ENGLISH Form Number : Paper Code : 1001CT102116063 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) LEADER COURSE (SCORE-I) & ENTHUSIAST COURSE (SCORE-II) Test Type : FULL SYLLABUS Test Pattern : JEE-Main TEST DATE : 12 - 03 - 2017 Important Instructions Do not open this Test Booklet until you are asked to do so. 1. Immediately fill in the form number on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited. 2. The candidates should not write their Form Number anywhere else (except in the specified space) on the Test Booklet/Answer Sheet. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the question paper A,B,C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 6. One Fourth mark will be deducted for indicated incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the Answer Sheet. 7. Use Blue/Black Ball Point Pen only for writting particulars/marking responses on Side–1 and Side–2 of the Answer Sheet. Use of pencil is strictly prohibited. 8. No candidate is allowed to carry any textual material, printed or written, bits of papers, mobile phone any electronic device etc, except the Identity Card inside the examination hall/room. 9. Rough work is to be done on the space provided for this purpose in the Test Booklet only. 10. On completion of the test, the candidate must hand over the Answer Sheet to the invigilator on duty in the Room/Hall. However, the candidate are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet. Your Target is to secure Good Rank in JEE (Main) 2017 Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PART A - PHYSICS 1. A right circular cylinder has a mass m, radius r, 3. A uniform magnetic field B that is and a height h. The cylinder is completely submerged in a fluid of density , as shown in perpendicular to the plane of the page now the diagram and is released. What is the passes through the loops, as shown. The field magnitude of the net force on the cylinder is confined to a region of radius a, where a < b, initially? and is changing at a constant rate. The induced emf in the wire loop of radius b is . What is the induced emf in the wire loop of radius 2b :- (1) rgh + mg (2) | rgh – mg| (3) r2gh + mg (4) r2gh – mg| 2. The distribution of relative intensity I () of blackbody radiation from a solid object versus the wavelength is shown in the figure. If the (1) 0 (2) /2 (3) (4) 2 Wien displacement law constant is 2.9 × 10–3 mK, 4. If the maximum and minimum voltage of an what is the approximate temperature of the object? amplitude modulated wave are V and V max min respectively, then modulation factor is :- (1) m Vmax (2) m Vmin Vmax Vmin Vmax Vmin (1) 10K (2) 50K (3) 250K (4) 1500K (3) m Vmax Vmin (4) m Vmax Vmin Vmax Vmin Vmax Vmin SPACE FOR ROUGH WORK 1001CT102116063 E-1/16
Target : JEE (Main + Advanced) 2017/12-03-2017 5. During which time interval is the particle 8. If the B-H curves of two samples of P and Q of described by these position graphs at rest? iron are as shown below, then which one of the xy following statements is CORRECT ? 0 1 2 3 t(s) 0 t(s) B B 0 4 0 1 2 34 H H (1) 0 – 1 s (2) 1 – 2 s (3) 2 – 3 s (4) 3 – 4 s 6. Consider telecommunication through optical Sample P Sample Q fibres. Which of the following statements is (1) Both P and Q are suitable for making NOT true ? permanent magnet (1) Optical fibres can be of graded refractive (2) P is suitable for making permanent magnet index and Q for making electromagnet (2) Optical fibres are subjected to (3) P is suitable for making electromagnet and electromagnetic interference from outside Q is suitable for permanent magnet (3) Optical fibres have extremely low (4) Both P and Q are suitable for making transmission loss electromagnets (4) Optical fibres may have homogeneous core 9. Statement-1 : A photodiodes operates in with a suitable cladding reverse bias. 7. Two polaroids are placed in the path of Statement-2 : The fractional change due to the photo-effects on the minority carrier dominated unpolarized beam of intensity I such that no 0 reverse bias current is more easily measurable than the fractional change in the forward bias light is emitted from the second polaroid. If a current. (1) Statement-1 is true, Statement-2 is true, third polaroid whose polarization axis makes an angle with the polarization axis of first Statement-2 is the correct explanation of polaroid, is placed between these polaroids, Statement-1. then the intensity of light emerging from the last polaroid will be :- (1) I0 sin 2 2 (2) I0 sin 2 2 (2) Statement-1 is true, Statement-2 is true, 8 4 Statement-2 is not the correct explanation of Statement-1. (3) I0 cos2 (4) I0 cos4 (3) Statement-1 is true, Statement-2 is false. 2 (4) Statement-1 is false, Statement-2 is true. SPACE FOR ROUGH WORK E-2/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 10. A particle of mass 5 × 10–5 kg is placed at lowest 13. All wires have same resistance and equivalent point of smooth parabola x2 = 40y resistance between A and B is R. Now keys (x and y in m). If it is displaced slightly such are closed, then the equivalent resistance will that it is constrained to move along parabola, become :- angular frequency of oscillation (in rad/s) will be approximately:- K1 y B g A x K2 (1) 2 (2) 10 (3) 1 (4) 5 7R 7R (3) 7R R (1) (2) (4) 2 3 9 3 11. The average degree of freedom per molecule 14. The Davisson-Germer experiment that first of a gas is 6. The gas performs 25 J work, while demonstrated the wave nature of matter used expanding at constant pressure. The heat electron accelerated to 54 V. Determine the absorbed by the gas is :- wavelength of the electrons in the Davisson- (1) 75 J (2) 100 J (3) 150 J (4) 125 J Germer experiment. 12. The potential energy of a diatomic molecule is (1) 1.67 Å (2) 1.21 Å given by U A B . A and B are positive (3) 3.45 Å (4) 5.21 Å r12 r6 constants. The distance r between them at 15. Liquid cools from 50ºC to 45ºC in 5 minutes equilibrium is :- and from 45ºC to 41.5ºC in the next 5 minutes. (1) A 1/ 6 (2) 2A 1/ 6 The temperature of the surrounding is :- B B [Assume newton's law of cooling is applicable] A 1/ 6 (1) 27ºC (2) 40.3ºC (3) 23.3ºC (4) 33.3ºC (3) 2B (4) None of these SPACE FOR ROUGH WORK 1001CT102116063 E-3/16
