ANSWERS, HINTS & SOLUTIONS

AIITS-PT-IV-PCM(SOL)-JEE(Main)/6 FIITJEE Students From All Programs have bagged in Top 00, 77 in Top 00 and 05 in Top 500 All India Ranks. FIITJEE Performance in JEE (Advanced), 05: 455 FIITJEE Students from All Programs have qualified in JEE (Advanced), 05. FIITJEE ALL INDIA INTEGRATED TEST SERIES ANSWERS, HINTS & SOLUTIONS PART TEST IV (MAIN) JEE(Main)-06 Q. No. PHYSICS CHEMISTRY MATHEMATICS. D B A. B C C. C A B 4. D B C 5. B D B 6. C D A 7. B C C 8. B B B 9. C A A 0. A C A. C A B. D D C. B A A 4. D A A 5. C A D 6. C C A 7. C B A 8. C A A 9. C D A 0. D A D. C D D. C D D. C D D 4. C C B 5. C D B 6. B C C 7. B D A 8. D C B 9. C D B 0. A C C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 Physics HINTS AND SOLUTION PART I. / v 0 v = + 90 cm 4 (4 / ) ( / ) f 90 60 f 69 cm.. Refractive index of bubble with respect to water is less than. So shifting of the fringe pattern would be towards B.. R = = A d / 4 4 R = d R d 00 00 R d = 4.05 % 4. use ev = hc W 6. V U f y = V, x = U, c = f y x = c y = x + c Intercept c = 0. /cm from graph f = 5cm c 7. Time at which no. of un-decayed nuclei become 0.7 N 0 is equal to the mean life of the sample. t av 7.5 = 8. = 7.5 / min. 0 N = N 0 e -t 7.5 = N 0 e = N 0 e -4 N0 N = 4 e sini sin90.5 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV-PCM(SOL)-JEE(Main)/6 sin i = =.5 0.8A (A x) (0.8A).5 0.64A (A x) (0.8A).5 x + Ax 0.44 A = 0 x = 0. A x x i 0.8A i 9. v 5 0 v = 0 cm. v h m = u Shift of image = II = ( + ) = cm. cm O O I I h 0. sin 0 = sin r = and tan r = sin r = 5 = / 5 / 5 0 0 R/ A R r r B. u v R 5 v 0 v = 0 cm. dh v dh v v I =.v = 0 4mm / s 0 dt u dt u 5.. Resultant intensity is I R = 4I cos / I max = I 0 = 4I I R = I 0 cos I (/6) = 0 4. (n n )(n n ) Number of lines obtained = n = 7 = (7 )(5 ) 5 n = 4. Number of nuclei of z will increase till the rates will become equal FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 4 i.e. k = N max or k N = dn dt, N max = k 5. Maximum K.E. of the photoelectron is KE max = hc = 4eV nm.5ev =.9 ev 80nm since 0 k.e..9 ev K.E..4 ev. 6. X X n He m 4n = 8 n m = n =, m = A A 8 4 0 z z 7. Angular fringe width = = d If & d d/ then new fringe width = 4 4 d 8. 0. hc z Rhc (n ) n n A b A b 0. 0.0 A = (0.0.00)cm 0.0.00 = 0. cm. + = 90 0 = 90 0 = 45 0 M 0 0 M 0 0 60 0 60 0 0 0 M ks 5. Energy received at photo sensitive surface per unit time = 4a FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

5 AIITS-PT-IV-PCM(SOL)-JEE(Main)/6 Number of photons = Maximum energy = ks ks 4 a hc 4 hca hc 6. F = du he dr r... (i) mv he F = r r... (ii) Also mvr = h h solving equation we get r =. me... (iii) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 6 Chemistry PART II. Total pressure of a mixture of volatile liquids is given by P Total = PA X A + PB X B = 00 80 = (5 + 60) = 85 mm Hg 4 4 Mole fraction of A in the vapour phase ( X A ) is given by 00 PA 4 X 5 A = = = 0.94 P 85 85 T X B = 0.706 These mole fractions will go into the distillate. The total vapour pressure of the distillate would be P Total = PA XA PB XB = (00 0.94) + (80 0.706) = (9.4 + 56.48) = 85.88 mm Hg.. At the same temperature, isotonic solutions will have same concentrations. Na SO 4 Na + + SO 4 C = C glu cos e Na SO 4 0.04( + ) = 0. = 0.75. KI + HgI K [HgI 4 ] K + + [HgI 4 ] Thus, there is net decrease in number of ions present in solution and freezing point is raised 4. T f = i K f m 0.0054 = i.80 0.00 i = i = + (n ) n = The correct formula of the compound is [Pt(NH ) 4 Cl ]Cl As S (NH ) S (NH ) AsO 5. 4 4 soluble 9. The number of units of AB in a unit cell (fcc) = 4 Now, r for octahedral co-ordination = 0.44 r 0 0 r 00 0 0 cm r r 0 0 So, r 4.5 pm Hence, (A). FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

