ALL INDIA TEST SERIES

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1 From Classroom/Integrated School Programs 7 in Top 0, 3 in Top 00, 54 in Top 300, 06 in Top 500 All India Ranks & 34 Students from Classroom /Integrated School Programs & 373 Students from All Programs have been Awarded a Rank in JEE (Advanced), 03 FIITJEE ALL INDIA TEST SERIES JEE(Advanced)-04 ANSWERS, HINTS & SLUTINS FULL TEST II PAPER- Q.N PHYSICS CHEMISTRY MATHEMATICS. A A C. C A D 3. B A A 4. C A B 5. C A B 6. A A C 7. A, B, C, D A, B, C, D A, C, D 8. A, B A, C, D A, C 9. C, D A, B, C A, B 0. B, C A, D A, D. B B C. B A B 3. B A A 4. C C A 5. A B C 6. B C D 7. A C C 8. A B B 9. C D C 0. D C D. C D B. D B D 3. C A A FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

2 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/4 Physics PART I. (v cos)t = (v cos)t = () (v sin)t gt +h = (v sin)t from () and () h = 60 m gt... () 3. = tan g u cos g tan g = tan tan 4 4 tan + = 0 If particle will not hit the target. (b 4ac) < < 0 4 > 3 mv B 4. At B, mg sin = r Using energy considerations... () Smooth u g A B (, ) N v B mvb = mgr(cos sin )... () From () and () mg sin = mg(cos sin ) cos = 3 sin 7. Here the position y on the screen will correspond to maima. nd y d 30,000 when n =,, 3, 4,.. n 8. a = k da k d tan 60 = k 3 k a dv v d 9 C 3 v v C at = 0, v = 3 m/s 6 hence a =.5 and v 3 9 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

3 3 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/4 T cos 9. Capillary height h h = 59.6 mm rg here 59.6 mm is greater than the protruding part hence water will rise in the capillary of insufficient height 5 mm. T Now, R 0.6 mm hrg Fluid particles passing through the bend are in circular motion. The centripetal acceleration is provided by the variation in pressure. In the section as shown P A P A ma C P P applying Bernoulli theorem at and i.e., P v gh P v gh P v P v as h h P P v v ma C P A P A 5. mag sin = 5 0 sin 37 = = 30 N f ma AB(mC m B )gcos 48 N hence f = 30 N 3 6. mcgsin 0 N 5 4 fma mcgcos N 5 hence f =.6 N 7. T fa fc mbgsin37 = 9.6 N T f C f a m Bg sin ptical path difference between the beans arriving at P ( ) dsin for maima ( ) dsin n sin n ( ) d n sin 40 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

4 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/ sin n 40 or 0 n 60 Hence number of maima = 60 0 = At C, phase difference ( ) = 80 Hence for maimum intensity will appear at C. Now for minimum intensity at C ( )t t t 500 nm ( ) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

5 5 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/4 Chemistry PART II 3. Neither N nor F contain d-orbitals. Further in N F 4 N N bond is shorter than in N H 4 due to s-character (Bent s Rule). 4. Due to poor shielding effect of 3d electron the size of Br ion nearly same as that of Cl ion. 5. N C 8 H 7 Cl 87 m Molar mass of C 8 H 7 Cl = = 54.5 g Molar mass of N = = 44g According to Graham s Law of diffusion rn MC8H 7Cl : r M 44 C8H7Cl N.87 dn 87 = 87 th row from N side.87.0 dc8h 7Cl 87 = 00 th row from weeping gas side.87 Therefore, the spectator from the side of N in the 87 th row will be laughing and weeping simultaneously Alternatively, the spectator from the side of weeping gas in 00 th row will laugh and weep. 6. H attacks at the more reactive (C=) group, (containing more EWG or less EDG) Et is more EDG than (due to +I effect here no H.C.). Therefore H attack (C=) with () group. H H C H H3 C H H Et Et Et NH N aq N H NHH HN N H Hence, (A) and (D) are the correct answers. 3. Velocity = 6 Z.8 0 n 4. (i) (ii) E Br 3 M/ 5 E Br 3 M/ 6 ratio = 6/5 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

