Paper Code : 100 1CT103516001 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 : 31 - 07 - 2016 Time : 3 Hours PAPER – 1 Maximum Marks : 186 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 24 pages and that all the 18 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) 2017/31-07-2016/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-2/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PHYSICS PART-1 : PHYSICS SECTION–I(i) : (Maximum Marks : 15) This section contains FIVE questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases 1. A ball is thrown vertically upward under the influence of gravity. Suppose observer A films this motion and play the tape backwards (so the tape begins with the ball at its heighest point and ends with it reaching the point from which it was released), and another observer B observes the motion of the ball from a frame of reference moving at constant velocity which is equal to the initial velocity of the ball. The ball has a downward acceleration according to observer(s) : (A) (A) and (B) (B) only (A) (C) only (B) (D) neither 2. A car is traveling on a highway at a speed 25 m/s as shown. A passenger in a car throws a ball at an angle 37° with horizontal in a plane perpendicular the motion of the car. The ball is projected with a speed of 10 m/s relative to car. What will be the initial velocity of the ball in unit vector notation ? y x z (A) 25ˆi 8ˆj 6kˆ (B) 10iˆ 8ˆj 6kˆ (C) 10iˆ 25ˆj 6kˆ (D) 25ˆi 6ˆj 8kˆ Space for Rough Work 1001CT103516001 E-3/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 PHYSICS3. Different particles are projected horizontally (in same plane) with different speeds from the same point. The point of projection is taken as origin. Determine the locus of the points in space where their velocities are same in magnitude. (A) circle (B) ellipse (C) hyperbola (D) parabola 4. A force = (3tiˆ 5ˆj) N acts on a body due to which its displacement varies as (2t2iˆ 5ˆj) . F s Work done by this force in t = 0 to 2 sec is :- (A) 23 J (B) 32 J (C) zero (D) can’t be obtained. 5. A grass hopper jumps a maximum distance of 1.6m. It spends negligible time on the ground. How far will it go in 10 seconds ? (A) 5 2 m (B) 10 2 m (C) 20 2 m (D) 40 2 m Space for Rough Work E-4/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 SECTION–I(ii) : (Maximum Marks : 32) PHYSICS This section contains EIGHT 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. Partial Marks : +1 For darkening a bubble corresponding to each correct option, Provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. for example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened 6. The vertical height of point P above the ground is twice that of point Q. A particle is projected downward with a speed of 5 m/s from P and at the same time another particle is projected upward with the same speed from Q. Both particles reach the ground simultaneously, if PQ lies on same vertical line then :- (A) PQ = 30 m (B) PQ = 60 m (C) time of flight of the stones = 3s (D) time of flight of the stones = 1/3s 7. The system shown is in equilibrium. Positive value of x1 & x2 denote extensions of the springs, while negative values denote compressions. Which of the following is not possible ? µ=0.5 4 kg 500 N/m 2 kg µ=0.6 5 kg µ=0.4 x1 300 N/m x2 (A) x1 = 13cm, x2 = 2cm (B) x1 = 8cm, x2= 3 cm (C) x1 = 1cm, x2 = –8cm (D) x1 = –1cm, x2 = –2cm Space for Rough Work 1001CT103516001 E-5/24
PHYSICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 8. A gymnast of mass m climbs a vertical rope attached to the ceiling. You can ignore the weight of the rope. (A) If the gymnast climbs at a constant velocity, then tension in rope is greater than his weight. (B) If the gymnast hangs motionless on the rope, then tension in rope is equal to his weight. (C) If the gymnast accelerates up the rope, then tension in rope is greater than his weight. (D) If the gymnast slides down the rope with a downward acceleration, then tension in rope is less than his weight. 9. A man of mass 40 kg is standing on a trolley A of mass 140 kg. While standing on trolley A he pushes another trolley B of mass 60 kg, so that they are set in motion. Then: (Assume the friction between ground and trolley A and B is zero) (A) speed of trolley A is 3 times that of trolley B immediately after the interaction. (B) speed of trolley B is 3 times that of trolley A immediately after the interaction. (C) distance travelled by trolley B is 3 times that of trolley A in same time. (D) distance travelled by trolley B is 9 times that of trolley A in same time. Space for Rough Work E-6/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 10. To give trainee astronauts experience of weightlessness, inside of a large plane. How can the PHYSICS plane be flown so that it has an acceleration of 9.8 ms–2 vertically down for some time? (A) vertically up with decreasing speed. (B) on a parabolic path with a constant speed (C) on a parabolic path with increasing speed while moving upwards and decreasing speed when moving downwards (D) on a parabolic path with decreasing speed while moving upwards and increasing speed when moving downwards 11. A model rocket rests on a frictionless horizontal surface and is joined by a string of length to a fixed point so that the rocket moves in a horizontal circular path of radius . The string will break if its tension exceeds a value T. The rocket engine provides a thrust F of constant magnitude along the rocket's direction of motion. The rocket has a mass m that does not change appreciably with time (Ignore any air resistance). Starting from rest at t = 0 at later time t1 the rocket is travelling so fast that the string breaks. Mark the correct option(s) : mT 1/ 2 (A) t1 = F2 (B) The magnitude of instantaneous net acceleration at time t1 [T2 16F2 ]1/2 is . 2 4m (C) The magnitude of instantaneous net acceleration at time t1 [T2 16F2 ]1/2 is . 2 2m 2mT 1 / 2 (D) t1 = F2 Space for Rough Work 1001CT103516001 E-7/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 12. A force of constant magnitude F acts on a particle (mass m) moving in a plane such that it is PHYSICS always perpendicular to the velocity v(|v| v) of the body. Select the correct option(s). (A) The time taken by the particle to cover a distance S is t = S/v FS (B) The angle turned by the velocity vector of the particle during a distance S is mv2 . Ft (C) The magnitude of change in velocity vector during the time t is 2v sin mv . (D) The magnitude of change in velocity vector during the time t is 2v sin Ft . 2mv 13. A cable stretched between the fixed supports A and B is under a tension T of 1000 N. Express the tension as a vector using the unit vectors ˆi and ˆj , first, as a force TA acting on A and second, as a force TB acting on B. 3m A y 4m x (A) TA (600iˆ 800ˆj)N B (C) TA (800iˆ 600ˆj)N (B) TB (600iˆ 800ˆj)N (D) TB (800iˆ 600ˆj)N 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 E-8/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 SECTION–IV : (Maximum Marks : 15) This section contains FIVE questions. PHYSICS 