Home Explore 15- Question Report (15)

# 15- Question Report (15)

## Description: Question Report (15)

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 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-2/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 HAVE CONTROL  HAVE PATIENCE  HAVE CONFIDENCE  100% SUCCESS BEWARE OF NEGATIVE MARKING PHYSICS PART-1 : PHYSICS SECTION–I(i) : (Maximum Marks : 24)  This section contains EIGHT 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 very long straight conductor and isosceles triangular conductor lie in a plane and are separated from each other as shown in the figure. If a = 10 cm, b = 20 cm and h = 10 cm, find the coefficient of mutual induction. I b ah (A) 4.8 × 10–6 H (B) 3.6 × 10–8 H (C) 2.4 × 10–8 H (D) 1.2 × 10–8 H 2. A uniform magnetic field B increasing with time exists in a cylindrical region of centre O and radius R. The direction of magnetic field is inwards the paper as shown. The work done by external agent in taking a unit positive charge slowly from A to C via paths APC, AOC and AQC be WAPC, WAOC and WAQC respectively. Then : PB R COA Q (A) WAPC = WAOC = WAQC (B) WAPC > WAOC > WAQC (C) WAPC < WAOC < WAQC (D) WAPC = WAQC < WAOC Space for Rough Work 1001CT103516014 E-3/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 3. D.C. supply of 120V is connected to a large resistance X. A voltmeter of resistance 10 k placed in series in the circuit reads 4V. This is an unusual use of voltmeter for measuring very high resistance. The value of X is : X 10k V 120V () k (A) 390 k (B) 290 k (C) 190 k (D) 300 k 4. A loop is formed by two parallel conductors connected by a solenoid with inductance L and a conducting rod of mass m, which can freely (without friction) slide over the conductors. The conductors are located in a horizontal plane in a uniform vertical magnetic field with induction B. The distance between the conductors is equal to . At the moment t = 0 a constant force F starts acting on the rod. Angular frequency of motion of rod is . If F and m are doubled then: B F=constant L m, 1 (B)  become 2 times (A)  become 2 times (C)  become 2 times 1 (D)  become 2 times 5. In a photoelectric-effect experiment, if 'f' is the frequency of incident radiations on the metal surface and I is the intensity of the incident radiations, then mark the INCORRECT satement :- (A) If 'f' is increases, keeping I and work function constant, then stopping potential and maximum kinetic energy of photoelectron increases. (B) If distance between cathode and anode increases, stopping potential remains same. (C) If 'I' is increased keeping 'f' and work function constant, saturation current decreases but stopping potential remains same. (D) Work function is decreased keeping f and I constant. Then, stopping potential and maximum kinetic energy of photoelectrons increases. Space for Rough Work E-4/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 PHYSICS 6. An isolated parallel-plates capacitor has circular plates of radius 4.0 cm. Its gap is filled with a partially conducting material of dielectric constant K and conductivity 5.0 × 10–4 –1 m–1. When the capacitor is charged to a surface charge density of 15 µC/m2, the initial current between the plates is 1.0 A. Determine the value of dielectric constant K (approximately). (A) K = 1.32 (B) K = 4.26 (C) K = 8.74 (D) K = 2.1 7. An uncharged cubical conducting block has a spherical cavity within it. The block is placed in a region permeated by a uniform electric field which is directed upwards. Which of the following is a correct statement describing conditions in the interior of the block's cavity? (A) The electric field in the cavity is directed upwards (B) The electric field in the cavity due to charges induced on surface is directed downwards (C) There is non-uniform electric field in the cavity (D) The electric field in the cavity is of varying magnitude and is zero at the exact center. 