FIITJEE Talent Reward Eam PAPER Time: Hours Instructions: for student presently in Class 11 CODE A11 A Maimum Marks: 70 Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results. 1. You are advised to devote 1 Hour on Section-I and Hours on Section-II and Section-III.. This Question paper consists of sections. All questions will be multiple choice single correct out of four choices with marking scheme in table below: Section Question no. Marking Scheme for each question SECTION I (IQ) SECTION II (PCM) SECTION III (PCM) correct answers wrong answers Q. 1 to 8 + 1 Q. 9 to 16 +6 Part A Physics Q. 17 to 5 + 1 Part B Chemistry Q. 6 to + 1 Part C Mathematics Q. 5 to + 1 Part A Physics Q. to 8 +6 Part B Chemistry Q. 9 to 5 +6 Part C Mathematics Q. 5 to 58 +6. Answers have to be marked on the OMR sheet.. The Question Paper contains blank spaces for your rough work. No additional sheets will be provided for rough work. 5. Blank papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any form, are not allowed. 6. Before attempting paper write your Name, Registration number and Test Centre in the space provided at the bottom of this sheet. Note: Check all the sheets of this question paper. Please ensure the same SET is marked on header of all the sheets inside as indicated above Maimum Marks of this page. In case SET marked is not the same on all pages, immediately inform the invigilator and CHANGE the Questions paper. Registration Number : Name of the Candidate Test Centre : : FIITJEE Ltd. ICES House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1100 16, Ph : 65699, 651599 Fa : 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)- Section-I IQ Directions (Q. 1 to ): Read the following information carefully and answer the given questions. L$M means L is son of M, L*M means L is wife of M, L#M means L is mother of M and L@M means L is sister of M. 1. What does A$B*C mean? (A) C is husband of A (C) B is father of A. What does A@B#C mean? (A) C is nephew of A (C) A is Aunt of C (B) C is father of A (D) C is brother of A (B) C is son of B (D) B is father of C Directions (Q. to 5): On the basis of the information given below, answer the given questions: Eight friends Shobhit, Abhishek, Shaurya, Abhinav, Shubham, Saurav, Animesh and Kartik are sitting in a circle facing the centre. Abhishek is between Animesh and Abhinav. Kartik is third to the left of Abhishek and Second to the right of Shobhit. Shaurya is sitting between Shobhit and Animesh and Abhishek and Shubham are not sitting opposite to each other.. Who is third to the left of Abhinav? (A) Shaurya (C) Saurav (B) Animesh (D) Kartik. Which of the following statement is not correct? (A) Shaurya is third to the right of Abhinav (B) Shobhit is sitting between Shaurya and Saurav (C) Abhinav and Shobhit are sitting opposite to each other (D) Shubham is sitting between Abhishek and Shobhit. 5. Who is second to the left of Kartik? (A) Shobhit (C) Shubham (B) Animesh (D) Abhishek FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)- 6. In a row of girls, Seema is eighth from the left and Reema is seventeenth from the right. If they inter-change their places, Seema becomes fourteenth from the left. How many girls are there in that row? (A) 0 (B) 5 (C) 9 (D) 7 7. There are two clocks on a wall. One clock gains minutes in one hour while another clock loses minutes in one hour. Both clocks show correct time at 10 am. Find the time difference of the two clocks at 8 pm net day? (A) hours 16 minutes (B) hours minutes. (C) hours minutes (D) hours 8 minutes 8. A earns more than B and less than C. D earns more than A and C. E earns more than only B. Who earns the least among the five of them? (A) E (B) D (C) B (D) C 9. A s house is located to the north of B s house. C s house is located to the west of B s house. M s house is located to the south-west of B s house and to the south-east of C s house while D s house is located to the north-west