Paper Code : 1001CT103516015 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 PHASE-III to VII (SCORE-I) Test Type : PART TEST Test Pattern : JEE-Advanced TEST DATE : 05 - 03 - 2017 Time : 3 Hours PAPER – 1 Maximum Marks : 234 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) 2017/05-03-2017/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 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PHYSICS PART-1 : PHYSICS SECTION–I(i) : (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. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases. 1. A circular conducting loop of radius r0 and having resistance per unit length as shown in the figure is placed in a magnetic field B which is constant in space and time. The ends of the loop are crossed and pulled in opposite directions with a velocity v such that the loop always remains circular and the radius of the loop goes on decreasing, then (A) Radius of the loop changes with r as r = r0 – vt/ (B) EMF induced in the loop as a function of time is e = 2Bv[r0 – vt/] Bv (C) Current induced in the loop is I = 2 Bv (D) Current induced in the loop is I = Space for Rough Work 1001CT103516015 E-3/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 PHYSICS2. Current in R3 in the shown circuit :- L E R3 R1 C R2 (A) just after closing the switch is zero (B) long after closing the switch is zero (C) just after closing the switch is E E R3 (D) long after closing the switch is R3 3. A vessel of uniform cross-section open at the top with an orifice at its bottom contains oil (relative density 0.8) on top of water. It is immersed vertically in a large open tank of same oil as shown in figure. In which of following configuration will liquid level remain same. (Water does not come out or oil does not enter vessel) :- (Take : area of orifice = 1 mm2 and area of vessel as 100 cm2) oil 5m oil Water 10m oil (A) 15m Oil 5m Oil 12m (C) 15m Oil 8m Oil 12.5m Water Water Water Oil Oil Oil (B) 15m Oil Oil Oil (D) 15m 8m 2.4m 2m Oil Water 4m Oil Space for Rough Work E-4/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 PHYSICS 4. A series RLC circuit is activated by a frequency () variable ac source of voltage VS volt as shown in the circuit. VRL and VC are the potential drops across RL and C respectively. Select the incorrect statement(s) :- VS ~ C RL (A) At low frequency limit, both VRL and VC are proportional to 1 (B) At high frequency limit, VRL approaches VS but VC is proportional to 2 1 (C) At high frequency limit, both VRL and VC are proportional to 2 (D) At low frequency limit, VRL is proportional to , whereas VC approaches VS 5. A long straight cylindrical shell has inner radius Ri and outer radius R0. It carries current i, uniformly distributed over its cross section. A wire is to be placed parallel to the cylinder axis, in the hollow region (r < Ri ). (A) The magnetic field is zero everywhere in the hollow region. We conclude that the wire is on the cylinder axis and carries current i in the same direction as the current in the shell (B) The magnetic field is zero everywhere outside the shell (r > R0). We conclude that the wire is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell. (C) The magnetic field is zero everywhere in the hollow region. We conclude that the wire may be anywhere in the hollow region but must be carrying current i in the same direction as the current in the shell (D) The magnetic field is zero everywhere in the hollow region. We conclude that the wire does not carry any current 6. A soap film is formed on a vertical equilateral triangular frame ABC. Side BC can move vertically always remaining horizontal. If m is the mass of rod BC, and T is surface tension of soap film. Mark the correct statement(s) :- mg (A) The length of BC in equilibrium is 2T mg (B) The length of BC in equilibrium is T (C) The time period of small oscillations about the equilibrium position is 3m T m (D) The time period of small oscillations about the equilibrium position is 3T Space for Rough Work 1001CT103516015 E-5/32
PHYSICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 7. The arrow OA in the PV diagram shown in figure represents a reversible adiabatic expansion of an ideal gas. The same sample of gas, starting from the same state O, now undergoes a process to the same final volume. Select the appropriate combination of process and final state. PO B A CV (A) Process : isothermal ; final state : B (B) Process : free expansion ; final state : B (C) Process : isothermal ; final state : C (D) Process : free expansion ; final state : C 8. If you were to blow smoke into the space between the barrier of standard YDSE and the viewing screen of figure, the smoke would show S1 r1 P dQ r2 y O S2 L Viewing screen (A) No evidence of interference between the barrier and the screen (B) Evidence of interference everywhere between the barrier and the screen. (C) Maxima are localised and located on hyperboloids. (D) Maxima are non-localised and located on hyperbolic planes. Space for Rough Work E-6/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 SECTION–I(ii) : (Maximum Marks : 18) PHYSICS This section contains THREE paragraphs. Based on each paragraph, there are TWO 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 answer is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases. Paragraph for Questions 9 and 10 The optical properties of a medium are governed by its relative permittivity (r) and relative permeability (µr). For conventional materials like water or glass, which are usually optically transparent, both of their r and r are positive, and refraction phenomenon meeting Snell’s law occurs when light from air strikes obliquely on the surface of such kind of substances. In 1964, a Russia scientist V. Veselago rigorously proved that a material with simultaneously negative r and r would exhibit many amazing and even unbelievable optical properties. In early 21st century, such unusual optical materials were successfully demonstrated in some laboratories. Nowadays study on such unusual optical materials has become a frontier scientific research field. Through solving several problems in what follows, you can gain some basic understanding of the fundamental optical properties of such unusual materials. It should be noticed that a material with simultaneously negative r and r possesses the following important property. When a light wave propagates forward inside such a medium for a distance , the phase of the light wave will decrease, rather than increase an amount of rr k as what happens in a conventional medium with simultaneously positive r and r . Here, positive root is always taken when we apply the square-root calculation, while k is the wave vector of the light. In the questions listed below, we assume that both the relative permittivity and permeability of air are equal to 1. Space for Rough Work 1001CT103516015 E-7/32
PHYSICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 9. A light beam strikes from air on the surface of such an unusual material with relative permittivity r < 0 & relative permeability µr < 0. If 'd' denotes the length, then : air A1 B medium E 1 C D r (A) dCD d AE d BC (B) dCD dBC rr dAE rr (D) dCD dAE rr dBC (C) sin r rr sin 1 10. Choose the CORRECT option :- (A) (B) µr < 0 µr < 0 r < 0 r < 0 (C) (D) µr < 0 µr < 0 r < 0 r < 0 Space for Rough Work E-8/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 PHYSICS Paragraph for Questions 11 and 12 Let us do the Young's double slit experiment with electron rather than light waves. We make an electron gun which consists of a tungsten wire heated by an electric current and surrounded by a metal box with a hole in it. If the wire is at negative potential w.r.t. the box, electron emitted by the wire (thermionic emission) will be accelerated towards the walls and same will pass through the hole. All the electrons coming out of this gun will have the same energy, qV. In front of gun there is a wall with two slits in it as shown. Beyond the wall there is another plate which acts as a screen. The no. of electrons arriving at the screen/area/time can be measured. When the graph of this is plotted with y coordinate on the screen, we obtain the graph as seen in YDSE experiment. This can