Target : JEE (Main + Advanced) 2017/12-03-2017 16. A point charge +Q is positioned at the centre 18. The radioactive sources A and B have half lives of the base of a square pyramid as shown. The of 2hr and 4hr respectively, initially contain the flux through one of the four identical upper same number of radioactive atoms. At the end faces of the pyramid is :- of 2 hours, their rates of distintegration are in the ratio:- (1) 4 : 1 (2) 2 : 1 (3) 2 : 1 (4) 1 : 1 +Q 19. A satellite S moves around a planet P in an Q Q elliptical orbit as shown in figure. The ratio of (1) (2) the speed of the satellite at point a to that at point b is :- 160 40 Q (4) None of these (3) s 80 a rP 17. If the behavior of light rays through a convex 3r b lens is as shown in the adjoining figure, then; (1) 1 : 9 (2) 1 : 3 (3) 1 : 1 (4) 3 : 1 µ2 µ µ2 20. The maximum possible acceleration of a train moving on a straight track is 10 m/s2 and maximum possible retardation is 5 m/s2. If maximum achievable speed of train is 10m/s then minimum time in which train can (1) µ = µ2 (2) µ < µ2 complete a journey of 135m starting from rest (3) µ > µ2 (4) µ µ2 and ending at rest, is :- (1) 5s (2) 10s (3) 15s (4) 20s SPACE FOR ROUGH WORK E-4/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 21. A metal wire PQ slides on parallel metallic rails 23. Correct output for given logic circuit and inputs having separation 0.25 m, each having is :- negligible resistance. There is a 2 resistor and A C 10V battery as shown in figure. There is a B uniform magnetic field directed into the plane of the paper of magnitude 0.5 T. Aforce of 0.5N to the left is required to keep the wire PQ A B moving with constant speed to the right. With T0 2T0 t what speed is the wire PQ moving ? 2T0 (Neglect self inductance of the loop) 10V P output output 2 0.25 m (1) t (2) output output Q (1) 8 m/s (2) 16 m/s (3) 24 m/s (4) 32 m/s (3) T0 (4) t 22. A rod of length L is held vertically on a smooth 2T0 horizontal surface. The top end of the rod is 24. A transverse wave is passing through a given a gentle push. At a certain instant of time, stretched string with a speed of 20 m/s. The when the rod makes an angle with horizontal tension in the string is 20 N. At a certain point the velocity of COM of the rod is v0. The P on the string, it is observed that energy is velocity of the end of the rod in contact with being transferred at a rate of 40 mW at a given the surface at that instant is : instant. Find the speed of point P. (1) v0 cot (2) v0 cos (1) 40 cm/s (2) 20 cm/s (3) v0 sin (4) v0 tan (3) 2 mm/s (4) 20 mm/s SPACE FOR ROUGH WORK 1001CT102116063 E-5/16
Target : JEE (Main + Advanced) 2017/12-03-2017 25. In the circuit shown, power factor of circuit is 28. A hollow sphere of radius R is filled completely 3 with an ideal liquid of density . sphere is 1 and power factor of box is 5 . Find reading moving horizontally with an acceleration 2g, of ammeter :- where g is acceleration due to gravity in the C = 1mF space. If minimum pressure of liquid is P0, then Box pressure at the centre of sphere is :- A 2g 90V, 25 rad/s (1) 5A (2) 6A (3) 4A (4) 3A 26. When the gap is closed without placing any object in the screw gauge whose least count is (1) P0 + gR (2) P0 + g R 2 0.005 mm, the 5th division on its circular scale (3) P0 + g R 5 with the reference line on main scale, and when (4) P0 gR a small sphere is placed reading on main scale 5 advances by 4 divisions, whereas circular scale 29. Two steel wires of same length but radii r and reading advances by five times to the 2r are connected together end to end and tied corresponding reading when no object was to a wall as shown. The force stretches the placed. There are 200 divisions on the circular combination by 10 mm. How far does the scale. The radius of the sphere is :- midpoint A move :- (1) 4.10 mm (2) 4.05 mm Radius= r A Radius= 2 r (3) 2.10 mm (4) 2.05 mm LL 27. A household refrigerator with a coefficient of (1) 2 mm (2) 4 mm (3) 6 mm (4) 8 mm performance 1.2 removes heat from the refrigerated space at the rate of 60kJ/min.What 30. A closed organ pipe of length is sounded together with another closed organ pipe of would be cost of running this fridge for one length + x (x << ) both in fundamental mode. month (30 days) (assuming each day it is used for 4 hours and cost of one electrical unit If v = speed of sound, the beat frequency is 6Rs.) :- heard is: (1) 180 Rs. (2) 300 Rs. vx vx vx2 vx (3) 480Rs. (4) 600 Rs. (1) 22 (2) 42 (3) 4 (4) 2 SPACE FOR ROUGH WORK E-6/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 PART B - CHEMISTRY 31. Select correct statement regarding interstitial 35. PCl5 + P4O10 o xygen cont aining hydrides: compound (X). (1) They do not conduct electricity in solid How many moles of an oxy acid are formed state. on hydrolysis of 1 mole (X). (2) Their density is more than that of parent (1) Zero (2) 1 (3) 4 (4) 3 metal. (3) Their composit ion may vary with 36. Select correct statement about anionic part of temperature. Zeise’s salt, K[Pt (2–C2H4)Cl3] : (4) Each metal of d-block form such type of (1) C=C bond is coplanar with Pt and 3 Cl hydride. atoms. 32. Find incorrect comparison of second ionisation (2) oxidation state of Pt is +3. energy : (1) Te < Sb (2) In > Sr (3) HCH bond angle is not same as that in free (3) He > B (4) Fe < Fe+ ethylene molecule. 33. Which property is not typical of a refractory (4) Pt-Cl bond is not affected by synergic material? bonding between Pt and C2H4 (1) Chemically inert 37. Maximum number of P–H bond(s) are there in a molecule of (2) High thermal conductivity (1) Orthophosphorous acid (3) High melting point (2) Pyrophosphoric acid (4) High electrical conductivity (3) Pyrophosphorous acid 34. Potassium manganate is not formed in (4) Hypophosphoric acid (1) KMnO4 38. Which of the following species is paramagnetic ? (2) KMnO 4 KOH (1) K3[Co(C2O4)3] (2) NO[PF6] ( conc.) (3) [NMe4]O3 (4) H[BF4] (3) MnO2 KOH O2 39. How many different types of OBO angles are (4) KMnO4 H2SO4 there in sodium peroxoborate ? ( conc .) (1) 2 (2) 3 (3) 5 (4) 1 SPACE FOR ROUGH WORK 1001CT102116063 E-7/16
