant k. If sphere A be (2) electric field at point P is less than that given a charges Q, the potential at the at point P. outer surface of the dielectric is :- (3) electric fields at points P and P are equal V Q (4) electric field at point P is E = . (1) 40kb KD (2) Q 1 1 40 a k(b 5. Two identical capacitors 1 and 2 are connected in series. The capacitor 2 contains a dielectric slab of constant K as shown. They are connected to a battery of emf V0 volts. The dielectric slab is then removed. Let Q1 and Q2 be the charge stored in the capacitors before removing the slab and Q1, and Q2 be the values after removing the slab. Then : a) 1 Q 2 vo (3) 40b (4) None of these SPACE FOR ROUGH WORK E-2/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 7. Three resistances of magnitude R each are 9. A wire of mass m and length can slide connected in the form of an equilateral freely on a pair of smooth, vertical rails triangle of side a. The combination is placed (figure). A magnetic field B exists in the region in the direction perpendicular to the in a magnetic field B = B0 e–t perpendicular plane of the rails. The rails are connected to its plane. The induced current in the at the top end by a capacitor of capacitance circuit is given by : A C. The acceleration of the wire neglecting any electric resistance is : RR mg × × C × × BC (1) m CB22 ×× ×× R 2mg ×× ×× (1) a2 B0 et (2) a2 B0 et (2) m CB22 ×× ×× 2 3R 4 3R mg (3) a2 B0 e t (4) a2B0R et (3) CB22 ×× ×× 4 3R 4 3 mg ×× ×× 8. Statement-1 : When a magnet is made to (4) 2(m CB22 ) fall freely through a closed coil, its acceleration is always less than 10. A rectangular loop has a sliding connector PQ of length 2 m and resistance 10 and acceleration due to gravity. and it is moving with a speed 5 m/s as shown. Statement-2 : Current induced in the coil The set-up is placed in a uniform magnetic opposes the motion of the magnet, as per field 3T going into the plane of the paper. Lenz's law. The three currents I1, I2 and I are :- (1) Statement-1 is true, Statement-2 is true; P Statement-2 is a correct explanation 5m/s for Statement-1 (2) Statement-1 is true, Statement-2 is true, 10 10 10 Statement-2 is NOT a correct × 3T I I2 explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False I1 Q (4) Statement-1 is False, Statement-2 is (1) I1 = I2 = 3A, I = 1A (2) I1 = I2 = 5A, I = 2A True (3) I1 = I2 = 1A, I = 2A (4) I1 = I2 = I = 2A SPACE FOR ROUGH WORK 1001CT103516012 E-3/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 11. A conducting ring of radius R is placed in 13. A particle is projected with a velocity uniform imnwovairndgmwaitghnevteilcocfiiteyldvBinaistssphloawnne,. (10m/s) along y-axis from point (2, 3). If ring is Magnetic field of 3iˆ 4ˆj Tesla exist the induced emf across arc PQ will be :- (1) vBR 1 1 P uniformly in the space. Its speed when 2 Q particle passes through y-axis for the third 2 time is : (neglect gravity) 45° v (1) 20 m/s (2) vBR 1 1 B 2 (3) vBR 1 1 (4) vBR 1 1 (2) 60 m/s 2 2 2 12. A uniform but time varying magnetic field (3) 30 m/s is present in a circular region of radius R. (4) 10 m/s The magnetic field is perpendicular and into the plane of the loop and the 14. A rod of length is rotating with constant magnitude of field is increasing at a angular velocity and we marked the rod constant rate . There is a straight 0, 1, 2, .......... 8 at equal spacing. Potential conducting rod of length 2R placed as difference between consecutive points, if shown in figure. The magnitude of induced we go from left to right will be : emf across the rod is : × ×× ××××××× × × ××R×× × × × 0 1 2 3 4 5 6 7 8 ××××××× ×× ×× (1) in increasing arithmetic progression. ××××× (2) in increasing geometric progression. ××××× ×× 2R (1) R2 R2 R2 R2 (3) in increasing form 12, 22, 32 .......... 82 (2) (3) (4) (4) in decreasing form, 82, 72, 62 ......... 12 22 4 SPACE FOR ROUGH WORK E-4/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 15. A simple pendulum with a bob of mass 17. In a uniform electric field, a cube of side m = 1 kg, charge q = 5 C and string length 1 cm is placed. The total energy stored in = 1 m is given a horizontal velocity u in a the cube is 8.85µJ. The electric field is uniform electric field E = 2 × 106 V/m at its parallel to four of the faces of the cube. The bottom most point A, as shown in figure. electric flux through any one of the It is given a speed u such that the particle remaining two faces is. leave the circular path at its topmost point C. Find the speed u. (Take g = 10 m/s2) (1) 1 V/m (2) 100 2V / m 52 (1) 40 m/s C (2) 50 m/s DB (3) 5 2V / m (4) 10 2V / m (3) 35 m/s E 18. Figure shows electric field lines in which Au an electric dipole p is placed as shown. (4) None of these Which of the following statements is 16. Figure shows a solid metal sphere of radius correct? ‘a’ surrounded by a concentric thin metal shell of radius 2a. Initially both are having charges Q each. When the two are connected by a conducting wire as shown in the figure, then amount of heat produced in this process will be : Q Q (1) The dipole will not experience any force. a (2) The dipole will experience a force 2a towards right. KQ2 KQ2 (2) KQ2 KQ2 (3) The dipole will experience a force (1) (3) (4) towards left. 4a 6a 8a 2a (4) The dipole will experience a force upwards. SPACE FOR ROUGH WORK 1001CT103516012 E-5/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 19. Figure shows the variation of the moment 21. A circular hoop of mass m and radius R rests of inertia of a uniform rod, about an axis flat on a horizontal frictionless surface. A passing through its centre and inclined at bullet, also of mass m and moving with a an angle to the length. The moment of velocity v, strikes the hoop and gets inertia of the rod about an axis passing embedded in it. The thickness of the hoop through one of its ends and making an is much smaller than R. The angular velocity with which the system rotates after angle = 3 will be the bullet strikes the hoop is :- 0.6 I (kg-m2) (rad) (1) V/(4R) (2) V/(3R) (3) 2V/(3R) (4) 3V/(4R) (1) 0.45 kg–m2 (2) 1.8 kg–m2 22. The time period of oscillation of a magnet (3) 2.4 kg–m2 (4) 1.5 kg–m2 is 2 sec. When it is remagnetised so that 20. A 3.0 kg bobbin consists of a central its pole strength is 4 times, its period will cylinder of radius 5.0 cm and two end plates be - each of radius 6.0 cm. It is placed on a (1) 4 sec (2) 2 sec (3) 1 sec (4) 8 sec slotted