Paper Code : 100 1CT103516005 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) ENGLISH JEE (Main + Advanced) : LEADER COURSE DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR Test Type : MINOR PHASE : III, IV & V Test Pattern : JEE-Advanced TEST DATE : 25 - 09 - 2016 Time : 3 Hours PAPER – 1 Maximum Marks : 264 READ THE INSTRUCTIONS CAREFULLY GENERAL : 1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so. 2. Use the Optical Response sheet (ORS) provided separately for answering the questions. 3. Blank spaces are provided within this booklet for rough work. 4. Write your name, form number and sign in the space provided on the back cover of this booklet. 5. After breaking the seal of the booklet, verify that the booklet contains 32 pages and that all the 20 questions in each subject and along with the options are legible. If not, contact the invigilator for replacement of the booklet. 6. You are allowed to take away the Question Paper at the end of the examination. OPTICAL RESPONSE SHEET : 7. The ORS will be collected by the invigilator at the end of the examination. 8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work. 9. Write your name, form number and sign with pen in the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your form number. DARKENING THE BUBBLES ON THE ORS : 10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS. 11. Darken the bubble COMPLETELY. 12. The correct way of darkening a bubble is as : 13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way. 14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or \"un-darken\" a darkened bubble. 15. Take g = 10 m/s2 unless otherwise stated. Please see the last page of this booklet for rest of the instructions
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 Atomic No. SOME USEFUL CONSTANTS Atomic masses : H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17, Br = 35, Xe = 54, Ce = 58, H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127, Xe = 131, Ba=137, Ce = 140, Boltzmann constant k = 1.38 × 10–23 JK–1 Coulomb's law constant 1 = 9 ×109 Universal gravitational constant 4 0 Speed of light in vacuum Stefan–Boltzmann constant G = 6.67259 × 10–11 N–m2 kg–2 Wien's displacement law constant c = 3 × 108 ms–1 Permeability of vacuum = 5.67 × 10–8 Wm–2–K–4 b = 2.89 × 10–3 m–K Permittivity of vacuum µ0 = 4 × 10–7 NA–2 Planck constant 1 0 = 0c2 h = 6.63 × 10–34 J–s Space for Rough Work E-2/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PHYSICS PART-1 : PHYSICS SECTION–I : (Maximum Marks : 40) This section contains TEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) Zero Marks darkened. : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. 1. Consider a sphere of mass ‘m’ radius ‘R’ doing pure rolling motion on a rough surface having velocity v0 as shown in the Figure. It makes an elastic impact with the smooth wall and moves back and starts pure rolling after some time again. Then v0 O (A) Change in angular momentum about ‘O’ in the entire motion equals 2mv0R in magnitude. (B) Moment of impulse provided by the wall during impact about O equals 2mv0R in magnitude. (C) Final velocity of ball will be 3 7 v0 (D) Final velocity of ball will be – 3 v 0 7 Space for Rough Work 1001CT103516005 E-3/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 2. A point object of mass m is connected to a cylinder of radius R via a massless rope. At time t = 0 the object is moving with an initial velocity v0 perpendicular to the rope, the rope has a length L0, and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds Tmax. Express your answer in terms of Tmax, L0, R, and v0. Mark the CORRECT statement(s) : (A) Angular momentum of the object with respect to the axis of the cylinder at the instant that the rope breaks is m2v03 Tmax (B) The kinetic energy of the object at the instant that the rope breaks is mv 2 0 2 (C) When tension is Tmax, the length of the rope (not yet wound) is L0 – R (D) When tension is Tmax, the length of the rope (not yet wound) is mv02 Tmax Space for Rough Work E-4/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 3. Slender bar A is rigidly connected to a massless rod BC in Case 1 and two massless cords in Case 2 as shown. The vertical thickness of bar A is negligible compared to L. In both cases A PHYSICS is released from rest at an angle = 0. When = 0° Case 1 Case 2 B L L CA A Mark the CORRECT statement(s) :- (A) The kinetic energy will be the same in both the cases. (B) The speed of the centers of gravity will be same For Option (C) and (D) : Consider system to be at rest intially in vertical position. If bullet D strikes A with a speed v0 initially directed horizontally and becomes embedded in it (C) The speed of the center of gravity of A immediately after the impact in case-1 will be larger. (D) The speed of the center of gravity of A immediately after the impact in case-2 will be larger. 