7 AIITS-PT-IV-PCM(SOL)-JEE(Main)/6 0. o 0.059 (Cr ) E E log / 7 n (CrO 7 ) (H ).. Pt H H H H Pt 4 atm x 0 atm E 0.059 (0 ) cell log x x = 0 6 M 4 o o 4 Ag e Ag ; E 0.799 V; G 7.7 0 J 8 o 4 Ag(NH ) Ag NH ; K 6 0 ; G (.0 RT)logK 4. 0 J G ( 7.70 J) (4. 0 J).59 0 J E o o 4 4 4 net o G 0.7 V nf 4. The total charge of Fe + and Fe + should be. Let, the percentage of Fe + = x Hence, x + (0.9 x) = x 0.4 % of Fe + 0.4 00 5% 0.9 7. It does not dimerise due to delocalization of odd electrons. 9. HBr + SO H O + Br + SO ; 500K 0. Ti + I TiI 4 700K Ti + I Volatile stable Pure metal compound. Carnallite is an important ore of magnesium. It s formula is KCl.MgCl.6H O.. PbS + PbO Pb + SO, SnO + C Sn + CO, SnO + CO Sn + CO. 5. VOSO 4 VO + + SO 4 So oxidation state of V is +4. CO H O H CO 6. CrO H Cr O H O 4 7 orange red 7. Mn + : d 5 4s 0 In presence of weak field ligand, three will be no pairing of electrons, so it will form a high spin complex i.e. the number of unpaired electrons = 5. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 Mathematics 8 PART III. Let c (c, c, c ), then c a b c c c SECTION A It is given that the angles between the vectors are identical and equal to (say), then a.b 0 0 cos a b a.c c c b.c c c and a c b c Hence c c,c c and c c c On solving the equation, we get c, ;c 4 0, ;c, 4 Hence co-ordinates of c are (, 0, ) or,,. Trick : Obviously, length of the vector (, 0, ) i.e., ˆ i k ˆ is equal to length of a and b. Also it makes equal angle with a and b and equal to that of between a and b i.e.,.. Since the vectors a cxi ˆ 6ˆj kˆ and b xi ˆ j ˆ cxkˆ a.b 0 cx 6cx 0 for all x c 0 and Discriminant 0 c 0 and 6c 48c 0 c 0 and c 4 0 c 0 and c 4 0 4 c 0.. [pqr ] p.(q r ) b c c a a b. a.b c a.b c a.b c (b c). (c a) (a b) [a.(b c)] (b c). {(c a).b}a {(c a).a}b [a.(b c)] (b c) {(c a).b}a, (c a).a 0 a.(b c) (b c) {(a.(b c)}a [a.(b c)] [(b c).a] [a.(b c)] [a.(b c)] make an obtuse angle, therefore FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