of C s house. Then in which direction M s house is located with respect to D s house? (A) North-West (B) South-West (C) South-East (D) North-East Directions (Q. 10 to 1): Study the following information carefully and answer these questions: P, Q, R, S, T, U and V are students of a school and have different favourite subjects viz Mathematics, History, Biology, Chemistry, English, Physics and Economics but not necessarily in the same order. Each one of them also has a liking for different colours viz Black, Red, Blue, Green, Purple, Yellow and Pink not necessarily in the same order. Q likes Mathematics and Blue colour. T s favourite subject is Physics but his favourite colour is not Purple or Black. The one whose favourite subject is English likes Pink colour. P s favourite subject is Chemistry and he likes Yellow colour. U likes Red colour but his favourite subject is neither Biology nor Economics. The one whose favourite subject is Economics does not like Black colour. R likes Purple colour and V s favourite subject is English. 10. Which of the following is correct combination? (A) V-English-Pink (C) U-Chemistry-Black 11. S s favourite subject is? (A) History (C) Biology (B) R-Mathematics-Purple (D) S-Economics-Black (B) English (D) Economics 1. Who likes Green colour? (A) T (C) U (B) S (D) R FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)- Directions (Q. 1 to 1): Find the missing term in the following figures: 1. 8 5 119 9 7 7 8? 7 5 6 (A) 15 (B) 96 (C) 87 (D) 108 1. 17 1 15 1 80 8 10 6 5? 9 5 9 7 (A) 88 (B) 78 (C) 98 (D) 91 15. If = 5, and 6 7 = 85, then 8 5 = (A) 65 (B) 89 (C) 9 (D) 77 16. Sarita correctly remembers that her father s birthday is after 8 June but before 1 th June. Her brother correctly remembers that their father s birthday is after 10 th June but before 15 th June. On which day of June was definitely their father s birthday? (A) 10 th (B) 11 th (C) 1 th (D) 9 th FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-5 Section-II Science and Mathematics (PCM) Physics (Part A) 17. Displacement of an oscillating particle is given by y = A sin(b + Ct + D) where is position of the particle, t is time and A, B, C, D are constants. The dimensional formula for [ABCD] is (A) [M 0 L 1 T 1 ] (B) [M 0 L 1 T 0 ] (C) [M 0 L 0 T 1 ] (D) [M 0 L 1 T 1 ] 18. Let v and a be the instantaneous velocity and the acceleration respectively of a particle moving in a plane. The magnitude of rate of change of speed d v of the particle is dt (A) a (B) v a v (C) v a v (D) v a a 19. Two identical rings A and B are acted upon by torques A and B respectively. A is rotating about an ais passing through the centre of mass and perpendicular to the plane of the ring. B is 1 rotating about a chord at a distance times the radius of the ring. If the angular acceleration of the rings is the same, then (A) A = B (B) A > B (C) A < B (D) nothing can be said about A and B due to insufficient data. 0. A metal rod of uniform linear mass density is placed as shown in the figure. The co-ordinates of centre of mass of the rod is 1 1 1 1 (A), (B), 1 1 1 1 (C), (D), 8 8 16 16 y (0, ½) (½, 0) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-6 1. The maimum tension in the string of a simple pendulum is three times the minimum tension. If 0 be the angular amplitude, then cos 0 is (A) 1/ (B) / (C) /5 (D) /. A particle moves in one dimension in a conservative force field. The potential energy is depicted in the graph. If the particle starts to move form rest from the point A, then choose the INCORRECT Statement. U() (A) The speed is zero at the points A and E (B) The acceleration vanishes at the points A, B, C, D, E (C) The acceleration vanishes at the points B, C, D (D) The speed is maimum at point D A B C D E. Variation of momentum with time of one of the bodies in a two body collision is shown in the figure. The instantaneous force is maimum corresponding to the point (A) P (B) Q (C) R (D) S momentum