Target : JEE (Main + Advanced) 2017/12-03-2017 40. Anhydrous AlCl3 is formed in : 44. 2gm of benzoic acid dissolved in 25gm of (1) AlCl .6H O benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for 32 benzene is 4.9K kg mol–1. What is % degree of (2) Al(Red hot) + HCl(Moist) (3) Al O + C + Cl (dry) association of acid 23 2 (1) 99.2 (3) 90 (4) All of these (2) 95 (4) 0.8 41. The radii of two of the first four Bohr’s orbits of the hydrogen atom are in the ratio 1 : 4 The energy difference between them may be : 45. On addition of one ml solution of 10% NaCl (1) Either 12.09 eV or 10.2 eV to 10 ml gold sol in the presence of 0.0250 gm (2) Either 2.55 eV or 10.2 eV of starch the coagulation is just prevented. (3) Either 13.6 eV or 3.4 eV Starch has the following gold number : (4) Either 3.4 eV or 0.85 eV (1) 0.025 (2) 0.25 42. Pure ammonia is placed in a vessel at a (3) 2.5 (4) 25 temperature where its dissociation constant () is appreciable. At equilibrium : 46. Which does not have hexagonal crystal system (1) K does not change significantly with (1) Graphite (2) Zinc oxide p pressure. (2) does not change with pressure. (3) CdS (4) HgS (3) concentration of NH does not change with 3 pressure. 47. By how much will the potential of half (4) concentration of hydrogen is less than that cell Cu2+ / Cu change if the solution is diluted of nitrogen to 100 times at 298 K 43. A system of 100 kg mass undergoes a process in which its specific entropy increases from (1) Increases by 59 mV 0.3 kJkg–1K–1 to 0.4kJkg–1K–1. At the same (2) Decreases by 59 mV time, the entropy of the surrounding decreases from 80 kJK–1 to 75 kJK–1. (3) Increases by 29.5 mV Find the (S)universe in kJK–1. (4) Decreases by 29.5 mV (1) 0 (2) 5 (3) 10 (4) 15 SPACE FOR ROUGH WORK E-8/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 48. Solubility curves of four ionic salts X, Y, Z, W O CCH O are given below : H2/Pd ‘X’ 51. Cl–C–CH2– C–CH2–CH2 –C–CH3 BaSO4 CH2 CN In which case the value of Hsol. < 0 Which statement is correct for product ‘X’ ? (1) 1º alcohol functional group is present in ‘X’ (2) Ethyl group is present in ‘X’ (1) X (2) Y (3) Z (4) W. (3) 2º-alcohol functional group is present in ‘X’ 49. Match of following (where Urms = root mean (4) Alkene functional group is present in ‘X’ square speed, U = average speed, U = most 52. Which of the following can not show av mp probable speed) geometrical isomerism ? List I List II O (a) Urms / Uav (i) 1.22 O (1) (b) Uav / Ump (ii) 1.13 O O (c) U / U (iii) 1.08 (2) CH3 – CH = CH – CH3 rms mp (1) (a)-(iii), (b)-(ii), (c)-(i) CH3 CH3 (2) (a)-(i), (b)-(ii), (c)-(iii) (3) (3) (a)-(iii), (b)-(i), (c)-(ii) Br Cl (4) (a)-(ii), (b)-(iii), (c)-(i). Cl (4) Br 50. 27 Al is a stable element. 29 Al is expected to 53. Which one is an example of bacteriocidal 13 13 disintegrate by : antibiotics ? (1) -emission (2) -emission (1) sulfa drugs (2) Tetracyclin (3) Positron emission (4) Proton emission (3) Erythromycin (4) Pencillin ‘G’ SPACE FOR ROUGH WORK 1001CT102116063 E-9/16
Target : JEE (Main + Advanced) 2017/12-03-2017 54. In given reactions which one is correct ? OH CH3 55. CH3 – CH – CH3 Cu/300ºC Product (X) (1) CH2= CH – C – COOH CH3 Which of the following is correct about product 'X' ? NaOH + CaO/ CH2= CH – CH – CH 3 (1) Degree of unsaturation of product is two. CH3 (2) Product can reduce tollens reagent. COOH (3) Product is an aldehyde. (4) Product can undergo aldol condensation. H CH3 (i) Ag2O 56. In given options, which one is correct for rate (2) CH3 (ii) Br2/CCl4/ of hydrolysis ? H D O CH2–Cl CH2–Cl CH2–Cl CH2–Cl Br N > > NO2 O NO2 > (1) NO2 NO2 NO2 H CH3 Cl Cl Cl Cl H CH3 + Enantiomer (2) O O O O> > > O D (3) CH3–CH–CH2–NO2 OH–/H2O CH2=CH–CH2–NO2 O O O O C–Cl C–Cl C–Cl C–Cl (3) F > > > O CHO C NO2 CH3 O––CH3 O Cl Cl Cl Cl (4) (i) 50% KOH (4) > >> CH2 (ii) H+/ CHO SPACE FOR ROUGH WORK E-10/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 57. Which one is correct for heat of hydrogenation 59. Which of the following is a non-reducing sugar? (1) > (1) glucose (2) > (2) fructose (3) sucrose (3) > (4) maltose (4) CH2 = CH2 > CH3 – CH = CH – CH3 60. In which of the following reaction alcohol is formed as major product (1) N 58. Which of the following statement is false for CH3 NaNO2 HCl Biodegradable polymers. 3 (1) Biodegradable polymers are specific type NH2 of polymers that breaks down after it’s (2) NaN0O52ºCHCl intended purpose to result in natural byproducts. NH2 (3) NaNO2 HCl (2) These polymers are found both naturally and synthetically. (3) PHBV and Nylon2-Nylon6 are example of NH–CH3 Biodegradable polymers. (4) NaNO2 HCl (4) Biodegradable polymers can not be synthesized by condensation reactions. SPACE FOR ROUGH WORK 1001CT102116063 E-11/16
Target : JEE (Main + Advanced) 2017/12-03-2017 PART C - MATHEMATICS 61. The real value of for which the expression 6 5 . Value of 1 i cos 4sin 9sin 21sin 39sin 51sin 69sin 81 1 2i cos is a real number is (n I) sin 54 (1) (2n + 1) (2) (2n + 1)/2 is equal to (3) 2n (4) None of these 1 1 62. If m and 2 are the mean and variance of (1) 16 (2) 32 random variable x, whose distribution is 11 (3) 8 (4) 4 given by Xx 0 1 2 3 4 66. Number of common t angents of t he P(X x) 1 1 0 1 , then (x 2)2 (y 2)2 0 32 6 ellipse 1 and the circle (1) m = 2 = 2 94 x2 + y2 – 4x + 2y + 4 = 0 is - (2) m = 1, 2 = 2 (1) 0 (2) 1 (3) m = 2 = 1 (4) m = 2, 2 = 1 (3) 2 (4) more than 2 63. Mr. A has six children and atleast one child is a 67. Consider the equation x2 + x + = 0 having girl, then probability that Mr. A has 3 boys and roots , such that Also consider the inequality 3 girls, is - 20 1 |y – | – < , then (1) 63 (2) 6 (1) inequality is satisfied by exactly two integral 5 1 values of y (3) 11 (4) 32 (2) inequality is satisfied by all values of 64. log cos ec2 sin2 3 equals to : y (– 4, 2) 1 8 8 (3) Roots of the equation are of same sign 8 (1) 0 (2) 1/2 (4) x2 + x + > 0 x [–1, 0] (3) 1 (4) not defined SPACE FOR ROUGH WORK E-12/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 68. If a, b, c are pth, qth, rth terms of an H.P. and 71. = (q r) ˆi + (r p) ˆj + (p q) kˆ , = ˆi ˆj kˆ , (a b) (b c) (a b b c c a) is u a b c then: (1) [a b c][(b a a c)b (| b |2 b c)a] (2) [a b c][(b a a c)b (| b |2 b c)a] (1) u , are parallel vectors (3) [a b c][(b a a c)b (| b |2 b c)a] (2) u , are orthogonal vectors (4) [a b c][(a c b a)b (| b |2 b c)a] (3) . = 1 3 u 72. sin2 sin2 2d is equal to - (4) = ˆi + ˆj + kˆ 3 u 69. The image of the point (1, 2, –1), on the plane 3 5 x1 y 3 z 2 (1) (2) (3) (4) 6 3 2 1 2 2 containing the line and the 73. If rate of change of area of a square S is equal point (0, 7, –7), is. to its side length, if rate of change of side of S (1) 1 , 7 , 1 (2) 1 , 2 , 7 is same as that of a cube C, then rate of change 3 3 3 3 3 3 of volume of C, at the time when its side length is 2 units, will be- 1, 7 1 2 7 (3) 3 0, 3 (4) 3 , 3 , 3 (1) 24 units/sec. (2) 12 units/sec. 70. The bisectors of the angles of a parallelogram (3) 6 units/sec. (4) 3 units/sec. enclose a: 74. Let ƒ is a differentiable function satisfying D C ƒ(x + 2y) = 2yƒ(x) + xƒ(y) – 3xy + 1 x, y R such that ƒ'(0) = 1, then ƒ(2) is equal to - S PR (1) 4 (2) 1 (3) 5 (4) 3 Q 75. Minimum distance between the curves AB x2 + y2 + 4x + 16y + 66 = 0 and y2 = 8x is - (1) Rhombus (2) Rectangle (1) 3 2 units (2) 5 2 units (3) Square (4) none of these (3) 4 2 2 units (4) 4 2 2 units SPACE FOR ROUGH WORK 1001CT102116063 E-13/16
