Paper Code : 100 1CT103316001 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) ENGLISH JEE (Main + Advanced) : LEADER COURSE DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR Test Type : MINOR PHASE : III Test Pattern : JEE-Advanced TEST DATE : 03 - 07 - 2016 Time : 3 Hours Maximum Marks : 210 READ THE INSTRUCTIONS CAREFULLY GENERAL : 1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so. 2. Use the Optical Response sheet (ORS) provided separately for answering the questions. 3. Blank spaces are provided within this booklet for rough work. 4. Write your name, form number and sign in the space provided on the back cover of this booklet. 5. After breaking the seal of the booklet, verify that the booklet contains 28 pages and that all the 20 questions in each subject and along with the options are legible. If not, contact the invigilator for replacement of the booklet. 6. You are allowed to take away the Question Paper at the end of the examination. OPTICAL RESPONSE SHEET : 7. The ORS will be collected by the invigilator at the end of the examination. 8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work. 9. Write your name, form number and sign with pen in the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your form number. DARKENING THE BUBBLES ON THE ORS : 10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS. 11. Darken the bubble COMPLETELY. 12. The correct way of darkening a bubble is as : 13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way. 14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or \"un-darken\" a darkened bubble. 15. Take g = 10 m/s2 unless otherwise stated. Please see the last page of this booklet for rest of the instructions
Target : JEE (Main + Advanced) 2017/03-07-2016 SOME USEFUL CONSTANTS Atomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Atomic masses : Cl = 17, Br = 35, Xe = 54, Ce = 58, H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127, Xe = 131, Ba=137, Ce = 140, Boltzmann constant k = 1.38 × 10–23 JK–1 Coulomb's law constant 1 = 9 ×109 4 0 Universal gravitational constant Speed of light in vacuum G = 6.67259 × 10–11 N–m2 kg–2 Stefan–Boltzmann constant c = 3 × 108 ms–1 Wien's displacement law constant = 5.67 × 10–8 Wm–2–K–4 Permeability of vacuum b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2 Permittivity of vacuum 1 Planck constant 0 = 0c2 h = 6.63 × 10–34 J–s Space for Rough Work E-2/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING PHYSICS PART-1 : PHYSICS SECTION–I(i) : (Maximum Marks : 18) This section contains SIX 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. 10 gm of ice at 0°C is mixed with ‘m’ gm of water at 50°C. What is minimum value of m so that ice melts completely. (Lf = 80 cal/gm and sw = 1 cal/gm-°C) (A) 32 gm (B) 20 gm (C) 40 gm (D) 16 gm 2. Three rods of identical cross sectional area and made from the same identical material form the sides of an equilateral triangle. The point A and B are maintained at T and 2T respectively. In steady state temperature of point C is TC. Assuming only heat conduction take place, TC equals C (A) T/2 (B) T AB (D) 3T/2 (C) 2/3 T Space for Rough Work 1001CT103316001 E-3/28