incline, where friction is sufficient 23. Curie's law states that to prevent sliding. A block of mass 4.5 kg (1) magnetic susceptibility is inversely is suspended from a cord wound around proportional to the absolute the bobbin and passing through the slot temperature under the incline. If the bobbin is in static (2) magnetic susceptibility is inversely equilibrium, what is the angle of tilt of the proportional the square root of the incline? absolute temperature (1) 30° (3) magnetic susceptibility is directly (2) 37° proportional to the absolute (3) 53° temperature (4) More information is required. (4) magnetic susceptibility is constant SPACE FOR ROUGH WORK E-6/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 24. The average dipole moment of Fe atoms is 26. The figure shows three situations when an 1.8 × 10–23 A-m2. The magnetic moment of electron with velocity v travels through a an iron rod of length 10 cm and diameter uniform magnetic field . In each case, 1cm is : (density and at. wt. of Fe are B 7.87 g/cm3 and 55.87) what is the direction of magnetic force on the electron? (1) 3.0 A-m2 (2) 4.0 A-m2 y x y y x (3) 6.0 A-m2 (4) 12.0 A-m2. v Bx B B v z v z 2 z 1 3 25. In the diagram, I1, I2 are the strength of (1) positive z–axis, negative x–axis, positive the currents in the loop and infinite long y–axis straight conductor respectively. OA = AB = R. The net magnetic field at the (2) negative z–axis, negative x–axis and centre O is zero. Then the ratio of the zero currents in the loop and the straight conductor is : (3) positive z–axis, positive y–axis and zero (4) negative z–axis, positive x–axis and zero O 27. A thin uniform rod with negligible mass I1 and length is attached to the floor by a frictionless hinge at point P. A horizontal R spring with force constant k connects the other end to wall. The rod is in a uniform A I2 magnetic field B directed into the plane of R paper. What is extension in spring in equilibrium when a current i is passed B through the rod in direction shown. Assuming spring to be in natural length (1) (2) 2 initially. 1 1 (3) (4) 2 SPACE FOR ROUGH WORK 1001CT103516012 E-7/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 29. A nucleus at rest undergoes -decay B according to equation X226 Y + At × 92 53° × B t = 0 the emitted -particle enters in a i × P region of space where a uniform magnetic P 53° Hinge Spring in relaxed position field B = B0 ˆi and electric field E E0iˆ Hinge exists. The -particle enters in a region In equilibrium position with velocity v = v0 ˆj at x = 0. 5iB 3iB 5iB 5iB 3 107 m (1) 8k (2) 8k (3) 4k (4) 6k q E0 time later the particle was 28. A disc of radius r and carrying positive found to have twice the initial speed. Initial charge q is rotating with an angular speed speed of the -particle is : in a uniform magnetic field B about a fixed (1) 106 ms–1 (2) 107 ms–1 axis as shown in figure, such that angle (3) 7 × 107 ms–1 (4) 3 × 107 ms–1 made by axis of disc with magnetic field is 30. A bar of mass m is suspended horizontally . Torque applied by axis on the disc is : on two vertical springs of spring constant k and 3k. The bar bounces up and down Disc while remaining horizontal. Find the time B period of oscillation of the bar (Neglect mass of springs and friction everywhere). Fixed axis Fixed pulley qr2B sin 3K K (1) , clockwise (1) 2 m (2) 2 m 2 k 3k qr2B sin (2) , anticlockwise 4 qr2B sin (3) , anticlockwise 2 qr2B sin 2m (4) 2 m (4) , clockwise (3) 3k 4k 4 SPACE FOR ROUGH WORK E-8/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 PART B - CHEMISTRY 31. If 1mL of 1M HCl solution is added to 99mL 34. 1 mole of H2C2O4 is oxidised by X mole of of aqueous solution of NaCl so that pH of MnO4– in strong basic medium and 1 mole NaCl solution change by X units. The value of NaHC2O4 is oxidised by y mole of MnO4– of X is : in acidic medium. Ratio of x/y is- (1) 2 : 1 (2) 5 : 1 (1) 2 (2) 5 (3) 3 : 1 (4) 1 : 3 (3) 7 (4) 1 35. The degree of hydrolysis of 0.1 M solution 32. Which of the following solutions will act of ArNH3+ Cl– is 0.01. Find the degree of as buffer solution - hydrolysis in 0.4 M solution of ArNH3+ Cl– (1) 10 ml of 0.1M NaOH + 5 ml of 0.1 M HCl (1) 0.1 (2) 0.01 (2) 15 ml of 0.1M NaOH + 10 ml of 0.1 M (3) 0.005 (4) 0.05 36. A real gas cannot be liquefied in which of CH3COOH the following conditions : (3) 10 ml of 0.1M NH4OH + 15 ml of 0.1 M (1) T < TC, P > PC (2) T < TC , P < PC HCl (3) T = TC, P > PC (4) V > VC, P > PC (4) 10 ml of 0.1M H2S + 15 ml of 0.1 M Na2S 37. Which of the following option is incorrect (1) Fraction of total molecules having energy 33. The pressure of a vander waal gas is less greater than or equal activation energy than the pressure of an ideal gas because is e–Ea/RT of : (2) Lesser the activation energy, faster will (1) Infinitesimal size of molecules be the reaction (2) The collisions with the wall become (3) In Arrhenius equation K = Ae–Ea/RT, inelastic Arrhenius factor (A) is the product of collision frequency and probability factor (3) Intermolecular attraction (4) Molecular movement is more random (4) Reaction having high value of activation energy is less temperature dependent SPACE FOR ROUGH WORK 1001CT103516012 E-9/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 38. For a certain reaction : (A)(g) B(g) 41. The formula of 3 membered discrete chain Half life for different initial pressures of A silicate is :- is given below : (1) Si3O96– (2) Si3O108– PA0 (atm) 0.1 0.025 (3) Si3O63– (4) Si3O1212– t1/2(sec) 100 50 42. Which of the following complex compound The correct statement about order of re- have highest number of geometrical action is : isomers ? (1) 1 (2) 2 (1) [PtCl2Br2]2– (2) [CoCl2Br2]2– (3) 3 (4) 0.5 (3) [Pt(NH3)(Gly)Br] (4) [Co(NH3)2(Gly)2]+ 39. One mole of an ideal monoatomic gas is 43. According to V.B.T., in which of the heated in a process PV5/2 = constant. following complex, transition of electron Amount heat absorbed in the process for occurs from one shell to other shell of 36ºC rise in temperature. central metal :- (1) 60 cal (2) 30 cal (1) [Fe(H2O)5(NO)]2+ (2) K3[Cu(CN)4] (3) 108 cal (4) 180 cal (3) [Ni(CN)6]4– (4) [Co(CN)6]4– 40. On heating a mixture of SO2Cl2 and CO, two equilibria are simultaneously 44. If Hund's rule is violated then select the established ; CORRECT statement regarding SO2Cl2(g) SO2(g) + Cl2(g) [Ni(NH3)6]2+ is :- CO(g) + Cl2(g) COCl2 (g) (1) sp3d2 , paramagnetic On adding more SO2 at