Target : JEE (Main + Advanced) 2017/12-03-2017 76. Let ƒ : R R, ƒ(x) = max.{|tan–1x|, cot–1x}. 80. The value of ‘p’ so that both the roots of the Consider the following statements : equation (p – 5)x2 – 2px + (p – 4) = 0 are I. Function is continuous and derivable x R positive, one is less than 2 and other is lying between 2 & 3, lies in the interval II. Range of function is , (1) 49 , 24 (2) (5, ) 4 4 III. ƒ(x) is many one-into. Identify the correct option - (3) (–, 4) U 49 , (4) None of these (1) All 3 statements are wrong. 4 (2) Exactly one of above statements is correct. 1 2 3 (3) Exactly two of above statements are correct. 81. Let A 2 2 1 and (4) All 3 statements are correct. 3 0 k 77. sin–1(sin 100) + cos–1(cos 100) + tan–1(tan 100) ƒ(x) = x3 – 2x2 – x + = 0. + cot–1(cot 100) equals to : If A satisfies ƒ(x) = 0, then- (1) 100 – 31 (2) 100 – 32 (1) k = 1, = 14 (2) = 13, = 22 (3) k = –1, = 22 (4) = –14, = –22 (3) 200 – 63 (4) None of these 82. The value of 78. The solution of y5 x + y – x dy = 0 is 1 2 3 n dx nC1 1Cr nC2 2Cr nC3 3Cr ......nCn nCr r0 r0 r0 r0 (1) x4/4 + 1/5 (x/y)5 = C is equal to (2) x5/5 + (1/4) (x/y)4 = C (3) (x/y)5 + x4/4 = C (1) 2n (2) 3n (3) (3n – 1) (4) (3n + 1) (4) none of these 83. Let B1 = 3x + 4y – 7 = 0 & B2 4x – 3y – 14 = 0 The area bounded by the curve y = ex and are angle bisectors of the angle between the 79. t he lines y = x 1, x = 2 is given by: lines L1 = 0 & L2 = 0 in which L1 is passes (1) e² + 1 through the point (1, 2) then (2) e² 1 (1) B1 is acute angle bisector (3) e² 2 (2) B2 is acute angle bisector (4) none (3) B1 & B2 both are right angle bisector (4) Data is insufficient SPACE FOR ROUGH WORK E-14/16 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 84. ABC is a variable triangle such that A is (1, 2), 88. The graph of y = f(x) is shown then number B and C lie on line y = x + (where is a of solutions of the equation f(f(x)) =2 is variable), then locus of the orthocenter of triangle ABC is (1) (x – 1)2 + y2 = 4 (2) x + y = 3 (3) 2x – y = 0 (4) none of these 85. If area bounded by the curve x2y + y2x = xy is 2 units, then possible values of is / are (1) 2 (2) 1 (1) 1 (2) 4 (3) 4 (4) 3 (3) 3 (4) none of these 86. If f(a) = a2 + a+ 1, then number of solutions of 89. If f(x) satisfies f(7 – x) = f(7 + x) x R such equation f(a2) = 3f(a) is that f(x) has exactly 5 real roots which are all (1) 0 (2) 1 distinct such that sum of the real roots is S then (3) 2 (4) more than 2 S/7 is equal to x3 x 2 10 x 5 , x 1 (1) 1 (2) 3 (3) 5 (4) 7 2 , x 1 Let f(x) = 87. 90. Let the matrix b2 2 x log2 the set of values of b for which f(x) has 1030 5 1020 4 1020 6 greatest value at x = 1 is given by : 108 7 A = 104 2 106 4 1010 2n , 104 8 1015 9 (1) 1 b 2 n N, then (2) b = {1, 2} (1) A is invertible for all n N (3) b (, 1) (2) A is not invertible for all n N (3) A may or may not be invertible depending (4) 130, 2 U 2 , 130 on value of n N (4) Data insufficient SPACE FOR ROUGH WORK 1001CT102116063 E-15/16
Target : JEE (Main + Advanced) 2017/12-03-2017 SPACE FOR ROUGH WORK SPACE FOR ROUGH WORK E-16/16 1001CT102116063
Form Number : Paper Code : 1001CT102116063 HINDI CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) LEADER COURSE (SCORE-I) & ENTHUSIAST COURSE (SCORE-II) Test Type : FULL SYLLABUS Test Pattern : JEE-Main TEST DATE : 12 - 03 - 2017 Important Instructions Do not open this Test Booklet until you are asked to do so. 1. Immediately fill in the form number on this page of the 1. Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited. 2. 2. The candidates should not write their Form Number anywhere else (except in the specified space) on the Test Booklet/Answer Sheet. 3. 3 4. 90 360 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum 5. A,B,C 30 marks are 360. 4 5. There are three parts in the question paper A,B,C consisting of Physics, Chemistry and Mathematics 6. having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 7. 6. One Fourth mark will be deducted for indicated incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the Answer Sheet. 8. 7. Use Blue/Black Ball Point Pen only for writting particulars/marking responses on Side–1 and Side 2 of the Answer Sheet. Use of pencil is strictly prohibited. 9. 8. No candidate is allowed to carry any textual material, 10. printed or written, bits of papers, mobile phone any electronic device etc, except the Identity Card inside the examination hall/room. 11. 9. Rough work is to be done on the space provided for this purpose in the Test Booklet only. 10. On completion of the test, the candidate must hand over the Answer Sheet to the invigilator on duty in the Room/ Hall. However, the candidate are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet. Your Target is to secure Good Rank in JEE (Main) 2017 Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PART A - PHYSICS 1. A right circular cylinder has a mass m, radius r, 1. m r and a height h. The cylinder is completely submerged in a fluid of density , as shown in h the diagram and is released. What is the magnitude of the net force on the cylinder initially? (1) rgh + mg (2) | rgh – mg| (1) rgh + mg (2) | rgh – mg | (3) r2gh + mg (4) r2gh – mg| (3) r2gh + mg (4) r2gh – mg | 2. The distribution of relative intensity I () of 2. blackbody radiation from a solid object versus I () the wavelength is shown in the figure. If the Wien displacement law constant is 2.9 × 10–3 mK, what is the approximate temperature of the 2.9 × 10–3 mK object? (1) 10K (2) 50K (3) 250K (4) 1500K (1) 10K (2) 50K (3) 250K (4) 1500K 1001CT102116063 H-1/31
Target : JEE (Main + Advanced) 2017/12-03-2017 3. A uniform magnetic field B that is 3. B perpendicular to the plane of the page now a passes through the loops, as shown. The field a<b is confined to a region of radius a, where a < b, b and is changing at a constant rate. The induced 2b emf in the wire loop of radius b is . What is the induced emf in the wire loop of radius 2b :- (1) 0 (2) /2 (1) 0 (2) /2 (3) (4) 2 (3) (4) 2 4. If the maximum and minimum voltage of an 4. amplitude modulated wave are V and V VmaxVmin max min respectively, then modulation factor is :- (1) m Vmax (2) m Vmin (1) m Vmax (2) m Vmin Vmax Vmin Vmax Vmin Vmax Vmin Vmax Vmin (3) m Vmax Vmin (4) m Vmax Vmin (3) m Vmax Vmin (4) m Vmax Vmin Vmax Vmin Vmax Vmin Vmax Vmin Vmax Vmin H-2/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 5. During which time interval is the particle 5. described by these position graphs at rest? xy xy 0 1 2 3 t(s) 0 t(s) 0 12 3 t(s) 0 t(s) 0 4 0 1 2 34 0 4 0 1 2 34 (1) 0 – 1 s (2) 1 – 2 s (3) 2 – 3 s (4) 3 – 4 s (1) 0 – 1 s (2) 1 – 2 s (3) 2 – 3 s (4) 3 – 4 s 6. Consider telecommunication through optical 6. fibres. Which of the following statements is NOT true ? (1) (1) Optical fibres can be of graded refractive index (2) (2) Optical fibres are subjected to electromagnetic interference from outside (3) (3) Optical fibres have extremely low transmission loss (4) (4) Optical fibres may have homogeneous core with a suitable cladding 7. Two polaroids are placed in the path of 7. I0 unpolarized beam of intensity I such that no 0 light is emitted from the second polaroid. If a third polaroid whose polarization axis makes /kzqo.k v{k izFke /kzqod dh /kqzo.kv{k ls an angle with the polarization axis of first polaroid, is placed between these polaroids, then the intensity of light emerging from the last polaroid will be :- I0 I0 8 4 I0 I0 (1) sin 2 2 (2) sin 2 2 8 4 (1) sin 2 2 (2) sin 2 2 (3) I0 cos2 (4) I0 cos4 (3) I0 cos2 (4) I0 cos4 2 2 1001CT102116063 H-3/31