Target : JEE (Main + Advanced) 2017/03-07-2016 PHYSICS3. A wire can support a maximum load of W. If the wire is cut in 2 equal parts and joined in parallel, what is the maximum load it can now support? (A) W (B) 2W (C) W/2 (D) 4W 4. The emissivity of a 1 m long filament of a 100 W bulb is 0.4. If its radius be 40 m, the approximate temperature at which it operates at correct rating is approximately : (Assume surrounding to be at 0 K) (A) 625 K (B) 815 K (C) 1025 K (D) 2100 K Space for Rough Work E-4/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 PHYSICS 5. In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper ( = 4.5eV). The emitter is illuminated by a source of monochromatic light of wavelength 200nm. (A) the minimum kinetic energy of the photoelectrons reaching the collector is 0. (B) the maximum kinetic energy of the photoelectrons reaching the collector is 3.7eV. (C) if the polarity of the battery is reversed then answer to part A will be 0. (D) if the polarity of the battery is reversed then answer to part B will be 1.7eV. 6. If the de Broglie wavelength of an particle and a neutron are same, then the velocity of (A) particle is greater than that of neutron (B) neutron is greater than that of particle (C) both neutron and -particle is same (D) none of the above Space for Rough Work 1001CT103316001 E-5/28
Target : JEE (Main + Advanced) 2017/03-07-2016 SECTION–I(ii) : (Maximum Marks : 16) PHYSICS This section contains FOUR questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks : 0 In all other cases. 7. A thin cylindrical metal rod is bent into a ring with a small gap as shown in figure. On heating the system :- r d (A) decreases, r and d increases (B) increases (C) d & r increases (D) is constant, 8. What happens when the light intensity incident on a photoelectric surface is doubled? (A) the frequency of the incident photons may be doubled (B) the number of photons may be doubled (C) the frequency of the emitted photons is tripled (D) the number of photons is tripled. Space for Rough Work E-6/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 PHYSICS 9. Two separate segments of equal area are isolated in the energy distribution of blackbody radiation. Are the emissive powers over the respective wavelength intervals the same? What about the number of emitted photons in each segment? —ddE– S1 S2 0 (A) The spectral emissive power of S1 > S2 (B) The spectral emissive power of S1 < S2 (C) Number of photons in segment S1 > S2 (D) Number of photons in segment S1 < S2 10. A bar of uniform crosssection and of length 90cm is made of three materials A, B & C. Their lengths are 40cm, 30cm and 20cm respectively. The coefficients of their thermal conductivities are in the ratio 2: 3: 4. The ends are maintained at 100°C & 30°C & there is no loss of heat from the sides of the bar. At steady state: ' 100° A B C 30° 40cm 30cm 20cm (A) = 60°C (B) = 40°C (C) ’ = 60°C (D) ’ = 40°C Space for Rough Work 1001CT103316001 E-7/28
Target : JEE (Main + Advanced) 2017/03-07-2016 SECTION–I(iii) : (Maximum Marks : 12) PHYSICS This section contains TWO 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 11 and 12 The energy radiated by a black body at 2300 K is found to have the maximum at a wavelength 1260 nm, its emissive power being 8000 Wm–2. When the body is cooled to a temperature T K, the emissive power is found to decrease to 500 Wm–2. 11. Find the temperature T (A) 1150 K (B) 2300 k (C) 1725 k (D) 2000 K 12. Find the wavelength at which intensity of emission is maximum at the temperature T. (A) 1260 nm (B) 1870 nm (C) 2000 nm (D) 2520 nm Space for Rough Work E-8/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 PHYSICS Paragraph for Questions 13 and 14 A rectangular piece of elastic material is shown in the following diagram. The material is compressed parallel to (i) the x-axis, (ii) the y-axis and (iii) the z-axis by 1 mm. z 5cm 1cm 3cm y x 13. Young’s modulus Y is (B) highest parallel to the y-axis. (A) highest parallel to the x-axis. (D) same in all directions. (C) highest parallel to the z-axis. (B) highest parallel to the y-axis. 14. The force constant k in Hooke’s law is (D) same in all directions. (A) highest parallel to the x-axis. (C) highest parallel to the z-axis. Space for Rough Work SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III 1001CT103316001 E-9/28