4. A block A (5 kg) rests over another block B (3 kg) placed over a smooth horizontal surface. There is friction between A and B. A horizontal force F1 gradually increasing from zero to a maximum is applied to A so that the blocks move together without having motion relative to each other. Instead of this, another horizontal force F2 gradually increasing from zero to a maximum is applied to B so that the blocks move together without relative motion. The magnitude of friction forces between the blocks in the two cases are f1 & f2 respectively during the variation of F1 and F2 respectively. Then :- (A) f1max > f2max (B) F1max : F2max = 3 : 5 (C) F1max : F2max = 5 : 3 (D) f1 < F1 Space for Rough Work 1001CT103516005 E-5/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 5. Consider a hollow cylinder of radius R rolling without slipping on an inclined plane. The mass of the cylinder is M and its moment of inertia is I = MR2. The axis of the cylinder is connected to a fixed wall at the top of the inclined plane by a light spring of spring constant k as shown in the figure. In equilibrium, the spring has an extension x0 and system has same kinetic energy in all the situations quoted below at equilibrium. Mark the CORRECT statement(s): k (A) The angular frequency 0 of small oscillations around this equilibrium is 2m 2k (B) The angular frequency 0 of small oscillations around this equilibrium is m (C) If instead of being hollow cylinder, another solid cylinder of same mass and same radius is used then the cylinder oscillates with more angular frequency than before but with same amplitude. (D) If instead of pure rolling the cylinder were sliding on a smooth surface then the cylinder oscillates with more angular frequency than before but with same amplitude. 6. A 20 gm particle is subjected to two simple harmonic motions x1 = 2 sin 10t, x2 = 4 sin 10t 3 along same straight line. Where x1 and x2 are in metre and t is in sec. (A) the displacement of the particle at t = 0 will be 2 3 m (B) maximum speed of the particle will be 20 7 m/s (C) magnitude of maximum acceleration of the particle will be 200 7 m/s2. (D) Energy of the resultant motion will be 28 J. Space for Rough Work E-6/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 PHYSICS 7. There is a vertical spring on a railroad car at a station. The spring obeys Hooke's Law with spring constant 6000 N/m. You stand at rest on a light platform attached to the spring and it depresses by a distance 0.1 meter in equilibrium. Now if someone else displaces you vertically and then release at t = 0 such that your vertical displacement is represented by y = Acos(t) meters, where = 10s–1. Mark the CORRECT statement(s): (Assume g = 10 m/s2.) (A) Your mass is 60 kg (B) The largest value of A for which you do not lose contact with platform is 0.1 m (C) Now suppose you are oscillating with amplitude A = 0.05 m and the railroad car is moving to the right with constant velocity. Treating your equilibrium position at station as origin, as seen by someone standing still at the railroad station, your path is (D) Now suppose you are oscillating with amplitude A = 0.05 m and the railroad car is moving to the right with constant velocity. Treating your equilibrium position at station as origin, as seen by someone standing still at the railroad station, your path is 8. A charge q is revolving around another charge q as shown in a conical pendulum. The motion is in a horizontal plane. l qq (A) Tension in the string is greater than the weight of the ball. (B) The tension in the string is greater than the electrostatic repulsive force (C) If the charge is removed, the speed of ball has to be increased to maintain the angle. (D) If the charge is removed, the speed of ball has to be decreased to maintain the angle. Space for Rough Work 1001CT103516005 E-7/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 9. The spacecraft is spinning with a constant angular velocity about the z-axis