9 AIITS-PT-IV-PCM(SOL)-JEE(Main)/6. a.b c 4. Let P be any point in the plane of the triangle ABC. Then A O G S B D PA PB PC PA (PB PC) PA PD ( )PG PG Since G divides AD in the ratio :. SA SB SC SG SG SG GO SG SO, ( OG SG). 5. [ abc ] 0 Also [ abc ] = [ bc ca ab] C = [ abc ] or t = t t = t, t = 0,, - but [ abc ] 0, [abc] = and [ abc ] = - a b a b a a b b a b a.b a a b b a a.b b a.b a b b a ( a b ) a.b a b a b a b is parallel to a b 6. Hence 7. Let x y z 5 5 7...(i) Then x, y 5,z 7 5 General point on this line is (,5,7 5) Again let x y 4 z 6...(ii) 5 Then x,y 4,z 5 6 A general point on this line is (, 4,5 6) For intersection, they have a common point, for some values of and, we must have bcx cay abz 0, (5 ) ( 4), (7 5) (5 6) From first two we have, (iii) and 5 7..(iv) From (iii), put the values of in (iv), we have ( ) 5 7 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 0 9 9 5 7 or 4 or Put in (iii), we get (Putting The required point of intersection is 5 7,, 5,,. ) 8. sin 6 4 sin 4. 9. 406 9. The equation of plane, in which the line x 5 y 7 z 4 4 5 lies, is A(x 5) B(y 7) C(z ) 0 (i) Where 4A 4B 5C 0 (ii) Also, since line x 8 y 4 z 5 lies in this plane 7 7A B C 0 (iii) By (ii) and (iii), we get A B C 7 47 4 The required plane is 7(x 5) 47(y 7) ( 4)(z ) 0 7x 47y 4z 7 0. 0. Equation of a plane passing through origin is ax by cz 0, but it also passes through (,, ) and (,,), therefore a b c 0 and a b c 0. c a c and b c Plane is cx y cz 0 4x y z 0. 4 Hence direction cosines of plane are,, 9 9 9. Let A (,, 0), B (, 0, ) direction ratios of segment AB are (,, ). If '' be the acute angle between segment AB and normal to plane,...6 cos = 4 6. 4 46 Length of projection = (AB) sin = 6. 9 = 46 7 6 units.. Given line passes through (,, 4) and this point also lies on the given plane. Thus, required line will be in the form of x y z 4. l m n Any point on the given line is ( r +, r +, r + 4). If r =, this point becomes P = (0, 5, 5) Let Q = (a, b, (C) be the reflection of P in the given plane, then FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV-PCM(SOL)-JEE(Main)/6 a b 5 5 c... 7 i.e., a + b + c = 4, and a b 5 c 5 = (say) a =, b = 5 +, c = 5 + 0 + = 4 = Thus, Q (,, ) Hence, direction ratios of reflected line are (,, ) Thus, it's equation is x y z 4. In first trial he get s a head = ½ probability that say in first trial he get s a trial then in dice throwing we should get ( to 4) and then 4 head comes. Probability of that 6 6 / So total probability will become... 6 8 / 4 4. No one dies probability = ( p) n atleast one dies = ( p) n now A, A, A n can make n! cases between them if the fix A to die first then rest A A n will make n! (n )! cases between them so probability of A dies first n! n n So, total probability of A dies and dies first p n. 5. We have or 00 50 50 00 49 C50 p ( p) C 5( p) p 00! 50!. 50! 50 p 5!. 49! 00! 5 or 5 5p 50p 5 p. 0 6. The man will be one step away from the starting point if (i) either he is one step ahead or (ii) one step behind the starting point. The required probability P(i) P(ii) The man will be one step ahead at the end of eleven steps if he moves six steps forward and five steps 6 5 backward. The probability of this event C (0.4) (0.6). 6 The man will be one step behind at the end of eleven steps if he moves six steps backward and five steps forward. 6 5 6 The probability of this event C (0.6) (0.4). Hence the required probability 6 5 6 5 C 6(0.4) (0.6) C 6(0.6) (0.4) 5 5 5 6 6 C (0.4) (0.6) (0.4 0.6) C (0.4). 7. P(x.5) P(x 0) P(x ) : expected minimum positive integer FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 k P(x k) e e : Euler number. k! P(x.5). e e e 8. n n n n n n 4n ( ) ( ) ( ) n n n n n [( ) ] 0 [ ( ). ] n n n 0 [ ] 0. 9. As we have A A I A.A (A I) A IA A and A A I A A IA (A I) A A A I 4 5 Similarly, A 4A I, A 5A 4I.. Hence b m c d n y e,x y e. 0. We have a,a,a... a n in G.P. a an an then r i.e., r..... a a a n n Hence r n n n n Now log log(a ) log(a ) log(a ) log(a )... loga loga loga n n n loga loga loga n n4 n5 loga loga loga n6 n7 n8 Operate C C C and C C C loga n (logan loga n ) (logan loga n) = loga (loga loga ) (loga loga ) = n n4 n n5 n4 loga (loga loga ) (loga loga ) n6 n7 n6 n8 n7 loga logr logr n loga logr logr 0. n loga logr logr n6. c n (). But we have to find the value of the square of the determinant, so required value is () 464.. Since (AB) B A.. For the given matrix to be singular, we must have, x 0 x x 0 x x 0, [R R R,R R R ] x x 0 x FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV-PCM(SOL)-JEE(Main)/6 x 4 x 0 x x 0 0 x 0 (x 4)(0 x ) 0 x 4,0 Note that only 4 [ 4, ]., [C C C C ] 4. sin x sin y sin x sin y so for sin x sin y sin x & sin y x = and y = area required (x ) dx 0 sq. units. 5. The curves given by y = x + sin x and y = f (x) are images of each other in the line y = x. Hence required area = 0 0 x sin x x dx cos x 6. Circle has (, ) as its centre and radius of this circle is. Thus if P(x, y) be any point on it, then x [0, 4]. x x Let g(x) 00 50 7. Since x g'(x) 0 x (0, 4] 00 50 The g(x) is increasing in [0, 4] 8 g(0) 0, g(4) 5 8 g(x) 0, x [0,4] 5 [g(x)] = 0 x [0, 4] x x y, represents the x-axis. 00 50 CA = CB =, CD = CD cos CA ACB, ACB sin Area of sector ACB = 4 sq. units a 0 a a f(x)dx sina cosa Differentiating both sides w.r.t. a, we get FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594

AIITS-PT-IV PCM(SOL)-JEE(Main)/6 4 a f(a) a cosa sina sina f 0.. 8. For x, 4 /, y sin 0 sin So required area 4/ sin dx sin Sq. units. 9. f x x 6 A x dx / 6 0 5 0. y (c c )cos(x c ) c e xc 4 4 x c 5 y (c c )sin(x c ) c e x C5 xc5 4 4 y (c c )cos(x c ) c e y c e xc5 4 y y c e y y y Hence the differential equation is y y y y 0, which is of order. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fax 6594