P Q R time S. A stone of mass m, tied to the end of string is whirled around in a horizontal circle. The length of string is reduced gradually keeping the angular momentum of the stone about the centre of circle constant. The tension in the string is proportional to r n where r is the instantaneous radius of circle. What is the value of n. (neglect force of gravity)? (A) (B) (C) (D) 5. If a lighter body (mass M 1 and velocity V 1 ) and a heavier body (mass M and velocity V ) have the same kinetic energy then (A) M V < M 1 V 1 (B) M V = M 1 V 1 (C) M V 1 = M 1 V (D) M V > M 1 V 1 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-7 Chemistry (Part B) 6. The bond length of O, O, O & O in the decreasing order is (A) O O O O (B) O O O O (C) O O O O (D) O O O O 7. Which of the following statement is true? (A) Higher the value of critical temperature, easier to liquify a real gas. (B) Higher the value of b (vander Wall s constant), easier to liquify a real gas. (C) Lower the value of a (vander Wall s constant), easier to liquify a real gas. (D) All are correct 8. Enthalpy of combustion of C(graphite), H (g) and CH (g) are, y and z kcal/mol respectively. Enthalpy of formation of CH in kcal/mol is (A) y + z (B) y + z (C) + y z (D) + y + z 9. A miture of 50 ml of H and 50 ml of O is allowed to effuse through an effusometer till the residual gas miture occupies 90 ml. The volume of H gas diffused is (A) ml (B) 8 ml (C) 5 ml (D) ml 0. A miture of CO and CO has molar mass 0 gm. 100 gm of this miture contains (A) 0.65 mole of CO (B) 1.875 mole of CO (C) 0.75 mole CO (D) equal mole of CO and CO 1. How is the following equilibrium affected by the addition of a catalyst and change in temperature? SO g O g SO g H = 6 kj (A) With the addition of catalyst, equilibrium constant increases. (B) With the increase in temperature, equilibrium constant increases. (C) With the addition of catalyst, equilibrium constant is not affected but with the increase in temperature equilibrium constant decreases. (D) Equilibrium constant is a universal constant and hence can not be affected by any change.. The bond energy of B F bond in BF is 66 kj/mole, while bond energy of N F bond in NF is 80 kj/mole, this is because (A) N is more electronegative than B (B) The atomic mass of N is higher than that of B (C) B F bond gets a partial double bond character due to P P overlap (D) N has a lone pair of electron while B does not have FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-8. 10 ml of a gaseous hydrocarbon is eploded with 100 ml of oygen. The residual gas on cooling is found to measure 95 ml. On passing through caustic soda solution, volume decreases by 0 ml and remaining gas is absorbed by alkaline pyrogallol. The formula of the hydrocarbon is (A) CH (B) C H 6 (C) C H (D) C H. 50 ml of an aqueous solution of H O was treated with an ecess of KI solution and dil H SO. The liberated iodine required 10 ml of 0.1 N. Na S O solution for complete reaction. The concentration of H O in g/l is (A) 0.68 g/l (B) 0. g/l (C) 0.17 g/l (D) 1.6 g/l / 1/ 1/ 1/ 5. If y = 1 1 1 Mathematics (Part C), then dy is equal to d (A) 1 (B) 1 (C) 1/ (D) none of these 6. The circumcentre of the triangle formed by the line + y = 1 and the co ordinate aes will be (A) (, ) (B) (6, ) (C) (, ) (D) (0, 0) q 7. log p m is equal to (where m = n k ) n (A) pk q (C) pq k (B) qk p (D) p k FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-9 8. e d is (A) e e c (C) e e c (B) (D) e e c e e c 9. If a, b R, a 0 and the quadratic equation a b 1 0 has imaginary roots, then a+ b+ 1 is (A) positive (B) negative (C) zero (D) depends on the sign of b. 0. If tan ( ) = a and tan ( + ) = b then tan equals to (A) a b (B) b a 1 ab 1 ab (C) a b (D) none of these 1 ab 1. The number of numbers from the set of first 500 natural numbers which are multiples of or 5 but not of 7, is (A) 10 (B) 00 (C) 190 (D) 71. The number of values of for which sin + cos = is (A) 0 (B) 1 (C) (D) Infinite. Largest number out of the following numbers, which divides 60 C 0 1 is given by (A) 7 (B) 1 (C) 5 (D) 61 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-10 Section-III Science & Mathematics (PCM) Physics (Part A). A man is coming down an incline of angle 0. When he walks with speed m/s, he has to keep his umbrella vertical to protect himself from rain. The actual speed of rain is 5 m/s. at what angle with vertical should he keep his umbrella when he is at rest so that he does not get drenched? (A) 0 (B) 0 (C) 7 (D) 5 0 5. An elevator of mass M is accelerated upwards by applying a force F. A mass m initially situated at a height of 1 m above the floor of the elevator is falling freely. It will hit the floor of the elevator after a time equal to: (A) (C) M F mg M F (B) (D) M F Mg m F FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-11 6. A wooden cube is placed on a rough horizontal table. A force is applied to cube. Gradually the force is increased. Whether the cube slides before toppling or topples before sliding is independent of (A) the position of point of application of the force (B) the length of the edge of the cube (C) mass of the cube (D) coefficient of friction between the cube and the table. 7. A boy is running at a constant speed v = 1 m/s leftward without slipping on a rough horizontal surface. He is holding string which is connected to a block of mass m = kg as shown in the figure. Find out the power developed by internal forces of the boy in the situation shown in the figure. (Take g = 10 m/s ) v = 1 m/s 5 m m (A) 1. W (C) 1.8 W (B) 1 W (D) 11.8 W 8. As shown in diagram, two wedges of same height are placed on a smooth surface, and moving in opposite direction with constant velocity. A sphere is placed in such a way that the contact between wedges and sphere is maintained throughout motion. Velocity of sphere will be m 9m 8 m/s m/s 5 60 (A) 10 (B) 1 (C) (D) 10 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-1 Chemistry (Part B) 9. In a sample of hydrogen atoms, all atoms are originally in th ecited state. The minimum number of H atoms that should be present in this sample in order to show all possible spectral lines are (A) 0 (B) 05 (C) 06 (D) 0 50. A solution is saturated with respect to SrF and SrCO. If the fluoride ion concentration in the solution is 10 M, the concentration CO ion in the solution is 10 10 Ksp SrF 8 10, K sp SrCO 7 10 (A) 1. 10 M (B).5 10 (C).10 M (D) 1.8 10 M 51. If the radius of 1 st Bohr orbit of hydrogen atom is r, the de Broglie wavelength of an electron in the nd Bohr orbit of hydrogen atom is (A) r (B) r (C) r (D) r 5. Structure of N(SiH ) is trigonal planar. Which of the statement is correct regarding the geometry of N(SiH ) (A) Nitrogen atom in N(SiH ) has sp hybridization and in bonding p orbital of N and either d y or d z orbital of Si are involved. (B) Nitrogen atom in N(SiH ) has sp hybridization and in bonding p z orbital of N and d z orbital of Si are involved. (C) Nitrogen atom in N(SiH ) has sp hybridization and there is no lateral overlapping. (D) In N(SiH ), atomic orbitals of nitrogen do not involve in hybridization at all. 5. The ph of a miture obtained by miing 100 ml, 0.1 M H PO with 100 ml of 0. M Na PO, is (For H PO ; Ka 1 = 1 10, Ka = 1 10 8, Ka = 1 10 11 and log10 0.7 ) (A) 7.5 (B) 10.5 (C) 11.7 (D).7 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-1 Mathematics (Part C) 5. Let f () = + b + c; where b, c R. If f () is a factor of both + 6 + 5 and + + 8 + 5, then the least value of f () is (A) (B) (C) (D) 6 55. r1 r r 1 r r 1 r r is equal to (A) (B) 1 (C) (D) / 56. If AB is chord of contact of point P (5, 5) to the circle of triangle PAB, then 1 1 is equal to (A) (B) 5 (C) 5 (D) y 5 and, is the orthocentre 57. If H is the orthocentre of acute angled triangle ABC, R is circumradius and P = AH + BH + CH, then (A) R < P R (B) R < P R (C) R < P 5R (D) none of these 