be explained using de broglie's hypothesis which states that electrons also behave like waves. y detector 1 d 2 ELECTRON D GUN Screen WALL (b) (a) 11. Which of following would increase fringe width, assuming that << d ? (A) Increase temperature of wire (B) Decrease D (C) Increase d (D) Decrease voltage V 12. Suppose we go on increasing the voltage from a very low value.Beyond a certain voltage, we obtain 3 maxima on the screen. What is the voltage? h2 5h2 9h2 7h2 (A) 2mqd2 (B) 2mqd2 (C) 2mqd2 (D) 2mqd2 Space for Rough Work 1001CT103516015 E-9/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 Paragraph for Questions 13 and 14 PHYSICS The first nuclear reaction ever observed was by ernest Rutherford in 1919. It was triggered by -particles incident on an isotope of nitrogen 14 N . He observed a proton was emitted 7 along with another element x. Let us assume that 14 N nucleus was initially stationary. For 7 this reaction to occur, -particle must touch the nitrogen nucleus. The distance between their centres at this moment is d. For this problem, we will neglect the effect of outer electrons in 14 N . Symbols have their usual meanings. 7 13. Value of d is (A) R0(21/3 + 71/3) (B) R0(22/3 + 72/3) (C) R0(22/3 + 142/3) (D) R0(22/3 + 141/3) 14. The minimum initial kinetic energy of -particle so that reaction can occur is :- 18ke2 14ke2 18ke2 14ke2 (A) R0 (22 / 3 142/ 3 ) (B) R0 (22/ 3 141/ 3 ) (C) R0 (22/ 3 141/ 3 ) (D) R0 (22 / 3 72 / 3 ) Space for Rough Work E-10/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 PHYSICS SECTION–II : (Maximum Marks : 8) This section contains ONE question. 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 shows certain system at t = 0. Column-II shows its behavior there after. Neglect dissipative forces in all cases. Column-I Column-II (A) System experiences net torque (P) A charged bead is constrained to move along intially about C. a non conducting ring in gravity free space. A dipole is fixed at the centre of ring. Charge is released from perpendicular bisector of dipole at t = 0. System is bead. C +Q p charged bead Space for Rough Work 1001CT103516015 E-11/32
PHYSICS (B) Angular velocity about C first Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 increases then decreases (Q) A spherical ball made of wood is released inside water from rest in the configuration shown at t = 0. Point C is geometric centre of ball. System is ball. C.M. of ball, nonuniform mass distribution C (C) System executes periodic motion (R) Disc hangs with an elastic wire of constant torsional stiffness. Uniform magnetic field is suddenly switched on at t = 0. System is disc. B uniformly C charged disc (D) Acceleration of centre of mass (S) A current carrying coil is placed in a uniform of system is zero initially. magnetic field parallel to the plane of coil at t = 0, coil lies in a gravity free space. System is coil. B C (T) A boy on a swing of small angular amplitude. When swing gets to mean position (t = 0) boy instantaneously stands up. System is swing & boy. C Space for Rough Work SECTION–III : Integer Value Correct Type No question will be asked in section III E-12/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 SECTION–IV : (Maximum Marks : 20) 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 : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. A point like sound source emits sound in all direction uniformly. An observer who is at a distance of 50 m from the source detects sound of intensity 10–8 watt/m2. If the bulk modulus of air is 1.6 × 105 N/m2 and velocity of sound is 340 m/s. Find pressure amplitude at the position of observer. If your answer is x × 10–3 N/m2 fill value of [x] (nearest integer) 2. One mole of an ideal monoatomic gas is taken from state A to state B through a polytropic process in which gas obeys law P = 3 T1/ 2 . It is found that its temperature increases by 100 K 2 in this process. Now it is taken from state B to C through another process for which internal energy is related to volume as U = 1 V1/ 2 , where volume at B is 100 m3 and at C it is 1600 m3. 