Target : JEE (Main + Advanced) 2017/12-03-2017 8. If the B-H curves of two samples of P and Q of 8. PQ B -H iron are as shown below, then which one of the following statements is CORRECT ? BB BB HH HH Sample P Sample Q Sample P Sample Q (1) P Q (1) Both P and Q are suitable for making permanent magnet (2) P Q (2) P is suitable for making permanent magnet and Q for making electromagnet (3) P Q (3) P is suitable for making electromagnet and Q is suitable for permanent magnet (4) P Q (4) Both P and Q are suitable for making electromagnets 9. Statement-1 : A photodiodes operates in 9. -1 : reverse bias. Statement-2 : The fractional change due to the -2 : photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. (1) Statement-1 is true, Statement-2 is true, (1) -1 -2-2-1 Statement-2 is the correct explanation of Statement-1. (2) -1 -2-2-1 (2) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1. (3) -1 -2 (3) Statement-1 is true, Statement-2 is false. (4) -1 -2 (4) Statement-1 is false, Statement-2 is true. H-4/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 10. A particle of mass 5 × 10–5 kg is placed at lowest 10. 5× 10–5 kg point of smooth parabola x2 = 40y x2 = 40y (x y; m) (x and y in m). If it is displaced slightly such that it is constrained to move along parabola, angular frequency of oscillation (in rad/s) will (rad/s) be approximately:- y y g g xx (1) 2 (2) 10 (3) 1 (4) 5 (1) 2 (2) 10 1 (4) 5 2 (3) 2 11. The average degree of freedom per molecule 11. of a gas is 6. The gas performs 25 J work, while 6 expanding at constant pressure. The heat 25 J absorbed by the gas is :- (1) 75 J (2) 100 J (3) 150 J (4) 125 J (1) 75 J (2) 100 J (3) 150 J (4) 125 J 12. The potential energy of a diatomic molecule is 12. U A B r12 r6 given by U A B . A and B are positive A B r12 r6 r constants. The distance r between them at equilibrium is :- A 1/ 6 2A 1/ 6 (1) B (2) B A 1 / 6 2A 1/ 6 B (1) (2) B A 1/ 6 (4) (3) 2B (3) A 1/ 6 (4) None of these 2B 1001CT102116063 H-5/31
Target : JEE (Main + Advanced) 2017/12-03-2017 13. All wires have same resistance and equivalent 13. A resistance between A and B is R. Now keys B R are closed, then the equivalent resistance will become :- K1 K1 B B A A K2 K2 7R 7R (3) 7R R 7R 7R (3) 7R R (1) (2) (4) (1) (2) (4) 3 9 3 3 9 3 14. The Davisson-Germer experiment that first 14. demonstrated the wave nature of matter used 54V electron accelerated to 54 V. Determine the wavelength of the electrons in the Davisson- Germer experiment. (1) 1.67 Å (2) 1.21 Å (1) 1.67 Å (2) 1.21 Å (3) 3.45 Å (4) 5.21 Å (3) 3.45 Å (4) 5.21 Å 15. Liquid cools from 50ºC to 45ºC in 5 minutes 15. 50ºC 45ºC 5 and from 45ºC to 41.5ºC in the next 5 minutes. 5 45ºC 41.5ºC The temperature of the surrounding is :- [Assume newton's law of cooling is applicable] [] (1) 27ºC (2) 40.3ºC (1) 27ºC (2) 40.3ºC (3) 23.3ºC (4) 33.3ºC (3) 23.3ºC (4) 33.3ºC H-6/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 16. A point charge +Q is positioned at the centre 16. +Q of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is :- +Q +Q Q Q Q Q (1) (2) (1) (2) 160 40 160 40 Q (4) None of these Q (4) (3) (3) 80 80 17. If the behavior of light rays through a convex 17. lens is as shown in the adjoining figure, then; µ2 µ µ2 µ2 µ µ2 (1) µ = µ2 (2) µ < µ2 (1) µ = µ (2) µ < µ (3) µ > µ (4) µ µ 2 2 2 2 (3) µ > µ2 (4) µ µ2 1001CT102116063 H-7/31
Target : JEE (Main + Advanced) 2017/12-03-2017 18. The radioactive sources A and B have half lives 18. AB2hr of 2hr and 4hr respectively, initially contain the 4hr same number of radioactive atoms. At the end 2 of 2 hours, their rates of distintegration are in the ratio:- (1) 4 : 1 (2) 2 : 1 (1) 4 : 1 (2) 2 : 1 (3) 2 : 1 (4) 1 : 1 (3) 2 : 1 (4) 1 : 1 19. A satellite S moves around a planet P in an 19. S P elliptical orbit as shown in figure. The ratio of a the speed of the satellite at point a to that at b point b is :- s 3r b s 3r b a rP a rP (1) 1 : 9 (2) 1 : 3 (1) 1 : 9 (2) 1 : 3 (3) 1 : 1 (4) 3 : 1 (3) 1 : 1 (4) 3 : 1 20. The maximum possible acceleration of a train 20. moving on a straight track is 10 m/s2 and 10 m/s25m/s2 maximum possible retardation is 5 m/s2. If maximum achievable speed of train is 10m/s 10m/s then minimum time in which train can complete 135m a journey of 135m starting from rest and ending at rest, is :- (1) 5s (2) 10s (3) 15s (4) 20s (1) 5s (2) 10s (3) 15s (4) 20s H-8/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 21. A metal wire PQ slides on parallel metallic rails 21. PQ 0.25 m having separation 0.25 m, each having negligible resistance. There is a 2 resistor and 2 10V 10V battery as shown in figure. There is a uniform magnetic field directed into the plane 0.5T of the paper of magnitude 0.5 T. Aforce of 0.5N PQ to the left is required to keep the wire PQ moving with constant speed to the right. With 0.5N what speed is the wire PQ moving ? PQ (Neglect self inductance of the loop) () 10V P 10V P 2 0.25 m 2 0.25 m QQ (1) 8 m/s (2) 16 m/s (3) 24 m/s (4) 32 m/s (1) 8 m/s (2) 16 m/s (3) 24 m/s (4) 32 m/s 22. A rod of length L is held vertically on a smooth 22. L horizontal surface. The top end of the rod is given a gentle push. At a certain instant of time, when the rod makes an angle with horizontal v0 the velocity of COM of the rod is v0. The velocity of the end of the rod in contact with the surface at that instant is : (1) v0 cot (2) v0 cos (1) v0 cot (2) v0 cos (3) v0 sin (4) v0 tan (3) v0 sin (4) v0 tan 1001CT102116063 H-9/31