equilibrium what (2) d2sp3 , diamagnetic will happen ? (3) sp3d2 , diamagnetic (1) Amount of CO will decrease (4) d2sp3 , paramagnetic (2) Amount of SO2Cl2 and COCl2 will 45. When aqueous solution of cupric bromide increase is electrolyzed, the product obtained at (3) Amount of CO will remain unaffected cathode will be :- (4) Amount of SO2Cl2 and CO will increase (1) H2 (2) Cu (3) Br2 (4) O2 SPACE FOR ROUGH WORK E-10/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 46. Which of the following complex ion is 50. Select the INCORRECT statement :- (1) Concentration of 'Cerrusite' ore is done expected to absorb light in visible region ? by froth floatation process (2) Roasting process is involved for (1) [Ti(en)3]4+ (2) [Sc(NH3)4(H2O)2]3+ extraction of iron (3) Byproduct of Serpeck's process is silicon (3) [CrO4]2– (4) [Zn(CN)4]2– (4) Cassiterite and Rutile are oxide ores of the metals 47. In which of the following pair, the molecule 51. Which of the following reaction will not has higher IE1 value compared to its atom :- produce \"Acetaldehyde\" ? (1) F2 and F (2) O2 and O (3) N2 and N (4) All of these 48. Select CORRECT statement :- (1) Acidic strength : H2O > H2S (2) p–p bonding M.O. has (gerade) (i) AlH(i-Bu)2 (ii) H2O symmetry (1) CH3–C N (3) NCl3 on hydrolysis gives HOCl but NF3 O gives HF in ordinary condition H2–Pd (4) Bond angle order : OF2<H2O<Cl2O<ClO2 (2) CH3–C–Cl BaSO4 49. Select the CORRECT IUPAC name of complex compound :- (3) HC CH Hg(OAc)2 – NaBH4 / OH (1) Triamminetricyanidochromium(III) hexanitrito-N-irridate(III) N (i) LiAlH4 (ii) H2O (2) Pentaamminecyanidochromium(III) (4) CH3–C hexanitrito-N-irridium(III) 52. Which of the following acyclic compound (3) Hexanitrito-N-irridium(III) pentaammine will react fastest with Br2 gas ? cyanidochromate(II) (1) C2H6 (2) C3H6 (4) Pentaamminecyanidochromium(III) (3) C2H4 (4) C2H2 hexanitrito-N-irridate(III) SPACE FOR ROUGH WORK 1001CT103516012 E-11/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 53. How many optically active (Lactide) 56. Which of the following will not produce CH4 product(s) are formed in given reaction ? gas with CH3MgBr ? OH CH3–CH–COOH O (1) 3 (2) 4 (1) S–OH (2) CH3–NH2 (3) 2 (4) 1 O 54. Which of the following on reaction with excess RMgX will give only 2º alcohol ? (3) CH3–CH2–OH (4) O CH3 57. Which of the following reaction will not (1) CH3–C–O–CH CH3 produce CO2 ? O (1) CH3 –CH2–SO3H NaHCO3 (2) H–C–O–CH2–CH3 OO O (3) H–C–O–CH(CH3)2 O (4) CH3–C–O–CH3 (2) HO OH 55. Best reagent to carry out following (3) CH3–CH2–COONa Electrolysis conversion is : CHO CH3 OH COOH O (1) Red P / I2 (3) Zn-Hg / HCl OH (4) (2) LiAlH4 , Et2O – (4) (i) N2H4 ; (ii) OH, SPACE FOR ROUGH WORK E-12/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 58. The relative reactivities for replacement of 60. If (P) & (Q) are major products, then find –H are 3º : 2º : 1º = 6 : 4 : 1. Relative amount incorrect statement for given sequence : of given product is : Cl OH Cl2 (+– ) CH3–CH2–CH–CH3 H3PO4 P hv (1) 3/31 (2) 6/31 Ni D2 (3) 16/31 (4) 16/29 Q 59. In which of the reaction carbanion is not formed as one of the intermediate : (1) Formation of (Q) is syn addition O phenomenon (1) NaOH (2) Formation of (P) is unimolecular OH + CaO – elimination reaction (2) OH (3) Formation of (Q) is redox reaction N (4) (Q) is optically inactive due to internal NH2 compensation Cl Na (3) Et2O OH (i) AgOH (4) (ii) Br2 / CCl4 , O SPACE FOR ROUGH WORK 1001CT103516012 E-13/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 PART C - MATHEMATICS 61. Let a, b, c, d, e be five numbers satisfying 65 . lim 1 12 sin 1 22 sin 2 32 sin 3 .... n2 sin n n3 n n n n the system of equations n 2a + b + c + d + e = 6 equals - a + 2b + c + d + e = 12 (1) cos1 + 2sin1 (2) 2sin1 – 2 a + b + 2c + d + e = 24 (3) cos1 – 2sin1 – 2 (4) cos1 + 2sin1 – 2 a + b + c + 2d + e = 48 66. Function ƒ x x2 2 - a + b + c + d + 2e = 96, 1 x2 (1) is always increasing then |c| is equal to (2) is always decreasing (1) 6 (2) 7 (3) 8 (4) 25 (3) has exactly one point of minima (4) has exactly one point of maxima 62. If domain of function ƒx nmsinx 4 is R, then number of possible integral x3 Hx values of m is 67. Let Hx x 1 sin t3dt , then lim x 1 (1) 3 (2) 4 x2 x 1 (3) 6 (4) 7 equal to - 63. If function y = ƒ(x) satisfy the differential (1) sin 1 (2) –sin1 (3) 2sin1 (4) 0 equation (x3 + 1)dy = x(1 – 3xy)dx and x8 x3 x 68. The integral dx is 3x11 8x6 24x4 1/ 3 ƒ(0) = 0, then lim x2 is equal to x0 ƒ(x) equal to- (1) 0 (2) 1 (1)2 2/3 3 3x11 8x6 24x C (3) 2 (4) 4 3x8 8x3 24x 2/ 3 C 64. Identify the correct statement about 1 (2) function ƒ(x) = max(x2 – 1, 7 – x2, 5) 16 (1) ƒ(x) is not differentiable at 4 points (3)2 2/3 3x8 8x3 24x 3 C (2) range of ƒ(x) is [3, ) (3) ƒ(x) is an injective function (4) 1 3x8 8x3 24x 2/3 C 16 (4) ƒ(x) is discontinuous at 4 points. (where C is constant of integration) SPACE FOR ROUGH WORK E-14/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 69. Locus of a point which moves so that its 1 distance from the origin is thrice its distance from y = 2x is- 73. Let (x) exet(t)dt x and (n(e2 3)) (1) a straight line 0 (2) a pair of straight lines (3) a circle is equal to A, then - (4) a parabola (1) A = n(e2 – 3) – 2 70. Triangle formed by the lines 3x + y + 4 = 0, (2) A (3,4) 3x + 4y – 15 = 0 and 24x – 7y = 3 is a/an (3) A = e2 – 3 (1) equilateral triangle (4) A = n(e2 – 3) + 2 (2) isosceles triangle 74. If the sum of the first 11 terms of the series (3) acute angle triangle (4) scalene triangle 1 4 2 1 5 2 1 6 2 22 2 1 2 ...... is 7 7 7 7 30 11 , then is equal to (4) 39 7 71. x 42 ex2 dx x 42 ex2 dx is equal (1) 36 (2) 37 (3) 38 75. Least value of 03 to- (2) 8(e3 – 1) x2y2 2x2y 2x2 2xy 2x 1 is , then (1) 8e3 x2y x (3) 3 e4 1 (4) 3 e8 1 {where x,y R+, x2y + x 0} (1) (0,1) (2) [1,3) 72. Let a, b, c are roots of equation x3 + 8x + 1 = 0, (3) [3,4] (4) (4,7) then the value of 76. Let P1 : y = –x2 + 4x + 2 and P2 : x2 + 5x + bc ac ab 17 = y are two parabolas, then number of (8b 1)(8c 1) (8a 1)(8c 1) (8a 1)(8b 1) 8 is equal to common tangents of P1 and P2 is - (1) 0 (2) –8 (1) 0 (2) 1 (3) –16 (4) 16 (3) 2 (4) 3 SPACE FOR ROUGH WORK 1001CT103516012 E-15/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 77. Let S = 0 is the locus of centre of a 81. P and Q are two points on the parabola variable circle which intersect the circle y2 = 8x and S is its focus. PS and QS meet x2 + y2 – 4x – 6y = 0 orthogonally at (4, 6). the