ENGLISH Paper Code : 100 1CT103516009 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE : III, IV & V Test Pattern : JEE-Advanced Test Type : MINOR TEST DATE : 20 - 11 - 2016 Time : 3 Hours PAPER – 1 Maximum Marks : 180 READ THE INSTRUCTIONS CAREFULLY 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 24 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 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. COMPLETELY. 15. Take g = 10 m/s2 unless otherwise stated. Please see the last page of this booklet for rest of the instructions DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR GENERAL : Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 SOME USEFUL CONSTANTS Atomic No. 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 ×10 9 4 0 Universal gravitational constant Speed of light in vacuum Stefan–Boltzmann constant Wien's displacement law constant Permeability of vacuum G = 6.67259 × 10–11 N–m2 kg–2 c = 3 × 108 ms–1 = 5.67 × 10–8 Wm–2 –K–4 b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2 Permittivity of vacuum 0 = Planck constant h = 6.63 × 10–34 J–s 1 0 c2 Space for Rough Work E-2/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 PHYSICS HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PART-1 : PHYSICS SECTION–I : (Maximum Marks : 30) 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 : +3 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks 1. : 0 In all other cases. Electric field, due to an infinite line of charge, as shown in figure, at a point P at a distance r from the line is E. If one half of the line of charge is removed from right side of point A, thenY P r X A (A) Electric field at P will have magnitude E 2 (B) Electric field at P in x direction will be E 2 (C) Electric field at P in y direction will be E 2 (D) None of these Space for Rough Work 1001CT103516009 E-3/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS 2. The figure shows, two point charges q1 = 2Q(>0) and q2 = –Q. The charges divide the line joining them in three parts I, II and III. I III II +2Q –Q (A) Region III has a local maxima of electric field. (B) Region I has a local minima of electric field. (C) Equilibrium position for a test charge lies in region II. (D) The equilibrium for constrained motion along the line joining the charges is stable for a negative charge. 3. Which of the following statement(s) are CORRECT ? (A) Friction acting on a cylinder without sliding on an inclined surface is always upward along the incline irrespective of any external force acting on it. (B) Friction acting on a cylinder without sliding on an inclined surface may be upward or may be downwards depending on the external force acting on it. (C) Friction acting on a cylinder rolling without sliding may be zero depending on the external force acting on it. (D) Nothing can be said exactly about it as it depends on the friction coefficient on inclined plane. Space for Rough Work E-4/24 1001CT103516009 4. Consider a Gaussian surface as shown. A and C are the circular faces while B is the curved surface. A charge q is moved along the axis of cylinder with constant velocity v. Choose the correct variation of flux through the surface/cylinder. B A C q t (A) (B) t Through the cylinder Through surface A (C) t Through surface C t (D) Through surface B Space for Rough Work 1001CT103516009 E-5/24 PHYSICS Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS 5. Figure shows the net power dissipated in R versus the current in a simple circuit shown. P(W) 5 4 3 E 2 r 1 0 6. 