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 : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. You are shown a photo of a car driven on a vertical inside wall of a huge cylinder with a radius of 50 m. The coefficient of static friction between the car tires and the cylinder is s = 0.8. The minimum speed in (m/s), at which the car can be driven like that is 5x. Find x. 2. A block slides down an inclined plane of angle 53° with constant velocity. It is then projected up the same plane with an initial velocity of 8m/s. How far up (in m) the incline will it move before coming to rest? Space for Rough Work 1001CT103516001 E-9/24
PHYSICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 3. The system shown in the figure is released from rest. After the 60 kg block strikes the ground, it does not rebound. Find the total distance (in meter) moved by the 30 kg block on the incline before coming to rest. 30 kg 60 kg µ = 0.5 6m 37° 4. The figure shows an accelerometer, a device for measuring the horizontal acceleration of cars and airplanes. A ball is free to roll on a parabolic track described by the equation y = x2, where both x and y are in meters. A scale along the bottom is used to measure the ball's steady position x. What is the acceleration of vehicle (in m/s2) if x = 20 cm ? y = x2 –0.5m 0m 0.5m 5. A car is travelling in steady rain with constant acceleration in a straight line. When it begins to move the driver sees that the raindrops make an angle of 37° with the vertical. After 20 s, the raindrops make an angle of 53° with vertical in same direction. Find the acceleration of the car in cm/s2. Rain is falling at 3 m/s. Space for Rough Work E-10/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 PART-2 : CHEMISTRY CHEMISTRY SECTION–I(i) : (Maximum Marks : 15) This section contains FIVE questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases 1. According to the equipartition principle, the predicted high temperature limiting value of the molar heat capacity at constant volume of C2H2 is - (A) 5.5 R (B) 6.0 R (C) 9.0 R (D) 9.5 R 2. The following reaction is carried out at 1 atm and 300 K 2H2(g) + O2(g) 2H2O(l) U = –550kJ/mol Assuming ideal behaviour for H2 and O2, the value of H of the above reaction is - [R = 25/3 J mol–1K–1] (A) –550 kJ/mol (B) –557.5 kJ/mol (C) –542.5 kJ/mol (D) –567.5 kJ/mol 3. For an ideal gas undergoing reversible carnot cycle, the plot of enthalpy (H) versus entropy (S) is - H H H H (A) (B) (C) (D) SSS (D) sp2 S 4. What is the hybridisation of the anionic part of Cl2O6(s) :- (A) sp3 (B) sp3d (C) sp3d2 5. NaNH2 Conjugate base Acid How many contributing resonating structures of monoanion conjugate base are possible. (A) 3 (B) 2 (C) 5 (D) Conjugate base do not show resonance Space for Rough Work 1001CT103516001 E-11/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 CHEMISTRY SECTION–I(ii) : (Maximum Marks : 32) This section contains EIGHT 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. Partial Marks : +1 For darkening a bubble corresponding to each correct option, Provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. for example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened 6. Which of the following statement(s) is/are correct - (A) A closed system is one whose volume is constant (B) For a closed gaseous system , the value of PdV for the change of the gas from one given state to another is independent of the path so long as all process are reversible (C) Every process that has T = 0, is an isothermal process (D) For every process Usys = – Usurr. 