Target : JEE (Main + Advanced) 2017/03-07-2016 SECTION–IV : (Maximum Marks : 24) PHYSICS This section contains SIX questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. The figure below shows a vaccum tube containing electrodes made of different metals, 1 and 2 whose work functions are 1 and 2. The electrodes are illuminated simultaneously. The maximum kinetic energy of photoelectrons reaching plate 2 is 1 eV and maximum kinetic energy of photoelectrons reaching plate 1 is 3 eV. Assume that photoelectron emitted from either plate do not interact with each other. 1 = 1.5 eV and 2 = 0.7 eV. Find wavelength (in nm) of the electromagnetic wave used. Write 100 in OMR sheet. Light source 12 +– A V0 Space for Rough Work E-10/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 PHYSICS 2. The force exerted by a light beam of intensity I, incident on a symmetric cone (height h and base radius R) as shown in figure if surface of cone is perfectly absorbing is I Rh . Fill the c value of in OMR sheet. I 3. A glass vessel is partially filled with mercury and when both are heated together, the volume of the unfilled part of vessel remain constant at all temperature. Find the initial volume (in cm3) of mercury if the empty part measures 34 cm3. [Given : vessel = 9 x 10–6/°C; Hg = 18 × 10–5/°C] Space for Rough Work 1001CT103316001 E-11/28
Target : JEE (Main + Advanced) 2017/03-07-2016 4. A composite heavy rope of two materials is suspended vertically from a high ceiling. The PHYSICS ratios of different quantities for upper to lower rope are: length Lu 1 , cross sectional area L1 2 Au 2 , density du 2 . The ratio of maximum stress in the two ropes is A1 1 d1 3 6 . Fill in OMR sheet. Lu L Space for Rough Work E-12/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 PHYSICS 5. A 2 m long light metal rod AB is suspended from the ceiling horizontally by means of two vertical wires of equal length, tied to its ends. One wire is of brass and has cross-section of 0.3 × 10–4 m2 and the other is of steel with 0.1 × 10–4 m2 cross-section. In order to have equal stresses in the two wires, a weight is hung from the rod. Let the position of the weight along the rod from end A is x cm. Write x/10 in OMR sheet. Brass Steel AB W 6. The atmospheric temperature above a lake is ‘’ below 0°C. Neglect heat inflow from the bottom of the lake and assume initial temperature of whole lake water at 0oC. It is found that 2 cm of ice layer from top is formed in four days. In how many more days will the thickness increase to 3cm. Space for Rough Work 1001CT103316001 E-13/28
Target : JEE (Main + Advanced) 2017/03-07-2016 CHEMISTRY PART-2 : CHEMISTRY SECTION–I(i) : (Maximum Marks : 18) This section contains SIX 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. 