2 4 5 6 8 I(A) 10 R (A) The internal resistance of battery is 0.2 (B) The emf of battery is 2V (C) R at which power is 5W is 2.5 (D) At i = 2A, power is 3.2 W A parallel plate capacitor of capacitance 'C' has charges on its plates initially as shown in the figure. Now at t = 0, the switch 'S' is closed. Select the CORRECT alternative(s) for this circuit diagram. S t=0 A B –2C C (A) In steady state the charges on the outer surfaces of plates 'A' and 'B' will be same in magnitude and sign. (B) In steady state the charges on the outer surfaces of plates 'A' and 'B' will be same in magnitude and opposite in sign. (C) In steady state the charges on the inner surfaces of plates 'A' and 'B' will be same in magnitude and opposite in sign. 52C (D) The work done by the battery till time steady state is reached is 2 Space for Rough Work E-6/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 In the figure shown. (All batteries are ideal) :10V 5V 20V PHYSICS 7. 10V 25V 5 8. 10 10 5 (A) Current through 25 V cell is 20 A (B) Current through 25 V cell is 10.5 A (C) Power supplied by 20 V cell is 20W (D) Power supplied by 20V cell is –20W In the shown circuit, the power dissipated in resistor R3 is 30W. The reading of ammeter and voltmeter are 500 mA and 10V respectively. Ammeter, voltmeter and battery are ideal. 10=R2 R3 R1 A V (A) the resistance R1 is 20 (B) the resistance R3 is 7.5 (C) emf of battery is 15V (D) power supplied by battery is 37.5 W 9. A dipole of dipole moment 8 × 10–10 î C-m is kept at the origin of coordinate system :(A) the flux through the y-z plane passing through origin is zero (B) the potential at (1, 1, 1) m is 0.8 volt (C) the potential at (2, 3, –6) m is positive (D) the electric field at any point on the z-axis is in the negative x-direction 10. A particle moving with kinetic energy = 3 joule makes an elastic head on collision with a stationary particle. Mass of the stationary particle is double of the moving particle. (A) The minimum kinetic energy of the system is 1 joule. (B) The maximum elastic potential energy of the system is 2 joule. (C) Momentum and total kinetic energy of the system are conserved at every instant. (D) The ratio of kinetic energy to potential energy of the system first decreases and then increases. Space for Rough Work SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III 1001CT103516009 E-7/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS SECTION–IV : (Maximum Marks : 30) This section contains TEN 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 : 1. Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. A charged particle is in equilibrium at a height h = 1m from a horizontal infinite line charge with uniform linear charge density. The charge lies in the vertical plane containing the line charge. If the particle is displaced slightly in vertical direction, find the time period of SHM in second. (= 10 & g = 10m/s2) 2. