ALL INDIA TEST SERIES

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1 1 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 FIITJEE Students From lassroom / Integrated School Programs have secured to Zonal, 6 State & 18 ity Topper Ranks. in Top 1, 78 in Top and 5 in Top 5 All India Ranks bagged by FIITJEE Students from All Programs have qualified in JEE (Advanced), 15. FIITJEE ALL INDIA TEST SERIES ANSWERS, INTS & SLUTINS FULL TEST I (PAPER-) No. PYSIS EMISTRY MATEMATIS 1. A B. A B A. B 4. A A 5. B B 6. A 7. D A A 8. A, B A, B A, B 9. B, B, D A, 1., D B, B, D 11. A, D A, B, A, B 1. D D B B D A 15. A B 16. D 1.. (A) p,q, (B) p,q, () p,q,s, (D) r,s (A) r, (B) t, () q, (D) r JEE(Advanced)-16 (A) p,r,s, (B)p,r,s,t, () q, r, (D) p, r, s (A)p,s, (B) q,r,t, () p,s, (D) q,r,t (A) p, q, r, s (B) p, r, s () t (D) p, s (A) q, (B) p, () r, s, (D) r FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

2 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 Physics PART I SETIN A 1. charge in flux Q R resistance B A n I 1 A n R 4 r R IAn rr. Relative acceleration of rod with respect to wedge is parallel to the inclined plane.. v v D fapp f v fs Simply use the above relation 4. for B to part A B h 1 v = gt h liquid u = gt v v g a v T v a gt g 1 T 5. alculate the initial and final distance of object from the concave mirror. 6. To avoid t toppling FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

3 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 F N net f x mg a F a mg mg F Now apply Newton s laws of motion. 7. As we know L = L 1 + L L 1 = due to rotational motion L = due to linear motion. 8. If temperature of a body is constant that means amount of hear energy received and amount of hear energy emitted are equal. 9. If direction of field is along the x-axis then its magnitude will be V/cm other wise greatest then hat. 1. Apply conservation of linear momentum KE mass 11. alculate the time constant R R a Battery is replaced by a wire L L R R ab b 1 1. Now apply newton s law of motion for the particle B. N B T 6 o T m r r mg mg FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

4 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/ Initially current will pass through the resistance only. I R 1 = 4 k 1 E R 7 k Power in R = 7 (I ) =.8 W Power in R 1 = 4 I = 4 (4) = 1.6 harge in 1 = = 4. SETIN B 1. In such type of situation, we have to use these basic facts. (a) charge will flow between the two body if there is potential difference between the two. (b) during this process energy will lose due to sparking.. At equilibrium F net will be zero and at the extreme point, particle will be in state of rest. 1. in this situation block B will not move at all. N f k a 1 T T = T 1 mag Now apply newton s laws of motion. T 1 SETIN mg mg a 1. asin v R o o v = 8 m/sec o asin a = m/sec Where R is radius of curvature. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

5 5 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16. apply newton s laws of motion for the body A and body B N M A Mgcos T Mg Mgsin B m 4. length of rod is R ollow sphere v v P Velocity of point P v R v KE of rod = rotational KE t translatory KE 5. simply use this concept After earthing a conductor its potential will become zero. If conductor are connected together both will be at the same potential. 6. from P T diagram, write the P T equation. By using PV = nrt Find P in terms of volume w Pdv. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

6 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 6 hemistry PART II SETIN A 1. Reimer Tiemann s reaction. B The cell reaction is: (g) + I (s) + (aq) + I - (aq). 4. A E cell = o E cell.7714 =.55 p =.591 I log P log 1 Mosley s equation is : az b = az ab Given that, ab = 1, a = tan45 o = z S l Sl4 S S sp 1 ybridisation of S in S = 5. B up down 6. A [Ag + 1 ] initial just after mixing = 1.1 M 5 5 [l - ] initial just after mixing M 5 5 I.P = [Ag + ] [l - ] = I.P K sp FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

7 7 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 Ag aq l aq Agl s.1m.5 initial conc. x.4 M conc. after precipitation Ksp Ag l 1 1 Ag.4 9 Ag.5 1 M 7. A Ph - = /Pt + + Ph Ph Ph + Ph + Ph Ph rearrangement Ph Ph + Ph - + Ph (Y) (X) 8. A, B All reducing sugars are muatrotating. Although (IV) is an hydroxy ketone and hence reducing but it can t mutarotate as it not a carbohydrate, can t form ring. In (II) the glycosidic linkage is in between two anomeric carbons and hence ring opening can t occur, thus non reducing as well as non mutarotating. 9. B, D In B and D compound after mixing undergoes bonding and hence boiling point will be raised and hence vapour pressure lowered. 1. B, Fact FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

8 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/ A, B, Redhot Fe tube n Al /r / l t - Buo - l N N 1. D N N N 4 A (B) () N P P N (D) 1. P l l P P 6 4 P D If k 1, k and k be the rate constants of the reaction along path I, II and III respectively, then overall rate constant of consumption of A will be k 1 + k + k. So, min k k k or k1 k k 81 min k1 41 min k 1 min 46.5 k min 1 k 1 % distribution of 1 k1 k k A d A k k k A 16. dt mol L min 1 1 d B d d D dt dt dt k1a k A k A 1 1 k1 k k A mol L min. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

9 9 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 E aoverall k1e a k 1 Ea k Ea k k k k k k k k k kj / mol SETIN B 1. (A p, r, s ) (B p, r, s, t ) ( q, r) (D p, r, s ) XeF4 6 Xe Xe F P P P 4 B 6 B 6 6 B B 4 Br Na 5NaBr NaBr. (A p, s ) (B q, r, t ) ( p, s) (D q, r, t ) Fact SETIN 1. U n U Np Pu o Effective no. of atoms of X = Effective no. of atoms of Y = 4 1 = Effective no. of atoms Z = 8 4 = 4 Therefore, required answer = FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