8. At t = 0 cylinder has only translational velocity v and plank is at rest as shown in figure. Choose the correct statement : Rv P rough Smooth Q (A) Angular momentum of cylinder is conserved about point P (B) Angular momentum of cylinder + block is conserved about point P (C) Linear momentum of cylinder + block is non-conserved. (D) Kinetic energy of cylinder + block is conserved. Space for Rough Work 1001CT103516014 E-5/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 PHYSICS SECTION–I(ii) : (Maximum Marks : 24)  This section contains FOUR 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 experimental setup is shown in figure. It consists of two identical metal disks D and D' of radius b mounted on the conducting shaft SS'. A motor rotates the set at an angular velocity , which can be adjusted for measuring R. Two identical coils C and C' (of radius a and with N turns each) surround the disks. They are connected in such a form that the current I flows through them in opposite directions. The whole apparatus serves to measure the resistance R. The disks are connected to the circuit by brush contacts at their rims 1 and 4. The galvanometer G detects the flows of current through the circuit 1-2-3-4.  4 R S G 3 C D 1 2 S' D' I C' 9. Assume that the current I flowing through the coils C and C' creates a uniform magnetic field B around D and D', equal to the one at the centre of the coil. Compute the electromotive force  induced between the rims 1 and 4, assuming that the distance between the coils is much larger than the radius of the coils and that a >> b. (A)   N 0b2I (B)   N 0b2I (C)   N 0b2I (D)   N 20b2I a 4a 2a a 10. The resistance R is measured when G reads zero. Give R in terms of the physical parameters of the system. (A) R  N 20b2 (B) R  N 0b2 (C) R  N 0b2 (D) R  N 0b2 a 2a a 4a Space for Rough Work E-6/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 Paragraph for Questions 11 and 12 PHYSICS How does a particle interacts with another particle. Have you ever thought, how does one electron get to know the presence of other? Obviously there must be some information exchange between the particles. This information exchange takes place with the help of exchange of some elementary particles, which are called force carriers and force between two particles should depend on number of forces carriers reaching either of the particles. Since particles are essentially point objects so force carriers starting from one particle and reaching other should be inversely proportional to square of distance between the particles. Yukawa explained the behaviour of force carriers for strong nuclear forces. According to his theory strong forces are carried by carriers which decay with time. So according to Yukawa theory the intensity of strong forces should not only diminish with distance as the inverse of square of distance but also get diminished because force carriers decay with time. Number of carriers remaining after time t is proportional to e–t where  is decay constant. 11. Let two nucleons are separated by a fixed disance r and speed of force carries is v. Which of the following is most appropriate expression of Yukawa force between two nucleons :- (Where k is a constant) k e  r k e    r2 k e   r k e 2 r v r2  v  r2 v  r2 v2 (A) F  (B) F (C) F (D) F r 12. Let a nucleon is moving in a circle of radius a0 around a fixed nucleon due to Yukawa force as mentioned in question no. 11. For circular motion to be stable for small perturbations in orbit (small change in radius) while keeping angular momentum of nucleon constant, which of the following is true :- v (A)   a0 v (B)   a0 (C) v  a0  (D) Circular motion is stable for both v  a0 and v  a0   Space for Rough Work 1001CT103516014 E-7/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 Paragraph for Questions 13 and 14 It is a matter of common knowledge that two electrons kept at a distance repel each other. Have you ever thought how does one electron get to know the presence of other? Obviously there must be some information exchange between the particles. This information exchange takes place with the help of exchange of some elementary particles, which are