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 2n resistance, resistance, it can be converted into a voltmeter of range 0 - 30 V. If connected to a 249 it becomes an ammeter of range 0 - 1.5 A. The value of n is :- 3. A pair of stars rotates about their common center of mass. One of the stars has a mass M which is twice as large as the mass m of the other. Their centres are at a distance d apart, d being large compared to the size of either star. Compare the kinetic energies of the two stars by calculating the ratio Km/KM. For what value of R (in in circuit, current through 4 resistance is zero. 4. R 4 2 4V 6V 10V Space for Rough Work E-8/24 1001CT103516009 5. In the circuit shown in figure the switch S is initially open and both the capacitors are initially uncharged. Find the ratio of current through 2 resistor, just after the switch S is closed and a long time after the switch S is closed. 4µF 2 S 6V 8 2µF 4 6. Current through 3 resistor is shown in graph. Find the heat produced in the resistor in first 2 sec in joule. i 1A t 1sec 7. A boy of mass 10 kg is standing on a 4kg block as shown. The whole system is on smooth level ground. The boy jumps to the right with a velocity of 7 m/s relative to the 4 kg block. Find the maximum velocity of 1 kg block (in m/sec) in the subsequent motion. k=100N/m 1kg 4kg Space for Rough Work 1001CT103516009 E-9/24 PHYSICS Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS 8. The spring block system lies on a smooth horizontal surface. The free end of the spring is being pulled towards right with constant speed v0 = 2m/s. At t = 0 sec,the spring, of constant k = 100 N/cm, is unstretched and the block has a speed 1 m/s to left. The maximum extension of the spring (in cm) is. 1m/s k = 100N/cm m 4kg 9. v0 = 2m/s 1 C & 10 mC respectively are connected through an 9 19 inelastic & light string of length 1m. Particle 'A' is fixed and 'B' has mass kg. A vertical 3 electric field 'E' is applied such that particle 'B' released from horizontal position with string remaining tight will move up and reaches to highest vertical position as shown in figure. When particle reaches the highest vertical position electric field is switch off but particle remains in circular motion all the time then find the least possible value of 'E' (in kN/C). g = 10m/s2 Two particles 'A' & 'B' having charge A (Fixed) B 10. Three very large identical metal plates are given charges as shown. The final charge on left surface of IInd plate is found to be NQ. Find out the value of N. Q –5Q 10Q I II III Space for Rough Work E-10/24 1001CT103516009 PART-2 : CHEMISTRY SECTION–I : (Maximum Marks : 30) 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 : +3 If only the bubble(s) corresponding to all the correct option(s) is (are) Zero Marks 1. : 0 darkened. In all other cases. Consider a first order reaction A B, with initial concentration of A being [A]0 and that of B being zero. Which of the following graphs decribes the variation of concentration with time : (A) [A] (C) [B] (B) [A] t 2. 3. (D) [B] t t t When you see the tip of a match flare the chemical reaction is likely to be : P4S3 + O2 P4O10 + SO2 Which of the following are correct regarding given reaction : (A) The number of electrons loose by one molecule of P4S3 is 32 (B) The number of electron gain by one molecule of O2 is 4 (C) It is an example of intermolecular redox reaction (D) P4S3 is oxidising agent Which of the following is/are correct for a vanderwaal real gas : (A) a 27 R 2 Tc 2 64 Pc (B) b RTc 8Pc (C) Pc 1 a 27 b 2 (D) Tc 1 a 27 bR Space for Rough Work 1001CT103516009 E-11/24 CHEMISTRY Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 CHEMISTRY 4. Which of the following heteroleptic complex(es) will show conductivity in it's aq. solution :(A) Bis(ethylenediamine)dichlorocobalt(III) chloride (B) Ziese's salt (C) Hexachloroplatinic(IV) acid (D) Triamminetrichlorochormium(III) III 5. Complex ion [Co(en)2 L]2 follow Sidwick E.A.N. rule then ligand 'L' may not be :(A) en 6. 