2 Find the total work performed by the gas (in Joule). Both the processes have propotionality constant unity with necessary dimensions. If your answer is x fill value of x 45 . 60 [Take : R = 8.3 J/mol-K] 3. A lizard of mass 4g is warming it self in the bright sunlight. It's surface area is 4.2 cm2, but its body casts a shadow of area 1.4 cm2 on a piece of paper held perpendicular to sun rays. The intensity of sun rays at earth is 1000 W/m2. Assume that specific heat capacity of lizard is same as that of water and its body behaves like a black body, also neglect the radiation from the lizard's body. How long (in minutes) must lizard lie in sun in order to raise it's temperature n by 5°C. Specific heat capacity of water is 4200 J/kgºC. If your answer is n fill value . 5 Space for Rough Work 1001CT103516015 E-13/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 PHYSICS4. The reaction 7 Li 11 H 7 Be 10 n is endothermic. Assuming that Li nuclei is free and at rest. 3 4 What is the minimum kinetic energy (in keV) of incident proton so that this reaction N occurs ? Take Q value of this reaction as –1645 keV. If your answer is N fill value 235 . 5. When the voltage applied to an X-ray tube increased from V1 = 15.5 KV to V2 = 31 KV, the wavelength interval between the K line and the short wavelength cut-off of the continuous X-ray spectrum increases by a factor of 1.3. Find the atomic number of the element of the N target. If your answer is N fill value of . (Take : hc = 1240 eV nm and R = 1 × 107 /m) 13 Space for Rough Work E-14/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 PART-2 : CHEMISTRY CHEMISTRY SECTION–I(i) : (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. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases. 1. Which of the following are correct regarding associated colloids : (A) They behave as normal electrolytes at low concentrations, but at higher concentrations exhibit colloidal behaviour (B) The formation of micelles takes below a particular temperature known as kraft temperature (Tk) (C) Associated colloids have both lyophobic and lyophillic parts (D) For soaps, the CMC is 10–4 to 10–3 mol L–1 2. Which of the following is/are correct regarding the given graph's for an ideal gas. dN Temp. is dN For the same Ndu same for both gas Ndu gas at two different temp. M1 M2 u T1 T2 u (A) M2 > M1 (B) M1 > M2 (C) T2 > T1 (D) T1 > T2 Space for Rough Work 1001CT103516015 E-15/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 CHEMISTRY3. Select the CORRECT order :- (A) Thermal stability : LiF > NaF > KF > RbF > CsF (B) Bond Angle : BF3 < BCl3 < BBr3 < BI3 (C) Rate of hydrolysis : SiF4 > SiCl4 > SiBr4 > SiI4 (D) Acidic nature : Mn2O7 > MnO2 > MnO 4. Select the CORRECT statement :- (A) The removal of impurity of Ag from the commercial lead is known as desilverization of lead (B) Nickel is purified by Van-Arkel method (C) For the isolation of metal from cinnabar no reducing agent is required (D) Tin stone consists of wolframite as non magnetic impurity 5. Which of the following complex ion(s) is/are not optically active :- (A) [Co(H2O)4(en)]3+ (B) [Sc(H2O)3(NH3)3]3+ (C) [Ti(en)2(NH3)2]4+ (D) [Co(gly)3] 6. If (Q) contains anomeric carbon at 2nd position then select correct option(s) ? C12H22O11 + H2O P+Q (Sucrose) (A) In sucrose, P & Q held together by -glucosidic linkage (B) Equimolar mixture of P & Q obtained in above reaction is known as invert sugar (C) Both P & Q on passing through Tollen's reagent followed by acidification gives gluconic acid (D) Sucrose is dextrorotatory but equimolar mixture of P & Q is laevorotatory 7. Tranquilizer obtained (P) is : CH2–CHO P (i) N2H4 (ii) H2, Ni (A) Iproniazid (B) Phenelzine (C) Meprobamate (D) Equanil 8. Which of the following amino acids moves towards cathod on elelctrophoresis at pH = 7 ? (A) Leucine (B) Lysine (C) Arginine (D) Histidine Space for Rough Work E-16/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 SECTION–I(ii) : (Maximum Marks : 18) CHEMISTRY This section contains THREE paragraphs. Based on each paragraph, there are TWO 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 answer is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases. Paragraph