Target : JEE (Main + Advanced) 2017/12-03-2017 23. Correct output for given logic circuit and inputs 23. is :- AA B CB C A 2T0 B A 2T0 B t T0 t T0 2T0 2T0 output output output output (1) t (2) (1) t (2) output output output output (3) T0 (4) t (3) T0 (4) t 2T0 2T0 24. A transverse wave is passing through a 24. 20m/s stretched string with a speed of 20 m/s. The 20N tension in the string is 20 N. At a certain point P P on the string, it is observed that energy is 40 mW P being transferred at a rate of 40 mW at a given instant. Find the speed of point P. (1) 40 cm/s (2) 20 cm/s (1) 40 cm/s (2) 20 cm/s (3) 2 mm/s (4) 20 mm/s (3) 2 mm/s (4) 20 mm/s H-10/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 25. In the circuit shown, power factor of circuit is 25. 1 3 35 1 and power factor of box is 5 . Find reading of ammeter :- C = 1mF C = 1mF Box Box A A 90V, 25 rad/s 90V, 25 rad/s (1) 5A (2) 6A (3) 4A (4) 3A (1) 5A (2) 6A (3) 4A (4) 3A 26. When the gap is closed without placing any 26. 0.005 mm object in the screw gauge whose least count is 0.005 mm, the 5th division on its circular scale 5 with the reference line on main scale, and when a small sphere is placed reading on main scale advances by 4 divisions, whereas circular scale 4 reading advances by five times to the corresponding reading when no object was placed. There are 200 divisions on the circular 200 scale. The radius of the sphere is :- (1) 4.10 mm (2) 4.05 mm (1) 4.10 mm (2) 4.05 mm (3) 2.10 mm (4) 2.05 mm (3) 2.10 mm (4) 2.05 mm 27. A household refrigerator with a coefficient of 27. 1.2 performance 1.2 removes heat from the 60kJ/min refrigerated space at the rate of 60kJ/min.What (30 ) would be cost of running this fridge for one month (30 days) (assuming each day it is used (4 for 4 hours and cost of one electrical unit 6Rs. ) is 6Rs.):- (1) 180 Rs. (2) 300 Rs. (1) 180 Rs. (2) 300 Rs. (3) 480Rs. (4) 600 Rs. (3) 480Rs. (4) 600 Rs. 1001CT102116063 H-11/31
Target : JEE (Main + Advanced) 2017/12-03-2017 28. A hollow sphere of radius R is filled completely 28. R with an ideal liquid of density . sphere is 2g moving horizontally with an acceleration 2g, g where g is acceleration due to gravity in the P0 space. If minimum pressure of liquid is P0, then pressure at the centre of sphere is :- 2g 2g (1) P0 + gR (2) P0 + g R 2 (1) P0 + gR (2) P0 + g R 2 (3) P0 + g R 5 (4) P0 gR (3) P0 + g R 5 (4) P0 gR 5 5 29. Two steel wires of same length but radii r and 29. r2r 2r are connected together end to end and tied to a wall as shown. The force stretches the 10mm combination by 10 mm. How far does the A midpoint A move :- Radius= r A Radius= 2 r Radius= r A Radius= 2 r LL LL (1) 2 mm (2) 4 mm (3) 6 mm (4) 8 mm (1) 2 mm (2) 4 mm (3) 6 mm (4) 8 mm 30. A closed organ pipe of length is sounded 30. together with another closed organ pipe of + x (x << ) length + x (x << ) both in fundamental mode. If v = speed of sound, the beat frequency v heard is: vx2 (3) 4 vx vx vx2 vx vx vx vx (1) 22 (2) 42 (3) 4 (4) 2 (1) 22 (2) 42 (4) 2 H-12/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 PART B - CHEMISTRY 31. Select correct statement regarding interstitial 31. hydrides: (1) (1) They do not conduct electricity in solid state. (2) Their density is more than that of parent (2) metal. (3) (3) Their composition may vary with temperature. (4) d- (4) Each metal of d-block form such type of hydride. 32. Find incorrect comparison of second ionisation 32. energy : (2) In > Sr (1) Te < Sb (2) In > Sr (1) Te < Sb (4) Fe < Fe+ (3) He > B (4) Fe < Fe+ (3) He > B 33. Which property is not typical of a refractory 33. material? (1) Chemically inert (1) (2) High thermal conductivity (2) (3) High melting point (3) (4) High electrical conductivity (4) 34. Potassium manganate is not formed in 34. (1) KMnO4 (1) KMnO4 (2) KMnO 4 KOH (2) KMnO 4 KOH ( conc.) ( conc.) (3) MnO2 KOH O2 (3) MnO2 KOH O2 (4) KMnO4 H2SO4 (4) KMnO4 H2SO4 ( conc .) ( conc .) 1001CT102116063 H-13/31
Target : JEE (Main + Advanced) 2017/12-03-2017 35. PCl + PO oxygen containing 35. PCl5 + P4O10 (X) 5 4 10 compound (X). 1 (X ) How many moles of an oxy acid are formed on hydrolysis of 1 mole (X). (1) Zero (2) 1 (3) 4 (4) 3 (1) (2) 1 (3) 4 (4) 3 36. Select correct statement about anionic part of 36. K[Pt (2–C2H4)Cl3] Zeise’s salt, K[Pt (2–C2H4)Cl3] : (1) C=C bond is coplanar with Pt and 3 Cl (1) C=C Pt3C l atoms. (2) Pt +3 (2) oxidation state of Pt is +3. (3) HCH bond angle is not same as that in free (3) HCH ethylene molecule. (4) Pt-Cl bond is not affected by synergic (4) Pt-Cl ,Pt C2H4 bonding between Pt and C2H4 37. Maximum number of P–H bond(s) are there in 37. P–H a molecule of (1) Orthophosphorous acid (1) (2) Pyrophosphoric acid (2) (3) Pyrophosphorous acid (3) (4) Hypophosphoric acid (4) 38. Which of the following species is 38. paramagnetic ? (2) NO[PF ] (1) K3[Co(C2O4)3] (2) NO[PF6] 6 (3) [NMe ]O (4) H[BF ] (1) K [Co(C O ) ] 3 2 43 (4) H[BF4] 43 4 (3) [NMe4]O3 39. How many different types of OBO angles are 39. there in sodium peroxoborate ? OBO (1) 2 (2) 3 (3) 5 (4) 1 (1) 2 (2) 3 (3) 5 (4) 1 H-14/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 40. Anhydrous AlCl3 is formed in : 40. AlCl3 (1) AlCl3.6H2O (1) AlCl3.6H2O (2) Al(Red hot) + HCl(Moist) (2) Al + HCl (3) Al O + C + Cl (dry) (3) Al O + C + Cl () 23 2 23 2 (4) All of these (4) 41. The radii of two of the first four Bohr’s orbits 41. of the hydrogen atom are in the ratio 1 : 4 The 1: 4 energy difference between them may be : (1) Either 12.09 eV or 10.2 eV (1) 12.09 eV 10.2 eV (2) Either 2.55 eV or 10.2 eV (2) 2.55 eV 10.2 eV (3) Either 13.6 eV or 3.4 eV (3) 13.6 eV 3.4 eV (4) Either 3.4 eV or 0.85 eV (4) 3.4 eV 0.85 eV 42. Pure ammonia is placed in a vessel at a 42. temperature where its dissociation constant () () is appreciable. At equilibrium : (1) Kp (1) Kp does not change significantly with (2) pressure. (3) NH3 (4) (2) does not change with pressure. (3) concentration of NH3 does not change with pressure. (4) concentration of hydrogen is less than that of nitrogen 43. A system of 100 kg mass undergoes a process 43. 100 kg in which its specific entropy increases from 0.3kJkg–1K–1 0.3 kJkg–1K–1 to 0.4kJkg–1K–1. At the same 0.4kJkg–1K–1 time, the entropy of the surrounding decreases 80kJK–1 75 kJK–1 from 80 kJK–1 to 75 kJK–1. (S)universe kJK–1 Find the (S)universe in kJK–1. (1) 0 (2) 5 (3) 10 (4) 15 (1) 0 (2) 5 (3) 10 (4) 15 1001CT102116063 H-15/31