7. (B) ox–2 (C) Br– (D) dmg– Complex compound [Cr(NH3)6]Cl3 have following properties :(A) paramagnetic nature (B) primary valency = 3 (C) Low spin complex (D) Shows synergic bonding A homodiatomic molecule 'X2' have bond order '1', then which of the following prediction(s) will be CORRECT for 'X2' :(A) The difference of total number of bonding and antibonding e– is 2 (B) Molecule can not be odd electron molecule (C) Total number of bonding electrons is more than antibonding electrons (D) Last e– of 'X2' molecule must be entered into bonding molecular orbital Space for Rough Work E-12/24 1001CT103516009 8. Which of the following compound can be obtained by Wurtz reaction from single organic monohalide / single organic dihalide in good yield : (A) 9. (B) H3C–CH–CO 2H (C) (i) Aq. KOH + (ii) H (D) Product Cl R-2-chloropropanoic acid Product is : (A) S-2-Hydroxy propanoic acid (B) R-2-Hydroxy propanoic acid (C) A mixture of S-2-Hydroxy propanoic acid and R-2-Hydroxy propanoic acid (D) S-1-carboxy ethanol 10. The compound which can show faster rate of SN2 mechanism than CH3–CH2–CH2–Cl in identical condition is/are : (A) CH3–CH–Cl Cl (B) O (C) CH2–Cl (D) CH3–Br CH3 Space for Rough Work SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III 1001CT103516009 E-13/24 CHEMISTRY Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 CHEMISTRY Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 SECTION–IV : (Maximum Marks : 30) This section contains TEN 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 Zero Marks 1. : +3 If only the bubble corresponding to the correct answer is darkened. : 0 In all other cases. A carnot engine has a cycle as given in figure. If W1 and W2 represent work done by one mole of monoatomic and diatomic gas respectively in the complete cyclic process, then calculate W1 ? W2 1 4T0 P 2 4 T0 V0 2. V 3 64V0 If the volume of a real gas molecule is 25 × 10–23 cm3 then find the free volume in litres 10 mole of this gas is present in 10 lt container. Use NA = 6 × 1023. 6 if 3. Heats of combustion of graphite and hydrogen at 298 K are –395 kJ mol–1 and –285 kJ mol–1 respectively. If on combustion of 1 gm of glucose releases 16 kJ of energy, calculate the magnitude of heat of formation of glucose in kJ at 298 K. Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer. Space for Rough Work E-14/24 1001CT103516009 4. Calculate the volume in lt of 0.1 M KMnO4 required to completely oxidised 5 mol of Fe0.9O in acidic medium. 5. One mole SOxCly produces two moles of HCl and one mole H2SO4 on complete hydrolysis. Then value of (x + y) will be :- 6. In the high spin complex [Cr(H 2O)5L]n total number of inner subshell 'd' orbitals used in hybridisation will be :- 7. Total number of complex compounds in which all geometrical isomers are optically active :[M(AB)3] ; [M(AA)abcd] ; [M(AA)b3c] ; [Ma3b3] Space for Rough Work 1001CT103516009 E-15/24 CHEMISTRY Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 CHEMISTRY Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 8. How many compounds when undergo Kolbe's electrolysis gives unsaturated hydrocarbon as major product ? (i) Sodium maleate (ii) Sodium fumerate (iii) Sodium ethanoate CO2Na (iv) Sodium succinate (v) Sodium propanoate (vi) CO 2Na 9. Find out number of reactions, where during product formation from reactant, reduction-oxidation phenomenan takes place : (i) H 3C–C–CH 3 1. Mg / Hg (ii) CH3–CH2–Cl + Na + 2. H Dry ether O (iii) H3C–CH–CH–CH3 Cl Zn dust (iv) H3C–CH=CH 2 CH 3 Cl (v) H3C–CH=CH2 (i) Hg(OAc)2, H2O (ii) NaBH4 (i) B2H6 (THF) (vi) CH3–CH2–Cl – Aq. KOH (ii) H2O2 + OH (vii) CH3–CH2–OH 10. SOCl2 Find the number of -hydrogen present the major alkene product : – + OH N Space for Rough Work E-16/24 1001CT103516009 PART-3 : MATHEMATICS SECTION–I : (Maximum Marks : 30) 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 : +3 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks : 0 /2 1. 0 xdx 1 sin 2x cos2 2x (A) 6 (C) 2. /2 0 In all other cases. is equal to- 1 1 sin 2x 1 2 sin 2x dx 2 1 12 3 3 (B) 12 (D) /2 0 1 1 sin 2x 1 2 sin 2x dx 4 1 12 3 3 Let ƒ(x) and g(x) are two differentiable functions such that ƒ ' x ƒ x g2 x g x g ' x , g ' x g x ƒ2 x ƒ x ƒ ' x , ƒ 0 1 2,g 0 1 2 , then - (A) Range of ƒ(x) is (0,) (B) Range of g(x) is (–,0) (C) There is exactly one solution of equation ƒ(x) = g(x) (D) ƒ(x) > g(x) x R 3. If ƒ(x) is a differentiable function on R such that ƒ(2) = ƒ(6) = 0. Then there exist a real number c in (2,6) for which ƒ'(c) = ƒ(c) where can be(A) 21 (B) –100 (C) –73 (D) 1010 Space for Rough Work 1001CT103516009 E-17/24 MATHEMATICS Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 MATHEMATICS Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 4. 5. If x1,x2 are real solutions to the equation x2 + a2x + b = 0 while x3,x4 are real solutions to x2 + 5ax + 7 = 0 such that x1 – x3 = x2 – x4 = 2, then (A) x1,x2,x3,x4 Q (B) x1,x2,x3,x4 Q (C) a + b = 2 (D) a + b = –25 Let ƒ : A B ƒ(x) = tan–1x + tan–12x + tan–13x is a real valued onto function and A,B are nonempty subsets of R, then- 6. 7. (A) If A = R, then B = R 3 3 (B) If A = R, then B , 2 2 (C) If A = [–1,1], then B = [–,] (D) If A = [–1,1], then B , 2 2 If function ƒ(x) satisfy ƒ(–tanx) + 2ƒ(tanx) = sin2x for all x , , then 2 2 (A) ƒ" is an odd function (B) ƒ'(0) = 2 (C) ƒ'(1) = 0 (D) ƒ(1) + ƒ(–1) = 0 Positive integers a,b and c are chosen so that a < b < c and the system of equations 2x + y = 2003 and y = |x – a| + |x – b| + |x – c| has exactly one solution, then(A) minimum value of c is 1002 (B) maximum value of c is 1002 (C) maximum value of b is 1001 (D) minimum value of b is 1001 Space for Rough Work E-18/24 1001CT103516009 8. Total number of real solutions of equation tan2x + cos6x = 1 – tan2x cos6x in [0,n] is(A) 4 if n is 1 9. (B) 7 if n is 2 (C) 2 if n is 1 2 (D) 3 if n is 3 4 Let combine equation of straight lines L1 and L2 is 6x2 + y2 + 5xy – 5x – 2y + 1 = 0 and is acute angle between them. If d1,d2 are distances of the lines from origin and () are their point of intersection, then- 10. 6 (B) 0, (A) + = 1 (C) d1d 2 1 5 2 (D) 3 + 2 = 2 Let y = ƒ(x) is a solution of differential equation cos2xdy – cos4xdx = y tan2x dx, x and 4 3 3 ƒ , then8 6 (A) lim ƒ x x 0 x 1 3 3 (B) ƒ 8 6 (C) equation of tangent to y = ƒ(x) at origin is y = x (D) graph of ƒ(x) is symmetric about origin Space for Rough Work SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III 1001CT103516009 E-19/24 MATHEMATICS Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 MATHEMATICS Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 SECTION–IV : (Maximum Marks : 30) This section contains TEN 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 Zero Marks 1. : +3 If only the bubble corresponding to the correct answer is darkened. : 0 In all other cases. If ƒ(x) is a differentiable function such that ƒ(0) = 0 and ƒ(1) = 1, then minimum value 1 of ƒ ' x 2 dx is 0 2. 