for Questions 9 and 10 The standard emf of the cell Pt(s) | H2(g) | HBr(aq)| AgBr(s) | Ag(s) was measured over a range of temperature, and the data were fitted to the following polynomial (F = 96500 C mol–1) Eº (volt) 0.07 5 104 T 298 3.5 106 T 298 2 K K 9. The standard Gibbs energy of reaction for 1 mole electron at 298 K is : (A) –6.755 kJ / mol (B) –5 kJ / mol (C) –13.51 kJ / mol (D) +6.755 kJ / mol 10. The standard entropy of reaction for 1 mole electron at 298 K is : (A) –48.25 JK–1mol–1 (B) –96.5 JK–1mol–1 (C) +48.25 JK–1mol–1 (D) +96.5 JK–1mol–1 Space for Rough Work 1001CT103516015 E-17/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 CHEMISTRY Paragraph for Questions 11 and 12 Due to presence of -acid lignads synergic bonding takes place in coordination compounds. 11. Which of the following complex synergic bonding does not take place :- (A) [Fe(-C5H5)2] (B) [Fe(H2O)5NO]2+ (C) [Ni(PF3)(CO)3] (D) None of these 12. The CORRECT order of metal-carbon bond order is :- (A) [Mn(CO)6]+ > [Cr(CO)6] > [V(CO)6]– (B) [Cr(CO)6] > [Mn(CO)6]+ > [V(CO)6]– (C) [V(CO)6]– > [Mn(CO)6]+ > [Cr(CO)6] (D) [V(CO)6]– > [Cr(CO)6] > [Mn(CO)6]+ Space for Rough Work E-18/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 Paragraph for Questions 13 and 14 CHEMISTRY C4H4O4 represent dicarboxylic acid with degree of unsaturation is 3. It can show G.I. and its gemoetrical isomer are P & Q. P forms anhydride (R) on heating. 13. Major product of following reaction is : Q Bayer's reagent (A) Meso (B) Pair of diastereomer (C) Pair of enantiomers (D) Pair of G.I. 14. Correct statement about major product (S) is : +R Anhydrous S AlCl3 (A) It is acetophenone (B) It can show Tautomerism (C) It can consume only 1 mole of RMgX (D) Its degree of unsaturation is 7 Space for Rough Work 1001CT103516015 E-19/32
CHEMISTRY Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 SECTION–II : (Maximum Marks : 8) This section contains ONE question. 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) Rock salt type crystal (P) Cation is present in tetrahedral void (B) Antifluorite type crystal (Q) Cation is present in octahedral void (C) Zinc blende type crystal (R) Coordination number of cation and anion are same (D) Corundum type crystal (S) Coordination number of anion is 8 (T) Coordination number of anion is 4 Space for Rough Work SECTION–III : Integer Value Correct Type No question will be asked in section III E-20/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 SECTION–IV : (Maximum Marks : 20) 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 : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. If 0.025 M NaCl is mixed with 0.1 M Na2SO4 in 2 : 3 volume ratio at 273 K , then find the osmotic pressure of resulting solution in atomosphere upto nearest integer : 2. Find out the total number of stereoisomer of [Zn(gly)2] :- 3. The moles of NaOH required for complete reaction with one mole hydrolysis product of SF4 :- Space for Rough Work 1001CT103516015 E-21/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 CHEMISTRY4. How many of following compounds are less reactive than benzene for sulphonation by conc. H2SO4 : CH3 NO2 Cl OH COOPh + O D NMe3 NH–C–CH3 DD DD D 5. If P = number of compounds which gives (+ve) dye azo test on treatment with NaNO2, HCl followed by -napthol Q = number of compounds which gives (+ve) isocynide test Then find value of (P + Q) ? CH2–NH2 NH2 NH–CH3 NH2 NH NH2 O NH2 N N NH2 H CH3 Space for Rough Work E-22/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 PART-3 : MATHEMATICS MATHEMATICS SECTION–I(i) : (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. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases. 1. A circle (x – 3)2 + (y – 6)2 = r2 touches parabola y2 = 4x at P(a,b). If the slope of common tangent at P is m, then (where b,r > 0) - (A) r = 5 (B) r2 10 (C) (a + b)m = 4 (D) a b 8 m 2. A hyperbola intersects an ellipse x2 + 9y2 = 9 orthogonally. The eccentricity of the hyperbola is reciprocal of that of ellipse. If the axes of the hyperbola are along coordinate axes, then- (A) vertices of hyperbola are 8 ,0 3 (B) y coordinate of point of intersection of ellipse and hyperbola is either 1 or 1 3 3 2 (C) latus rectum of hyperbola is 3 4 (D) latus rectum of hyperbola is 3 3. Let are real numbers for which the system of linear equations 2x + y + z = 0 x + y + 2z = 1 2x + y – 3z = –2 never posses unique solution then point P() lies on a conic whose- (A) center is 1 ,0 (B) eccentricity is less than 3 2 (D) (–1,0) lies on it (C) one vertex is origin 4. Let A = {z1; z116 = 1, z1 C}, B = {z2 ; z272 = 1, z2 C} and P = {z1z2; z1 A, z2 B} are three sets of complex roots of unity (where C denotes set of complex numbers), then (A) n(A B) = 8 (B) n(A B) = 4 (C) n(P) = 144 (D) n(P) = 72 Space for Rough Work 1001CT103516015 E-23/32
MATHEMATICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 5. Let A is a four order diagonal matrix whose entries are complex numbers such that A4 = I4. If trace of A is zero then (here I4 is four order unit matrix) (A) There will be 24 distinct matrices A (B) There will be 36 distinct matrices A (C) Determinant value of all such matrices is 1 (D) Determinant value of all such matrices is –1 1 1 1 6. Let matrix B 0 1 1 and A is 3 ordered square matrix such that AB = BA. If all entries of 0 0 1 matrix A are whole numbers whose sum is 6, then- (A) If there are exactly 3 zero's in matrix A, then det(A) = 1 (B) If traceA = 6, then det(A) = 8 (C) A is always an invertible matrix (D) there are 4 such matrices A. 7. Let aiˆ bˆj ckˆ is another vector in R3 such that A be a unit vector and B AB 1, C 1 2iˆ 2ˆj kˆ and A B .C 1 , then which of the following statement(s) is (are) correct ? 3 (A) If A lies in plane x + y + z = 10, then there are exactly 2 choices for A . (B) If A lies in plane x + y + z = 10, then there are exactly 4 choices for A . (C) If a,b,c I, then there is no such vector A . (D) If a,b,c I, then there are infinitely many choices for A . 8. Let there are n planes in R3 such that any three have exactly one point in common and no four of them have a point in common. If ƒ(n) represent the number of parts in which these n planes will divide the space, then- (A) ƒ(3) = 8 (B) ƒ(4) = 15 (C) ƒ(4) = 12 (D) ƒ(10) = 176 Space for Rough Work E-24/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 SECTION–I(ii) : (Maximum Marks : 18) MATHEMATICS This section contains THREE paragraphs. Based on each paragraph, there are TWO 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 answer is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases. Paragraph for Questions 9 and 10 Let S(x1,0) for x1 > 0 is foci of the ellipse E : x2 y2 1. Suppose a parabola whose vertex is 9 4 V x1 5,0 touches the ellipse at points A and B in I and IV quadrant respectively. Axis of parabola is x-axis. On the basis of above information, answer the following questions : 9. Equation of normal to the ellipse E at A is- (A) 9x 2 3y 5 2 0 (B) 9x 2 3y 5 2 0 (C) 9x 2 3y 5 2 (D) 9x 2 3y 5 10. Tangent to the ellipse E at A and B meet at C, then area of ABC is- (A) 32 2 (B) 32 2 (C) 12 2 (D) 16 2 3 3 Space for Rough Work 1001CT103516015 E-25/32
MATHEMATICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 Paragraph for Questions 11 and 12 Let L1 : 4x – 3y + 13 = 0, L2 : 4x – 3y = 37, L3 : 3x + 4y = 34 are three lines in xy plane and L4:(1 + )x + (1 – )y = 24 is a variable line. P(a,b) is centre of circle which touches lines L1,L2 and L3. On the basis of above information, answer the following questions : 11. Maximum value of a + b is- (A) 13 (B) 15 (C) 17 (D) greater than 17 12. If L1,L2,L3 and L4 form a quadrilateral, then the value of for which slope of line L4 takes least positive integral value is- 4 (B) 2 7 5 (A) (C) (D) 3 5 4 Space for Rough Work E-26/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 MATHEMATICS Paragraph for Questions 13 and 14 A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer the following questions : 13. If n = 4 and p = 6, then the probability that there will be at least one passenger in each carriage is- 195 1564 (C) 6 C4 4 !42 (D) 5 C3 4 ! (A) 512 (B) 4096 46 46 14. n 1P n 2P n 3P ...... 1n 1 n n P is equal to (where n nCr ) - 1 2 3 n r (A) (–1)n–1n!, if P < n (B) 0, if P = n (C) n!, if P = n (D) (–1)n–1n!, if P = n Space for Rough Work 1001CT103516015 E-27/32
MATHEMATICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 SECTION–II : (Maximum Marks : 8) This section contains ONE question. 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) In a triangle ABC sinA,sinB,sinC are in A.P, then (P) Latus rectum of sin B ellipse x2 y2 1 2 16 4 sin A sin C is less than or equal to (Q) eccentricity of 22 hyperbola (B) In ABC if 2 + b2 + c2 = 2bc + a2, then value of x 2 which satisfy x2 – 5(sinA + cosA)x + 25sinAcosA = 0 x 2 y2 1 is equal to 2 16 (C) In ABC if a = 7, b = 3, c 131 and median AD (R) Radius of director 2 circle of x2 y2 1 meets circumcircle at E, then AE is greater than 36 13 (D) The point of contact of an inscribed circle of a right (S) Minimum value of angle triangle divides the hypoteneous in two parts of |z1 – z2| where lengths 4 and 7, then area of triangle is divisible by |z1| = 2 and |z2 – 9|= 3 E-28/32 (T) Half of maximum value of |z1 – z2| where |z1|= 2 and |z2 – 9| = 3 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 SECTION–III : Integer Value Correct Type MATHEMATICS No question will be asked in section III SECTION–IV : (Maximum Marks : 20) 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 : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. Let (x + 3)2 (x + 4)3 (x + 5)4 = (x + 1)9 + a1(x + 1)8 + a2(x + 1)7+..... + a9, then a2 – 365 is equal to 2. A plane P intersects 4 lines L1,L2,L3 and L4 (given below) at A,B,C,D L1 : x 3 y 3 z 3 , L2 : x 3 y 3 z 2 1 2 2 1 2 L3 : x y 3 z ,L4 : x y 3 z 3 , then the 2 1 2 2 1 2 minimum area of quadrilateral ABCD is 3. Coefficient of x17 in the expansion of (1 + x5 + x7)20 is , then is equal to 380 Space for Rough Work 1001CT103516015 E-29/32
MATHEMATICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 4. Number of ways in which two distinct natural numbers can be selected out of first 100 natural N numbers so that sum of their cubes is multiple of 8 is N, then 200 is equal to (where [.] denotes greatest integer function) 5. A die is rolled four times. If the probability that product of first 3 outcomes is equal to fourth 1 outcome is p, then is equal to (where [.] denotes greatest integer function) p Space for Rough Work E-30/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 Space for Rough Work 1001CT103516015 E-31/32
Target : JEE (Main + Advanced) 2017/05-03-2017/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 –1 — One or more If only the bubble(s) If none In all I(i) correct 8 corresponding of the other 32 option(s) to all the correct bubbles is cases option(s) is(are) darkened darkened Paragraph +3 0 –1 If none In all Based If only the bubble of the other bubbles is cases I(ii) (Single 6 corresponding to darkened 18 correct the correct option option) is darkened +8 +2 0 –1 II Matrix If only the bubble(s) For darkeninga bubble If none In all Match Type 1 corresponding to corresponding to each of the other 8 20 all the correct correct match is bubbles is cases match(es) is(are) darkened darkened darkened +4 0 Single digit If only the bubble In all IV Integer 5 corresponding — other — (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 1001CT103516015
Paper Code : 1001CT103516015 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) HINDI JEE (Main + Advanced) : LEADER COURSE PHASE-III to VII (SCORE-I) Test Type : PART TEST Test Pattern : JEE-Advanced TEST DATE : 05 - 03 - 2017 Time : 3 Hours PAPER – 1 Maximum Marks : 234 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) 2017/05-03-2017/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 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1PHYSICS HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING -1: –I(i) : ( : 32) (A), (B), (C) (D) : +4 : 0 : –1 1. r0 B v (A) r r =r0– vt/ (B) e=2Bv[r0 – vt/] (C) I=2Bv (D)I=Bv 1001CT103516015 H-3/32