called force carriers. Scientists have found these exchange particles for all types of interaction except gravitational forces. These particles are called gauge bosons. For strong nuclear forces these gauge bosons are called gluons, for electromagnetic forces these gauge bosons are photon and for weak interaction the gauge bosons are called W± gauge bosons. The great Scientist Richard Feynman explained the interaction between particles by now legendry Feynman diagram. The rules of these diagram is that time always goes up, So we start with incoming particle at bottom and outgoing particle at top while interaction takes place at the middle, as shown below. outgoing Interaction incoming In an actual situation, lets consider a case when a neutron gets converted into proton, an electron and an antineutrino, Feynman diagram will look like as shown below. (Outgoing) e– p W– v n (Boson) exchange particle (incoming) So conversion of neutron into proton involves emission of W– boson which quickly decays into electron and antineutrino. Let us consider a reaction in which a neutron and neutrino combine to given proton and electron. n + v  p + e– (Outgoing) e— v p+ W+ n (Boson) Space for Rough Work E-8/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 PHYSICS N 13. For a nucleus having Z ratio considerably less than 1, which of the Feynman diagram is most suitable, when it regains its stability. p e— p e+ W— v v (B) (A) W— n n n e+ n e+ v (C) v (D) p p e+ W— W+ e— 14. The Feynman diagram representing K-capture is given by : n W+ v n W— (A) p+ e— (B) p+ np nv (C) p+ W+ n (D) p+ W+ e— Space for Rough Work 1001CT103516014 E-9/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 Paragraph for Questions 15 and 16 Black box is a device having many terminals. Inside the box there is an intricate circuit which can be guessed by connecting the terminals to battery and measuring devices. Here we have such a block box with 4 terminals (ABCD). We connect an ideal battery in series with an ideal ammeter to terminals A and B. Terminals C and D are left open, the reading of ammeter is plotted against time here. The battery emf of 10 V. i t A Black C 1A 2s 3s B Box D 1s During this connection, an ideal voltmeter was connected across the terminals C and D. The graph of voltage against time was. V 2/ (in volt) 1s 2s 3s t 15. The possible circuit is : A (1/10H A (10H A (10/H A (10F C C (A) C (B) C (C) (D) (10/F (1/10F (5 (1/10H B B B D B D D D 16. If the possible circuit is as found in Q. 15 and if an ideal battery was connected to C and D and an ideal ammeter between A and B, its reading would be : iiii tt (A) (B) (C) (D) tt Space for Rough Work E-10/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 SECTION–I(iii) : (Maximum Marks : 12)  This section contains FOUR questions. PHYSICS  Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct  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 17. Match the list–I with list–II. List–II (Based on) List–I (Instruments) (1) Heating effect of current (2) Mutual induction (P) Hot wire ammeter (3) Magnetic effect of current (Q) Moving coil galvanometer (4) Wheatstone bridge (R) Transformer (S) Post office box S Codes : 4 4 P QR 3 (A) 1 2 3 2 (B) 1 3 2 (C) 2 4 1 (D) 3 1 4 Space for Rough Work 1001CT103516014 E-11/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 18. There is sufficient friction between object A & B (of same mass) to avoid any slipping between PHYSICS them and ground is smooth :- List-I List-II System Net force on object A Solid F (1) 2F sphere A 9 (P) B Solid F 7F sphere A 9 (Q) (2) B A F F (3) 2 (R) B F B 3F (S) Solid A (4) cylinder 4 Codes : P Q 2 3 RS (A) 2 3 14 (B) 1 2 41 (C) 2 1 34 (D) 34 Space for Rough Work E-12/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 19. Match the physical quantity of List-I with value given in List-II. List-I List-II PHYSICS 2M 2µF (P) 2M (1) 100 V time constant,  = .... sec (Q) 2M 2µF (2) 1.33 (3) 4 2M V 2M time constant,  = .... sec (R) C C C L L L C = 100 µF, L = 1H angular frequency  = ....... rad/sec (S) 2R (4) 6 