50 ml of HNO3(d = 1.2 gm/ml, 80% w/w) is mixed with 100 ml of HNO3(d =1.1gm/ml, 29%w/v), the mixture is diluted further to obtain 1100 ml HNO3. Calculate final molarity of HNO3 (A) 1/3 M (B) 9/4 M (C) 3/4 M (D) 10/9 M 2. Nitric acid can be produced NH3 in three steps- Process : (I) 4NH3(g) + 5O2(g) 4NO(g) + 6 H2O(g) (II) 2NO(g) + O2(g) 2NO2(g) (III) 3NO2(g) + H2O(l) 2HNO3(aq.) + NO(g) Percentage yield of 1st , 2nd and 3rd steps are 20%, 50% and x% respectively. If 630gm of HNO3 is produced by 600 moles of NH3 at 1 atm and 273K temperature , then the value of x will be - (A) 25 (B) 50 (C) 75 (D) 100 3. Molarity of 44.8V (at NTP)labelled H2O2 solution is - (A) 2M (B) 4M (C) 8M (D) 0.5M 4. The element which have highest magnitude of I.E. and electron gain enthalpy both amongst the following (Consider only one electronic change) (A) Br (B) O (C) F (D) I 5. An element 'Y' have following properties :- (I) higher I.E1 than 'Cl' (II) Bigger size than 'Cl' The element 'Y' may be :- (A) F (B) Br (C) S (D) None of these 6. The CORRECT decreasing order of electropositive character among the following elements is :- K , Ca , Sc , Fe , Br (A) Fe > Sc > Ca > K > Br (B) K > Ca > Sc > Fe > Br (C) Ca > Sc > K > Fe > Br (D) K > Sc > Fe > Ca > Br Space for Rough Work E-14/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 SECTION–I(ii) : (Maximum Marks : 16) CHEMISTRY This section contains FOUR questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Zero Marks : 0 In all other cases. 7. Two bulbs A and B contains 16gm O2 and 16 gm O3 respectively. Which of the statement/s is/are true - (A) Both bulb contain same number of molecules (B) Both bulb contain same number of total atoms (C) Bulb A contains NA molecule while bulbs B contains NA/3 molecules (NA = Avogadro's 2 numbers) (D) Both bulb contain 16 gm- molecule of respective gases 8. Two solution A and B of respective volume and concentrations are mixed together to form final solution C. The statement/s correct regarding solution C is /are - + 2 litre of 2M 2 litre of 2M Final NaOH solution H2SO4 solution Solution (Solution B) (Solution A) (C) (A) Molarity of Na+ is 1M in solution C (B) The solution is acidic (C) The solution is neutral (D) Molarity of SO42– ion in final solution C is 1M 9. Which of the following order is CORRECT for the property indicated against it :- (A) O < S > Cl > Br (Atomic size) (B) S < N < O < F (Electronegativity) (C) Pb < Sn < Si < C (I.E1) (D) O < S < Cl (E.A1) 10. Which of the following are isoelectronic species out of given options :- (A) Mg2+ (B) Al3+ (C) N–3 (D) F– Space for Rough Work 1001CT103316001 E-15/28
Target : JEE (Main + Advanced) 2017/03-07-2016 CHEMISTRY SECTION–I(iii) : (Maximum Marks : 12) This section contains TWO 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 11 and 12 Pressure exerted by a particular gas in the mixture having two or more than two different ideal gases at particular temperature is called its partial pressure which is equals to the product of its mole fraction & total pressure in gaseous state. According to Dalton's law of partial pressure, the total pressure exerted by the different gases in the container is equal to the sum of their respective partial pressure. 11. Total pressure exerted in a container having 2 moles of A and 3 moles of B is 15 atm. The partial pressure of gas B will be - (A) 3 atm (B) 6 atm (C) 9 atm (D) 12 atm 12. The correct graph between PT vs T for a given gas at constant volume - PT PT PT PT (A) (B) (C) (D) TT T Space for Rough Work T E-16/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 CHEMISTRY Paragraph for Questions 13 and 14 The first ionisation energy (H1), second ionisation energy (H2) and electron gain enthalpy Heg1 in KJ/mol of few elements are given below :- Element H1 H2 Heg1 (X) 520 7300 –60 (Y) 1681 3374 –328 (Z) 1008 1846 –295 (P) 2372 5251 +52 13. Which of the above element may belongs to alkali metals :- (A) X (B) Y (C) Z (D) P (D) X 14. Which of the above element may have inert gas configuration :- (A) Z (B) Y (C) P Space for Rough Work SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III 1001CT103316001 E-17/28