Target : JEE (Main + Advanced) 2017/12-03-2017 44. 2gm of benzoic acid dissolved in 25gm of 44. 2gm 25gm benzene shows a depression in freezing point 1.62 K equal to 1.62 K. Molal depression constant for 4.9K kg mol–1 benzene is 4.9K kg mol–1. What is % degree of % association of acid (1) 99.2 (2) 95 (1) 99.2 (2) 95 (3) 90 (4) 0.8 (3) 90 (4) 0.8 45. On addition of one ml solution of 10% NaCl 45. 0.0250 gm 10%NaCl to 10 ml gold sol in the presence of 0.0250 gm 1 ml 10 ml of starch the coagulation is just prevented. : Starch has the following gold number : (1) 0.025 (2) 0.25 (1) 0.025 (2) 0.25 (3) 2.5 (4) 25 (3) 2.5 (4) 25 46. Which does not have hexagonal crystal system 46. (1) Graphite (2) Zinc oxide (1) (2) (3) CdS (4) HgS (3) CdS (4) HgS 47. By how much will the potential of half 47. 298K 100 cell Cu2+ / Cu change if the solution is diluted Cu2+ / Cu to 100 times at 298 K (1) Increases by 59 mV (1) 59 mV (2) Decreases by 59 mV (2) 59 mV (3) Increases by 29.5 mV (3) 29.5 mV (4) Decreases by 29.5 mV (4) 29.5 mV H-16/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 48. Solubility curves of four ionic salts X, Y, Z, W 48. X, Y, Z, W are given below : : In which case the value of Hsol. < 0 Hsol. <0 (1) X (2) Y (3) Z (4) W. (1) X (2) Y (3) Z (4) W. 49. Match of following (where Urms = root mean 49. (U = , rms square speed, U = average speed, U = most Uav = ,U mp = ) av mp probable speed) List I List II I II (a) Urms / Uav (i) 1.22 (a) U / U (i) 1.22 rms av (b) U / U (ii) 1.13 (b) Uav / Ump (ii) 1.13 av mp (c) U / U (iii) 1.08 (c) U / U (iii) 1.08 rms mp rms mp (1) (a)-(iii), (b)-(ii), (c)-(i) (1) (a)-(iii), (b)-(ii), (c)-(i) (2) (a)-(i), (b)-(ii), (c)-(iii) (2) (a)-(i), (b)-(ii), (c)-(iii) (3) (a)-(iii), (b)-(i), (c)-(ii) (3) (a)-(iii), (b)-(i), (c)-(ii) (4) (a)-(ii), (b)-(iii), (c)-(i). (4) (a)-(ii), (b)-(iii), (c)-(i). 50. 27 Al is a stable element. 29 Al is expected to 50. 27 Al 2193 Al 13 13 13 disintegrate by : : (1) -emission (2) -emission (1) - (2) - (3) Positron emission (4) Proton emission (3) (4) 1001CT102116063 H-17/31
Target : JEE (Main + Advanced) 2017/12-03-2017 O CCH O O CCH O 51. Cl–C–CH2– C–CH2–CH2 –C–CH3 H2/Pd ‘X’ 51. Cl–C–CH2– C–CH2–CH2 –C–CH3 H2/Pd ‘X’ BaSO4 BaSO4 CH2 CH2 CN CN ‘X’ ? Which statement is correct for product ‘X’ ? (1) ‘X’ 1º (1) 1º alcohol functional group is present in ‘X’ (2) ‘X’ (2) Ethyl group is present in ‘X’ (3) ‘X’ 2º (3) 2º-alcohol functional group is present in ‘X’ (4) Alkene functional group is present in ‘X’ (4) ‘X’ 52. Which of the following can not show 52. geometrical isomerism ? ? OO O O O O (1) (1) O O (2) CH3 – CH = CH – CH3 (2) CH – CH = CH – CH 33 (3) CH3 (3) CH3 CH3 CH3 Br Cl Br Cl Cl Cl (4) Br (4) Br 53. Which one is an example of bacteriocidal 53. antibiotics ? ? (1) sulfa drugs (2) Tetracyclin (1) (2) (3) Erythromycin (4) Pencillin ‘G’ (3) (4) ‘G’ H-18/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 54. In given reactions which one is correct ? 54. ? CH3 CH3 (1) CH2= CH – C – COOH (1) CH2= CH – C – COOH CH3 CH3 NaOH + CaO/ NaOH + CaO/ CH2= CH – CH – CH 3 CH2= CH – CH – CH 3 CH3 CH3 COOH COOH H CH3 (i) Ag2O H CH3 (i) Ag2O (2) CH3 (ii) Br2/CCl4/ (2) CH3 (ii) Br2/CCl4/ H D H D Br Br H CH3 H CH3 H CH3 + Enantiomer H CH3 + Enantiomer D D (3) CH3–CH–CH2–NO2 OH–/H2O CH2=CH–CH2–NO2 (3) CH3–CH–CH2–NO2 OH–/H2O CH2=CH–CH2–NO2 F F O O CHO C CHO C O O (4) (i) 50% KOH (4) (i) 50% KOH CH2 CH2 (ii) H+/ (ii) H+/ CHO CHO 1001CT102116063 H-19/31
Target : JEE (Main + Advanced) 2017/12-03-2017 OH OH 55. CH3 – CH – CH3 Cu/300ºC Product (X) 55. CH3 – CH – CH3 Cu/300ºC (X) Which of the following is correct about product 'X' ? 'X' ? (1) (2) (1) Degree of unsaturation of product is two. (3) (4) (2) Product can reduce tollens reagent. (3) Product is an aldehyde. (4) Product can undergo aldol condensation. 56. In given options, which one is correct for rate 56. of hydrolysis ? ? O CH2–Cl CH2–Cl CH2–Cl CH2–Cl O CH2–Cl CH2–Cl CH2–Cl CH2–Cl N > > NO2 N > > NO2 O NO2 NO2 > (1) O > (1) NO2 NO2 NO2 NO2 NO2 NO2 Cl Cl Cl Cl Cl Cl Cl Cl (2) O O O (2) O O O O> > > O> > > O O O O O O O O O O C–Cl C–Cl C–Cl C–Cl C–Cl C–Cl C–Cl C–Cl (3) > > (3) > >> > NO2 CH3 O––CH3 NO2 CH3 O––CH3 Cl Cl Cl Cl Cl Cl Cl Cl (4) > >> >> (4) > H-20/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 57. Which one is correct for heat of hydrogenation 57. (1) > (1) > (2) > (2) > (3) > (3) > (4) CH = CH > CH – CH = CH – CH3 223 (4) CH = CH > CH – CH = CH – CH 223 3 58. Which of the following statement is false for 58. Biodegradable polymers. (1) Biodegradable polymers are specific type (1) of polymers that breaks down after it’s intended purpose to result in natural byproducts. (2) (2) These polymers are found both naturally and synthetically. (3) PHBV 2- 6, (3) PHBV and Nylon2-Nylon6 are example of Biodegradable polymers. (4) (4) Biodegradable polymers can not be synthesized by condensation reactions. 1001CT102116063 H-21/31
Target : JEE (Main + Advanced) 2017/12-03-2017 59. Which of the following is a non-reducing 59. ? sugar? (1) glucose (2) fructose (1) (2) (3) sucrose (4) maltose (3) (4) 60. In which of the following reaction alcohol is 60. formed as major product (1) CH3 N (1) N NaNO2 HCl CH3 NaNO2 HCl 3 3 NH2 NH2 (2) NaN0O52ºCHCl (2) NaNO2HCl 05ºC NH2 NH2 (3) NaNO2 HCl (3) NaNO2 HCl NH–CH3 NH–CH3 (4) NaNO2 HCl (4) NaNO2 HCl H-22/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 PART C - MATHEMATICS 61. The real value of for which the expression 61. 1 i cos 1 i cos (n I) 1 2i cos is a real number is (n I) 1 2i cos (1) (2n + 1) (2) (2n + 1)/2 (1) (2n + 1) (2) (2n + 1)/2 (3) 2n (4) None of these (3) 2n (4) 62. If m and 2 are the mean and variance of 62. m 2 x random variable x, whose distribution is given by Xx 0 1 2 3 4 Xx 0 1 2 3 4 1 1 1 0 1 1 1 , then P(X x) 0 0 32 6 P(X x) 0 32 6 (1) m = 2 = 2 (1) m = 2 = 2 (2) m = 1, 2 = 2 (2) m = 1, 2 = 2 (3) m = 2 = 1 (3) m = 2 = 1 (4) m = 2, 2 = 1 (4) m = 2, 2 = 1 63. Mr. A has six children and atleast one child is a 63. A girl, then probability that Mr. A has 3 boys and A 3 3 3 girls, is - - 20 1 20 1 (1) 63 (2) 6 (1) (2) 5 1 63 6 (3) 11 (4) 32 5 1 (3) 11 (4) 32 64. log sin2 3 equals to : 64. log cos ec2 sin2 3 : 1 8 1 8 8 8 cos ec2 8 8 (1) 0 (2) 1/2 (1) 0 (2) 1/2 (3) 1 (4) not defined (3) 1 (4) 1001CT102116063 H-23/31
Target : JEE (Main + Advanced) 2017/12-03-2017 6 5 . Value of 4sin 9sin 21sin 39sin 51sin 69sin 81 4sin 9sin 21sin 39sin 51sin 69sin 81 65. sin 54 sin 54 is equal to 11 11 (1) 16 (2) 32 (1) 16 (2) 32 11 (3) 8 (4) 4 11 (3) 8 (4) 4 66. Number of common tangents of the 66. (x2)2 (y 2)2 1 94 ellipse (x 2)2 (y 2)2 1 and the circle x2 + y2 – 4x + 2y + 4 = 0 94 - x2 + y2 – 4x + 2y + 4 = 0 is - (1) 0 (2) 1 (1) 0 (2) 1 (3) 2 (4) more than 2 (3) 2 (4) 2 67. Consider the equation x2 + x + = 0 having 67. x2 + x + = 0 , roots , such that Also consider the inequality |y – | – < , |y – | – < , then (1) inequality is satisfied by exactly two integral (1) y values of y (2) inequality is satisfied by all values of (2) y (– 4, 2) y (– 4, 2) (3) Roots of the equation are of same sign (3) (4) x2 + x + > 0 x [–1, 0] (4) x2 + x + > 0 x [–1, 0] H-24/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 68. If a, b, c are pth, qth, rth terms of an H.P. and 68. a, b, c p , q, r = (q r) ˆi + (r p) ˆj + (p q) kˆ , ˆi ˆj kˆ , u = (q r) ˆi + (r p) ˆj + (p q) kˆ , u = a b c ˆi ˆj kˆ then: = : abc (1) u , are parallel vectors (1) , u (2) u , are orthogonal vectors (2) , u (3) . = 1 u (3) u . = 1 (4) = ˆi + ˆj + kˆ = u u ˆj (4) ˆi + + kˆ 69. The image of the point (1, 2, –1), on the plane 69. x31 y 3 z 2 2 1 containing the line x1 y 3 z 2 and the (0, 7, –7) (1, 2, –1) 3 2 1 point (0, 7, –7), is. (1) 1 , 7 , 1 (2) 1 , 2 , 7 (1) 1 , 7 , 1 (2) 1 , 2 , 7 3 3 3 3 3 3 3 3 3 3 3 3 (3) 1 , 0, 7 (4) 1 , 2 , 7 (3) 1 , 0, 7 (4) 1 , 2 , 7 3 3 3 3 3 3 3 3 3 3 70. The bisectors of the angles of a parallelogram 70. enclose a: : DC DC S R S R P P Q Q AB AB (1) Rhombus (2) Rectangle (1) (2) (3) Square (4) none of these (3) (4) 1001CT102116063 H-25/31
