ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 FIITJEE Students From ll Programs have bagged in Top, 66 in Top and 7 in Top 5 ll India Ranks. FIITJEE Performance in JEE (dvanced), : 5 FIITJEE Students from assroom / Integrated School Programs & 579 FIITJEE Students from ll Programs have qualified in JEE (dvanced),. FIITJEE LL INDI TEST SERIES NSWERS, HINTS & SOLUTIONS FULL TEST V (Paper-) Q. No. PHYSICS CHEMISTRY MTHEMTICS.... 5. 6. 7. 8. 9...... 5. 6. 7. 8..... 5. JEE(dvanced)-5 B B D C C D B C D D D C C D B, D, B, C, C, B, D, B, C B, C, C, B, C, D, B C, D, B B C D D D D D B C C C B B B D 7 6 9 9 5 7 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 Physics PRT I SECTION. pply conservation of energy and conservation of angular momentum between points at infinite separation and on the surface of the sphere.. To prevent reflection µt 7 5. t.µm... ll the corners of the hexagonal faces will be at same potential. So wires joining the corners can be removed. Then in the remaining circuit we will have six R resistors connected in parallel between and B.. E due to rod. 6L = Q dx. 6L 8L x = Q 6 L E net = E cos. x dx Q, 6L 5L E 5. px qv r dx dv px qv dt dt p a x q Particle is executing S.H.M. with angular frequency p. q 6. Particle will move in helical path having circular projection in y-z plane. In the given time it will complete half revolution. So, x = v t, y =, z = R 7. E P.R R r P will be maximum when r = 8. dv D dh. dt dt dh 6.5 dt. (6.).59 cm / s.5 5T a T. Where T = sec a = 5.75 m/s s we can not find velocity of bullet with respect to ground, we can not calculate distance from the given data. 9.. QBC ncpt Q nc T D V W Q Q Q absorbed released absorbed ncv TD T TD T ncp TC TB TC TB FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 a. a. a.. Dipole moment will be angle of to x axis of magnitude E will be. at angle of with +x axis. a. For E P = q Q q Q E Q Q q E Q Q E inside the gap = E Q Q E on the left side of plates = E Q Q E on the right side of plates = E q P Q q (Q q) Q + Q q. pply conservation of ngular momentum about COM, to calculate final speed of the astronauts. Work = K.E. = Mv. Torque about COM is zero hence angular momentum remains conserved 5. K d / d / D= k d / d / 6. K d K d d d 7-8. First case : = i +. i =.8 E = i +. (r + ) E =. +. r E r (i.). i V FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 Second case : = i 6 6 i 6 E i' r 6r E..() Solving () and () E = 8V, r = i E r 6 i' V 6 SECTION C. B v t v t b when identical balls collide obliquely and collision is elastic and one of them at rest, they move mutually perpendicular to each other after collision. From energy conservation mu mv mv so b = u t = b = m = m rea = b = m. B v b 9 v B U=m/s B. E n m : E m : E 5 : 7 : E Transition corresponding to visible region. m E n : E 77 : : 77 : n n = 9.. For lens B according to the figure u = + v = 6 f = m. f v u. Conservation of angular momentum 5 mv sin (k )R e m ve Re conservation of M.E. GMem GMem 5 mv m v e (k )R e Re 6 solving we get k = 5 or 9. KR e R e v FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
5 ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 5. Given : Floop i B 8N F i B 9N B FBC Floop Force on CD F CD = i B = i B = i B i B i B F F F F F F CD = = loop B loop = loop B 9 9 = 7 N. B C D FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 6 Chemistry PRT II SECTION. On removing atoms along the plane cutting two opposite edges. In new Na lattice; No. of Na + ion : No. of atoms Na + atom : 5 atom. In new fluorite structure: No. of Ca + ion : No. of F ion 5 Ca : F New cationic ratio 6 : 5 5 New anionic ratio = 5 : 8. O O CN NaCN HOEt H /HO MeMgBr excess C O C H H OH C OH C. 5. Cu 8HNO NO Cu NO H O NO has valence e with an unpaired. e in * p orbital. is diamagnetic in solid state due to dimerisation. shows resonance. S S SN N N N N five membered ring S S 6. If PF is better accepter than CO, than C O bond order in II complex is more than I complex. 7. HNO pale blue in colour due to dissolution of N O. Reaction with: () Br water : HNO HO Br HNO HBr (B) KMnO : KMnO HSO 5HNO KSO MnSO HO 5HNO FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