PHYSICS2. R3 Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 EL R1 R3 C R2 (A) (B) E (C) R3 E (D) R3 3. 0.8 ( )=1mm2 100 cm2 oil 5m oil Water 10m oil (A) 15m Oil 5m Oil 12m (C) 15m Oil 8m Oil 12.5m Water Water Water Oil Oil Oil (B) 15m Oil Oil Oil (D) 15m 8m 2.4m 2m Oil Water 4m Oil H-4/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1PHYSICS 4. () V S acRLCVRL VC RL C VS ~ C RL (A) VR L V C (B) VRL, VS VC, 12 (C) VRL V C 12 (D) VR L, VC,VS 5. Ri R0 i (r<Ri ) (A) i (B) (r>R0) i (C) i (D) 6. ABCBC m BC T (A) BC m2Tg (B) BC mTg (C) 3m T (D) m 3T 1001CT103516015 H-5/32
PHYSICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 7. PVOAO PO B A CV (A) : ; :B (B) : ; :B (C) : ; :C (D) : ; :C 8. Y DSE (Ba rrier) S1 r1 P dQ r2 y S2 L O Viewing screen (A) (B) (C) (D) H-6/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1PHYSICS –I(ii) : ( : 18) (A), (B), (C) (D ) : +3 : 0 : –1 9 10 (r) (µr) r r 1964V.Veselago r r 21 rr rrrr k k 1 1001CT103516015 H-7/32
PHYSICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 9. r<0 µr< 0 'd' air A1 B medium E 1 C D r (A) dCD d AE d BC (B) dCD dBC rr dAE rr (D) dCD dAE rr dBC (C) sin r rr sin 1 10. (A) (B) µr < 0 µr < 0 r < 0 r < 0 (C) (D) µr < 0 µr < 0 r < 0 r < 0 H-8/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1 11 12 PHYSICS qV y YDSE y detector 1 d 2 ELECTRON D GUN Screen WALL (b) (a) 11. <<d? (A) (B) D (C) d (D) V 12. h2 5h2 9h2 7h2 (A) 2mqd2 (B) 2mqd2 (C) 2mqd2 (D) 2mqd2 1001CT103516015 H-9/32
PHYSICS Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 13 14 1919147 N -x147N - d147N 13. d (A) R0(21/3 + 71/3) (B) R0(22/3 + 72/3) (C) R0(22/3 + 142/3) (D) R0(22/3 + 141/3) 14. - 18ke2 14ke2 18ke2 14ke2 (A) R0(22 / 3 142/ 3 ) (B) R0 (22/ 3 141/ 3 ) (C) R0 (22/ 3 141/ 3 ) (D) R0(22 / 3 72 / 3 ) H-10/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1PHYSICS –II : ( : 8) -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. -It= 0 -II -I -II (A) C (P) t =0 C +Q p charged bead 1001CT103516015 H-11/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 PHYSICS (B) C (Q) t=0 C C.M. of ball, nonuniform mass distribution C (C) (R) t= 0 B uniformly C charged disc (D) (S) t=0 B C (T) t = 0 C –III : III H-12/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1PHYSICS –IV : (: 20) 09 : +4 : 0 1. 50 m 10–8watt/m2 1.6 × 105N/m2 340 m/s x×10–3N/m2[x] 2. AB P=3T1/2 2 100K B C U=1V1/2 B 2 100 m3 C 1600 m3 xx6045 [:R = 8.3 J/mol-K] 1001CT103516015 H-13/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 PHYSICS3. 4g 4.2cm2 1.4cm2 1000W/m2 5°C nn5 4200J/kgºC 4. 73L i 11 H 7 Be 10 n Li 4 (keV) Q –1645 keV N2N35 5. X-V1=15.5 KV V2 = 31 KV K X-1.3 N N(hc = 1240 eV nm R = 1 × 107 /m ) 13 H-14/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1CHEMISTRY -2: –I(i) : ( : 32) (A), (B), (C) (D) : +4 : 0 : –1 1. (A) (B) (Tk) (C) (D) CMC 10–4 10–3 mol L–1 2. dN Temp. is dN For the same Ndu same for both gas Ndu gas at two different temp. M1 M2 u T1 T2 u (A) M2 > M1 (B) M1 > M2 (C) T2 > T1 (D) T1 > T2 3. :- (A) :LiF > NaF > KF > RbF > CsF (B) :BF3 < BCl3 < BBr3 < BI3 (C) : SiF4 > SiCl4 > SiBr4 > SiI4 (D) :Mn2O7 > MnO2 > MnO 1001CT103516015 H-15/32
Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 CHEMISTRY4. :- (A) Ag(desilverization) (B) (C) (D) 5. :- (A) [Co(H2O)4(en)]3+ (B) [Sc(H2O)3(NH3)3]3+ (C) [Ti(en)2(NH3)2]4+ (D) [Co(gly)3] 6. (Q) 2nd C12H22O11 + H2O P+Q (A) PQ - (B) PQ (C) P Q (D) PQ 7. (Tranquilizer) (P) CH2–CHO P (i) N2H4 (ii) H2, Ni (A) (B) (C) (D) 8. pH=7 (A) (B) (C) (D) H-16/32 1001CT103516015
Leader Course/Phase-III to VII/Score-I/05-03-2017/Paper-1CHEMISTRY –I(ii) : ( : 18) (A), (B), (C) (D ) : +3 : 0 : –1 9 10 Pt(s) | H2(g) | HBr(aq)| AgBr(s) | Ag(s) emf (polynomial) (F = 96500 C mol–1) Eº (volt) 0.07 5 104 T 298 3.5 106 T 298 2 K K 9. 298 K 1 (A) –6.755 kJ / mol (B) –5 kJ / mol (C) –13.51 kJ / mol (D) +6.755 kJ / mol 10. 298 K 1 (A) –48.25 JK–1mol–1 (B) –96.5 JK–1mol–1 (C) +48.25 JK–1mol–1 (D) +96.5 JK–1mol–1 1001CT103516015 H-17/32
CHEMISTRY Target : JEE (Main + Advanced) 2017/05-03-2017/Paper-1 11 12 - 11. :- (A) [Fe(-C5H5)2] (B) [Fe(H2O)5NO]2+ (C) [Ni(PF3)(CO)3] (D) 12. :- (A) [Mn(CO)6]+ > [Cr(CO)6] > [V(CO)6]– (B) [Cr(CO)6] > [Mn(CO)6]+ > [V(CO)6]– (C) [V(CO)6]– > [Mn(CO)6]+ > [Cr(CO)6] (D) [V(CO)6]– > [Cr(CO)6] > [Mn(CO)6]+ H-18/32 1001CT103516015
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