Codes : R S (A) 3 (B) RC 2 (C) 2 (D) I +– 1 V Space for Rough Work I(0) = ..... I() I(0) and I() are current drawn from the battery at t = 0 and t  P QR 214 314 341 423 1001CT103516014 E-13/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 PHYSICS20. List-I shows different configurations of charge distribution. List-II gives the corresponding expressions of physical quantities mentioned in list-I :- List-I List-II (P) 6kq2 Four point charges are kept at the vertices of a regular (1) tetrahedron of side R.Total electrostatics energy of the configuration. R 6kq2  9  (Q) (2) R  40  Two thin spherical uniformly charged shells having charge q are separated by distance 4R. Total electrostatics energy of the configuration. (R) 6kq2 A solid non-conducting sphere of radius R having uniform (3) charge q is kept inside a thin spherical non-conducting shell of radius 2R having uniform charge q. Total electrostatics 6R energy of the configuration. (S) 6kq2 (4) (3) Work done in contracting the sphere having uniform charge q from radius 4R to R 40R Codes : 1001CT103516014 P QRS (A) 1 3 4 2 (B) 4 2 3 1 (C) 3 2 1 4 (D) 1 3 2 4 E-14/36

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 PART-2 : CHEMISTRY CHEMISTRY SECTION–I(i) : (Maximum Marks : 24)  This section contains EIGHT 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. In the galvanic cell shown below, which arrow indicates the spontaneous electron flow- AB CD Ag(s) Zn(s) Ag+(aq.)(1M) Zn2+(aq.)(1M) (A) A (B) B (C) C (D) D 2. What is the pH of a 0.2M solution of C6H5COONa ? (The Ka of C6H5COOH is 6.4 × 10–5) [log 2 = 0.3] (A) 5.25 (B) 5.4 (C) 8.6 (D) 8.75 3. The active ingredient in commericial bleach is sodium hypochlorite, NaOCl, which can be determined by iodometric analysis as indicated in these equations - OCl– + 2H+ + 2I–  I2 + Cl– + H2O l2 + 2S2O3–  S4O62– + 2I– If 1.356 g of a bleach sample requires 19.5 ml of 0.1 M Na2S2O3 solution, what is the percentage by mass of NaOCl in the bleach - (A) 2.68 % (B) 13.7 % (C) 5.35 % (D) 10.7 % 4. Which of the following oxide can be reduced by Al in thermite process ? (A) TiO2 (B) Cr2O3 (C) Fe2O3 (D) All of these Space for Rough Work 1001CT103516014 E-15/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 CHEMISTRY5. Which of the following reaction produce colourless and odourless gas :- (A) S2O3–2 + dil. H2SO4 (B) Zn + Conc. H2SO4 (C) Zn + dil. HCl (D) Zn + very dil. HNO3 6. A compound C5H10O3 shows positive iodoform test and (+) Tollen's test the compound is O OCH3 O OH (A) CH3–C–CH OCH3 (B) CH3–C–CH2–CH OCH3 O OH O OH (C) H–C–CH2–CH2–CH OCH3 (D) Me PH OH Q 7. H OH Me P & Q are respectively : – – (A) Bayer's reagent and OsO4 / OH (B) OsO4 / OH and CF3CO3H (C) Bayer's reagent and CF3CO3H / H3O (D) CF3CO3H / H3O and OsO4 / – OH 8. Consider the following reaction sequence : Ketone (A) PhMgBr (B) H+ / (C) (i) O3 O O The ketone (A) is : (ii) Zn / H2O + CH3–C–Ph C H H3C (A) (B) O O O O (C) C (D) H Space for Rough Work E-16/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 SECTION–I(ii) : (Maximum Marks : 24) CHEMISTRY  This section contains FOUR 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 Corrosion of metals is associated with electrochemical reactions. This also applies for for the formation of rust on iron surface, where the initial electrode reactions usually are- (1) Fe(s)  Fe2+(aq.) + 2e– ; E0  0.44V 298 K (2) O2(g) + 2H2O(l) + 4e–  4OH–(aq.) ; E0  0.4V 298 K An electrochemical cell in which these electrode reactions take place is constructed. The temperature is 25ºC. The cell is represented by the following cell diagram : Fe(s) |Fe2+(aq.) || OH–(aq.) | O2(g) | Pt(s) Nernst factor : RT ln10  0.06V at 298K F Faraday constant : F = 96500 Cmol–1 9. The standard electromotive force at 25ºC (A) –0.04 V (B) 0.04 V (C) 0.84 V (D) –0.84 V 10. The equilibrium constant at 25ºC for the overall cell reaction is (A) 1020 (B) 1028 (C) 1030 (D) 1048 Space for Rough Work 1001CT103516014 E-17/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 CHEMISTRY Paragraph for Questions 11 and 12 Solubility is