CHEMISTRY Target : JEE (Main + Advanced) 2017/03-07-2016 SECTION–IV : (Maximum Marks : 24) This section contains SIX questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. Hardness of water is defined as weight in grams of CaCO3 in 106 ml of water (dH2O = 1gm /ml). If hardness of water is 200, molarity CaCO3 in water is x ×10–3 M, then value of x is. Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer. 2. A sample of hydrocarbon CxHy on complete combustion give 88 gm CO2 and 54 gm H2O. If vapour density of hydrocarbon is 15, then calculate (x + y). Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer. 3. A gas mixture contain He and CH4 in molar ratio of 1 : 1. After n steps if it is get enriched to 8 : 1 molar ratio, then the value of n will be. Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer. Space for Rough Work E-18/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 CHEMISTRY 4. Total number of electrons in Zn(atomic number = 30) for which n + l + m = 0 5. If '' for the last electron of an element is given as '' = 18 × 0.85 + 10 × 1 then atomic number of this element will be ? Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer. 6. Total number of elements of 2nd period which have higher 2nd ionisation energy than the I.E2 of Li. Space for Rough Work 1001CT103316001 E-19/28
Target : JEE (Main + Advanced) 2017/03-07-2016 MATHEMATICS PART-3 : MATHEMATICS SECTION–I(i) : (Maximum Marks : 18) This section contains SIX 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 x 0 1. Let ƒ x 0 1 0 , then (ƒ(x))n (n N) is equal to- 0 0 1 (A) ƒ(x) (B) nƒ(x) (C) ƒ(nx) (D) ƒ(xn) 2. If where > are roots of x2 – 4x + 2 = 0, then value of determinant 1 2 2 2 1 2 is- 2 2 1 (A) 125 (B) 64 (C) 25 (D) 16 3. Let A and B are two non-singular square matrices of order n with real entries such that adjA = adjB, then which of the following is necessarily true- (A) A = B if n is even (B) A = –B if n is even (C) A = – B if n is odd (D) A = B if n is odd Space for Rough Work E-20/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 MATHEMATICS 4. In a triangle ABC, A – B = 60º and 642 = 3abc (a + b + c) {where a,b,c are sides of ABC opposite to vertices A,B,C respectively and is area of the triangle}, then cosC is equal to- 1 1 5 7 (A) 2 (B) (C) 8 (D) 8 2 5. If 2 ; , 0 , then tan( + ) – 2tan is equal to- 2 (A) tan (B) tan (C) –tan (D) 2 tan tan 6. Let the angles of a triangle ABC satisfy cos3A + cos3B + cos3C = 1. If two sides of the triangle are 1 and 2 then maximum possible length of third side is- (A) 2 2 (B) 7 3 (D) 5 (C) 2 Space for Rough Work 1001CT103316001 E-21/28
Target : JEE (Main + Advanced) 2017/03-07-2016 MATHEMATICS SECTION–I(ii) : (Maximum Marks : 16) This section contains FOUR questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) Zero Marks darkened. : 0 In all other cases. 7. Let there are two triangle ABC and ABD such that AB = 17, AC = AD = 13, BC < BD and ABC = ABD = sin1 5 , then - 17 (A) Area of triangle ACD = 60 (B) CD = 24 12 (D) Inradius of ACD 6 (C) Inradius of ACD 5 5 x2 4 4 8. ƒ(x) is a polynomial defined as ƒ x 4 x2 4 , then ƒ(x) is divisible by 4 4 x2 (A) x2 + 4 + 4x (B) x2 + 4 – 4x (C) x2 + 8 (D) x2 – 8 9. Let ƒ(x) is a polynomial in x defined as ƒ(x) = (1 + x + x2)(1 + x)5(1 + x2)5, then - (A) coefficient of x17 in ƒ(x) is 1 (B) coefficient of x16 in ƒ(x) is 6 (C) coefficient of x15 in ƒ(x) is 