7. For reversible melting of solid benzene at 1 atm and its normal melting point which of the following is/are correct - (A) w = – ve (B) q = + ve (C) U = + ve (D) H = + ve Space for Rough Work E-12/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 8. An ideal gas is subjected to following cyclic process. Which of the following are correct regarding CHEMISTRY this process - 50 1 2 3 P (kPa) 20 25 V(m3) (A) w12 = –105 kJ (B) w23 = +60 kJ (C) w1231 = –45 kJ (D) q1231 = +45 kJ 9. One mole of ideal gas N2 undergo following process State A Onestep State B 0ºC irrevers ible process 2atm 1atm 0ºC Choose the correct statement(s) (A) H = 0 (B) S = – 0.0821 ln2 atm litre (C) q = – 44.8 atm-litre K (D) W = 22.4 atm-litre 10. Which of the following statement is/are CORRECT for BCl3 :- (A) One unhybridised 2p orbital is on B (B) Act as lewis acid (C) All bond angles are equal (D) Boron is sp2 hybridised 11. Identify the chemical species in which all A–O bond lengths are equal (where A is underlined atom) (A) CO32– (B) HCO3– (C) NO2– (D) SO42– 12. In which of the following compound inductive effect, mesomeric effect and hyperconjugation all three are present : : CH3 O–CH3 OCH3 NH2 (B) (C) (A) (D) O H3C Space for Rough Work 1001CT103516001 E-13/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 13. Find out the correct statement(s) : CHEMISTRY : : : CH3 + + > Ph–CH2 (A) H3C +> CH3 (Order of stability of carbocation) CH3 H3C –– – CH2 CH2 CH2 NO2 > NO2 (Order of stability of carbanion) CO2H (B) > CO2H > (Order of acidic strength) NO2 OH OH HO CO2H CO2H (C) OH OH > > OO (D) NH > NH > NH (Order of basic strength) O 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 E-14/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 SECTION–IV : (Maximum Marks : 15) CHEMISTRY This section contains FIVE 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 : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. During adiabatic reversible compression of ideal gas He its volume become 1/8 times of its initial volume. Calculate its molar enthalpy change in (kcal) if initial temperature of gas is 400K 2. A triatomic non-linear ideal gas is subjected to isobaric expansion to double its volume. Calculate the value of H for the process. Assume vibrational degree of freedom is inactive. |W| 3. Find the maximum number of identical bond angle in ClO3– . Space for Rough Work 1001CT103516001 E-15/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 CHEMISTRY4. Find the number of chemical species having p–p bond. ClO – ; NO – ; CO ; CO 2– ; PO 3– ; XeO 22 34 3 5. If Number of Aromatic compounds are X. Number of Non Aromatic compounds are Y. and number of Anti aromatic compounds are Z. Then value of (X + 2Y – Z) is : OH ::NH2 : B N ::O NO (a) (b) (c) H (d) O +H – – : :N N H (e) (f) (g) (h) Space for Rough Work E-16/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 PART-3 : MATHEMATICS MATHEMATICS SECTION–I(i) : (Maximum Marks : 15) This section contains FIVE questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases 1. Number of real number(s) which satisfy the equation |x – 2| cosecx = 1 – |x – 3| cosecx is- (A) 0 (B) 1 (C) 2 (D) more than 2 2. Number of integers which satisfy the inequation log2 x log x 2 is 2 4 (A) 4 (B) 6 (C) 8 (D) 10 3. The sum of distinct solutions of the equation (4 tanx + tan2x + 1) 2 2 sin x 4 1 tan2 x in the interval [0,] is- 4 (C) 2 7 (A) 2 (B) 3 (D) 6 4. If are the solutions of equation 2 x20log22 x12 such that < , then is- (A) 4 (B) 64 (C) 256 (D) 512 1 b2 b3 b4 8b5 b6 b7 b13 5. If b > 0, then minimum value of b5 is equal to- (A) 89/8 (B) 7 (C) 64 (D) 15 Space for Rough Work 1001CT103516001 E-17/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 MATHEMATICS SECTION–I(ii) : (Maximum Marks : 32) This section contains EIGHT 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. Partial Marks : +1 For darkening a bubble corresponding to each correct option, Provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. for example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened 6. If are roots of equation 3x3 + 7x + 3 = 0, then the number(s) which satisfy the equation is(are)- (A) 1 (B) 1 (C) 1 1 (D) 1 7. Let Sn n 3r1 2r 3 , then r r 1 r 2 (A) S9 is divisible by 4 (B) S9 is divisible by 21 (C) 10.S10+3 = 310 (D) 7(S7 + 3), 10(S10 + 3), 13(S13+ 3) are in G.P. Space for Rough Work E-18/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 MATHEMATICS 8. For the quadratic equation x2 + (2 – m)x – 4 = 0 in x identify the correct statement(s)- (A) if m < 5 then exactly one root is smaller than – 1 (B) if m < 5 then both roots are smaller than – 1 (C) if m > 5 then both roots are greater than –1 (D) roots are distinct real numbers for each m R 9. Let x1,x2,x3,x4 are four roots of equation x4 + ax3 + bx2 + cx + d = 0, then- (A) x12x2x3 ac (B) x12x2x3 ac 4d 12 terms 12 terms (C) x12x22x3 3ad bc (D) x12x22x3 bc 3ad 12 terms 12 terms 10. If is a real number from 0 to such that tan 4x2 6x 9 4x2 6x 9 where x R, then- (A) sin cos (B) sin cos 6 6 6 6 (C) sin 1 , 3 (D) cos 1 , 3 2 2 2 2 Space for Rough Work 1001CT103516001 E-19/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 MATHEMATICS11. If a,b,c are three distinct real numbers in A.P., then- (A) a,2b,3c can not be in A.P. (B) 2a, b, c + 2 can be in G.P. (C) a,c,b can be in G.P. (D) a2,b2,c2 can be in A.P. 12. Let a = (sin)sin, b = (sin)cos c = (cos)sin, d = (cos)cos, then (A) b < c for 0, (B) b > c for , 4 4 2 (C) a < d for 0, (D) ab > cd for , 4 4 2 13. If 6x+1 = 23x+1, then x is equal to- log2 3 (B) log4/33 (C) log3/43 1 (A) 2 log2 3 (D) 1 2 log3 2 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 E-20/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 SECTION–IV : (Maximum Marks : 15) MATHEMATICS This section contains FIVE 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 : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. If a,b,c are real numbers greater than 1 and x 1 1 1 , then 2x is equal to 1 loga3b2 c2 / a 1 logb3c2 a2 / b 1 logc3a2 b2 / c 2. If tan 5 2 tan tan n , then smallest positive integer n is equal to 16 8 16 3. Let a(x) = 1 – 2sinx + 3sin2x – 4sin3x......... b(x) = 1 + 2cosx + 3cos2x + 4cos3x........ where x 0,4 x|x n , n I , then number of values of x for which a(x) = b(x) 2 Space for Rough Work 1001CT103516001 E-21/24
MATHEMATICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 4. Number of different integral values of for which the equation |x + 5| – |x – 3| = 4, (where x R) posses integral solutions is 5. If cos() = 1 and 16cos2( + ) = 2 where [0,], then number of orderd pairs of which satisfy both equations is Space for Rough Work E-22/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 Space for Rough Work 1001CT103516001 E-23/24