3. m Let A(0,0) B(1,2) C(3,3) D(4,0) are vertices of a quadrilateral ABCD. A line y x divide n the quadrilateral in two parts of equal area then n–m is equal to (where m,n are relatively prime no.) Let line 4x + 5y = 20 intersect x-axis at A and the y-axis at B. A line L intersect AB and OA at points C and D respectively. The least value of CD2 for which line L divides the area of OAB b into two regions of equal area is a 41 b where a,b N, then is equal to a Space for Rough Work E-20/24 1001CT103516009 4. Let y = x and y = –2x are internal angle bisectors of angle B and C of triangle ABC. If A is (1,2) and r is its inradius then 5. 6. 1 r2 is equal to Value of tan1º tan5º + tan5º tan9º + tan9º tan13º........+ tan177º tan181º is equal to L, then L 5 is equal to A cylindrical container is to be made from certain solid material with following constraints. It has a fixed inner volume of V mm3, has a 3mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 3 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 7mm then the value of V is 49 Space for Rough Work 1001CT103516009 E-21/24 MATHEMATICS Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 MATHEMATICS Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 7. The number of distinct real roots of equation 2x4 – 8x3 + 8x2 – 1 = 0 is 8. min ƒ t , t x , x 1 Consider ƒ(x) = |x2 – 2x| and g x . If number of points where max ƒ t ,1 t x , x 1 g(x) is not differentiable is 9. 1 tan 1 n 2 tan n 1 , then [–L] is equal to Let L lim n 2 n n 1 n (where [.] denotes greatest integer function) 10. 2 Let the area of region bounded by y sin x and 2 y x is 1 , then 'n' is equal to n Space for Rough Work E-22/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 Space for Rough Work 1001CT103516009 E-23/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 QUESTION PAPER FORMAT AND MARKING SCHEME : 16. The question paper has three parts : Physics, Chemistry and Mathematics. 17. Each part has two sections as detailed in the following table. Section I IV Que. Type No. of Que. Category-wise Marks for Each Question Full Partial Marks Marks +3 One or more If only the bubble(s) correct corresponding 10 to all the correct — option(s) option(s) is(are) darkened +3 Single digit If only the bubble Integer 10 corresponding — (0-9) to correct answer is darkened Zero Marks 0 In all other cases 0 In all other cases Maximum Negative Marks of the Marks section — 30 — 30 NAME OF THE CANDIDATE ................................................................................................ FORM NO. ............................................. I have read all the instructions and shall abide by them. I have verified the identity, name and Form 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 E-24/24 info@allen.ac.in www.allen.ac.in Your Target is to secure Good Rank in JEE 2017 1001CT103516009 Paper Code : 1001CT103516009 HINDI CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE PHASE : III, IV & V Test Pattern : JEE-Advanced Test Type : MINOR TEST DATE : 20 - 11 - 2016 Time : 3 Hours PAPER – 1 Maximum Marks : 180 2. (ORS) 3. 4. 5. 24 20 1. 6. 7. 8. 9. : 10. 