Frm Classrm/Integrated Schl Prgrams 7 in Tp, in Tp, 5 in Tp, 6 in Tp 5 ll India Ranks & Students frm Classrm /Integrated Schl Prgrams & 7 Students frm ll Prgrams have been warded a Rank in JEE (dvanced), FIITJEE LL INDI TEST SERIES Q. N. NSWERS, HINTS & SLUTINS FULL TEST I (Paper-) PHYSICS CHEMISTRY MTHEMTICS. B B B. B C. D. B D 5. C 6. D B 7. C B 8. D B C 9. B B.. D B C. B B D. B C. B 5. B C C 6. C C 7. B 8. C B 9. C C C... () (r, s), (B) (r, s), (C) (q, s), (D) (p, s) () (p, s), (B) (p, s), (C) (q, s), (D) (p, s) () (p, q, r, s), (B) (q, s) (C) (q, s), (D) (p, q, r, s) JEE(dvanced)- (q, r) B (q, r) C (p, s) D (p, s) (r, s) B (r, s) C (p, r) D (p, q) (q) B (r) C (s) D (p) () (s), (B) (p), (C) (q), (D) (r) () (q), (B) (s), (C) (p), (D) (r) () (q), (B) (s), (C) (p), (D) (r) FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ Physics PRT I. Nw dwnward frce n the right blck is mre. T T mg mg. I C = I + M(C) = I + M(B + BC ) = I B + M(BC). Ptential difference acrss C is zer as I C = 5 I = I =.5 Let the resistance f part BC be r pplying KVL + 5 I Ir I =.5r = 7.5 r = s resistance f part B = 9 Length C = 66.7 cm G E = V r = C E = 5V r = B 5. Initial energy f electrn = ev Energy after frmatin f hydrgen atm in the grund state =.6 ev. Energy released = (.6) = 5.6 ev hc 79 Å 5.6 6. Threshld wavelength = 5Å Wrk functin = hc =.8 ev K.E. = ev = ev hc 58 Å 5.8 7. V V P R = 9 R P V current i = R 9 Let L be the required inductance, then 5 5 r = z (9) x X L L H W f 5 FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ 8. In steady state ptential difference acrss each capacitr = E 9. H = E 6t dt dt = 6 J R. q + q = kq kq kq v = R R R kq kq kq v C = R R R v = v C q = Q/ and q = Q/ Q Q Q Q 5. v = k R R R 6 R Q Q Q 5Q 6. v B = k 6R R R 8 R Q q q B C 7-9. When current is maximum di dt emf acrss L = and ptential difference acrss the capacitr will be same. Frm cnservatin f charge V + CV + CV = 6CV CV 5CV V = Lss in energy f capacitr = energy stred in inductr I max = V C L V + C C S L FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ Chemistry PRT II. g I gi e.. g e g E E.5 V.8 V g I gi E cell =. 95 V.RT t eqm, E g I K SP.RT.95 lgksp.59 lgk F lgk 6. SP t SP n differentiating the equatin we get: d d dt r dt Hence rder is -. r.5. s it lies in the range. t.7. B has ctahedral structure like that f NaCl. r B a r r 6 pm B Vlume = a = (6 pm) = 6 pm. BaF s Ba aq F aq 5. sp sp K Ba F F gain K sp Ba F K Ba K sp F Ba C s s p 6 s p 6 d 6 6 d s p d sp hybridisatin 6. Bnd rder f N,N,N,N,CN and CN are,.5,,.5,,.5 respectively. Higher is bnd rder smaller is bnd length. Bnd rder f C and C + are and.5. 7. Br C N CH 8. CH CH Cl Cl /h MgCl Br /h KCN H H C 6 H 5 Br CHCH FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
5 ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ CH Br Br 9. C 6 H 5 /Zn H. H H CaS. H s H H g 5. 6. H CaS.H s = + 8 kj fr kg H H S = 79 J ml - G.RTlgK p 8 kj ml G lgkp 7. p.rt p 8. atm H p, K H G at eqm. p H S = 7 C H T S T 8 K H Y C 6 H 5 H KH Cannizzar reactin H H K + C 6 H 5 FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ 6 Mathematics PRT III. x + y x y = x + y = Cmmn chrd is x y = x + y = y = x Pints f intersectin f circles are,. k 5 P(E) 8 7 k 8 C. The line can be written as y = mx and curve as x + y = Let, C(h, k) be a pint n the circles and, be given pint then, h h k m k = m Nw, this pint (h, k) lies n the circle, C(h, k) B (, m) (, ) y = mx m 9 9m m 9 m m m m 6 m m m m m m m m m m m m,. Suppse m is an integer rt f x ax bx cx d = as d m (I) m > m am bm cm = d d = m(m am bm c) d m ls, m am = bm + cm + d m (m a) = bm + cm + d m > a cntradictin (II) m < m = n n + an bn + cn d = FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