curve again in T and R respectively. If P is a variable point of S = 0, then least value of OP is (where O is origin) - If PQ passes through a fixed point (–2, 3), (1) 13 (2) 2 13 then TR also passes through a fixed point (3) 10 (4) 13 whose coordinates are (1) (2, –3) (2) (3, – 2) 78. Number of integral values of a for which (3) (–2, 3) (4) (–3, 2) smaller root of quadratic equation 82. Slope of common tangents of parabola x2 – 2ax – 4 + a2 = 0 is smaller than 1 and (x – 1)2 = 4(y – 2) and ellipse (x –1)2 + (y 2)2 1 bigger root is greater than 6 is - 2 (1) 0 (2) 1 are m1 and m2, then m12 m22 is equal to - (3) 2 (4) infinite values (1) 2 (2) 3 79. Let a circle S = 0 touches both the circles (3) 4 (4) 6 x2 + y2 = 400 and x2 + y2 – 10x – 24y + 120 = 0 83. If x, y, z R+ are such that z > y > x > 1, externally and also touches x-axis. The 5 10 radius of circle S = 0 is - logyx + logxy = 2 and logzy + logyz = 3 , (1) 200 (2) 33 then logxz is equal to - (3) 120 (4) 240 (1) 2 (2) 3 (3) 6 (4) 12 80. Length of latus rectum of the parabola 84. The equation of normal to the curve 9x2 + 16y2 + 24xy – 4x + 3y = 0 is - 1 1 y3 + 2xy + x3 = (x – 1)3 at point (1, –1) is - (1) 20 (2) 4 (1) 5x + y = 4 (2) 5x – y = 6 1 (4) 1 (3) x + 5y + 4 = 0 (4) x – 5y = 6 (3) 5 SPACE FOR ROUGH WORK E-16/18 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 x2 88. Let ƒ(x) = min{sin–1x, cos–1x}, then area bounded by ƒ(x) and x-axis is - 85. Let ƒ(x) = (t 1)(t 4)(t 9)dt , then 0 (1) 2 (2) 2 1 (1) ƒ''(x) = 0 have 4 distinct positive solutions (2) ƒ'''(x) = 0 have 2 distinct positive solutions 1 (4) 2 1 (3) 2 (3) ƒ'''(x) = 0 have 3 distinct positive solutions (4) ƒ(x) have 6 critical points. 89. Let S = 0 is an ellipse whose vartices are the extremities of minor axis of the ellipse 86. Let ƒ(x) = a sin( x b) x0 E: x2 y2 1, a b . If S = 0 passes through 6x7 x 1 is differentiable a2 b2 x0 for all real x. If a R and b [0,2] , the foci of E, then its eccentricity is (considering the eccentricity of E as e) then number of ordered pair(s) of (a, b) 1 2e2 1 is - (1) 1 e2 (2) 1 e2 (1) 1 (2) 2 (3) 4 (4) more than 4 1 2e2 e2 (3) 1 e2 (4) 1 e2 87. Number of solutions of the equation 2tan–1(cos2x) = tan–1(2cosec2x) in [0,5] is m, 90. Let L is distance between two parallel then - normals of x2 y2 1 , a > b, then maximum (1) m < 1 a2 b2 (2) m {2, 3, 4} (3) m = 5 value of L is - (4) m > 5 (1) 2a (2) 2b (3) a + b (4) 2(a – b) SPACE FOR ROUGH WORK 1001CT103516012 E-17/18
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 SPACE FOR ROUGH WORK SPACE FOR ROUGH WORK E-18/18 1001CT103516012
Form Number : Paper Code : 1001CT103516012 HINDI CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE : III, IV & V Test Type : MINOR Test Pattern : JEE-Main TEST DATE : 18 - 12 - 2016 PAPER – 2 Important Instructions Do not open this Test Booklet until you are asked to do so. 1. Immediately fill in the form number on this page of the 1. Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited. 2. 2. The candidates should not write their Form Number anywhere else (except in the specified space) on the Test Booklet/Answer Sheet. 3. 3 4. 90 360 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum 5. A,B,C 30 marks are 360. 4 5. There are three parts in the question paper A,B,C consisting of Physics, Chemistry and Mathematics 6. having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 7. 6. One Fourth mark will be deducted for indicated 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. 8. 7. Use Blue/Black Ball Point Pen only for writting particulars/marking responses on Side–1 and Side 2 of the Answer Sheet. Use of pencil is strictly prohibited. 9. 8. No candidate is allowed to carry any textual material, 10. printed or written, bits of papers, mobile phone any electronic device etc, except the Identity Card inside the examination hall/room. 11. 9. Rough work is to be done on the space provided for this purpose in the Test Booklet only. 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 candidate are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet. Your Target is to secure Good Rank in JEE (Main) 2017 Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PART A - PHYSICS 1. A smooth sphere A of mass m collides 1. m A elastically with an identical sphere B at B rest. The velocity of A before collision is A 60° 8 m/s in a direction making 60° with the 8 line of centres at the time of impact. (i) A (i) The sphere A comes to rest after collision. (ii) B,8 (ii) The sphere B will move with a speed of 8 m/s after collision. (iii) AB (iii) The directions of motion A and B after collision are at right angles. (iv) B 4 (iv) The speed of B after collision is 4 m/s. The correct option is :- (1) (i), (ii) (2) (ii), (iii), (iv) (1) (i), (ii) (2) (ii), (iii), (iv) (3) (iii), (iv) (4) (ii), (iii) (3) (iii), (iv) (4) (ii), (iii) 2. A block of mass M is tied to one end of 2. M massless rope. The other end of rope is in the hands of a man of mass 2M as show in figure. Initially the block and the man are 2M resting on a rough plank of mass 2M as 2M shown in figure. The whole system is resting on a smooth horizontal surface. The man pulls the rope. Pulley is massless and frictionless. What is the magnitude of displacement of the plank when the block meets the pulley (Man does not leave his ( position on the plank during the pull). ) 1001CT103516012 H-1/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 2M 2M M M 3m 2M 3m 2M (1) 0.5m (2) 0.6m (1) 0.5m (2) 0.6m (3) 1.5m (4) 2m (3) 1.5m (4) 2m 3. In a parallel plate air capacitor, a cathode 3. beam comprising n = 106 electrons is v0=108 m/s n=106 emitted with a velocity v0 = 108 m/s into the space between the plates. The =400 V potential difference between the plate is = 400 V, the seperation between the d= 2cm. plates is d = 2 cm and the area of each 2 = 100 cm2 plate is 2 = 100 cm2. The deflection of the electron beam, is :- (1) 1.6 mm (1) 1.6 mm (2) 1.76 mm (2) 1.76 mm (3) 0 mm (3) 0 mm (4) 5 mm (4) 5 mm H-2/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 4. A parallel plate capacitor, partially filled 4. K with a dielectric slab of dielectric constant V K, is connected with a cell of emf V volt, as shown in the figure. Separation between D the plates is D. Then V V D P P' D P P' (1) electric field at point P is less than that (1) P P at point P. (2) electric field at point P is less than that (2) P'P at point P. (3) electric fields at points P and P are (3) PP equal V (4) PE= V (4) electric field at point P is E = KD . KD 5. Two identical capacitors 1 and 2 are 5. 