58. From a point t on the parabola y = a, a focal chord and a tangent are drawn. Two circles are drawn in which one circle is drawn taking focal chord as diameter and other is drawn by taking intercept of tangent between point t and directri as diameter. Then the locus of mid point of common chord of the circles is (A) y = + a (B) 9a ay y + 6a + a = 0 (C) 9a + ay + y 6a a = 0 (D) None of these FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-1 (FTRE-01) CLASS XI HINTS (SET-A) PAPER- 1. (B) Clearly, A$B*C means A is the son of B and B is wife of C. Hence, Answer is (B) i.e. C is father of A.. (C) Clearly A@B#C means A is sister of B and B is mother of C. Hence, Answer is (C) i.e. A is Aunt of C. Directions (Solutions for Q. to 5): Saurav Kartik Shobhit Shubham Shaurya Abhinav Animesh Abhishek. (C) Saurav is third to the left of Abhinav.. (D) Shubham is sitting between Kartik and Abhinav. So, it is wrong statement. 5. (A) Shobhit is second to the left of Kartik. 6. (A) Since Seema and Reema interchange position, so Seema s new position is the same as Reema s earlier position. This position is 1 th from the left (Seema s new position). Number of girls in the row = (1 + 1 + 16) = 0. So, Answer is (A). 7. (C) One clock is gaining min and other clock is losing minutes. So, difference between two clocks = 6 minutes. It shows correct time at 10 am. So, time difference 10 am 8 pm (net day) = hours. Differences of two clocks = hours 6 minutes = 0 minutes hours and minutes. 8. (C) In terms of earning, we have: D > C > A > E > B. So, least among five is B. Hence, Answer is (C). FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-15 9. (C) From the figure we can see that M s house is in the south east of D s house. D s House C s House A s House B s House Directions (Q. 10 to 1): From the given information we can prepare table: Subject Colour P Chemistry Yellow Q Mathematics Blue R Economics Purple S Biology Black T Physics Green U History Red V English Pink 10. (A) 11. (C) S s favourite subject is Biology. Hence Answer is (C). 1. (A) T likes Green colour. Hence Answer is (A). 1. (B) 7 5 1, 8 5, 1 5 119 (5 ) 9, (9 6), 9 7 8 11, 7 5, 11 5 96 1. (D) 17 5 85 1 85 5 80 15 7 105 9 5 105 1 91 15. (B) 1 9 117 8 6 117 1 10 5, 6 7 85, 8 5 89. M s House 16. (B) Sarita remembers that her father s birthday is after 8 th June but before 1 th June, while her brother remembers that it falls after 10 th June but before 15 th June. So, definitely it s falls on 11 th June. 17. B + Ct + D must be dimensionless so dimension of B is [L 1 ], C is [T 1 ] and D is a dimensionless constant. A is having dimension [L 1 ] [ABCD] is having dimensional formula [T 1 ] 18. d v dt is component of a along v. So d v Concept Involved 1. tangential acceleration. Dot product dt v a v 19. Moment of inertia of the ring A, above the ais = mr Moment of inertia of the ring B, above the ais = mr FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-16 mr R m mr because angular acceleration of both the rings are equal. So, A = B Concept Involved 1. moment of inertia. = I 1 m m m 1 1 co-ordinate of C.M. is, 8 8 Concept Involved (i) centre of mass 0. Xcom Ycm 1. Tension is maimum at the bottom most point and T ma = mg mg cos 0 Tension is minimum at the etreme position and T min = mg cos 0 T ma = T min Cos 0 = /5 Concept (i) relative velocity (ii) vector diagram. At B, C and D, dv 0 d So, force on the particle is zero Concept involved (i) Potential energy (ii) Conservation of mechanical energy. dv is increasing upto point R. So at point R acceleration is maimum i.e., force is maimum. dt Concept involved (i) fundamental collision. mvr = C (constant) v = C mr mv mc C T r mr mr T r Concept involved (i) Centripetal force (ii) Angular momentum 5. 1 1 M V M V M 1 < M M V M V 1 1 1 1 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-17 M 1 V 1 < M V Concept involved (i) linear momentum (ii) kinetic energy 6. (D) The Bond length is inversely proportional to bond order. 