at the same time that its mass centre O is traveling with a velocity O in the y-direction. If a tangential hydrogen- peroxide jet is fired when the craft is in the position shown. At the instant the jet force is F and directed along – ˆj . The radius of gyration of the craft about the z-axis is k, and its mass is m. Distance of point of application of force is also r. z O v0 y A F x r (A) Instantaneous velocity of centre of mass is v0ˆj (B) Instantaneous angular acceleration is Fr kˆ mk2 (C) The expression for the absolute instantaneous acceleration of point A on the spacecraft rim aA Fr2 iˆ F r2 ˆj mk2 m (D) The expression for the absolute instantaneous acceleration of point A on the spacecraft rim aA Fr2 iˆ F r2 ˆj mk2 m Space for Rough Work E-8/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 PHYSICS 10. Figure-(a) shows a frictionless roller coaster track. Figure-(i), (ii), (iii) shows potential energy and total energy for a car on roller coaster. Figure-(i), (ii), (iii) shows drawing of actual roller coaster track because gravitational potential energy is directly proportional to height it is also U vs x graph. Car AB C Figure-(a) U U (i) ETotal A (ii) ETotal U D D x BC x A B C (iii) B ETotal A D x C (A) In case (i) the car can negotiate the hill at C and reach D. (B) In case (ii) the motion is confined between two turning points where the total energy and potential energy curves intersect. (C) In case (iii) motion of car will be confined in part of valley formed between A and B. (D) The turning points are attained where Etotal = U for confined motion. Space for Rough Work 1001CT103516005 E-9/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 PHYSICS SECTION–II : (Maximum Marks : 16) This section contains TWO questions. Each question contains two columns, Column-I and Column-II. Column-I has four entries (A), (B), (C) and (D) Column-II has five entries (P), (Q), (R), (S) and (T) Match the entries in Column-I with the entries in column-II. One or more entries in Column-I may match with one or more entries in Column-II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below : (A) (P) (Q) (R) (S) (T) (B) (P) (Q) (R) (S) (T) (C) (P) (Q) (R) (S) (T) (D) (P) (Q) (R) (S) (T) For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry (A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). For each question, marks will be awarded in one of the following categories : For each entry in Column-I Full Marks : +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened Zero Marks : 0 In none of the bubbles is darkened Negative Marks : –1 In all other cases 1. Column-I Column-II (A) Periodic motion (P) Simple pendulum with small amplitude. (B) Simple harmonic motion A is mean position of bob. (Q) Block released on a smooth fixed inclined plane. Collision with wall is elastic. A is starting point. m e=1 (C) Body crosses point A when its acceleration is zero. (R) Conical pendulum. A is a point on circular path. (D) Centripetal acceleration is zero atleast at some point of the motion. A (S) ball tied to a string and projected horizontally with v = 6g . A is top most point of circular path. l (T) A block connected to a spring is pulled to right and released. A is position corresponding to natural length of the spring. m µ=0 E-10/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 PHYSICS 2. A variable force whose variation is shown in force position graph below. For each of the position indicated in Column-I, match appropriate description in Column-II. The force is acting on a body of mass 1kg initially at rest at x = 0. (Assume no other forces acting accept this one) F 10N x 34 01 2 –10N Column–I Column–II (A) 0 < x < 1 (P) Speed is increasing (B) 1 < x < 2 (Q) Power supplied by force is constant (C) 2 < x < 3 (R) Body moves in positive x direction (D) 3 < x < 4 (S) Speed is decreasing (T) Power of the force is increasing Space for Rough Work SECTION –III : Integer Value Correct Type No question will be asked in section III 1001CT103516005 E-11/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 PHYSICS SECTION–IV : (Maximum Marks : 32) This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. The bob of a pendulum has mass m = 1 kg and charge q = +40 C. Length of pendulum is l = 0.9m. The point of suspension also has the same charge +40 C. What the minimum speed u (ms–1) should be imparted to the bob so that it can complete vertical circle? (g = 10 m/s2) 