12 connected in series. The capacitor 2 2 K V0 contains a dielectric slab of constant K as shown. They are connected to a battery of Q1Q2 Q1Q2 emf V0 volts. The dielectric slab is then removed. Let Q1 and Q2 be the charge stored in the capacitors before removing the slab and Q1, and Q2 be the values after removing the slab. Then : 1 1 2 vo 2 vo 1001CT103516012 H-3/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 (1) Q1 ' K 1 (2) Q2 ' (K 1) (1) Q1 ' K 1 (2) Q2 ' (K 1) Q1 K Q2 2 Q1 K Q2 2 (3) Q2 ' K 1 (4) Q1 ' K (3) Q2 ' K 1 (4) Q1 ' K Q2 2K Q1 2 Q2 2K Q1 2 6. A and C are concentric conducting 6. AC a c spherical shells of radius a and c A respectively. A is surrounded by a a, b k concentric dielectric radius a, outer radius b and dielectric constant k. If sphere A be A Q given a charges Q, the potential at the outer surface of the dielectric is :- Q Q (1) 40kb (1) 40kb (2) Q 1 1 (2) Q 1 1 a) 40 a k(b 40 a k(b a) Q Q (3) 40b (3) 40b (4) None of these (4) H-4/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 7. Three resistances of magnitude R each are 7. R connected in the form of an equilateral a triangle of side a. The combination is placed B=B0e–t in a magnetic field B = B0 e–t perpendicular to its plane. The induced current in the circuit is given by : A A RR RR B C R B C R a2 a2 (1) a2 B0 e t (2) a2 B0 et 2 3R et 4 3R et 2 3R (1) B0 (2) B0 4 3R a2 B0 a2B0R a2B0 a 2 B0R 4 3R 4 3 e t et (3) e t (4) e t (3) (4) 4 3R 4 3 8. Statement-1 : When a magnet is made to 8. -1: fall freely through a closed coil, its acceleration is always less than acceleration due to gravity. and Statement-2 : Current induced in the coil -2 : opposes the motion of the magnet, as per Lenz's law. (1) Statement-1 is true, Statement-2 is true; (1) 1 2 2 Statement-2 is a correct explanation for Statement-1 1 (2) Statement-1 is true, Statement-2 is true, (2) 12; 2 Statement-2 is NOT a correct explanation for Statement-1 1 (3) Statement-1 is True, Statement-2 is False (3) 1 ,2 (4) Statement-1 is False, Statement-2 is (4) 1 2 True 1001CT103516012 H-5/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 9. A wire of mass m and length can slide 9. m freely on a pair of smooth, vertical rails (figure). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected B at the top end by a capacitor of capacitance C C. The acceleration of the wire neglecting :- any electric resistance is : mg × × C× × mg × × C× × (1) m CB22 ×× ×× (1) m CB22 ×× ×× 2mg ×× ×× 2mg ×× ×× (2) m CB22 ×× ×× (2) m CB22 ×× ×× mg ×× ×× mg ×× ×× (3) CB22 (3) CB22 ×× ×× mg ×× ×× (4) 2(m CB22 ) mg (4) 2(m CB22 ) 10. A rectangular loop has a sliding connector 10. PQ PQ of length 2 m and resistance 10 and 2m10 it is moving with a speed 5 m/s as shown. 5 m/s The set-up is placed in a uniform magnetic 3T field 3T going into the plane of the paper. The three currents I1, I2 and I are :- I1, I2 I P P 5m/s 5m/s 10 10 10 × 3T I I2 10 10 10 I1 Q × 3T I I2 (1) I1 = I2 = 3A, I = 1A (2) I1 = I2 = 5A, I = 2A I1 Q (3) I1 = I2 = 1A, I = 2A (4) I1 = I2 = I = 2A (1) I1 = I2 = 3A, I = 1A (2) I1 = I2 = 5A, I = 2A (3) I1 = I2 = 1A, I = 2A (4) I1 = I2 = I = 2A H-6/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 11. A conducting ring of radius R is placed in 11. R uniform imnwovairndgmwaitghnevteilcocfiiteyldvBinaistssphloawnne,. If ring is B v the induced emf across arc PQ will be :- PQ (1) vBR 1 1 P vBR 1 1 P 2 Q 2 Q 2 (1) 2 45° 45° v v (2) vBR 1 1 B vBR 1 1 B 2 (2) 2 (3) vBR 1 1 vBR 1 1 (3) vBR 1 1 (4) vBR 1 1 2 2 2 (4) 2 2 2 12. A uniform but time varying magnetic field 12. R is present in a circular region of radius R. The magnetic field is perpendicular and into the plane of the loop and the magnitude of field is increasing at a 2R constant rate . There is a straight conducting rod of length 2R placed as shown in figure. The magnitude of induced emf across the rod is : ×× ×× × × ××R×× × × × × × × ××R×× × × × × ×× ×× ×× ×× ××××× ××××× ××××× ××××× ×× ×× 2R 2R R2 R2 R2 R2 R2 R2 (4) (1) R2 (2) (3) 2 (4) (1) R2 (2) (3) 2 2 2 4 4 1001CT103516012 H-7/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 13. A particle is projected with a velocity 13. (10 m/s) y- (10m/s) along y-axis from point (2, 3). (2, 3) 3iˆ 4ˆj Magnetic field of 3iˆ 4ˆj Tesla exist y - uniformly in the space. Its speed when particle passes through y-axis for the third time is : (neglect gravity) (1) 20 m/s (1) 20 m/s (2) 60 m/s (2) 60 m/s (3) 30 m/s (3) 30 m/s (4) 10 m/s (4) 10 m/s 14. A rod of length is rotating with constant 14. angular velocity and we marked the rod 0,1, 2,.......... 8 0, 1, 2, .......... 8 at equal spacing. Potential difference between consecutive points, if we go from left to right will be : ××××××× ××××××× 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 ××××××× ××××××× (1) in increasing arithmetic progression. (1) (2) in increasing geometric progression. (2) (3) in increasing form 12, 22, 32 .......... 82 (3) 12, 22, 32 .......... 82 (4) in decreasing form, 82, 72, 62 ......... 12 (4) 82, 72, 62 ......... 12 H-8/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 15. A simple pendulum with a bob of mass 15. E=2×106 m = 1 kg, charge q = 5 C and string length m=1 = 1 m is given a horizontal velocity u in a q =5C uniform electric field E = 2 × 106 V/m at its = 1 bottom most point A, as shown in figure. It A u is given a speed u such that the particle leave the circular path at its topmost point u C C. Find the speed u. (Take g = 10 m/s2) ug=10 2 ) (1) 40 m/s C C (1) 40 m/s (2) 50 m/s DB (2) 50 m/s DB (3) 35 m/s (3) 35 m/s E E (4) None of these Au (4) A u 16. Figure shows a solid metal sphere of radius 16. a 2a ‘a’ surrounded by a concentric thin metal shell of radius 2a. Initially both are having Q charges Q each. When the two are connected by a conducting wire as shown in the figure, then amount of heat produced in this process will be : Q Q Q Q a a 2a 2a KQ2 KQ2 KQ2 KQ2 KQ2 KQ2 KQ2 KQ2 (1) (2) (3) (4) (1) (2) (3) (4) 2a 4a 2a 4a 6a 8a 6a 8a 1001CT103516012 H-9/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 17. In a uniform electric field, a cube of side 1 17. 