11 (D) coefficient of x in ƒ(x) is 6 10. Numerically greatest term is the expansion of (3 + 7x)25 is (A) 19th term if x = 1 (B) 18th term if x = 1 (C) 19th term if x = –1 (D) seventh term if x 1 7 Space for Rough Work E-22/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 SECTION–I(iii) : (Maximum Marks : 12) MATHEMATICS This section contains TWO 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 11 and 12 9 Let ABC is a triangle such that AB = 5, AC = 6 and cosC = 16 , BC N. CM is angle bisector of angle C meeting AB at M, r is its inradius, R is its circumradius ; LA,LB are length of medians passing through A and B respectively then 11. CM is equal to - (A) 3 (B) 3 2 3 (D) 6 (C) 2 12. Which of the following is incorrect- (A) r 7 (B) R 8 (C) LA > 5 (D) LB 7 2 7 3 Space for Rough Work 1001CT103316001 E-23/28
Target : JEE (Main + Advanced) 2017/03-07-2016 MATHEMATICS Paragraph for Questions 13 and 14 Consider system 'S' of linear equations in x, y, z ax + 2y – z = 1, 2x + y + z = 1, x – 3y + bz = 4, a, b N. 13. Number of ordered pair (a, b) for which 'S' posses infinite solutions (A) 0 (B) 1 (C) 2 (D) more than 2 14. Number of ordered pair (a, b) for which 'S' posses no solutions (A) 0 (B) 1 (C) 2 (D) more than 2 Space for Rough Work SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type No question will be asked in section II and III E-24/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 SECTION–IV : (Maximum Marks : 24) MATHEMATICS This section contains SIX questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +4 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases. 1. Let A 6 4 ,B 2 3 and C = A + A2B + A3B2+......A100B99. If sum of elements of matrix 6 3 9 4 CB is then 36 + is 2. Let matrix A 1 3 and b,c R are such that A2 + bA = cI2, then number of ordered pair(s) 2 1 of (b,c) is {here I2 is 2 × 2 unit matrix} 3. Let (3 + 2x + 3x2)n = a0 + a1x + a2x2 + ...... + a2nx2n, then a0 2 a1 5 a2 is equal to a2n a2n1 a2n2 Space for Rough Work 1001CT103316001 E-25/28
Target : JEE (Main + Advanced) 2017/03-07-2016 MATHEMATICS 40 36 40 36 40 36 40 36 4. 25 24 23 0 25 Let 0 1 2 ...... is equal to C36 C36 , then is 12 11 here n nCr r 2 1 3 5. If inverse of matrix A is A1 0 4 5 , then determinant value of adj(3A) is 1 2 0 6. In the figure below PQRS is a square of side 2 units. With SR as P Q T diameter, a semicircle is drawn. PTR and QTS are quadrants of R circles of radius 2 units. If the shaded area is n 3, n 1 then n is S Space for Rough Work E-26/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 Space for Rough Work 1001CT103316001 E-27/28
Target : JEE (Main + Advanced) 2017/03-07-2016 QUESTION PAPER FORMAT AND MARKING SCHEME : 16. The question paper has three parts : Physics, Chemistry and Mathematics. 17. Each part has two sections as detailed in the following table. Que. No. Category-wise Marks for Each Question Maximum Section Type of Full Partial Zero Negative Marks of the Que. Marks Marks Marks Marks section +3 0 –1 If none In all I(i) Single If only the bubble of the other bubbles is cases correct option 6 corresponding to — darkened 18 the correct option 0 In all is darkened other cases +4 0 One or more If only the bubble(s) If none of the I(ii) correct 4 corresponding — bubbles is — 16 option(s) to all the correct darkened 12 –1 option(s) is(are) 0 In all In all other darkened other cases cases I(iii) Paragraph +3 — Based If only the bubble (Single 4 corresponding to correct the correct option option) is darkened +4 Single digit If only the bubble IV Integer 6 corresponding — — 