6 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/ H N A H HN N N + H N H C C4 bond breaks Ring ep ansion H H (B) RC3H Baeyer Villiger (C) (Cyclic ester) i LiAlH4 ii H H (D) H (Lactone) KI Pb N Pb I KN mol of Pb(N 3 ) reacts with mol of KI m mol of Pb(N 3 ) reacts = 5.5 = 0.5 m mol m mol of KI 3 left = 0.5 =.5 Hence Pb(N 3 ) is the limiting reagent.5 m mol I M 5 35 ml 0. Due to common ion sp S = K = = I left in the solution the solubility 9 Pb I decreases.. ph range of all these indicators lies in between -. Hence, all are suitable indicators.. Since curve is given for strong acid and strong base hence, ph at the end point is 7. 3 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

7 7 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/4 Mathematics PART III. y = a + b + c, verte is (4, ) b 4ac b 4 =, b = 8a, c = + a 4a Now = abc = 8a ( + 6a) = 6(a + 8a 3 ) d = 6(a + 4a ) < 0 a [, 3] da b 4a = + 6a ma = 44, min = 3600 Difference = [] = {} [ + ] = {} 4 f d = 6.. = t dt lim 0 = sin a t = 0 lim 0 sin a = 4. a = lim 0 0 t dt a t sin = (given) a = lim 0 a cos 4. (3) 6 = 79 < 900 and (3) 7 = 87 > Now by property of triangles A RQ = BC = a R Q Similarly PQ = c, PR = b Area of ABC = abc 4R a b c That of PQR = 4R' Also area of ABC = 4 (area of PQR) abc abc R 3R' 6R R' B P C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

8 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/ Clearly solutions B y = 3 y =log 3 7. f() = cos ( + []) f(0) =, f(0 0) = cos ( ) = f(/) = cos = 0, f 0 cos f 0 cos 0 for (0, ) f() = cos for (, 0) f() = cos ( ) = cos = 0 8. P = 5 sec P = 5 cosec 00 PP P = sin ( PP P ) min = 00 = /4 P =0 P = (0, 0), ( 0, 0) P P P 9. Let P(, ) goes units along + y = upto A and 5 units along y = 4 up to B Slope of PA = = tan 35, slope of PB = = tan sin =, cos = 5 5 A ( + r cos 35, y = r sin 35 ) = ( +, + ) = (, ) B ( + r cos, y + r sin ) = ( + 5 5, + ) = (5 +, 5 ) Slope of tangent = a b a b FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

9 9 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/4. dy d dy 4 t dt < 0 d 4 t dt b So, 0 a a b > 0 A PA B P P B B or is positive 5 3. a b c d a b c d e 6 e e 6e 64 4e 5e 6e 0 0 e Let A(a, P(a)), B(b, P(b)), then slope of AB = P(a) = P(b) from LMVT c (a, b) Where P(c) = slope of AB 5. Given QT = QA = Let PQ =, then PT Then TQP and AP are similar triangles Then T A ( ) 8 = 5 3 T P A Q 6. From above, A = and AQ = coordinate of Q (, ) equation is ( ) + (y ) = 7. RQ is a right angled triangle. Then and R are the etremities of the diameter then the coordinates of R( 3, ) Equation of circle ( 0)( 3) (y 0)(y ) 0 y 3 y 0 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

10 AITS-FT-II (Paper-)-PCM(Sol)-JEE(Advanced)/ Perpendicular tangents intersect on the director circle of hyperbola and director circle of rectangular hyperbola is a point circle. Hence centre of hyperbola is (, ) and equation of asymptotes are ( ) = 0 and y = 0 9. Equation of hyperbola is y y + + = 0 It passes through (3, ) hence = Equation of hyperbola is y = + y + 0. From the centre of hyperbola we can draw two real normals to the rectangular hyperbola. Taking point (r cos, r sin ), we get r = 3 3 sin 4 3 r ma =, r min = 3 Ma eist when 4 8. Required equation is 3 3 y 3. Centre of circle be origin and its radius is the length of semi minor ais Hence area = FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

ALL INDIA TEST SERIES

ALL INDIA TEST SERIES AITS-FT-I (Paper-)-PCM(Sol)-JEE(Advanced)/5 FIITJEE Students From All Programs have bagged 4 in Top, 66 in Top and 74 in Top 5 All India Ranks. FIITJEE Performance in JEE (Advanced), 4: 5 FIITJEE Students