11. 12. : 13. 14. 15. g = 10 m/s2 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 SOME USEFUL CONSTANTS Atomic No. 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 ×10 9 4 0 Universal gravitational constant Speed of light in vacuum Stefan–Boltzmann constant Wien's displacement law constant Permeability of vacuum G = 6.67259 × 10–11 N–m2 kg–2 c = 3 × 108 ms–1 = 5.67 × 10–8 Wm–2–K–4 b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2 Permittivity of vacuum 0 = Planck constant h = 6.63 × 10–34 J–s 1 0 c2 H-2/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 PHYSICS HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING -1 : –I 1. : ( : 30) (A), (B), (C) (D) : +3 : 0 r P E A Y P r X A E (A) P 2 E (B) P x 2 E (C) P y 2 (D) 1001CT103516009 H-3/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS 2. q1 = 2Q (>0) q2= –Q I, II III I III II +2Q –Q (A) III (local) (B) I (local) (C) II (D) 3. (A) (B (C (D) H-4/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 A C B q v PHYSICS 4. B A C q t (A) (B) t A (C) t C t (D) B 1001CT103516009 H-5/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS 5. R P(W) 5 4 3 E 2 r 1 0 2 4 5 6 8 (A) 0.2 I(A) 10 R (B) 2V (C) R = 2.5 5W 6. (D) i = 2A 3.2 W t = 0 'S' 'C' S t=0 A B –2C C (A) 'A' 'B' (B) 'A' 'B' (C) 'A' 'B' 52C (D) 2 H-6/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 10V 5V 20V PHYSICS 7. 10V 25V 10 5 8. 10 5 (A) 25 V 20 A (B) 25 V 10.5 A (C) 20 V 20W (D) 20V –20W R3 30W 500 mA 10V 10=R2 R3 R1 A V (A) R1 20 (B) R3 7.5 (C) 15V (D) 37.5 W 9. 8 × 10–10 î C-m (A) y-z (B) (1, 1, 1) m 0.8 (C) (2, 3, –6) m (D) z-x- 10. 3J (A) 1J (B) 2J (C) (D) –II : & –III : II III 1001CT103516009 H-7/24 PHYSICS Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 –IV : (: 30) 0 9 : : +3 0 1. h = 1m (= 10 & g = 10m/s2) 2. 0.006 A 4990 2n 249 0 - 30 V 0 - 1.5 A n : 3. M m d d Km/KM 4. R ( 4 R 4 2 4V 6V 10V H-8/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 S S 2 PHYSICS 5. 4µF 2 S 6V 8 2µF 4 6. 3 2 i 1A t 1sec 7. 10 kg 4kg 4kg 7 m/s 1 kg (m/sec ) k=100N/m 1kg 4kg 1001CT103516009 H-9/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 PHYSICS 8. v0 = 2m/s t = 0 sec k = 100 N/cm 1 m/s (cm ) 1m/s k = 100N/cm m 4kg 9. v0 = 2m/s 1 9 A C ) B 10 mC) 'A' 19 kg 3 'B' 'E' 'B' E (kN/C ) g = 10m/s2 A (Fixed) B 10. NQ N Q –5Q 10Q I II III H-10/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 –I : ( : 1. 30) (A), (B), (C) (D) : +3 : 0 A B A [A]0 B : (A) [A] (C) [B] (B) [A] t 2. CHEMISTRY -2 : (D) [B] t t t : P4S3 + O2 P4O10 + SO2 : (A) P4S3 32 4 (B) O2 (C) (D) P4S3 3. (A) a 27 R 2 Tc 2 64 Pc (B) b RTc 8Pc (C) Pc 1 a 27 b 2 (D) Tc 1 a 27 bR 1001CT103516009 H-11/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 CHEMISTRY 4. heteroleptic :(III) (B) Ziese's) (A) (IV) (D) (III) (C) III 5. [Co(en)2 L]2 E.A.N. 'L' :(B) ox–2 (A) en 6. 7. (C) Br– (D) dmg– [Cr(NH3)6]Cl3 :(A) (B) =3 (C) (D) 'X2' '1' 'X2' : (A) 2 (B) (C) (D) 'X2' 8. (A) (B) (C) (D) H-12/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 H3C–CH–CO2H (i) Aq. KOH + (ii) H CHEMISTRY 9. Cl R-2- (A) S-2- (B) R-2- (C) S-2- R-2- (D) S-1- 10. CH3–CH2 –CH2–Cl S 2 N (A) CH3–CH–Cl Cl (B) O (C) CH2–Cl (D) CH3–Br CH3 –II : & –III : II III 1001CT103516009 H-13/24 CHEMISTRY Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 –IV : (: 30) 0 9 : : 1. +3 0 W1 W2 W1 W 2 1 4T0 P 2 4 T0 V0 2. 25 × 10 V 10–23 cm3 3 64V0 10 NA 6 3. 