2. A flat disc of mass 0.6 kg and radius 0.2 m lies on a frictionless horizontal table. A string wound around the cylindrical surface of the disc exerts a force of F = 3 N in the direction shown in (fig.). Calculate length (in m) of the string that is unwound in 2 second. If your answer is N fill value of N/4. F BOA Space for Rough Work E-12/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 3. A thin rectangular plate of mass M, sides a and b is suspended from four vertical wires of the same length . Determine the period of small oscillations of the plate when it is given a small PHYSICS horizontal displacement in a direction perpendicular to AB for a = b = . If your answer is T 2 x fill value of x + y. yg D C b G A aB 4. A ball of mass m = 100 gm is given a charge of 100 µC is connected to a spring of spring constant 1000 N/m and kept on a smooth level ground. An electric field E = 100 N/C exists towards the right. Ball is released when spring is in a natural length. What is the maximum speed of the ball in subsequent motion ? (in mm/sec). 5. In a 2-D region of x-y space, a uniform electric field exists, given by: E = (2 V/m)(3i + 4j). A charged particle of mass 10–2 g and charge 10–5 C is fired from the origin with initial velocity 20 m/s, directed along the negative y-direction. Find the minimum speed (in m/s) of the particle during its subsequent motion. Neglect gravity. If your answer is N fill value of N/3 Space for Rough Work 1001CT103516005 E-13/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 6. A tennis ball launcher is placed on the floor at the front end of a train that has a uniform forward acceleration of 2 m/s2. The launcher projects a ball at an initial speed of 20 m/s with respect to the train toward its rear. The ball achieves a maximum height of 10.0 m. Find how far (in m) from the front end of the train the ball lands inside the train. If your answer is N fill value of N/8. (Ignore air resistance, g = 10 m/s2) 7. A man hangs from the midpoint of a 10m long rope, the ends of rope are tied to two light rings which are free to move on a horizontal fixed rod (see fig.). What is the maximum possible separation d (in m) of the rings when the man is hanging in equilibrium, if the relevant coefficient of static friction is 0.75? d d/2 5m 5m 8. The combined frictional and air resistance on a bicyclist has the force F = aV, where V is his velocity and a = 4 newton-sec/m. At maximum effort, the cyclist can generate 400 watts propulsive power. What is his maximum speed (in m/s) on level ground with no wind ? If your answer is N fill value of N/5. Space for Rough Work E-14/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 PART-2 : CHEMISTRY CHEMISTRY SECTION–I : (Maximum Marks : 40) This section contains TEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. 1. The reduction of iron oxide in the blast furnace proceeds according to the following reactions: 3Fe2O3(s) + CO(g) 2Fe3O4(s) + CO2(g) ; H0298 –12kcal / mol Fe O (s) + CO(g) 3FeO(s) + CO (g) ; H0298 9kcal / mol 34 2 FeO(s) + CO(g) Fe(s) + CO2(g) ; H2098 4.4kcal / mol then H2098 for the reaction is : Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) (A) –8.8 kcal/ mol (B) –7.8 kcal/ mol (C) –6.8 kcal/ mol (D) –5.8 kcal/ mol 2. Half a mole of an ideal gas expands isothermally and reversibly at 300K from a volume of 10 litre to 20 litre. Which of the following are correct [ ln2 = 0.7] (A) Ssys=0.7 cal K–1 (B) Stotal=0 cal K–1 (C) q = –210 cal K–1 (D) q = –210 cal K–1 sys surr 3. The proposed mechanism for Br– catalysed aqueous reaction (H+ + HNO2 + C6H5NH2 Br C6H5N2+ + 2H2O) is I. H+ + HNO2 kk11 H2NO2+ (rapid eq.) II. H NO + + Br– k3 ONBr + HO (slow) 22 2 III. ONBr + C6H5NH2 k4 C6H5N2+ + H2O + Br– (fast). Which of the following are correct for given data : (A) Step I has more effect of temperature than II and III (B) Step II has more effect of temperature than I and III (C) Overall order of reaction is 3 (D) H2NO2+ and ONBr are reaction intermediate Space for Rough Work 1001CT103516005 E-15/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 CHEMISTRY4. Which of the following order INCORRECT between PCl3F2 and PCl2F3 :- (A) PCl3F2 > PCl2F3 (Dipole moment) (B) PCl3F2 > PCl2F3 (Total number of lone pair) (C) PCl3F2 > PCl2 F3 (Maximum number of atom in one plane) (D) PCl3F2 > PCl2F3 (P–Cl Bond length) 5. Dibrorane is hydride of boron and it exist in gaseous state and poisonous in nature. Which of the following statement is/are CORRECT for diborane :- (A) It is e– deficient species (B) Banana bond is present (C) Boron is sp3 hybridised (D) Bridge bond is longer than B–Ht (t = terminal) 6. Identify the INCORRECT match(es) in the following :- (A) XeF3+ : sp3d (Bent T-shape) (B) XeF5– : sp3d3 (Pentagonal planar) (C) XeO3F2 : sp3d2 (square planar) (D) XeO2F2 : sp3d (see-saw) 7. Which of the following overlapping results ungerade bonding molecular orbital :- (A) – (B) + (C) + (D) – Space for Rough Work E-16/32 1001CT103516005