1cm cm is placed. The total energy stored in the 8.85 µJ cube is 8.85µJ. The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is. (1) 1 V/m (2) 100 2V / m (1) 1 V/m (2) 100 2V / m 52 52 (3) 5 2V / m (4) 10 2V / m (3) 5 2V / m (4) 10 2V / m 18. Figure shows electric field lines in which 18. an electric dipole p is placed as shown. p Which of the following statements is correct? (1) The dipole will not experience any force. (1) (2) The dipole will experience a force (2) towards right. (3) The dipole will experience a force (3) towards left. (4) (4) The dipole will experience a force upwards. H-10/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 19. Figure shows the variation of the moment 19. of inertia of a uniform rod, about an axis passing through its centre and inclined at an angle to the length. The moment of inertia of the rod about an axis passing =3 through one of its ends and making an angle = 3 will be 0.6 0.6 I (kg-m2) I (kg-m2) (rad) (rad) (1) 0.45 kg–m2 (2) 1.8 kg–m2 (1) 0.45 kg–m2 (2) 1.8 kg–m2 (3) 2.4 kg–m2 (4) 1.5 kg–m2 (3) 2.4 kg–m2 (4) 1.5 kg–m2 20. A 3.0 kg bobbin consists of a central 20. 3 kg 5cm cylinder of radius 5.0 cm and two end plates 6cm each of radius 6.0 cm. It is placed on a slotted incline, where friction is sufficient to prevent sliding. A block of mass 4.5 kg 4.5k g is suspended from a cord wound around the bobbin and passing through the slot under the incline. If the bobbin is in static equilibrium, what is the angle of tilt of the incline? (1) 30° (1) 30° (2) 37° (2) 37° (3) 53° (3) 53° (4) More information is required. (4) 1001CT103516012 H-11/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 21. A circular hoop of mass m and radius R rests 21. mR flat on a horizontal frictionless surface. A bullet, also of mass m and moving with a mv velocity v, strikes the hoop and gets embedded in it. The thickness of the hoop is much smaller than R. The angular R velocity with which the system rotates after the bullet strikes the hoop is :- (1) V/(4R) (2) V/(3R) (1) V/(4R) (2) V/(3R) (3) 2V/(3R) (4) 3V/(4R) (3) 2V/(3R) (4) 3V/(4R) 22. The time period of oscillation of a magnet 22. 2sec is 2 sec. When it is remagnetised so that its pole strength is 4 times, its period will 4 be - (1) 4 sec (2) 2 sec (3) 1 sec (4) 8 sec (1) 4 sec (2) 2 sec (3) 1 sec (4) 8 sec 23. Curie's law states that 23. (1) magnetic susceptibility is inversely (1) proportional to the absolute temperature (2) (2) magnetic susceptibility is inversely proportional the square root of the absolute temperature (3) (3) magnetic susceptibility is directly proportional to the absolute temperature (4) (4) magnetic susceptibility is constant H-12/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 24. The average dipole moment of Fe atoms is 24. Fe 1.8×10–23A-m 2 1.8 × 10–23 A-m2. The magnetic moment of 10cm1cm an iron rod of length 10 cm and diameter (Fe 1cm is : (density and at. wt. of Fe are 7.87 g/cm3 55.87 ) 7.87 g/cm3 and 55.87) (1) 3.0 A-m2 (1) 3.0 A-m2 (2) 4.0 A-m2 (2) 4.0 A-m2 (3) 6.0 A-m2 (3) 6.0 A-m2 (4) 12.0 A-m2. (4) 12.0 A-m2. 25. In the diagram, I1, I2 are the strength of 25. I1I2 the currents in the loop and infinite long OA=AB= R O straight conductor respectively. OA = AB = R. The net magnetic field at the centre O is zero. Then the ratio of the currents in the loop and the straight conductor is : O O I1 I1 R R A A I2 R R B B I2 (1) (2) 2 (1) (2) 2 1 1 1 1 (3) (4) 2 (3) (4) 2 1001CT103516012 H-13/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 26. The figure shows three situations when an 26. Bv electron with velocity v travels through a uniform magnetic field . In each case, B what is the direction of magnetic force on the electron? y x y y x y x y y x v Bx B v Bx B B v B v z v z v z 2 z z 2 z 1 1 3 3 (1) positive z–axis, negative x–axis, positive (1) z-, x-, y- y–axis (2) z-, x- (2) negative z–axis, negative x–axis and zero (3) z-, y- (3) positive z–axis, positive y–axis and zero (4) z-, x- (4) negative z–axis, positive x–axis and zero 27. A thin uniform rod with negligible mass 27. and length is attached to the floor by a frictionless hinge at point P. A horizontal P spring with force constant k connects the k other end to wall. The rod is in a uniform magnetic field B directed into the plane of B paper. What is extension in spring in equilibrium when a current i is passed i through the rod in direction shown. Assuming spring to be in natural length initially. H-14/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 B × B B × B × i × × i × 53° P 53° 53° P 53° P Hinge P Hinge In equilibrium position In equilibrium position Hinge Hinge Spring in relaxed position Spring in relaxed position 5iB 3iB 5iB 5iB 5iB 3iB 5iB 5iB (1) (2) (3) (4) (1) 8k (2) 8k (3) 4k (4) 6k 8k 8k 4k 6k 28. A disc of radius r and carrying positive 28. qr charge q is rotating with an angular speed B in a uniform magnetic field B about a fixed axis as shown in figure, such that angle made by axis of disc with magnetic field is . Torque applied by axis on the disc is : Disc Disc B B Fixed axis Fixed axis qr2B sin (1) qr2Bsin , (1) , clockwise 2 2 qr2B sin (2) qr2Bsin , (2) , anticlockwise 4 4 qr2B sin (3) qr2B sin (3) , anticlockwise , 2 2 qr2B sin (4) qr2Bsin , (4) , clockwise 4 4 1001CT103516012 H-15/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 29. A nucleus at rest undergoes -decay 29. according to equation X226 Y + At 92 X226 Y + t=0 92 t = 0 the emitted -particle enters in a - region of space where a uniform magnetic field B = B0 iˆ and electric field E E0iˆ EE 0ˆi B-=Bx0 iˆ exists. The -particle enters in a region = 0 v = v0 ˆj with velocity v = v0 ˆj at x = 0. 3107 qmE 0 3 107 m time later the particle was qE0 found to have twice the initial speed. Initial - speed of the -particle is : (1) 106 ms–1 (2) 107 ms–1 (1) 106 ms–1 (2) 107 ms–1 (3) 7 × 107 ms–1 (4) 3 × 107 ms–1 (3) 7 × 107 ms–1 (4) 3 × 107 ms–1 30. A bar of mass m is suspended horizontally 30. m k3k on two vertical springs of spring constant k and 3k. The bar bounces up and down while remaining horizontal. Find the time period of oscillation of the bar (Neglect mass of springs and friction everywhere). Fixed pulley Fixed pulley 3K K 3K K (1) 2 m (2) 2 m (1) 2 m (2) 2 m k 3k k 3k 2m (4) 2 m (3) 2m (4) 2 m (3) 3k 4k 3k 4k H-16/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 PART B - CHEMISTRY 31. If 1mL of 1M HCl solution is added to 99mL 31. 