an important factor for the measurement of the environmental pollution of salts. The solubility of a substance is defined as the amount that dissolves in a given quantity of solvent to form a saturated solution. This solubility varies greatly with the nature of the solute and the solvent, and the experimental conditions, such as temperature and pressure. The pH and the complex formation also may have influence on the solubility [log6 = 0.78] Ksp(BaSO4) = 10–10 Ksp[Mn(OH)2] = 1.6 × 10–13 Ksp(SrSO4) = 3 × 10–7 Ksp[Mg(OH)2] = 1.2 × 10–11 11. An aqueous solution contains BaCl2 and SrCl2 both in a concentration of 0.01M. Solid Na2SO4 is added slowly to this solution. The concentration of Ba2+ ion left when Sr2+ ion starts precipitating (B) 3.33 × 10–5 M (C) 3 × 10–5 M (D) 3 × 10–7 M (A) 3.33 × 10–6 M 12. To what pH must a solution be buffered if you wish to permit the maximum precipitation of Mn(OH)2 with no precipitation of Mg(OH)2, if the solution is 0.2M in both Mg2+ and Mn2+. (A) 5.32 (B) 7.89 (C) 8.89 (D) 6.78 Space for Rough Work E-18/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 CHEMISTRY Paragraph for Questions 13 and 14 'x' dil. HCl 'y' SO2 (aq.) 'z' (salt) (yellowish white) HgCl2(aq.) 'I' conc.(3HCl + HNO3) soluble Na2[Fe(CN)5NO] (Black) Purple colour 'J' 13. Which of the following statement is INCORRECT :- (A) 'y' produce black ppt. with Pb(OAc)2 solution (B) When 'y' react with acidic solution of FeCl3 produce 'z' (C) 'y' = SO2 , 'x' = SO3–2 (D) Al2S3 produce 'y' when react with water 14. Which of the following statement is CORRECT for 'J' :- (A) It is high spin complex (B) Oxidation number of Fe is +3 (C) It's magnetic moment is 24 B.M. (D) It have t 6 and eg0 electronic configuration 2g Space for Rough Work 1001CT103516014 E-19/36

CHEMISTRY Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 Paragraph for Questions 15 and 16 O O 2 Ph C Conc. NaOH Ph C + Ph–CH2–OH ONa H For the above reaction the following mechanism is suggested : O –O C – Ph—C—OH + OH (1) Ph H H –O –O – Ph—C—O– + H2O (2) Ph—C—OH + OH H H –O O rds O –O Ph–C (3) Ph—C—O– + Ph–C–H O– + Ph—C—H H H (4) Ph–CH2–O– + H2O Ph–CH2–OH 15. Order of above reaction is : (A) 2 (B) 3 (C) 4 (D) 1 16. A mixture of benzaldehyde and formaldehyde on heating with conc. NaOH solution gives: (A) benzyl alcohol and sodium formate (B) sodium benzoate and methyl alcohol (C) sodium benzoate and sodium formate (D) benzyl alcohol and methyl alcohol Space for Rough Work E-20/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 SECTION–I(iii) : (Maximum Marks : 12) CHEMISTRY  This section contains FOUR questions.  Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct  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 17. 10 ml of 0.1M H S solution is taken, to which the following quantities of 1M NaOH solution 2 have been added. Given : K (H S) = 10–7, K (H S) = 10–14. Match the following (neglect the volume of NaOH added 12 22 in final solution's volume to calculate pH.) List-I (Volume of NaOH added) List-II (pH of final solution) (P) 0 ml (1) 7 (Q) 0.5 ml (2) 14 (R) 1.0 ml (3) 4 (S) 1.5 ml (4) 10.5 Codes : P QRS (A) 4 3 4 4 (B) 3 1 4 2 (C) 1 2 3 4 (D) 3 1 2 4 Space for Rough Work 1001CT103516014 E-21/36

CHEMISTRY18. Match the following list :- Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 List - I (Reaction) List - II (Characteristic odour/product) (P) NO2 conc. HCl  (1) Coloured odd e species (Q) S2O3–2 dil.H2SO4 (2) Vinegar smell (R) HCO3 dil.H2SO4  (3) Yellowish-white turbidly (S) CH3COO dil.HCl  (4) Colourless gas comes out with brisk effervescence Codes : S P QR 2 3 (A) 4 3 1 2 (B) 2 4 1 2 (C) 1 3 4 (D) 3 1 4 Space for Rough Work E-22/36 1001CT103516014