8. The following reactions are as follows. Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 CHEMISTRY CH=O .....(1) + CN– X1 CH=O X2 .....(2) + NO2+ Which of the following statement(s) are correct : (A) During X1 & X2 formation, CH=O acting as electrophile & nucelophilic respectively (B) NO2+ acting as nucleophile (C) CN– acts as nucleophile CH=O (D) in both reactions acts as electrophile 9. The correct nomenclature of following compound is : CH3 H Cl H Br (A) 2R , 3R (B) 2S, 3R CH=O (D) 2S, 3S (C) 2R, 3S O 10. Which of the following molecules are dissymmetric : O H CH3 H CH3 Cl (D) (A) Cl (B) H Cl O H Cl O Cl H (C) CH3 CH3 Space for Rough Work 1001CT103516005 E-17/32
CHEMISTRY Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 SECTION–II : (Maximum Marks : 16) This section contains TWO questions. Each question contains two columns, Column-I and Column-II. Column-I has four entries (A), (B), (C) and (D) Column-II has five entries (P), (Q), (R), (S) and (T) Match the entries in Column-I with the entries in column-II. One or more entries in Column-I may match with one or more entries in Column-II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below : (A) (P) (Q) (R) (S) (T) (B) (P) (Q) (R) (S) (T) (C) (P) (Q) (R) (S) (T) (D) (P) (Q) (R) (S) (T) For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry (A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). For each question, marks will be awarded in one of the following categories : For each entry in Column-I Full Marks : +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened Zero Marks : 0 In none of the bubbles is darkened Negative Marks : –1 In all other cases 1. Column-I Column-II (A) Mixing of ideal gas at constant (P) S = + ive temperature and pressure sys (B) Adiabatic reversible expansion (Q) S = – ive of a real gas sys (C) Melting of ice at 1 atm and 273K (R) S = + ive (D) Isothermal reversible compression of surr ideal gas (S) S = – ive surr (T) S = 0 total Space for Rough Work E-18/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 2. Column-I Column-II CHEMISTRY CO2H (P) Ten stereoisomers are possible out of these, two are optically (A) (CHOH)4 inactive & eight are optically active. CO2H (Q) Four stereoisomers are possible out of these, two are optically CH3 inactive & two are optically active. CHCl (B) CHCl CHCl CH3 OH OH (R) Six stereoisomers are possible out of these, four are optically (C) active & two are optically inactive. Cl (S) Three stereoisomers are possible out of these, two are optically active and one is optically inactive. (D) Cl (T) At least one chiral centre is present in any one of the stereoisomers of the compound. Space for Rough Work SECTION –III : Integer Value Correct Type No question will be asked in section III 1001CT103516005 E-19/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 CHEMISTRY SECTION–IV : (Maximum Marks : 32) This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 5 BI C 1. P(N/m2)3 II A 1 2 V(m3) If a real gas is subjected to change in state from A to C by path I and II, If U for path I is 2J then find the heat absorbed by gas in path II. 2. If k is rate constant and A is collision frequency factor then find (Ea –Ea ) for reaction 21 2 and 1 in kcal/mol at 227ºC k = 100 k 12 10A1 = A2 ln10 = 7 3 Space for Rough Work E-20/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 CHEMISTRY 3. A(aq.) 2B(aq.) Decomposition of A to B follow first order kinetics. A andB are optically active with specific rotation of 20º & –10º per mol of A and B respectively. If the solution become optically inactive after 4 minutes if starting with 4 mol of A then find time in minutes for 75% completion of reaction. A(aq.) 2B(aq.) 4. Find the number of chemical species(s) which are planar and nonpolar in the following :- NO2 , O2F2 , XeF5+ , HCN , XeF5– , B3N3H6 5. Find the number of paramagnetic chemical specie(s) having bond order greater than two :- O22– , O2– , N2 , C22– , NO+ , CO Space for Rough Work 1001CT103516005 E-21/32