1mL , 1M HCl N aCl 99mL of aqueous solution of NaCl so that pH of NaC l NaCl solution change by X units. The value of X is : pH, x X (1) 2 (2) 5 (1) 2 (2) 5 (3) 7 (4) 1 (3) 7 (4) 1 32. Which of the following solutions will act 32. as buffer solution - - (1) 10 ml of 0.1M NaOH + 5 ml of 0.1 M HCl (1) 10 ml , 0.1M NaOH + 5 ml , 0.1 M HCl (2) 15 ml of 0.1M NaOH + 10 ml of 0.1 M (2) 15 ml , 0.1M NaOH + 10 ml , 0.1 M CH3COOH CH3COOH (3) 10 ml of 0.1M NH4OH + 15 ml of 0.1 M (3) 10 ml , 0.1M NH4OH + 15 ml , 0.1 M HCl HCl (4) 10 ml of 0.1M H2S + 15 ml of 0.1 M Na2S (4) 10 ml , 0.1M H2S + 15 ml , 0.1 M Na2S 33. The pressure of a vander waal gas is less 33. than the pressure of an ideal gas because : of : (1) (1) Infinitesimal size of molecules (2) (2) The collisions with the wall become (3) inelastic (4) (3) Intermolecular attraction (4) Molecular movement is more random 1001CT103516012 H-17/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 34. 1 mole of H2C2O4 is oxidised by X mole of 34. 1 H 2C2O4 X MnO4– in strong basic medium and 1 mole MnO4– 1 of NaHC2O4 is oxidised by y mole of MnO4– NaHC2O4 y M nO4– in acidic medium. Ratio of x/y is- x/y- (1) 2 : 1 (2) 5 : 1 (1) 2 : 1 (2) 5 : 1 (3) 3 : 1 (4) 1 : 3 (3) 3 : 1 (4) 1 : 3 35. The degree of hydrolysis of 0.1 M solution 35. ArNH3+ Cl– 0.1M of ArNH3+ Cl– is 0.01. Find the degree of hydrolysis in 0.4 M solution of ArNH3+ Cl– 0.01 ArN H3+ Cl–0.4 M (1) 0.1 (2) 0.01 (1) 0.1 (2) 0.01 (3) 0.005 (4) 0.05 (3) 0.005 (4) 0.05 36. A real gas cannot be liquefied in which of 36. the following conditions : : (1) T < TC, P > PC (2) T < TC , P < PC (1) T < TC, P > PC (2) T < TC , P < PC (3) T = TC, P > PC (4) V > VC, P > PC (3) T = TC, P > PC (4) V > VC, P > PC 37. Which of the following option is incorrect 37. - (1) Fraction of total molecules having (1) e–Ea/RT energy greater than or equal activation energy is e–Ea/RT (2) (2) Lesser the activation energy, faster will (3) K = Ae–Ea/RT be the reaction (A) (3) In Arrhenius equation K = Ae–Ea/RT, Arrhenius factor (A) is the product of (4) collision frequency and probability factor (4) Reaction having high value of activation energy is less temperature dependent H-18/34 1001CT103516012
38. For a certain reaction : (A)(g) B(g) Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 Half life for different initial pressures of A is given below : 38. : (A)(g) B(g) A : PA0 (atm) 0.1 0.025 PA0 (atm) 0.1 0.025 t1/2(sec) 100 50 t1/2(sec) 100 50 The correct statement about order of re- : action is : (1) 1 (2) 2 (1) 1 (2) 2 (3) 3 (4) 0.5 (3) 3 (4) 0.5 39. One mole of an ideal monoatomic gas is 39. heated in a process PV5/2 = constant. PV5/2 = 3 6ºC Amount heat absorbed in the process for 36ºC rise in temperature. (1) 60 cal (2) 30 cal (1) 60 cal (2) 30 cal (3) 108 cal (4) 180 cal (3) 108 cal (4) 180 cal 40. On heating a mixture of SO2Cl2 and CO, 40. SO2Cl2 CO two equilibria are simultaneously established ; ; SO2Cl2(g) SO2(g) + Cl2(g) CO(g) + Cl2(g) COCl2 (g) SO2Cl2(g) SO2(g) + Cl2(g) On adding more SO2 at equilibrium what CO(g) + Cl2(g) COCl2 (g) will happen ? SO2 ? (1) Amount of CO will decrease (1) CO (2) Amount of SO2Cl2 and COCl2 will increase (2) SO2Cl2 COCl2 (3) CO (3) Amount of CO will remain unaffected (4) SO2Cl2 CO (4) Amount of SO2Cl2 and CO will increase 1001CT103516012 H-19/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 41. The formula of 3 membered discrete chain 41. 3 (discrete) silicate is :- :- (1) Si3O96– (2) Si3O108– (1) Si3O96– (2) Si3O108– (3) Si3O63– (4) Si3O1212– (3) Si3O63– (4) Si3O1212– 42. Which of the following complex compound 42. have highest number of geometrical ? isomers ? (1) [PtCl2Br2]2– (2) [CoCl2Br2]2– (1) [PtCl2Br2]2– (2) [CoCl2Br2]2– (3) [Pt(NH3)(Gly)Br] (4) [Co(NH3)2(Gly)2]+ (3) [Pt(NH3)(Gly)Br] (4) [Co(NH3)2(Gly)2]+ 43. According to V.B.T., in which of the 43. V.B.T. following complex, transition of electron occurs from one shell to other shell of :- central metal :- (1) [Fe(H2O)5(NO)]2+ (2) K3[Cu(CN)4] (1) [Fe(H2O)5(NO)]2+ (2) K3[Cu(CN)4] (3) [Ni(CN)6]4– (4) [Co(CN)6]4– (3) [Ni(CN)6]4– (4) [Co(CN)6]4– 44. If Hund's rule is violated then select the 44. [Ni(NH3)6]2+ CORRECT statement regarding [Ni(NH3)6]2+ is :- :- (1) sp3d2 , paramagnetic (1) sp3d2 , (2) d2sp3 , (2) d2sp3 , diamagnetic (3) sp3d2 , diamagnetic (3) sp3d2 , (4) d2sp3 , (4) d2sp3 , paramagnetic 45. When aqueous solution of cupric bromide 45. is electrolyzed, the product obtained at cathode will be :- :- (1) H2 (2) Cu (1) H2 (2) Cu (3) Br2 (4) O2 (3) Br2 (4) O2 H-20/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 46. Which of the following complex ion is 46. expected to absorb light in visible region ? ? (1) [Ti(en)3]4+ (2) [Sc(NH3)4(H2O)2]3+ (1) [Ti(en)3]4+ (2) [Sc(NH3)4(H2O)2]3+ (3) [CrO4]2– (4) [Zn(CN)4]2– (3) [CrO4]2– (4) [Zn(CN)4]2– 47. In which of the following pair, the molecule 47. has higher IE1 value compared to its atom :- IE1 :- (1) F2 and F (2) O2 and O (1) F2 F (2) O2 O (3) N2 and N (4) All of these (3) N2 N (4) 48. Select CORRECT statement :- 48. :- (1) Acidic strength : H2O > H2S (1) :H2O > H2S (2) p –p ( ) (2) p –p bonding M.O. has (gerade) symmetry (3) NCl3 on hydrolysis gives HOCl but NF3 (3) NCl3HOCl gives HF in ordinary condition NF3 ,HF (4) Bond angle order : OF2<H2O<Cl2O<ClO2 (4) :OF2<H2O<Cl2O<ClO2 49. Select the CORRECT IUPAC name of 49. IUPAC :- complex compound :- (1) (III)- (1) Triamminetricyanidochromium(III) N-(III) hexanitrito-N-irridate(III) (2) (III)-N- (2) Pentaamminecyanidochromium(III) (III) hexanitrito-N-irridium(III) (3) -N-(III) (3) Hexanitrito-N-irridium(III) pentaammine cyanidochromate(II) (II) (4) Pentaamminecyanidochromium(III) (4) (III)-N- hexanitrito-N-irridate(III) (III) 1001CT103516012 H-21/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 50. Select the INCORRECT statement :- 50. :- (1) Concentration of 'Cerrusite' ore is done (1) by froth floatation process (2) Roasting process is involved for (2) extraction of iron (3) Byproduct of Serpeck's process is silicon (3) B y product (4) Cassiterite and Rutile are oxide ores of the metals (4) 51. Which of the following reaction will not 51. \"\" produce \"Acetaldehyde\" ? (1) CH3–C N (i) AlH(i-Bu)2 (1) CH3–C N (i) AlH(i-Bu)2 (ii) H2O (ii) H2O O O (2) CH3–C–Cl H2–Pd (2) CH3–C–Cl H2–Pd BaSO4 BaSO4 (3) HC CH Hg(OAc)2 (3) HC CH Hg(OAc)2 – – NaBH4 / OH NaBH4 / OH (4) CH3–C N (i) LiAlH4 (4) CH3–C N (i) LiAlH4 (ii) H2O (ii) H2O 52. Which of the following acyclic compound 52. Br2 will react fastest with Br2 gas ? (1) C2H6 (2) C3H6 (1) C2H6 (2) C3H6 (3) C2H4 (4) C2H2 (3) C2H4 (4) C2H2 H-22/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 53. How many optically active (Lactide) 53. product(s) are formed in given reaction ? OH OH CH3–CH–COOH CH3–CH–COOH (1) 3 (2) 4 (1) 3 (2) 4 (3) 2 (4) 1 (3) 2 (4) 1 54. Which of the following on reaction with 54. RMgX excess RMgX will give only 2º alcohol ? 