298 K 1 gm = 6 × 10 23 . –1 –395 kJ mol –285 kJ mol–1 16 kJ 298K kJ ( ) H-14/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 5 mol Fe0.9 O 0.1 M KMnO4 CHEMISTRY 4. 5. HCl H2SO4 (x + y) ? SOxCly 6. 'd' [Cr(H2O)5 L]n ? 7. : [M(AB)3] ; [M(AA)abcd] ; [M(AA)b3c] ; [Ma3b3] 1001CT103516009 H-15/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 CHEMISTRY 8. (i) (ii) (iii) CO2Na (iv) 9. (v) (vi) CO 2Na (i) H 3C–C–CH 3 1. Mg / Hg (ii) CH3–CH2–Cl + Na + 2. H Dry ether O (iii) H3C–CH–CH–CH3 Cl Zn dust (iv) H3C–CH=CH 2 CH 3 Cl (v) H3C–CH=CH2 (i) Hg(OAc)2, H2O (ii) NaBH4 (i) B2H6 (THF) (vi) CH3–CH2–Cl – Aq. KOH (ii) H2O2 + OH (vii) CH3–CH2–OH 10. SOCl2 - – + OH N H-16/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 –I : ( : (A), /2 0 (C) +3 (D) 0 xdx 1 sin 2x cos2 2x (A) 6 2. 30) : : 1. (B), (C) MATHEMATICS -3 : /2 0 1 1 sin 2x 1 2 sin 2x dx (B) 12 2 1 12 3 3 (D) /2 0 1 1 sin 2x 1 2 sin 2x dx 4 1 12 3 3 ƒ(x) g(x) ƒ ' x ƒ x g2 x g x g ' x , g ' x g x ƒ2 x ƒ x ƒ ' x , ƒ 0 1 2,g 0 1 2 (A) ƒ(x) (0,) (B) g(x) (–,0) (C) ƒ(x) = g(x) (D) ƒ(x) > g(x) x R 3. R ƒ(x) ƒ(2) = ƒ(6) = 0 (2,6) c ƒ'(c) = ƒ(c) (A) 21 (B) –100 (C) –73 (D) 1010 1001CT103516009 H-17/24 MATHEMATICS Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 4. 5. 6. 7. x1,x2 x2 + a2x + b = 0 x3,x4 x2 + 5ax + 7 = 0 x1 – x3 = x2 – x4 = 2 (A) x1,x2,x3,x4 Q (B) x1,x2,x3,x4 Q (C) a + b = 2 (D) a + b = –25 ƒ : A B ƒ(x) = tan–1x + tan–12x + tan–13x A,B R (A) A = R B = R 3 3 (B) A = R B , 2 2 (C) A = [–1,1] B = [–,] B , (D) A = [–1,1] 2 2 x , ƒ(x) ƒ(–tanx) + 2ƒ(tanx) = sin2x 2 2 (A) ƒ" (B) ƒ'(0) = 2 (C) ƒ'(1) = 0 (D) ƒ(1) + ƒ(–1) = 0 a,b c a < b < c 2x + y = 2003 y = |x – a| + |x – b| + |x – c| 1002 (A) c (B) c 1002 (C) b 1001 (D) b 1001 H-18/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 [0,n] tan2x + cos6x = 1 – tan2x cos6x (A) 4,n = 1 (C) 2, n = 9. 1 2 MATHEMATICS 8. (B) 7, n = 2 3 4 (D) 3, n = L1 L2 6x2 + y2 + 5xy – 5x – 2y + 1 = 0 d1,d2 () 10. 6 (B) 0, (A) + = 1 (C) d1d 2 1 y = ƒ(x) cos2xdy – cos4xdx = y tan2x dx, (A) lim ƒ x x 0 x (D) 3 + 2 = 2 5 2 x 4 3 3 8 6 ƒ 1 3 3 (B) ƒ 8 6 y = x (D) ƒ(x) (C) y = ƒ(x) –II : & –III : II III 1001CT103516009 H-19/24 MATHEMATICS Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 –IV : (: 30) 0 9 : +3 : 0 1 1. ƒ(0) = 0 ƒ(1) = 1 ƒ(x) ƒ ' x 2 dx 0 m n 2. y x , A(0,0) B(1,2) C(3,3) D(4,0) ABCD 3. n–m ( m,n ) 4x + 5y = 20, x- A y- B L, AB OA C D CD2 L, OAB b a a,b N a 41 b H-20/24 1001CT103516009 4. MATHEMATICS Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 y = x y = –2x, ABC B C A (1,2) r 1 2 r 5. L tan1º tan5º + tan5º tan9º + tan9º tan13º........+ tan177º tan181º = L 5 6. V mm3 3mm 3 mm V 49 7mm 1001CT103516009 H-21/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 2x4 – 8x3 + 8x2 – 1 = 0 8. ƒ(x) = |x2 – 2x| g x MATHEMATICS 7. ƒ t , t x , x 1 ƒ t ,1 t x , x 1 g(x) 9. tan 1 n 1 L lim n 2 n n 1 tan 1 n 2 n [–L] ( [.] ) 10. y sin2 x 2 1 n y x 'n' H-22/24 1001CT103516009 Leader Course/Phase-III, IV & V/20-11-2016/Paper-1 1001CT103516009 H-23/24 Target : JEE (Main + Advanced) 2017/20-11-2016/Paper-1 16. 17. I IV (0-9) +3 10 0 — +3 10 — 30 0 — — 30 ................................................................................................ ............................................. ____________________________ ____________________________ Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 H-24/24 info@allen.ac.in www.allen.ac.in Your Target is to secure Good Rank in JEE 2017 1001CT103516009