24 (0-9) to correct answer is darkened NAME OF THE CANDIDATE ................................................................................................ FORM NO. ............................................. I have read all the instructions I have verified the identity, name and Form and shall abide by them. number of the candidate, and that question paper and ORS codes are the same. ____________________________ ____________________________ Signature of the Candidate Signature of the invigilator Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-5156100 [email protected] www.allen.ac.in E-28/28 Your Target is to secure Good Rank in JEE 2017 1001CT103316001
Paper Code : 1001CT103316001 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) HINDI JEE (Main + Advanced) : LEADER COURSE Test Type : MINOR PHASE : III Test Pattern : JEE-Advanced TEST DATE : 03 - 07 - 2016 Time : 3 Hours Maximum Marks : 210 1. 2. (ORS) 3. 4. 5. 2820 6. 7. 8. 9. : 10. 11. 12. : 13. 14. 15. g = 10 m/s2
Target : JEE (Main + Advanced) 2017/03-07-2016 SOME USEFUL CONSTANTS Atomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Atomic masses : Cl = 17, Br = 35, Xe = 54, Ce = 58, H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127, Xe = 131, Ba=137, Ce = 140, Boltzmann constant k = 1.38 × 10–23 JK–1 Coulomb's law constant 1 = 9 ×109 4 0 Universal gravitational constant Speed of light in vacuum G = 6.67259 × 10–11 N–m2 kg–2 Stefan–Boltzmann constant c = 3 × 108 ms–1 Wien's displacement law constant = 5.67 × 10–8 Wm–2–K–4 Permeability of vacuum b = 2.89 × 10–3 m–K µ0 = 4 × 10–7 NA–2 Permittivity of vacuum 1 Planck constant 0 = 0c2 h = 6.63 × 10–34 J–s H-2/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS PHYSICS BEWARE OF NEGATIVE MARKING -1: –I(i) : ( : 18) (A), (B), (C) (D) : +3 : 0 : –1 1. 0ºC 10 gm 50ºC m ‘m’ (Lf = 80 cal/gm sw = 1 cal/gm-°C) (A) 32 gm (B) 20 gm (C) 40 gm (D) 16 gm 2. A B T2T CTc Tc C (A) T/2 (B) T AB (D) 3T/2 (C) 2/3 T 1001CT103316001 H-3/28
Target : JEE (Main + Advanced) 2017/03-07-2016 PHYSICS3. W (A) W (B) 2W (C) W/2 (D) 4W 4. 100W 1m0.4 40m 0K (A) 625 K (B) 815 K (C) 1025 K (D) 2100 K H-4/28 1001CT103316001
Leader Course/Phase-III/03-07-2016PHYSICS 5. (=4.5eV) 2.0V 200nm (A) 0 (B) 3.7eV (C) A0 (D) B1.7eV 6. (A) (B) (C) (D) 1001CT103316001 H-5/28
PHYSICS Target : JEE (Main + Advanced) 2017/03-07-2016 –I(ii) : ( : 16) (A), (B), (C) (D) : +4 : 0 7. r d (A) r d (B) (C) d r (D) 8. (A) (B) (C) (D) H-6/28 1001CT103316001
Leader Course/Phase-III/03-07-2016PHYSICS 9. —ddE– S1 S2 0 (A) S1 S2 (B) S2 S1 (C) S1 S2 (D) S2 S1 10. 90cmA, BC 40cm, 30cm 20cm 2: 3:4100°C 30°C :- ' 100° A B C 30° 40cm 30cm 20cm (A) = 60°C (B) = 40°C (C) ’ = 60°C (D) ’ = 40°C 1001CT103316001 H-7/28
Target : JEE (Main + Advanced) 2017/03-07-2016 PHYSICS –I(iii) : ( : 12) (A), (B), (C) (D ) : +3 : 0 : –1 11 12 2300 K 1260 nm 8000 Wm–2TK500Wm–2 11. T:- (A) 1150 K (B) 2300 k (C) 1725 k (D) 2000 K 12. T (A) 1260 nm (B) 1870 nm (C) 2000 nm (D) 2520 nm H-8/28 1001CT103316001
Leader Course/Phase-III/03-07-2016PHYSICS 13 14 (i)x-,(ii)y-(iii) z- 1mm z 5cm 1cm 3cm y x 13. Y (B) y- (A) x- (D) (C) z- 14. k (B) y- (A) x- (D) (C) z- –II : & –III : II III 1001CT103316001 H-9/28
PHYSICS Target : JEE (Main + Advanced) 2017/03-07-2016 –IV : ( : 24) 09 : +4 : 0 1. 12 1 2 2 1 eV 13eV 1=1.5eV2=0.7eV (nm)100 Light source 12 +– A V0 H-10/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 2. h RI PHYSICS I Rh c I 3. (cm3 )34cm3 [= 9 x 10–6/°C; Hg = 18 × 10–5/°C] 1001CT103316001 H-11/28