2º O CH3 O CH3 (1) CH3–C–O–CH CH3 (1) CH3–C–O–CH CH3 OO (2) H–C–O–CH2–CH3 (2) H–C–O–CH2–CH3 O O (3) H–C–O–CH(CH3)2 (3) H–C–O–CH(CH3)2 O O (4) CH3–C–O–CH3 (4) CH3–C–O–CH3 55. Best reagent to carry out following 55. conversion is : CHO CH3 CHO CH3 OH OH OH OH (1) Red P / I2 (2) LiAlH4 , Et2O (1) P / I2 (2) LiAlH4 , Et2O– (3) Zn-Hg / HCl – (4) (i) N2H4 ; (ii) OH, (3) Zn-Hg / HCl (4) (i) N2H4 ; (ii) OH, 1001CT103516012 H-23/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 56. Which of the following will not produce 56. CH3MgBr CH4 CH4 gas with CH3MgBr ? O O (1) S–OH (1) S–OH O O (2) CH3–NH2 (2) CH3–NH2 (3) CH3–CH2–OH (3) CH3–CH2–OH (4) (4) 57. Which of the following reaction will not 57. CO2 produce CO2 ? (1) CH3 –CH2–SO3H NaHCO3 (1) CH3 –CH2–SO3H NaHCO3 OO OO (2) HO OH (2) HO OH (3) CH3–CH2–COONa Electrolysis (3) CH3–CH2–COONa Electrolysis COOH COOH O O (4) (4) H-24/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 58. The relative reactivities for replacement 58. H– of –H are 3º : 2º : 1º = 6 : 4 : 1. Relative 3º : 2º : 1º = 6 : 4 : 1 amount of given product is : Cl Cl Cl2 Cl2 hv hv (1) 3/31 (2) 6/31 (1) 3/31 (2) 6/31 (3) 16/31 (4) 16/29 (3) 16/31 (4) 16/29 59. In which of the reaction carbanion is not 59. formed as one of the intermediate : O O (1) NaOH (1) NaOH OH + CaO OH + CaO – – (2) OH OH (2) N NH2 N NH2 Cl Na Cl Na (3) Et2O (3) Et2O OH (i) AgOH OH (i) AgOH (4) (ii) Br2 / CCl4 , (4) (ii) Br2 / CCl4 , O O 1001CT103516012 H-25/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 60. If (P) & (Q) are major products, then find 60. (P) (Q) incorrect statement for given sequence : OH OH (+– ) CH3–CH2–CH–CH3 H3PO4 P (+– ) CH3–CH2–CH–CH3 H3PO4 P Ni D2 Ni D2 Q Q (1) Formation of (Q) is syn addition (1) (Q) phenomenon (2) (P) (3) (Q) (2) Formation of (P) is unimolecular (4) (Q) , elimination reaction (3) Formation of (Q) is redox reaction (4) (Q) is optically inactive due to internal compensation H-26/34 1001CT103516012
Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 PART C - MATHEMATICS 61. Let a, b, c, d, e be five numbers satisfying 61. a, b, c, d, e the system of equations 2a + b + c + d + e = 6 2a + b + c + d + e = 6 a + 2b + c + d + e = 12 a + 2b + c + d + e = 12 a + b + 2c + d + e = 24 a + b + 2c + d + e = 24 a + b + c + 2d + e = 48 a + b + c + 2d + e = 48 a + b + c + d + 2e = 96, a + b + c + d + 2e = 96, then |c| is equal to |c| (1) 6 (2) 7 (1) 6 (2) 7 (3) 8 (4) 25 (3) 8 (4) 25 62. If domain of function ƒx nmsinx 4 62. ƒ x n m sin x 4 R is R, then number of possible integral m values of m is (1) 3 (2) 4 (1) 3 (2) 4 (3) 6 (4) 7 (3) 6 (4) 7 63. If function y = ƒ(x) satisfy the differential 63. y = ƒ(x) equation (x3 + 1)dy = x(1 – 3xy)dx and (x3 + 1)dy = x(1 – 3xy)dx ƒ(0) = 0, then lim x2 is equal to ƒ(0) = 0xlim0ƒx2x x0 ƒ x (1) 0 (2) 1 (1) 0 (2) 1 (3) 2 (4) 4 (3) 2 (4) 4 64. Identify the correct statement about 64. ƒ(x) = (x2 – 1, 7 – x2, 5) function ƒ(x) = max(x2 – 1, 7 – x2, 5) (1) ƒ(x) is not differentiable at 4 points (1) ƒ(x) (2) range of ƒ(x) is [3, ) (2) ƒ(x) [3, ) (3) ƒ(x) is an injective function (3) ƒ(x) (4) ƒ(x) is discontinuous at 4 points. (4) ƒ(x) 1001CT103516012 H-27/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 65. lim 1 12 sin 1 22 sin 2 32 sin 3 .... n2 sin n 65. 1 12 sin 1 22 sin 2 32 sin 3 .... n2 sin n n3 n n n n lim n n n n n n n3 equals - - (1) cos1 + 2sin1 (2) 2sin1 – 2 (1) cos1 + 2sin1 (2) 2sin1 – 2 (3) cos1 – 2sin1 – 2 (4) cos1 + 2sin1 – 2 (3) cos1 – 2sin1 – 2 (4) cos1 + 2sin1 – 2 66. Function ƒ x x2 2 - 66. ƒ x x2 2 1 x2 1 x2 (1) is always increasing (2) is always decreasing (1) (3) has exactly one point of minima (2) (4) has exactly one point of maxima (3) (4) 67. Let Hx x3 x 1 sin t3dt , then lim Hx 67. H x x3 x 1 sin t3dt limH x x 1 x 1 x1 x 1 x2 x2 equal to - - (1) sin 1 (2) –sin1 (3) 2sin1 (4) 0 (1) sin 1 (2) –sin1 (3) 2sin1 (4) 0 x8 x3 x x8 x3 x 8x6 24x4 68. The integral dx is 3x11 8x6 24x4 1/ 3 68. 1/3 dx 3x11 equal to- - (1)2 2/3 3 3x11 8x6 24x 2 (1) 2/3 C 3 3x11 8x6 24x C 3x8 8x3 24x 2/ 3 C 1 3x8 8x3 24x 2/ 3 C 1 (2) 16 (2) 16 (3)2 2/3 3x8 8x3 24x 2 (3) 2/3 3 C 3x8 8x3 24x C 3 3x8 8x3 24x 2/ 3 C 1 (4) 1 3x8 8x3 24x 2/3 C 16 (4) 16 (C ) (where C is constant of integration) H-28/34 1001CT103516012
69. Locus of a point which moves so that its Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 distance from the origin is thrice its distance from y = 2x is- 69. (1) a straight line y=2x (2) a pair of straight lines (3) a circle (1) (4) a parabola (2) (3) 70. Triangle formed by the lines 3x + y + 4 = 0, (4) 3x + 4y – 15 = 0 and 24x – 7y = 3 is a/an (1) equilateral triangle 70. 3x+ y + 4 = 0, 3x + 4y – 15 = 0 (2) isosceles triangle 24x – 7y = 3 (3) acute angle triangle (1) (4) scalene triangle (2) (3) (4) 30 30 71. x 42 ex2 dx x 42 ex2 dx is equal 71. x 42 ex2dx x 42 ex2dx 03 03 to- (2) 8(e3 – 1) - (2) 8(e3 – 1) (1) 8e3 (1) 8e3 (3) 3 e4 1 (4) 3 e8 1 (3) 3 e4 1 (4) 3 e8 1 72. Let a, b, c are roots of equation x3 + 8x + 1 = 0, 72. a, b, c x3 + 8x + 1 = 0 then the value of bc ac ab bc ac ab (8b 1)(8c 1) (8a 1)(8c 1) (8a 1)(8b 1) (8b 1)(8c 1) (8a 1)(8c 1) (8a 1)(8b 1) is equal to (1) 0 (2) –8 (1) 0 (2) –8 (3) –16 (4) 16 (3) –16 (4) 16 1001CT103516012 H-29/34
Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 11 73. Let (x) exet(t)dt x and (n(e2 3)) 73. (x) exet(t)dt x (n(e2 3)) 00 is equal to A, then - A - (1) A = n(e2 – 3) – 2 (1) A = n(e2 – 3) – 2 (2) A (3,4) (2) A (3,4) (3) A = e2 – 3 (3) A = e2 – 3 (4) A = n(e2 – 3) + 2 (4) A = n(e2 – 3) + 2 74. If the sum of the first 11 terms of the series 74. 1 4 2 1 5 2 1 6 2 22 2 1 2 ...... is 1 4 2 1 5 2 1 6 2 22 2 1 2 ...... 7 7 7 7 7 7 7 7 11 , then is equal to (4) 39 11 11 7 (1) 36 (2) 37 (3) 38 7 (1) 36 (2) 37 (3) 38 (4) 39 75. Least value of 75. x2y2 2x2y 2x2 2xy 2x 1 x2y2 2x2y 2x2 2xy 2x 1 is , then x2y x x2y x {x,y R+, x2y + x 0} {where x,y R+, x2y + x 0} (1) (0,1) (2) [1,3) (1) (0,1) (2) [1,3) (3) [3,4] (4) (4,7) (3) [3,4] (4) (4,7) 76. Let P1 : y = –x2 + 4x + 2 and P2 : x2 + 5x + 76. P1 : y = –x2 + 4x + 2 17 = y are two parabolas, then number of P2 : x2 + 5x + 17 = y P1P2 8 8 common tangents of P1 and P2 is - (1) 0 (2) 1 (1) 0 (2) 1 (3) 2 (4) 3 (3) 2 (4) 3 H-30/34 1001CT103516012
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