7 ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ n + n (an b) + (cn d) > S, cntradictin Hence, equatin has n integral slutin 5. (I) x x x x (II) x x + + + + + + + + + + + x (III) x x set x can be, x x x x 6. Required area, / / dx y dt dt = a a sin t cs t dt sint = / a cs t dt a x y (, a) (, a) y x 7. Distance frm rigin, D x y z where P(x, y, z) is any pint n the curve D = x + y + z = x + y + xy xy xy D = and ccurs at pint(s),, and,,,,,,, -, 8. Differentiating w.r.t. r, r dr = a cs d, r rdr cs d sin d r tan dr r a sin FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ 8 S, differential equatin f rthgnal trajectry, dr Slving tan d r ln(cs ) = ln r + k r = c cs d r ct dr 9. Let f x x bx cx. f, f b c. S, tan csec tan sin sec S,, sin tan tan tan tan sin sin sin sin as sin. is cntinuus every where except where + z = z z = i s when z < then abve pints are excluded s f(z) is cntinuus. Cntinuity : fr any pint z, lim f z lim x x f z zz zz. Nn differentiable : fr any pint z, f z z f z x f ' z lim lim z z z x iy nw x lim and x x iy. y cs msin x. m x x y mcs msin x x y xy m y y n n x yn n x yn yn xyn n..y n m yn Simplifying we get, dy at t dt t t dx 6a dt t Nw dy at t ()/ dx x y n y n m y n n n x lim x iy y x.-6. Here, Pv P PK s P v will make acute angle with all the vectrs frm P, P,..., Pk P P k P m n P n P n P P P FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
9 ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ P P P... P S, v k gain, if n is dd n = k P Pk Pk Nw Pk will make acute angle with all the vectrs frm P, P... Pk Pk P P... Pk P k... Pk Pk Pk Nw, Pk P P... Pk Pk P... Pk = P P... Pn gain, V makes acute angle with all the vectrs P, P,... P7 V S P + P +...+ P V S S, 7 7. Eigen values are rts f the equatin X = ( I)X = I = 8 ( )( 6) = =, 6 8. When = 6 8 6 x x 6 x x x x = x = x X C 9. I = 6 + 6 = =,,, Fr = X C which is rthgnal SECTIN B. () ssume sphere as, x + y + z + ux + vy + wz = Nw P, Q, R are ( u,, ), (, v, ), (,, w) x y z S, equatin f plane is u v w Since it passes thrugh (,, ) u v w If centre is (x, y, z) ( u, v, w) lcus is x y z (B) ssume equatin f sphere as, x + y + ux + vy + wz = FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ It passes thrugh (,, ), (,, ), (, ) r u, v=, Radius = + + = if (x, y, z) are crdinates f centrid x, y, z x y z 9 (C) P (,, ), Q (,, ), PQ = 5º tan 5º =.. () Nw if sphere is made with PQ as diameter (x )x + (y )y + z = x + y + z = ax + by.. () plane thrugh PQ parallel t z axis is x y.. () Using (), (), () we get x + y + z x y = (x + y) z z = xy(tan 5º + ct 5º) xy (D) Using family f planes any plane thrugh line f intersectin is x + y 6 + (x z ) = 6 Nw, its distance frm centre f sphere is radius, =, Equatin is x + y + z = 9; x + y z = 9. () Let p = cs x sin y cs z. s, y z, sin (y z) p cs z sin x y sin x y cs z s, sin (x y) and sin (x + y) = cs z p cs z cs 6 8 8 7 (B) The equatin can be written as u 8u u u 6 8 7 6 Clearly, u, u are negative s x, x < x + x < as ct x tan x = = ct x ct x = ct x ct x 7 x x and anther pair, x ' x' x', x' x + x ' + x + x' = 5 FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659
ITS-FT-I-(Paper-)-PCM(S)-JEE(dvanced)/ (C) EF E F = a [ BF cs B + CF cs C] DE C [ E cs + BD cs B] FD b [ CD cs C + F cs ] DC + CE = E + F = FB + BD = a b c DE + EF + FD a b c a b ccs csb csc a b c B F E F D E C S, minimum value is (D) sin x + sin x +.. + sin x = cs xi sin xj j Fr each i we have cs x sin x sin xi cs x 9 j i i i ji i cs x cs x... cs x sinx sinx... sinx i ji sin x j ji sin x j. () Prperty f ellipse (B) Nrmal chrd is made between pints, and B8, s its length is 6 (C) Q (, ), S : (x ) + (y ) = 6 then required circle is (x ) + (y ) + (x y) = (D) Here, = 9 and radius = y x c 6 5 5 e (c > ) and e (c < ) FIITJEE Ltd., FIITJEE Huse, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 66, 65699, Fax 659