Target : JEE (Main + Advanced) 2017/03-07-2016 PHYSICS4. ;Lu 1 , AA1u 12dd1u32 2 L1 jfLl;ksa esa vf/kdre çfrcy dk vuqikr ;f6n Lu L H-12/28 1001CT103316001
Leader Course/Phase-III/03-07-2016PHYSICS 5. 2 m AB 0.3×10–4 m2 0.1×10–4m2 Axcm x/10 Brass Steel AB W 6. 0°C ‘’ 0°C 2cm 3cm 1001CT103316001 H-13/28
Target : JEE (Main + Advanced) 2017/03-07-2016 CHEMISTRY -2: –I(i) : ( : 18) (A), (B), (C) (D) : +3 : 0 : –1 1. 50 ml HNO3(d = 1.2 gm/ml, 80% w/w) 100 ml HNO3 (d = 1.1 gm/ml, 29%w/v) 1100ml HNO3 H NO3 (A) 1/3 M (B) 9/4 M (C) 3/4 M (D) 10/9 M 2. NH3 - : (I) 4NH3(g) + 5O2(g) 4NO(g) + 6 H2O(g) (II) 2NO(g) + O2(g) 2NO2(g) (III) 3NO2(g) + H2O(l) 2HNO3(aq.) + NO(g) 1st , 2nd 3rd 20%,50% x% 1atm 273K 600 N H3 630gm HNO3 x - (A) 25 (B) 50 (C) 75 (D) 100 3. NTP 44.8V H2O2 (A) 2M (B) 4M (C) 8M (D) 0.5M 4. Magnitude (A) Br (B) O (C) F (D) I 5. 'Y' :- (I) 'Cl' I.E1 (II) 'Cl' 'Y' :- (A) F (B) Br (C) S (D) 6. :- K , Ca , Sc , Fe , Br (A) Fe > Sc > Ca > K > Br (B) K > Ca > Sc > Fe > Br (C) Ca > Sc > K > Fe > Br (D) K > Sc > Fe > Ca > Br H-14/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 –I(ii) : ( : 16) CHEMISTRY (A), (B), (C) (D) : +4 : 0 7. A B 16g m O2 16 gm O3 (A) (B) (C) AN A B NA/3 (NA = ) 2 (D) 16 gm- 8. A B C C - + 2 litre of 2M 2 litre of 2M Final NaOH solution H2SO4 solution Solution (Solution B) (Solution A) (C) (A) C Na+ 1M (B) (C) (D) C SO42– 1M 9. :- (A) O < S > Cl > Br ( ) (B) S < N < O < F () (C) Pb < Sn < Si < C (I.E1) (D) O < S < Cl (E.A1) 10. :- (A) Mg2+ (B) Al3+ (C) N–3 (D) F– 1001CT103316001 H-15/28
Target : JEE (Main + Advanced) 2017/03-07-2016 CHEMISTRY –I(iii) : ( : 12) (A), (B), (C) (D ) : +3 : 0 : –1 11 12 11. 2 A 3 B 15atm B - (A) 3 atm (B) 6 atm (C) 9 atm (D) 12 atm 12. PT vs T - PT PT PT PT (A) (B) (C) (D) TT T T H-16/28 1001CT103316001
Leader Course/Phase-III/03-07-2016 CHEMISTRY 13 14 (H1), (H2) Heg1, KJ/mol :- H1 H2 Heg1 (X) 520 7300 –60 (Y) 1681 3374 –328 (Z) 1008 1846 –295 (P) 2372 5251 +52 13. :- (A) X (B) Y (C) Z (D) P 14. :- (A) Z (B) Y (C) P (D) X –II : & –III : II III 1001CT103316001 H-17/28
CHEMISTRY Target : JEE (Main + Advanced) 2017/03-07-2016 –IV : ( : 24) 09 : +4 : 0 1. 106 ml(dH2O = 1gm /ml)CaCO3 200 CaC O3 x× 10–3 M x ( ) 2. CxHy 88gm CO2 54 gm H2O 15 (x+ y) ( ) 3. H e CH4 1 : 1 n 8: 1 n ( ) H-18/28 1001CT103316001
Leader Course/Phase-III/03-07-2016CHEMISTRY 4. Zn(=30) n+l+m=0 5. '' =18×0.85 + 10 × 1 () 6. 2nd LiI.E22nd 1001CT103316001 H-19/28
MATHEMATICS Target : JEE (Main + Advanced) 2017/03-07-2016 -3: –I(i) : ( : 18) (A), (B), (C) (D) : +3 : 0 : –1 1 x 0 1. ƒx 0 1 0 (ƒ(x))n (n N) - 0 0 1 (A) ƒ(x) (B) nƒ(x) (C) ƒ(nx) (D) ƒ(xn) 1 2 2 2. x2 – 4x + 2 = 0 > 2 1 2 - 2 2 1 (A) 125 (B) 64 (C) 25 (D) 16 3. A B, n adjA=adjB - (A) A = B n (B) A = –B n (C) A = – B n (D) A = B n H-20/28 1001CT103316001
Leader Course/Phase-III/03-07-2016MATHEMATICS 4. ABC A– B = 60º 642 = 3abc (a + b + c) {a,b,c A,B,C ABC } cosC - 1 1 5 7 (A) 2 (B) (C) (D) 2 8 8 5. 2 ; , 0 tan( + ) – 2tan - 2 (A) tan (B) tan (C) –tan (D) 2 tan tan 6. ABC cos3A + cos3B + cos3C = 1 12 - (A) 2 2 (B) 7 3 (D) 5 (C) 2 1001CT103316001 H-21/28
MATHEMATICS Target : JEE (Main + Advanced) 2017/03-07-2016 –I(ii) : ( : 16) (A), (B), (C) (D) : +4 : 0 7. ABCAB D AB = 17, AC = AD = 13, BC < BD A BC = ABD = sin1 5 17 - (B) CD = 24 (A) ACD 60 (C) ACD 12 (D) ACD 6 5 5 x2 4 4 8. ƒ(x) ƒx 4 x2 4 ƒ(x) 4 4 x2 (A) x2 + 4 + 4x (B) x2 + 4 – 4x (C) x2 + 8 (D) x2 – 8 9. ƒ(x), x ƒ(x) = (1 + x + x2)(1 + x)5(1 + x2)5 - (A) ƒ(x) x17 1 (B) ƒ(x)x16 6 (C) ƒ(x)x15 11 (D) ƒ(x) x6 10. (3 + 7x)25 (A) 19 x= 1 (B) 18 x= 1 (C) 19 x= –1 (D) x 1 7 H-22/28 1001CT103316001
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