19. Match the following list :- Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 CHEMISTRY List - I (Compound) List - II (Colour) (1) Brown colour (P) Fe(SCN)3 (2) Blue colour (3) Green colour (Q) [Fe(H2O)5(NO)]SO4 (4) Blood red colour (R) [Cu(NH3)4]SO4 S 2 (S) Cr2O3 3 3 Codes : 3 P QR Space for Rough Work (A) 1 4 3 (B) 1 4 2 (C) 4 2 1 (D) 4 1 2 1001CT103516014 E-23/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 CHEMISTRY20. List-I represents pair of compound & List-II represents reagents / tests used to differentiate them : List-II List-I O , (1) Baeyer's reagent (P) O (Q) , (2) I2 / NaOH OH , (3) Tollen's reagent (R) (4) Na metal O O OH OH (S) OCH3 , OH Codes : Q RS P 1 42 1 43 (A) 3 3 14 (B) 2 2 34 (C) 2 (D) 1 Space for Rough Work E-24/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 PART-3 : MATHEMATICS MATHEMATICS SECTION–I(i) : (Maximum Marks : 24)  This section contains EIGHT 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. Let ƒ(x) = x3 + x2 + 1; g(x) = x2 – 1. If the roots of ƒ(x) are x1,x2 and x3, then the value of g(x1).g(x2).g(x3) + 17g(x1x2x3) is - (A) 3 (B) 7 (C) 17 (D) 20 2. Consider the quadratic equation (x – 1)2 + x – 3 = 0. If  is of the form k k  1 , k  Q, then 2 roots of equation are necessarily- (B) imaginary (A) integers (D) cannot be predicted (C) rational numbers y  x2 dx  ƒ y  C , where C is an arbitrary constant,   3. If y2 – x2y – 2x = 0, x,y > 0 and x2  y y2  x then the value of ƒ'(y) at x = 1 is - 1 1 1 1 (A) 10 (B) 20 (C) 30 (D) 40 4. If the plane passing through the points (,1,1), (1,2,1) and (2,3,4) is parallel to the line   r ˆi  ˆj  2kˆ , ( R), then  is equal to- (A) – 1 (B) –1 (C) 3 (D) 0 2 2 Space for Rough Work 1001CT103516014 E-25/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 MATHEMATICS5. Let ƒ, g are two continuous and twice derivable functions such that ƒ(0) = ƒ(3) = 0; ƒ(1).ƒ(2) < 0; g(0) = g(3) = 0. Number of roots of equation ƒ\"(x).g(x) + ƒ'(x).g'(x) = 0 in (–1,5) cannot be (A) 2 (B) 3 (C) 4 (D) 5 6. If for unit vectors aˆ , bˆ and non-zero  , aˆ  bˆ  aˆ   and bˆ.c  0 , then volume of parallelopiped c c with coterminous edges aˆ , bˆ and c will be (in cu.units)- (A) 6 (B) 4 (C) 1 1 (D) 2 7. Range of the function ƒ(x) = cos(tan–1(sin(cot–1x))) is- (A)  1 ,1 (B) 1 ,1 (C)  1 ,1  (D) 1 ,1  2  2  2   2 8. If a,b,c  R are distinct numbers in A.P. ; a,,b are in G.P.; b,,c are also in G.P., then 2,b2,2 will be in- (A) H.P. (B) A.P. (C) G.P. (D) None Space for Rough Work E-26/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 SECTION–I(ii) : (Maximum Marks : 24) MATHEMATICS  This section contains FOUR 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 L1 be a line 5x – 7y = 35 which cuts x and y axis at A & B respectively. Variable line L2, which is perpendicular to L1 cuts x and y axis at C & D respectively. Locus of point of intersection of lines joining AD and BC is the curve S. 9. Area enclosed by curve S is (in sq. units)- (A) 37 (B) 49 37 49 (C) 2 (D) 2 10. Coordinates of a point P, which is farthest from origin, on S is- (A) (5,–7) (B) (5,7) (C) (7,5) (D) (7,–5) Space for Rough Work 1001CT103516014 E-27/36

MATHEMATICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 Paragraph for Questions 11 and 12 Let y = ƒ(x) is the solution of differential equation  x tan  y   dx  sec 2  y   xdy  ydx   0 ,   x    x    where ƒ 1   4  11. Value of ƒ 3 is -   (C) 3 6 (A) 3 3 (B) 6 3 6 (D) 3 12. If g x   ƒ  1  , then area bounded by g(x), x-axis and the lines x1 and x = 3 is (in sq. units)  x  3 (A)  n3 (B)  n3 (C)  n3 (D)  n3 6 4 3 2 Space for Rough Work E-28/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 MATHEMATICS Paragraph for Questions 13 and 14 Consider a parallelopiped OAC'BA'CB'O' where OA = 2, OB = 3 and OC = 4. If areas of OPQ, OQR and OPR are least (as shown in figure), then Z R C A' B' O' O B A XP C' QY 13. Volume of tetrahedron OPQR, is- (C) 48 cubic units (D) 16 cubic units (A) 24 cubic units (B) 32 cubic units 3 14. Shortest distance between BA' & PR is- (D) 5 (A) 3 7 4 (B) (C) 5 5 Space for Rough Work 1001CT103516014 E-29/36

MATHEMATICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 Paragraph for Questions 15 and 16 Let P(z) is a variable point in argand plane which satisfies z  z   z  2i 1  0 15. If amp(z) is least, then |z| is equal to- (A) 1 (B) 2 (C) 3 (D) 2 16. If P(z) also satisfies arg(z + 1 – 2i)    , then number of such points P, is- 4 (A) 1 (B) 2 (C) 0 (D) infinite Space for Rough Work E-30/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 SECTION–I(iii) : (Maximum Marks : 12) MATHEMATICS  This section contains FOUR questions.  Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct  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 I f none of t he bubbles is dar k ened. Negative Marks : –1 In all other cases 17. Match List-I with List-II and select the correct answer using the code given below the list. List-I List-II (P) If  are three positive, consecutive terms of a G.P., (1) 8 harmonic mean of  being 12 and arithmetic mean 16 of  being 3 , then geometric mean of  is  r  (Q) If the line aˆi  2ˆj 12kˆ ,   R, a > 0 makes an (2) 4  isosceles  triangle with the planes r. 2ˆi  ˆj  3kˆ 1 and   r. iˆ  2ˆj  3kˆ 1, then ten's digit of a is (R) Value of x in (0,2) where ƒ(x) = [sin[x]] is (3) 6 discontinuous, is (4) 7 (where [.] denotes greatest integer function) (S) Let ƒ be a derivable function satisfying x ƒ x  x2   et ƒ x  t dt then degree of polynomial 0 function ƒ(x2) is Codes : P QRS (A) 1 2 4 3 (B) 3 4 2 1 (C) 3 2 4 1 (D) 1 4 2 3 Space for Rough Work 1001CT103516014 E-31/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 MATHEMATICS18. Match List-I with List-II and select the correct answer using the code given below the list. List-I List-II (P) If mutually perpendicular tangents can be drawn (1) 2 from A(0,–b) to the hyperbola x2  y2 1 which a2 b2 a2 touches the hyperbola at B and C, then value of b2 is (Q) If x – y + 2 = 0 is a tangent to parabola y2 = 4ax + b, (2) 4 then ka2 – 2ka + b = 0, where k is (R) Least distance of point (–4,7) from y2 = 4x is  2 , (3) 5 where  is (4) 10 (S) If tangents drawn at A(x1,y1), B(x2,y2) on the ellipse x2  y2  1 are mutually perpendicular, then 10 1 the value of x1x2 is y1y2 Codes : P QR S (A) 1 2 3 4 (B) 4 3 2 4 (C) 1 4 3 1 (D) 3 2 2 3 Space for Rough Work E-32/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 19. Match List-I with List-II and select the correct answer using the code given below the list. MATHEMATICS List-I List-II (P) If for a derivable function ƒ defined for all real (1) 0 numbers, ƒ(x + y), ƒ(x).ƒ(y) and ƒ(x – y) are in A.P.  x,y  R and ƒ(0)  0, then value of ƒ'(2) + ƒ'(–2) is    dx is  , where k is (2) 3 (3) 4 (Q) Value of (4) 6 0 x2  1 x5 1 k (R) Number of solutions of the equation sin4x + cos7x = 1 in (–) is (S) If 3  i  a  ib c  id  , (where a, b, c, d > 0 and ac > bd) then tan1  a   tan1  c  is equal to 5 where k is (where i  1 )  b   d  k, Codes : P QRS (A) 1 3 4 2 (B) 3 4 3 4 (C) 2 4 4 3 (D) 1 3 2 4 Space for Rough Work 1001CT103516014 E-33/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 MATHEMATICS20. If ƒ(x) = x3 + ax2 + bx + c where ƒ'(–1) = 0 and ƒ\"(2) = 0 Match List-I with List-II and select the correct answer using the code given below the list. List-I List-II (P) ƒ'(5) is (1) –15 (Q) If c > 0, then minimum value of ƒ(–1) is (2) –6 (R) a is (3) 0 (S) b is (4) 8 Codes : P QR S (A) 3 4 1 2 (B) 2 1 3 4 (C) 3 4 2 1 (D) 2 3 4 1 Space for Rough Work E-34/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 Space for Rough Work 1001CT103516014 E-35/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 QUESTION PAPER FORMAT AND MARKING SCHEME : 16. The question paper has three parts : Physics, Chemistry and Mathematics. 17. Each part has one section 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 8 corresponding to — of the 24 the correct option bubbles is is darkened darkened Paragraph +3 0 –1 In all Based If only the bubble If none other cases I(ii) (Single 8 corresponding to — of the 24 correct the correct option bubbles is option) is darkened darkened Matching +3 0 –1 Lists Type If only the bubble If none In all other I(iii) (Single 4 corresponding to — of the cases 12 correct the correct option bubbles is option) is darkened 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-36/36 Your Target is to secure Good Rank in JEE 2017 1001CT103516014