7 ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 8. t stage I: CH COOH NaOH CH COONa Left 5 meq 5 meq So, its is a buffer solution, [Salt] = [cid] ph = pk a ph I = 5 t statge II: CH COOH NaOH CH COONa Left meq in ml. So, c =.5 So, anionic hydrolysis takes place and ph is: & ph = 7 pka log C phii 7 5 log.5 8.8 Stage III: CH COOH NaOH CH COONa Left meq/ml meq/ ml So, ph will be according to NaOH. [NaOH] = [a - ] So, [OH ] = a - poh =. ph III =.95. 9. Beckmann rearrangement: CH N CH CH OH PO5 CH CH NH O C H N Isoquinoline X (B) is aromatic, more basic than aniline. HO CH HO C H CH N. CH NH CH NH PF PF o 8 NO NO NO NO B.O B.O..5 6 Cr O Cr O cidic nature FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 8. () (C) CD CH OH CH CD CH CH CD CH CH H C conc.h SO Major Minor H CH H H H C (D), diequitorial cis, axial and equatorial = trans So, groups are more stable at equatorial position.. Electrode reactions are: nodic reaction Pb s HSO aq. PbSO s H aq. e No rearrangement willoccur Cathodic reaction PbO s HSO aq. H aq. e PbSO s H O Net reaction: Pbs PbO s H SO aq. PbSO s H O 5. MeOK in chloroform KMnO 6. conc. HNO Solution for the Q. No. 7 to 8. Major NO NO NO ZnS H SO dil. ZnSO H S C B Eq. () H S K Cr O H SO K SO Cr SO 7H O S 7 D H S B E colourless liq. D D O SO H O S C CuSO.5H O Blue ZnSO NaOH Zn OH Na ZnO H O NaOH Soluble In Eq. () n eq (H S) = n eq (K Cr O 7 ) Moles(H S) =. 6 = 6 milli moles wt. of ZnS required is Co = (6 97.) mg = 58.8 mg. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
9 ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 SECTION C. OH Me OH Me conc. H So Me Me O HCN HO Me Me CN chiral centre lone pairs. q U w w w [-(6 ) + (- ( 6)] = kj.. Solid P P P 5 6 sp sp d Solid PBr PBr Br 5 sp Sum 5. 5 sp sp Solid N O NO NO en Co en en Co Co en B en en & B diastereomers & C B & C enantiomers FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 Mathematics PRT III SECTION. Six letters can be arranged in following ways:,,, or,,, (,,, ) selection of boxes can be done as C (,,, ) selection of boxes can be done as C C Total such selections = 6 + 9 = 5 The number of required arrangements = 5 6! =,8 b. Distance between two roots = ac Required b ac That means that the point P(b, c) lies on the intersection of the region in between two parabola x = y and x = y + and the square formed by x =, x =, y = and y = Required probability = x dx. x y x y z x y z x 5y z Line is parallel to x y z So, ( + ) + ( ) + = ( + ) + ( + 5) + ( ) =. p, q, r will be in H.P. p + r = pr pr pr pr 5. Put x t 6. tan tan tan O(, ) P(x, y) FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 dy dx dy dx x y dy dy y x x dx dx dy dy x y x dx dx 7. f() = and f () is negative So, equation has one real root in between (, ) 8. x f(x) is a one-one function. x x. Either tan x + tan x + tan x = or tan x = tan x = tan x. Major axis will be x + y = and centre at (, ). Foci will be at distance from centre on major axis. Circle is x + y + 8x + 8y + 6 + k(x + y + ) = From origin chord of contact is T = (8 + k )x + (8 + k )y + 8( + k )= Which is same as x + y + = 8 k 8 k 6 k s PQ is chord to x + y + 8x + 8y + 6 + 6 x y From mid-point of PQ(h, k) chord is T = S (h, k) = ( 6, 6).-. The equation of line L is y = x and the value of p =, q = 6 and r = 5.-6. = is singular B = B is singular C = C is non-singular 5 5 is idempotent B B is nilpotent C I C is involutary Now, check the options FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 7. Equation of the given circle can be written as (x + y y 8) ax = which represent the family of circle passing through points of intersection of x + y y 8 = and ax =. Now, ax = ; x = put in circle we get y y 8 = (y ) = 9 y = y =, points can (, ), (, ) 8. Let the tangent at P and Q to a member of this family intersect at (h, k), then PQ is the chord of contact of (h, k) and its equation is hx + ky a(x + h) (y + k) 8 = x(h a) + y( k ) (ah + k + 8) = Comparing this with equation x = of PQ. We get k = and ah + k + 8 = Since (h, k) lies on the given line x + y + 5 = h + + 5 = h = a + + 8 = a = Hence the equation of the required member c of this family is x + y 6x y 8 =. Let the number of rows in triangle is k + +... k = n kk n k k lso, 9 k k + k + 98 = k + k + 8 k + k 8 = k + 6k 5k 8 = k = 5 55 Number of balls, n = 5 SECTION C. P = P = cos x sin x dx x, put x = t x / sin t dt t. sinrº sinr º cosº Tr = cosrº cos r º cosrºcos(r )º cosº sin (r )º r = sinº cosrº cos(r )º = cotº tan r º tanrº T T...T88 cotº(tan89º tanº ) 88 = cot (º ) 89 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-V-(Paper-) PCM(Sol)-JEE(dvanced)/5 5. X iy r a r r i a ar X, Y a a r X + Y = Y X Y Y So all points lies on circle x + y y r rea of regular octagon = 8 r sin 8 = 8 6 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659