CHEMISTRY Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 6. Which oxy acid have M = O and M–O–H both type of bonds (M = Central atom) Sulphurous acid ; Arsenous acid ; Hypochlorous acid ; Perchloric acid ; Peroxynitric acid ; Pyrosulphuric acid 7. Find out number of plane of symmetry present in regular octahedral geometry : 8. How many carbocation undergoing rearrangement : (a) + (b) + (c) + + OH O (d) + + + CH3 (f) CH2–CH2–CH2 (g) CH3 (e) CH2–CH2–C NH2 Space for Rough Work E-22/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 PART-3 : MATHEMATICS MATHEMATICS SECTION–I : (Maximum Marks : 40) This section contains TEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. 1. Let ƒ(x) = x +|x – 100| –|x + 100| and g(x) = |ƒ(x)| – 1, then- (A) ƒ(x) is an odd function (B) g(x) is an even function (C) ƒ(x) is neither even nor odd (D) there are exactly six different values of x satisfying g(x) = 0 a 2. Let b ,ab, a – b and a + b are first 4 terms of an A.P. in order (a,b R0), then- 6 (A) common difference of A.P is 5 (B) common difference of A.P is 5 6 117 (C) fifth term of A.P. will be 40 (D) ab 27 40 3. Let ƒ(x) is a continuous function for all the real numbers and ƒ(x) = 10 have exactly three solutions x = 2,5,8. If ƒ(10) = 6, then which of the following is necessarily true- (A) ƒ(7) > 10 (B) ƒ(12) < 10 (C) ƒ(14) < 14 (D) ƒ(15) > 10 Space for Rough Work 1001CT103516005 E-23/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 MATHEMATICS4. Let ƒn x exex2 ex3 .......exn ,n N and g x lim ƒn x , then- n (A) g ' 1 4e (B) g ' 1 4e (C) g ' 1 9 e (D) g ' 2 9e2 2 2 3 4 3 5. If ƒ 3x 4 x 2 and ƒ xdx an x 1 bx c , then- 3x 4 (A) a + b = –2 (B) a < b (C) 3(b – a) = 10 (D) ab 4 3 6. Let y ex ex ex ....... , then dy is equal to- dx ex y2 y ex ex (A) 2y 1 (C) 1 4ex (D) 1 4ex (B) 2y 1 7. Identify the correct statement(s) (A) x5 x 1/5 6/ 5 x6 dx 5 1 1 c 24 x4 (B) x5 x 1/5 6/ 5 x6 dx 5 1 1 c 24 x4 (C) tan x 12 tan3 x tan2 x tan x 1 c dx 2 tan3 x tan2 x tan x 1 dx 1 sec2 x tan x c 2 (D) Space for Rough Work E-24/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 MATHEMATICS 8. Let is the solution of equation sin–1(2sin–1(cos–1(tan–1x))) = 0 and is the solution of equation sin–1x + sin–1x2 = , then- 2 (A) (B) = tan1 (C) 2 sin18 (D) 2sin18 4 9. Let 2x2 2x 1 ex2 dx ƒ x C where C is integration constant and ƒ(0) = 1, then- (A) 2 ƒ x ex2dx x 12 C (B) ƒ(1) = 2e 1 (D) lim ƒ x1/ x e x 0 (C) lim ƒ xx e2 x0 89 10. Let cos6 k A , then- k 1 /2 (B) A 20,30 (A) A cos6 d 0 (C) A 180 /2 cos6 d /2 0 (D) A cos6 d 0 Space for Rough Work 1001CT103516005 E-25/32
MATHEMATICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 SECTION–II : (Maximum Marks : 16) This section contains TWO questions. Each question contains two columns, Column-I and Column-II. Column-I has four entries (A), (B), (C) and (D) Column-II has five entries (P), (Q), (R), (S) and (T) Match the entries in Column-I with the entries in column-II. One or more entries in Column-I may match with one or more entries in Column-II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below : (A) (P) (Q) (R) (S) (T) (B) (P) (Q) (R) (S) (T) (C) (P) (Q) (R) (S) (T) (D) (P) (Q) (R) (S) (T) For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry (A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). For each question, marks will be awarded in one of the following categories : For each entry in Column-I Full Marks : +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened Zero Marks : 0 In none of the bubbles is darkened Negative Marks : –1 In all other cases 1. Column-I Column-II (A) If y = |n(x2)| then y' 1 is equal to (P) 4 2 (B) If ƒ x 1 x3 tan1 x and g(x) = ƒ–1(x), (Q) 5 24 then 16g \" 1 is equal to (R) 2 2 e4 nx dx (C) If ƒ x xxx , then ƒ\"(1) is equal to (S) 1 2x /2 (T) 3 (D) sec2 x 1 tan x sin4 2x dx is equal to Harmonic mean of 3 and 2 0 Space for Rough Work E-26/32 1001CT103516005