Target : JEE (Main + Advanced) 2017/12-03-2017 71. (a b) (b c) (a b b c c a) is 71. (a b) (b c) (a b b c c a) (1) [a b c][(b a a c)b (| b |2 b c)a] (1) [a b c][(b a a c)b (| b |2 b c)a] (2) [a b c][(b a a c)b (| b |2 b c)a] (2) [a b c][(b a a c)b (| b |2 b c)a] (3) [a b c][(b a a c)b (| b |2 b c)a] (3) [a b c][(b a a c)b (| b |2 b c)a] (4) [a b c][(a c b a)b (| b |2 b c)a] (4) [a b c][(a c b a)b (| b |2 b c)a] 3 3 72. sin2 sin2 2d is equal to - 72. sin2 sin2 2d - 3 3 3 5 3 5 (1) (2) 2 (3) 2 (4) 6 (1) (2) 2 (3) 2 (4) 6 73. If rate of change of area of a square S is equal 73. S to its side length, if rate of change of side of S S is same as that of a cube C, then rate of change C of volume of C, at the time when its side length C is 2 units, will be- 2 (1) 24 units/sec. (2) 12 units/sec. (1) 24 (2) 12 (3) 6 units/sec. (4) 3 units/sec. (3) 6 (4) 3 74. Let ƒ is a differentiable function satisfying 74. ƒ ƒ(x + 2y) = ƒ(x + 2y) = 2yƒ(x) + xƒ(y) – 3xy + 1 x, y R 2yƒ(x) + xƒ(y) – 3xy + 1 x, y R such that ƒ'(0) = 1, then ƒ(2) is equal to - ƒ'(0) = 1 ƒ(2) (1) 4 (2) 1 (3) 5 (4) 3 (1) 4 (2) 1 (3) 5 (4) 3 75. Minimum distance between the curves 75. x2 + y2 + 4x + 16y + 66 = 0 y2 = 8x x2 + y2 + 4x + 16y + 66 = 0 and y2 = 8x is - - (1) 3 2 units (2) 5 2 units (1) 3 2 (2) 5 2 (3) 4 2 2 units (4) 4 2 2 units (3) 4 2 2 (4) 4 2 2 H-26/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 76. Let ƒ : R R, ƒ(x) = max.{|tan–1x|, cot–1x}. 76. ƒ : R R, ƒ(x) = Consider the following statements : {|tan–1x|, cot–1x} I. Function is continuous and derivable x R I. x R II. Range of function is , II. , 4 4 III. ƒ(x) is many one-into. III. ƒ(x) Identify the correct option - - (1) All 3 statements are wrong. (1) 3 (2) Exactly one of above statements is correct. (2) (3) Exactly two of above statements are correct. (3) (4) All 3 statements are correct. (4) 3 77. sin–1(sin 100) + cos–1(cos 100) + tan–1(tan 100) 77. sin–1(sin 100) + cos–1(cos 100) + tan–1(tan 100) + cot–1(cot 100) equals to : + cot–1(cot 100) (1) 100 – 31 (2) 100 – 32 (1) 100 – 31 (2) 100 – 32 (3) 200 – 63 (4) None of these (3) 200 – 63 (4) 78. The solution of y5 x + y – x dy = 0 is 78. y5 x + y – x dy = 0 dx dx (1) x4/4 + 1/5 (x/y)5 = C (1) x4/4 + 1/5 (x/y)5 = C (2) x5/5 + (1/4) (x/y)4 = C (2) x5/5 + (1/4) (x/y)4 = C (3) (x/y)5 + x4/4 = C (3) (x/y)5 + x4/4 = C (4) none of these (4) 79. The area bounded by the curve y = ex and 79. y = ex y = x 1, x = 2 t he lines y = x 1, x = 2 is given by: : (1) e² + 1 (1) e² + 1 (2) e² 1 (2) e² 1 (3) e² 2 (3) e² 2 (4) none (4) 1001CT102116063 H-27/31
Target : JEE (Main + Advanced) 2017/12-03-2017 80. The value of ‘p’ so that both the roots of the 80. ‘p’ equation (p – 5)x2 – 2px + (p – 4) = 0 are (p – 5)x2 – 2px + (p – 4) = 0 positive, one is less than 2 and other is lying 2 bet ween 2 & 3, lies in t he int erval 2 3 (1) 49 , 24 (2) (5, ) (1) 49 , 24 (2) (5, ) 4 4 (3) (–, 4) U 49 , (4) None of these (3) (–, 4) U 49 , (4) 4 4 1 2 3 1 2 3 81. Let A 2 2 1 and 81. A 2 2 1 3 0 k ƒ(x) = x3 – 2x2 – x + = 0. 3 0 k If A satisfies ƒ(x) = 0, then- ƒ(x) = x3 – 2x2 – x + = 0 A, ƒ(x) = 0 - (1) k = 1, = 14 (2) = 13, = 22 (1) k = 1, = 14 (2) = 13, = 22 (3) k = –1, = 22 (4) = –14, = –22 (3) k = –1, = 22 (4) = –14, = –22 82. The value of 1 2 nC3 3 ......nCn n 1 2 3 ......nCn n r0 r0 r0 nC1 1Cr r0 nC1 1Cr nC2 2Cr 3Cr nCr 82. nC2 2Cr nC3 3Cr nCr r0 r0 r0 r0 is equal to (2) 3n (1) 2n (4) (3n + 1) (3) (3n – 1) (1) 2n (2) 3n (3) (3n – 1) (4) (3n + 1) 83. Let B1 = 3x + 4y – 7 = 0 & B2 4x – 3y – 14 = 0 83. B1 = 3x + 4y – 7 = 0 B2 4x – 3y – 14 = 0, are angle bisectors of the angle between the lines L1 = 0 & L2 = 0 in which L1 is passes L1 = 0 L2 = 0 1 L (1, 2) through the point (1, 2) then (1) B1 (1) B1 is acute angle bisector (2) B (2) B2 is acute angle bisector 2 (3) B1 & B2 both are right angle bisector (3) B B 12 (4) Data is insufficient (4) H-28/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 84. ABC is a variable triangle such that A is (1, 2), 84. ABC A(1, 2), B B and C lie on line y = x + (where is a C y = x + ( ) variable), then locus of the orthocenter of ABC triangle ABC is (1) (x – 1)2 + y2 = 4 (2) x + y = 3 (1) (x – 1)2 + y2 = 4 (2) x + y = 3 (3) 2x – y = 0 (4) none of these (3) 2x – y = 0 (4) 85. If area bounded by the curve x2y + y2x = xy 85. x2y + y2x = xy 2 is 2 units, then possible values of is / are (1) 2 (2) 1 (1) 2 (2) 1 (3) 4 (4) 3 (3) 4 (4) 3 86. If f(a) = a2 + a+ 1, then number of solutions of 86. f(a) = a2 + a+ 1 f(a2) = 3f(a) equation f(a2) = 3f(a) is (1) 0 (2) 1 (1) 0 (2) 1 (3) 2 (4) more than 2 (3) 2 (4) 2 Let f(x) = x3 x 2 10 x 5 , x 1 f(x) = x3 x2 10 x 5 , x 1 2 , x 1 2 87. 87. b2 b2 2 x log2 2 x log2 , x 1 the set of values of b for which f(x) has b f(x) greatest value at x = 1 is given by : x = 1 : (1) 1 b 2 (1) 1 b 2 (2) b = {1, 2} (2) b = {1, 2} (3) b (, 1) (3) b (, 1) (4) 130, 2 U 2 , 130 (4) 130, 2 U 2 , 130 1001CT102116063 H-29/31
Target : JEE (Main + Advanced) 2017/12-03-2017 88. The graph of y = f(x) is shown then number 88. y = f(x) of solutions of the equation f(f(x)) =2 is f(f(x)) =2 (1) 1 (2) 4 (1) 1 (2) 4 (3) 3 (4) none of these (3) 3 (4) none of these 89. If f(x) satisfies f(7 – x) = f(7 + x) x R such 89. f(x), f(7 – x) = f(7 + x) x R that f(x) has exactly 5 real roots which are all f(x) 5 distinct such that sum of the real roots is S then S S/7 is equal to S/7 (1) 1 (2) 3 (3) 5 (4) 7 (1) 1 (2) 3 (3) 5 (4) 7 90. Let the matrix 90. 1030 5 1020 4 1020 6 1030 5 1020 4 1020 6 108 7 A = 104 2 106 4 1010 2n , A = 104 2 108 7 1010 2n , 104 8 106 4 1015 9 104 8 1015 9 n N n N, then (1) n N A (1) A is invertible for all n N (2) A is not invertible for all n N (2) n N A (3) A may or may not be invertible depending on value of n N (4) Data insufficient (3) n N (4) H-30/31 1001CT102116063
Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 1001CT102116063 H-31/31
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