Paper Code : 1001CT103516014 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 : 14 - 01 - 2017 Time : 3 Hours PAPER – 2 Maximum Marks : 180   1.  2. (ORS) 3.  4.  5. 3620  6.   7.  8.   9.   : 10.  11.    12. : 13.  14.   15. g = 10 m/s2              

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 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-2/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2PHYSICS HAVE CONTROL  HAVE PATIENCE  HAVE CONFIDENCE  100% SUCCESS BEWARE OF NEGATIVE MARKING -1:   –I(i) : ( : 24)         (A), (B), (C) (D)                                    : +3             : 0            : –1     1.  a=10 cm, b = 20 cm h= 10 cm  I b ah (A) 4.8 × 10–6 H (B) 3.6 × 10–8 H (C) 2.4 × 10–8 H (D) 1.2 × 10–8 H 2. BR O  ACAPC,  AOC AQC WAPC,WAOCWAQC  PB R COA Q (A) WAPC = WAOC = WAQC (B) WAPC > WAOC > WAQC (C) WAPC < WAOC < WAQC (D) WAPC = WAQC < WAOC  1001CT103516014 H-3/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 3. X120V DC 10k 4V X:  X 10k V 120V () k (A) 390 k (B) 290 k (C) 190 k (D) 300 k 4. L m B  t=0 F  Fm :  B F=constant L m, 1 (B)   2  (A)   2  (C)  2  (D)  12  5. fI  (A) If  (B)  (C) fI (D) f I    H-4/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 6. 4.0cmK PHYSICS 5 .0 × 10–4 –1 m–1 15µC/m2 1.0AK  (A) K = 1.32 (B) K = 4.26 (C) K = 8.74 (D) K = 2.1 7.   (A)  (B)  (C)  (D)  8. t=0v  Rv P rough Smooth Q (A) P (B) P (C)  (D)   1001CT103516014 H-5/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 –I(ii) : (: 24)                (A), (B), (C) (D )                                   : +3             : 0            : –1     9 10    bDD'  SS' R CC' ( a N )I  R 14 G1-2-3-4  R 4 G S 3 1 C D 2 D' S' I C' 9. CC'I,DD' B  14 a>>b (A)   N 0b2I (B)   N 0b2I (C)   N 0b2I (D)   N 20b2I a 4a 2a a 10.  G R R  (A) R  N 20b2 (B) R  N 0b2 (C) R  N 0b2 (D) R  N 0b2 a 2a a 4a  H-6/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 PHYSICS 11 12                  (Yukawa)    te–t  11. r v (k)  k e  r k e    r2 k e   r k  2 r v r2  v  r2 v  r2 v2 (A) F (B) F  (C) F (D) F  e r 12. 11a0                  (A) v  a0  (B) v  a0  v (C)   a0 (D) va0 v  a0    1001CT103516014 H-7/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 13 14         (gauge bosons) (gluons) W±  RichardFeynman Feynman     outgoing Interaction incoming  Feynman  (Outgoing) e– p W– v (Boson) exchange particle n (incoming) W–(boson)     n + v  p + e– (Outgoing) e— v p+ W+ n (Boson)  H-8/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2PHYSICS 13. NZ Feynman  p e— p e+ W— v v (B) (A) W— n n n e+ n e+ (C) v (D) v p p W— W+ 14. Feynm an K- n W+ v n W— e+ (A) p+ e— (B) p+ e— np n W+ v (C) p+ W+ n (D) p+ e—  1001CT103516014 H-9/36

PHYSICS Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 15 16     (ABCD) ABCD 10V i t A Black C 1A 2s 3s B Box D 1s CD   V 2/ (in volt) 1s 2s 3s t 15.  A (1/10H A (10H A (10/H A (10F C C (A) C (B) C (C) (D) (10/F (1/10F (5 (1/10H B B B D B D D D 16. CD  A  B  iiii tt (A) (B) (C) (D) tt  H-10/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 –I(iii) : ( : 12)      PHYSICS   (A), (B), (C)(D )     lgh    :  : +3             : 0            : –1     17. –I  –II    –I () –II () (P)   (1)  (Q)  (2)  (R)  (3)  (S)  (4)  Codes : P QRS (A) 1 2 3 4 (B) 1 3 2 4 (C) 2 4 1 3 (D) 3 1 4 2  1001CT103516014 H-11/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 18. AB PHYSICS  –I –II  A  Solid F (1) 2 F sphere A 9 (P) B Solid F (2) 7 F sphere A 9 (Q) B A F F (3) 2 (R) B F B 3F (S) Solid A (4) cylinder 4 Codes : P Q 2 3 RS (A) 2 3 14 (B) 1 2 41 (C) 2 1 34 (D) 34  H-12/36 1001CT103516014

Leader Course/Phase-III, IV & V/14-01-2017/Paper-2 19. -I -II PHYSICS -I -II 2M 2µF (P) 2M (1) 100 (2) 1.33 V (3) 4 ,  = .... sec (Q) 2M 2µF 2M V 2M ,  = .... sec (R) C C C L L L C = 100 µF, L = 1H   = ....... rad/sec (S) 2R (4) 6 Codes : R (A) (B) RC (C) (D) I +– V I(0) = ..... I() I(0) I() t= 0  t   P QRS 2143 3142 3412 4231  1001CT103516014 H-13/36

Target : JEE (Main + Advanced) 2017/14-01-2017/Paper-2 20. -I -II  PHYSICS -I -II 6kq2 (P) (1) R R   6kq2  9  (Q) (2) R  40  q4R   6kq2 (R) (3) 6R R q2R q  (S) 6kq2 (4) (3) q4RR  40R 1001CT103516014 Codes : P QRS (A) 1 3 4 2 (B) 4 2 3 1 (C) 3 2 1 4 (D) 1 3 2 4 H-14/36