2. Column-I Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 MATHEMATICS Column-II (A) If ƒ x ax2 5 x2 (P) 1 a2 is differentiable for all real (Q) 2 2bx x2 values of x then possible value(s) of |a| is/are (B) If ƒ(x) is a twice differentiable function such that 1 (R) 4 (S) 5 ƒ(0) = ƒ(2) and ƒ'(2) = 8, then xƒ \"2x dx is 0 greater than (C) Let ƒ(x) is an even, continuous function for all real x. If ƒ(0) = 3, ƒ(4) = –4 and ƒ(–6) = 6 then number of solution of ƒ(x) = 0 can be (D) If lim n 1 x2 x sin x exist then n can be (T) 6 x 0 xn Space for Rough Work SECTION –III : Integer Value Correct Type No question will be asked in section III 1001CT103516005 E-27/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 MATHEMATICS SECTION–IV : (Maximum Marks : 32) This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. lim x100nx is equal to x e4x tan 1 sin2 x 3 2. Number of real solutions of equation x + 1 = 99 sin(x) is n then n – 190 is 3. If x,y and (in radians) are real numbers satisfying the system of equation x3 3x2 2019x 2017 sin2 1 2 , then |x + y| is equal to y3 3y2 cos2 1 2019y 2017 2 Space for Rough Work E-28/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 n 1 2 cos 2x MATHEMATICS 3k x lim , 4. + ||x – 2| – Let n k1 3 then number of points where y = |x (x)| 1| is not differentiable in (0,3) is equal to 5. Let ƒ(x) = ax2 – bx – 16 (where a,b are real constants and x R) doesn't have distinct real roots then maximum value of 4a – b is 6. Let x,y,z are three positive real numbers such that xyz = 1 then minimum value of 1 xy 1 yz 1 zx is equal to 1x 1y 1z Space for Rough Work 1001CT103516005 E-29/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 MATHEMATICS7. Number of integers in the domain of ƒ x sin1 log2 x tan 1 9 x2 is x 1 8. If 2x2 3x 3 ax x2 2x 2 bn x 1 x 12 1 c x2 2x 2 (C is integration constant) then a + b is equal to Space for Rough Work E-30/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 1001CT103516005 E-31/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 QUESTION PAPER FORMAT AND MARKING SCHEME : 16. The question paper has three parts : Physics, Chemistry and Mathematics. 17. Each part has three sections as detailed in the following table. Que. No. Category-wise Marks for Each Question Maximum Section Type of Full Partial Zero Negative Marks of the Que. Marks Marks Marks Marks section +4 0 –2 If none In all One or more If only the bubble(s) of the other bubbles is cases I correct 10 corresponding — darkened 40 option(s) to all the correct option(s) is(are) darkened +8 +2 0 –1 In all II Matrix If only the bubble(s) For darkeninga bubble If none other cases Match Type 2 corresponding to corresponding to each of the 16 all the correct correct match is bubbles is match(es) is(are) darkened darkened darkened +4 0 Single digit If only the bubble In all IV Integer 8 corresponding — other — 32 (0-9) to correct answer cases is darkened NAME OF THE CANDIDATE ................................................................................................ FORM NO. ............................................. I have read all the instructions I have verified the identity, name and Form and shall abide by them. number of the candidate, and that question paper and ORS codes are the same. ____________________________ ____________________________ Signature of the Candidate Signature of the invigilator Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in E-32/32 Your Target is to secure Good Rank in JEE 2017 1001CT103516005
Paper Code : 1001CT103516005 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) HINDI JEE (Main + Advanced) : LEADER COURSE Test Type : MINOR PHASE : III, IV & V Test Pattern : JEE-Advanced TEST DATE : 25 - 09 - 2016 Time : 3 Hours PAPER – 1 Maximum Marks : 264 1. 2. (ORS) 3. 4. 5. 3220 6. 7. 8. 9. : 10. 11. 12. : 13. 14. 15. g = 10 m/s2
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 SOME USEFUL CONSTANTS Atomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Atomic masses : Cl = 17, Br = 35, Xe = 54, Ce = 58, H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127, Xe = 131, Ba=137, Ce = 140, Boltzmann constant k = 1.38 × 10–23 JK–1 Coulomb's law constant 1 = 9 ×109 4 0 Universal gravitational constant Speed of light in vacuum G = 6.67259 × 10–11 N–m2 kg–2 Stefan–Boltzmann constant c = 3 × 108 ms–1 Wien's displacement law constant = 5.67 × 10–8 Wm–2–K–4 Permeability of vacuum b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2 Permittivity of vacuum 1 Planck constant 0 = 0c2 h = 6.63 × 10–34 J–s H-2/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1PHYSICS HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING -1: –I : ( : 40) (A), (B), (C) (D) : +4 : 0 : –2 1. ‘m’‘R’ v0 v0 O (A) ‘O’2mv0R (B) O2mv0R (C) 73v0 (D) –37v0 1001CT103516005 H-3/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 2. mRt=0 v0L0 Tmax Tmax, L0, R v0 (A) mTm2vax03 (B) mv20 2 (C) Tmax L0 –R (D) Tmax mTmvax02 H-4/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 3. 