7. A Higher the value of critical temperature easier to liquify a gas. 8. C C gr O g CO g, H C gr Kcal / mol.... (1) 9. (B) comb 1 H g O g HO l, HcombH g ykcal / mol. () CH g O g CO g H O l, H CH g zkcal / mol. () comb Eq(1) + Eq() Eq() rh M O r M O H Let the volume of H effused be ml. 10 = 8 ml C gr H g CH g, H CH,g y z Kcal / mol f 0. A Let n 1 mole of CO and n mole of CO are present in the miture 8n n n n 0 n 1 1 n 1 n1 1 n n 1 1 1 mole of CO in 100 gm sample is 100 0.65mole 0 1. C Addition of catalyst do not affect the equilibrium constant. Since the given reaction is eothermic and hence equilibrium constant decreases with increase of temperature.. C Due to p p bonding. D y y CHy O CO HO y 10 ml 10 O ml 10 ml FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-18 Volume of CO = 0 ml 10 = 0, so = Volume of O = 75 ml Volume of O reacted = 100 75 = 5 ml y 10 5 y 10 5 y Formula of hydrocarbon = C H. B meq of Na S O = meq of I = meq of H O meq of H O = 10 0.1 = 1 50 N = 1 1 N 50 conc. of H O in g/l = N E = 1 17 0.g / L 50 5. On solving we get, y = 1 dy d = 1. 6. Hint : We know that in a right angled triangle circumcentre is a middle point of hypotenuse 6 0 0 Solution :,, (0, ) + y = 1 (0, 0) (6, 0) q kq qk 7. We have log p m log p n. n n p 8. e d Hint : Method of substitution Solution : Let t 1 d dt d t dt t t t e t dt e t e dt e e c. 9. f() a b 1 f(0) = 1 (positive) i.e. f() a b 1 0 ( a b + 1 = 0 has imaginary roots) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-19 f(-1) = a + b + 1 0 0. tan = tan [( + ) ( - )] tan tan b a 1 tan tan 1 ab 1. Let A = {, 6, 9,, 98} B = {5, 10, 15,, 500} C = {7, 1, 1,, 97} Required number = n (A B C) n (C) = n (A) + n (B) + n (C) n (A B) n (B C) n (C A) + n (A B C) n (C) = 166 + 100 1 + = 00.. As sin + cos no solution.. We have 60 60! C 0 1 = 1 0!0! = 60 59... 1 0 9... 1 0! = 61 1 61... 61 0 0 9... 1 61k. 0!. ABC = 7 Vector degree B A 5 Concept: (i) Relative velocity (ii) Vector diagram F Mg 5. ae M a m = g F g ae M 1 F 1 t M M t F Concept (i) Laws of Motion (ii) Relative Motion 0 C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-0 6. Block of slide If F > mg Block will topple if a mga Fh > mg or F h mga The sliding will occur earlier if mg h < a/h and toppling will occur earlier if > a/h, so independent of mass Concept involved (i) Toppling (ii) Shifting of normal h f S F mg a q 7. d 1 Pf P s b Pw mvm dt 1 d 0 P mgv m b m vm dt 1 d P mgv m b m v cos dt 1 d P mgv mv cos ( ) b m dt v sin Pb mgvm mv sin cos mv mgv cos sin cos = () (10) (1) 6 = 1 + 0. = 1. Watt 5 15 5 Concept involved (i) Power (ii) Work done by internal force 8. v sin = v y cos v sin and v = u = m/s v sin = v y cos u sin (u + v) sin = v y cos v y = (u + v) tan v y = v + u v y = 1 m/s. Concept involved (i) velocity constraints 9m v = 8 m/s m v y cos v sin 5 v y v sin v 60 u = m/s 9. C 5 1 (1) 5 1 () 5 1 () 5 1 () 5 1 (5) 5 1 (6) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)-1 50. A 10 8 10 7 Sr 5 10 M 10 7 10 CO 1. 10 5 10 10 7 51. D r r n r r n 0 0 nh h mv r r h r mv 5. A In p d overlapping d and d z can not be involved. y 5. B PO + HPO HPO + HPO Initial m.mol 0 10 0 0 m.mol after neutralization 0 0 10 10 also PO + HPO HPO Initial m.mols 0 10 10 m.mol after neutralization 10 0 0 So it is a buffer so, PO ph pka log HPO 10 ph 11 log 0 ph 10.5 5. f () will also be a factor of ( + 6 + 5) ( + + 8 + 5); which is equal to 1 ( + 5) So, f () = + 5. 55. r r 1 r 1 r r 1 r r r r 1 r r 1 r1 r1 r1 1 1 1 1 1 = r1 r r 1 r r 1 r1 r r 1 Applying V n method we get, S = 1 1. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519
FTRE-01-IQ+PCM-XI-Paper-(Set-A)- 56., has image in AB as (0, 0). 57. P = R(cos A + cosb + cosc) Since 1 < cosa + cosb + cosc R < P R. 58. Locus is mid point of AD On solving, we get the locus as 9a ay y + 6a + a = 0 y A y = a S 1 P D B S(a, 0) S = a FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 6106000, Fa 6519