Target : JEE (Main + Advanced) 2017/31-07-2016/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. 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 +3 0 –1 In all I(i) Single If only the bubble If none other cases correct option 5 corresponding to — of the 15 the correct option bubbles is is darkened darkened +4 +1 0 –2 In all One or more If only the bubble(s) For darkening a bubble If none other cases I(ii) correct 8 corresponding corresponding toeach of the 32 option(s) to all the correct correct option, provided bubbles is option(s) is(are) NO incorrect option darkened darkened darkened +3 0 In all Single digit If only the bubble other cases IV Integer 5 corresponding — — 15 (0-9) to correct answer 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-24/24 Your Target is to secure Good Rank in JEE 2017 1001CT103516001
Paper Code : 1001CT103516001 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 : 31 - 07 - 2016 Time : 3 Hours PAPER – 1 Maximum Marks : 186 1. 2. (ORS) 3. 4. 5. 2418 6. 7. 8. 9. : 10. 11. 12. : 13. 14. 15. g = 10 m/s2
Target : JEE (Main + Advanced) 2017/31-07-2016/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 H-2/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS PHYSICS BEWARE OF NEGATIVE MARKING -1: –I(i) : ( : 15) (A), (B), (C) (D) : +3 : 0 : –1 1. A B (A) (A) (B) (B) (A) (C) (B) (D) 2. 2 5 m/s 37°10m/s y x z (A) 25ˆi 8ˆj 6kˆ (B) 10iˆ 8ˆj 6kˆ (C) 10iˆ 25ˆj 6kˆ (D) 25ˆi 6ˆj 8kˆ 1001CT103516001 H-3/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 PHYSICS3. (A) (B) (C) (D) 4. F = (3tˆi 5ˆj) N s (2t2iˆ5ˆj) t=0 2 (A) 23 J (B) 32 J (C) (D) 5. 1.6m 10 (A) 5 2 m (B) 10 2 m (C) 20 2 m (D) 40 2 m H-4/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 –I(ii) : ( : 32) PHYSICS (A), (B), (C) (D) : +4 : +1 : 0 : –2 (A),(C) (D ) +4 (A)(D ) + 2 (A) (B ) –2 6. P Q5m/sP Q PQ (A) PQ = 30 m (B) PQ = 60 m (C) =3s (D) =1/3s 7. x1x2 µ=0.5 4 kg 500 N/m 2 kg µ=0.6 5 kg µ=0.4 x1 300 N/m x2 (A) x1 = 13cm, x2 = 2cm (B) x1 = 8cm, x2= 3 cm (C) x1 = 1cm, x2 = –8cm (D) x1 = –1cm, x2 = –2cm 1001CT103516001 H-5/24
PHYSICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 8. m (A) (B) (C) (D) 9. 40 kg 140kg AA60kg BAB (A) AB3 (B) BA3 (C) BA 3 (D) BA 9 H-6/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1PHYSICS 10. 9.8ms–2 (A) (B) (C) (D) 11. T F mt=0 t1 mT 1/ 2 (A) t1 = F2 (B) t1 [T216F2 ]1/2 2 4m (C) t1 [T216F2 ]1/2 2 2m 2mT 1 / 2 (D) t1 = F2 1001CT103516001 H-7/24
PHYSICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 12. mF v(|v| v) (A) S t= S/v (B) S mFvS2 (C) t 2vsinmFtv (D) t 2vsin2Fmtv 13. AB T = 1000 N ˆiˆj TA,ATB, B 3m A y 4m x (A) TA (600iˆ 800ˆj)N B (C) TA (800iˆ 600ˆj)N (B) TB (600iˆ 800ˆj)N (D) TB (800iˆ 600ˆj)N –II : & –III : II III H-8/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1PHYSICS –IV : ( : 15) 09 : +3 : 0 1. 50m s=0.8 5x(m/s x 2. 53° 8m/s (m )? 1001CT103516001 H-9/24
PHYSICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 3. 60kg 30kg () 30 kg 60 kg µ = 0.5 6m 37° 4. y = x2 xy x x =20 cm (m/s2 ) y = x2 –0.5m 0m 0.5m 5. 37°2053° (cm/s2)3m/s H-10/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 -2: CHEMISTRY –I(i) : ( : 15) (A), (B), (C) (D) : +3 : 0 : –1 1. C2H2 (A) 5.5 R (B) 6.0 R (C) 9.0 R (D) 9.5 R 2. 1 atm 300 K 2H2(g) + O2(g) 2H2O(l) U = –550kJ/mol H2 O2 H [R = 25/3 J mol–1K–1] (A) –550 kJ/mol (B) –557.5 kJ/mol (C) –542.5 kJ/mol (D) –567.5 kJ/mol 3. (H) (S) H H H H (A) (B) (C) (D) SS S S 4. Cl2O6(s) (A) sp3 (B) sp3d (C) sp3d2 (D) sp2 5. NaNH2 Acid (A) 3 (B) 2 (C) 5 (D) 1001CT103516001 H-11/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 CHEMISTRY –I(ii) : ( : 32) (A), (B), (C) (D) : +4 : +1 : 0 : –2 (A),(C) (D ) +4 (A)(D ) +2(A ) (B) –2 6. (A) (B) PdV (C) T= 0 (D) Usys = – Usurr. 7. 1 atm (A) w = – ve (B) q = + ve (C) U = + ve (D) H = + ve H-12/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 8. CHEMISTRY 50 1 3 2 P (kPa) 20 25 V(m3) (A) w12 = –105 kJ (B) w23 = +60 kJ (C) w1231 = –45 kJ (D) q1231 = +45 kJ 9. N2 A irreveOrsnibelsetepprocess B 0ºC 2atm 1atm 0ºC (A) H = 0 (B) S = – 0.0821 ln2 atm litre K (C) q = – 44.8 atm-litre (D) W = 22.4 atm-litre 10. BCl3 :- (A) B 2p (B) (C) (D) sp2 11. A–O (A ) (A) CO32– (B) HCO3– (C) NO2– (D) SO42– 1001CT103516001 H-13/24