1 A BC2 PHYSICS A LA = 0 =0° Case 1 Case 2 B L L CA A (A) (B) (C) (D) D v0A (C) -1 A (D) -2 A 4. A(5 kg) B(3kg) A B F1 A F2 B F1 F2 f1 f2 (A) f1max > f2max (B) F1max : F2max = 3 : 5 (C) F1max : F2max = 5 : 3 (D) f1 < F1 1001CT103516005 H-5/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 5. R M I =MR2k x0 (A) 02km (B) 02mk (C) (D) 6. 20 gm x1 =2sin10t, x2 = 4 sin 10t 3 x1x2t (A) t = 0 23 m (B) 20 7 m/s (C) 200 7 m/s2 (D) 28J H-6/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1PHYSICS 7. 6000N/m 0.1 t=0 y = Acos(t) =10s–1 (g=10m/s2.) (A) 60kg (B) A0.1 m (C) A = 0.05 m (D) A = 0.05 m 8. q q l qq (A) (B) (C) (D) 1001CT103516005 H-7/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 9. z-OO y- F–ˆj z-k m r z O v0 y A F x r (A) v0ˆj (B) mFkr2kˆ (C) A aA mFrk22 iˆ F r2 ˆj m (D) A aA mFrk22 iˆ F r2 ˆj m H-8/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1PHYSICS 10. (a)(i), (ii),(iii) (i),(ii),(iii) Ux Car AB C Figure-(a) U U (i) ETotal A (ii) ETotal U D D x BC x A B C (iii) B ETotal A D x C (A) (i)CD (B) (ii) turning (C) (iii) AB (D) Etotal = U turning 1001CT103516005 H-9/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 PHYSICS –II : ( : 16) -I -II -I (A), (B), (C) (D) -II (P), (Q), (R), (S) (T) -I -II -I -II 4 ×5 : (A) (P) (Q) (R) (S) (T) (B) (P) (Q) (R) (S) (T) (C) (P) (Q) (R) (S) (T) (D) (P) (Q) (R) (S) (T) -I -I(A) (Q),(R) (T) (B), (C)(D) : -I : +2 : 0 : –1 1. -I -II (A) (P) A (B) (Q) A m e=1 (C) A (R) A A (D) (S) v=6g A l (T) A m µ=0 H-10/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1PHYSICS 2. -I -IIx=01kg F 10N x 34 01 2 –10N –I –II (A) 0 < x < 1 (P) (B) 1 < x < 2 (C) 2 < x < 3 (Q) (D) 3 < x < 4 (R) x (S) (T) –III : III 1001CT103516005 H-11/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 –IV : (: 32) 09 : +4 : 0 1. m=1kgq=+40 C l =0.9m +40Cu(ms–1) (g=10 m/s2) 2. 0.6 kg 0.2 m F=3N (m)2NN/4 F BA O H-12/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1PHYSICS 3. M ab a = b = AB T 2 x x+y yg D C b B G Aa 4. m= 100 gm 100 µC 1000N/m E=100N/C (mm/sec ) 5. x-y 2 -D E=(2V/m)(3i + 4j) 10–2g 10–5 C y- 20m/s (m/s) NN/3 1001CT103516005 H-13/32
PHYSICS Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 6. 2m/s2 20 m/s10.0m NN/8 (g=10m/s2) 7. 1 0m d0.75 d d/2 5m 5m 8. F=aV V a = 4 newton-sec/m 400 (m/s) N N/5 H-14/32 1001CT103516005
Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 -2: CHEMISTRY –I : ( : 40) (A), (B), (C) (D) : +4 : 0 : –2 1. : 3Fe O (s) + CO(g) 2Fe O (s) + CO (g) ; H2098 –12kcal / mol 23 34 2 Fe O (s) + CO(g) 3FeO(s) + CO (g) ; H0298 9kcal / mol 34 2 FeO(s) + CO(g) Fe(s) + CO2(g) ; H2098 4.4kcal / mol H0298 : Fe O (s) + 3CO(g) 2Fe(s) + 3CO (g) 23 2 (A) –8.8 kcal/ mol (B) –7.8 kcal/ mol (C) –6.8 kcal/ mol (D) –5.8 kcal/ mol 2. 300K10 20 [ ln2=0.7] (A) Ssys=0.7 cal K–1 (B) Stotal=0 cal K–1 (C) q = –210 cal K–1 (D) q = –210 cal K–1 sys surr 3. Br– (H+ + HNO2 + C6H5NH2 Br C6H5N2+ + 2H2O) I. H+ + HNO kk11 H NO + () 2 22 II. H NO + + Br– k3 ONBr + HO () 22 2 III. ONBr + C H NH k4 C H N + + H O + Br– (). 2 65 2 652 : (A) II III I (B) I III II (C) 3 (D) H2NO2+ ONBr 1001CT103516005 H-15/32
Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 CHEMISTRY4. PCl3F2 PCl2F3 :- (A) PCl3F2 > PCl2F3 () (B) PCl3F2 > PCl2F3 () (C) PCl3F2 > PCl2 F3 () (D) PCl3F2 > PCl2F3 (P–Cl ) 5. :- (A) (B) B anana (C) sp3 (D) B–Ht (t = ) 6. :- (A) XeF3+ : sp3d (T-) (B) XeF5– : sp3d3 () (C) XeO3F2 : sp3d2 ( ) (D) XeO2F2 : sp3d (sea-saw) 7. ungerade:- (A) – (B) + (C) + (D) – H-16/32 1001CT103516005
8. Leader Course/Phase-III, IV & V/25-09-2016/Paper-1 CHEMISTRY .....(1) CH=O + CN– X1 CH=O X2 .....(2) + NO2+ CH=O (A) X1 X2 (B) NO2+ , (C) CN– , CH=O (D) 9. CH3 H Cl H Br CH=O (A) 2R , 3R (B) 2S, 3R (C) 2R, 3S (D) 2S, 3S 10. dissym metric O O CH3 CH3 Cl Cl Cl H (C) (D) H H H Cl O (A) (B) O Cl H CH3 CH3 1001CT103516005 H-17/32
CHEMISTRY Target : JEE (Main + Advanced) 2017/25-09-2016/Paper-1 –II : ( : 16) -I-II -I (A), (B), (C) (D) -II (P), (Q), (R), (S) (T) -I -II -I -II 4 ×5 : (A) (P) (Q) (R) (S) (T) (B) (P) (Q) (R) (S) (T) (C) (P) (Q) (R) (S) (T) (D) (P) (Q) (R) (S) (T) -I -I(A) (Q),(R) (T) (B), (C)(D) : -I : +2 : 0 : –1 1. -I -II (A) (P) S = + ive sys (B) (Q) S = – ive sys (C) 1 atm 27 3K (R) S = +ive surr (D) (S) S = –ive surr (T) S = 0 total H-18/32 1001CT103516005
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