10 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/ U n T v v.4 5 v 5 cal 1 R 7 cal p v p P V The gas is thus diatomic No. of eq. of Na that can be produced theoretically for 1% current efficiency = no of equivalents of Nal decomposed = no of eq. of u deposited wt. of u Eq. wt. of u 6.5 / No. of eq. of Na produced experimentally = N V % yield 1 6% offmann s mustard oil reaction is a test of 1 o amine. R N S S S S NR meq. of Na 1 gl RNS (Alkyl isothiocyanote) smells like mustard oil + gs + l FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

11 11 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 Mathematics PART III SETIN A 1. f x n n x dx x dx n n 1 n n f x n 1 n. yy y y y dx dx y y x A y A y A 1. d A d bcosec.cot asec tan A asec y P a b bcosec B x b a b a a b A a b 1 1 b a 4. Let,,, be the four roots Then 4 and 1 AM = GM x 4x ax 5x 1 x 1 a 6, b 4 5. Let APB = PA PB AB cos PA.PB FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

12 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 1 PA PB AB PA.PB cos PA.PB as cos 1 PA PB AB Thus the max value of PA PB is AB This is possible only when P lies on AB but P lies on AB P is the point of intersection of x + y = and x + y + 1 =. 6! P!q!r! P q r 6 & q r 5 6. P x q x r coeff tan tan tan tan sin sin coscos cos cos sin sin sin sin or sin n n, n z or n is not for possible as n is odd tan is not define. ence n, n z is the only solution. 8. AB c, B a, A b In BD A B z y A x D E D cos A B x cos A R x Rcos A a a b c x cos A R sin A sin A sinb sin a a x cot A tan A x Similarly b tanb y c And tan z Also tan A tanb tan tana. tanb. tan (when A+ B + = ) FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

13 1 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 a b c abc x y z 8xyz a b c abc x y z 4xyz 9. (A) x 4 tanx cot x also ln sinx tanx lnsinx cot x lnsin x (B) in, cosecx 1 ln cosecx ln cosec x 4 5 lncos x () in,, cos x,1 1 1 also lncosx lncosx = 5m n 7(5m n) + 5m n = 6m 7mn + n = n m n or m If m = /n from (1) = 1/ n If m = n/.. = -n/ cos a b c d a a.b 4 d. a b c a,b,c are mutually r Let d a b c Then d.a d.b d.c cos cos Also 1 cos or cos 1 a b c d end point of latus rectum P(1, ) & Q(1, -) PAQ is isosceles right angled FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

14 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 14 slope of PA is 1 In equation y = - (x 1) x + y = similarly equation of line QB x y = solving x + y = with parabola y = 4x x 4x x 1,9 co-ordinate of B & are (9, -6) & (9, 6) respectively Area of trapezium x8 64 sq units 1. let the circumcenter of trapezium be T(n, ) Then PT = PB n 1 4 n 9 6 or n = 7 radius z 1 ase I when 1 1 We get < 1 z i 1 or x = - y = 1 x or x + y = 1 case II > 1 1 z 1 or x 1, y Roots are 1,, 1, ne root lies inside the unit circle and other will lies outside the unit circle ase III where is very large then z 1 z SETIN B 1. total number of function = 6 = 79 (a) total number of onto function = FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

15 15 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 (b) since f(a i ) b i It means that a 1, a, a cannot be assigned images b 1, b, b Number of function = = 16 (c) number of invertible function = as function is not one one (d) total many one function = 79 = (a) sin sin cos cos 4 1 (b) (c) 4 1 tan x tan if x x if x (d) n n 1 n n 1 sin 1 tan nn 1 1 nn 1 1 n n 1 n n 1 sin tan n1 nn 1 n1 1 nn 1 tan 1 n SETIN 1. let each friend has n sons. tickets can be distributed among n sons in n ways. The number of ways distributing tickets such that two tickets go to the sons of one and one tickets goes to sons of the other. n n n n n n probability that two tickets go to the sons of one and one tickets goes the sons of the other n n 1 n 6n n 1 n n n 1 n n 1 But from question 6 n 7 n 1 n 4 ence total number of boys = 8.. the equation of the tangents to the circle x + y = a at P and the hyperbola x y = a at Q are x my a 1 m or x my a 1m respectively Where y = mx is intersecting line through (, ) Let (h, k) be the point of intersection of these two lines FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

16 AITS-FT-I-(Paper-)-PM(Sol)-JEE(Advanced)/16 16 h mk a 1 m are h mk a 1 m h mk a 1 m and h mk a 1 m m K a mhk h a m K a mhk h a Elimination m from these two equation and we get a 4y x a. A B i i B i A ia Since e 1, e e e ia iab i A e e e iab ib ib e e e ia ib i e e e put x 1 t x 5. P P or y 6. x y x 6 x Area of region = = sq unit 9 y dy n f x lim n n n... n 1 1 n log f x lim log log... log n n n n n n 1 r lim log n n n 1 f x r 1 logxdx 1 e 1 1 e ence e f(x) = e e f x e 1 1 n e e FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fax 65194

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FIITJEE JEE(Main) AITS-FT-II-PCM-Sol.-JEE(Main)/7 In JEE Advanced 06, FIITJEE Students bag 6 in Top 00 AIR, 75 in Top 00 AIR, 8 in Top 500 AIR. 54 Students from Long Term Classroom/ Integrated School Program & 44 Students