CHEMISTRY Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 12. : CH3 O–CH3 OCH3 NH2 (A) (B) (C) (D) O H3C 13. CH3 + + > Ph–CH2 (A) H3C +> CH3 CH3 H3C –– – : : CH2 : CH2 CH2 NO2 > (B) > NO2 CO2H CO2H NO2 CO2H OH OH CO2H HO > > > (C) OH O OH O (D) NH > NH > NH O –II : & –III : II III H-14/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1CHEMISTRY –IV : ( : 15) 09 : +3 : 0 1. He1/8 400K (kcal) 2. H |W| 3. ClO3– 1001CT103516001 H-15/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 CHEMISTRY4. p–p ClO –; NO – ; CO ; CO 2– ; PO 3– ; XeO 22 3 43 5. X Y Z (X + 2Y – Z) OH ::NH2 : B N ::O NO (a) (b) (c) H (d) O +H – – : :N N H (e) (f) (g) (h) H-16/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 -3: MATHEMATICS –I(i) : ( : 15) (A), (B), (C) (D) : +3 : 0 : –1 1. |x–2| cosecx = 1 – |x – 3| cosecx - (A) 0 (B) 1 (C) 2 (D) 2 2. log2x2 log x 2 4 (A) 4 (B) 6 (C) 8 (D) 10 3. [0, ] (4 tanx + tan2x + 1) 2 2 sin x 4 1 tan2 x - 4 (C) 2 7 (A) 2 (B) 3 (D) 6 4. 2 20 log 2 x x12 < - 2 (A) 4 (B) 64 (C) 256 (D) 512 5. b > 0 1b2 b3 b4 8b5 b6 b7 b13 - b5 (A) 89/8 (B) 7 (C) 64 (D) 15 1001CT103516001 H-17/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 MATHEMATICS –I(ii) : ( : 32) (A), (B), (C) (D) : +4 : +1 : 0 : –2 (A),(C) (D ) +4 (A)(D ) +2(A ) (B) –2 6. 3x3 + 7x + 3 = 0 - 1 1 11 1 (A) (B) (C) (D) 7. Sn n 3r1 2r 3 r r 1 r 2 (A) S9, 4 (B) S9, 21 (D) 7(S7 + 3), 10(S10 + 3), 13(S13+ 3) (C) 10.S10+3 = 310 H-18/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 MATHEMATICS 8. x x2+(2 – m)x – 4 = 0 - (A) m < 5 –1 (B) m < 5 –1 (C) m > 5 –1 (D) m R 9. x1,x2,x3,x4 x4 + ax3 + bx2 + cx + d = 0 - (A) x12x2x3 ac (B) x12x2x3 ac 4d 12 12 (C) x12x 2 x3 3ad bc (D) x12x22x3 bc 3ad 2 12 12 10. 0 tan 4x2 6x 9 x R - 4x2 6x 9 (A) sin cos (B) sin cos 6 6 6 6 (C) sin 1 , 3 (D) cos 1 , 3 2 2 2 2 1001CT103516001 H-19/24
MATHEMATICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 11. a,b,c- (A) a,2b,3c (B) 2a, b, c + 2 (C) a,c,b (D) a2,b2,c2 12. a= (sin)sin, b = (sin)cos c = (cos)sin, d = (cos)cos (A) 0, b < c (B) , b > c 4 4 2 (C) 0, a < d (D) , ab > cd 4 4 2 13. 6x+1 = 23x+1 x- log2 3 (B) log4/33 (C) log3/43 1 (A) 2 log2 3 (D) 1 2 log3 2 –II : & –III : II III H-20/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 MATHEMATICS –IV : ( : 15) 09 : +3 : 0 1. a,b ,c 1 x 1 1 1 2x 1 loga3b2 c2 / a 1 logb3c2 a2 / b 1 logc3a2 b2 / c 2. tan 5 2 tan tan n n 16 8 16 3. a(x) = 1 – 2sinx + 3sin2x – 4sin3x......... b(x) = 1 + 2cosx + 3cos2x + 4cos3x........ x 0,4 x|x n ,n I xa(x)= b(x) 2 1001CT103516001 H-21/24
MATHEMATICS Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 4. |x+5|–|x– 3| = 4 (x R) 5. cos() = 1 16 cos2( + ) = 2 [0,] H-22/24 1001CT103516001
Leader Course/Phase-III, IV & V/31-07-2016/Paper-1 1001CT103516001 H-23/24
Target : JEE (Main + Advanced) 2017/31-07-2016/Paper-1 16. 17. I(i) I(ii) IV +3 0 –1 5 — 15 +4 +1 0 –2 8 32 +3 0 5 — — 15 (0-9) ................................................................................................ ............................................. ____________________________ ____________________________ Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in H-24/24 Your Target is to secure Good Rank in JEE 2017 1001CT103516001
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