INDIA Sec: Sr. IIT_IZ Jee-Advanced Date: 9--7 Time: 0:00 PM to 05:00 PM 0_P MODEL Max.Marks:80 KEY SHEET PHYSICS B ABC BCD 4 CD 5 BD D 7 ABD 8 ABC 9 C 0 B C C A 4 B 5 A C 7 A 8 B 9 D 0 B CHEMISTRY ABCD ABC ABCD 4 ABD 5 ACD A 7 ABCD 8 ABCD 9 B 0 B B C C 4 C 5 D A 7 A 8 C 9 A 40 B MATHS 4 ABCD 4 ABCD 4 ABC 44 ACD 45 ACD 4 ABC 47 AB 48 BCD 49 D 50 B 5 A 5 B 5 B 54 D 55 B 5 C 57 D 58 A 59 C 0 B
. q q SOLUTIONS PHYSICS ; Elines diverging from +ve charge and terminate at ve charge.. 0 ; L constant, collision is elastic hence kinetic energy will also conserve if the. point of collision is l distance below from the hinge them impulse imported by hing will be zero hence momentum may also conserve. d x d y 0; g dt dt 4. In the given situation image distance is greater then object distance, size of image increases. 5. The pulse is inverted both length wise and side ways from a rigid well and only length wise from a yielding surface. 7 From work energy theorem Wn W f Wgravity 0 8. Very for from the capacitor the electric field is determined by the total electric charge of the system and the electric field outside the capacitor is zero. Hence the component of the total dipole moment normal to the plates is also zero. d dt db dt 9. BA Bl b E l b 0. Rt E 0.5 L t I e e R 0 V. Ball has vertical componentvy and nonzero horizontal component of velocity as. well. Ndt mv Sec: Sr. IIT_IZ Page
Ndt R mr 5... Consider velocity of slab after collision is V then and or Ndt MV mv MV Solving we get 4 mv mr V M 5 M However this is not possible as friction will stop acting once relative motion ceases during collision. Where and 4. In the plank frame, As 5. Velocity of cylinder as it just comes out of water. by work energy theorem Hence height above the surface of liquid Total height Sec: Sr. IIT_IZ Page
. Time for Time for Hence total time 7. (A) I mr (B) I I I (perpendicular axis theorem) I I mr mr I I (C) I mr I (D) I4 I mr mr 8. d = mm d = m I I0 I0 I0 cos yd x D x 9. (A) 0 0. both have same acceleration in downward direction a = 0 m/s aa = ab = 0 m/s (B) 0. 0 Sec: Sr. IIT_IZ Page 4
for A for B 0 = 9 a = 0 m/s a = 9 m/s (C) 0..0 0 = 09 0 + f f = 9.5 9 a = 9.5 m/s f = 0.5 0 = 0.5 N 0 So assumptions are right. f (D).0 0. 0N 0 + 0 = 9A aa = m/s For B 0 f = ab 0 = ab ab = 9 m/s 0. F = (8 x) At eq. position F = 0 Sec: Sr. IIT_IZ Page 5
So, 8 x = 0 x = 4 (A) x = 4 (B) Amp. A = 4 A = m (C) Total time per T Time take x = 4 to x = is T 4 4 (D) a = 4 x vdv 4 x dx v 4 v dv (4 x) dx 0 v x 4x 4 v At equilibrium position, energy of S.H.M. = K.E. of parts = mv = = 4 J. Solution: PbO SO PbSO4 CHEMISTRY PbO Oxidises SO to SO then combine to form PbSO 4. Here PbO is basic and SO is acidic.. Solution: PF cannot form PF 4 ion by accepting F ion or cannot form adduct with any Lewis base. The bond angles in both OF and SCl are 0. Due to resonance O O length in OF decreases. Sec: Sr. IIT_IZ Page
F O O F O F SOCl is used in the preparation of anhydrous metal halides by removing water molecules. SOCl is used for chlorination or as sulphonating agent O F. Solution: Oxidation of hypo with iodine gives NaS4O a salt of tetrathionic acid. 4. 5. SOL: Ag S O H O Ag S H SO 4 White Black H SO 4 gives white ppt with BaCl O S π bonding is facilitated by the electronegative halogen atom. HSO4 NH CONH HNSOH CO NH4HSO4 Sulphamic acid is used to remove NO ion from a mixture of NO and NO V m f i 0. dv b W PdV RT.0RT log. Vm b V b 0.4.08.00log 4.58KJ. Conceptual 7. Conceptual 8. Conceptual 9. Solution: NiCl 4 and NiH O attracted into magnetic field 0. Solution: No of unpaired electrons in A) NiCO 4 0 contain two unpaired electrons and paramagnetic. So B) Mn CN 4 C) Cr NH D) CoF 4 Sec: Sr. IIT_IZ Page 7
. l K R A k 000 molarity K 000 00 0.0 K 40 00 40. SOL: l 0.4 A 0 Nace V K 000 5 0 V 5 k 0 V 5 V 8000 V 0 0.4 50 8000 0.4 50 l K R A l A. 0.4 4 0 000 d A A 5 C t t dt C0 dc C t 0 dt 0 Ct ln Ct l n C0 Ct t C t t Ct C0e e t Ct e N0e t t Sec: Sr. IIT_IZ Page 8
4. Conceptual 5. Conceptual. Conceptual 7. Conceptual 8. Conceptual 9. Conceptual 40. Conceptual 4. P, P 0 and Pn Pn... Pn pn... Let Qn Pn Pn... Qn Qn Pn Pn Pn Pn n Sec: Sr. IIT_IZ Page 9 Qn is GP with st term Qi P i Pn n Qi P i summationon 4. By LMVT n Pn 4 f f 0 ' f c Similarly f f ' f c 0 ' ' f c, f c and f c for at least one c by IVT. Also consider a quadratic g x a x b g 0 0, g, g Such that MATHS Q and common ratio. ' f is continuous.
Which gives gx x 9 8 H x f x g x H 0 H H 0 Let Applying Rolle s twice on 4. Centre of sphere will be,, 7, Centre of inscribed sphere is equation of Distance between P & P H x gives elmaining options. and at distance units from P,, P x y z 7 Volume 9 4. 9 4 7 89. 450 58 T T 44. A A;B I and CC I C C T T A I A A ;B B and C C (A) T T T ABCC B C BA AB C T (B) adjab C 4 adjc adjb adja 4 (C) C B A A B C T T C B BC BC CB adj A BC 9adj A B C T T (D) 45. Z, Z,.. Z9 are roots of equation z 0 = 0 other than z all are vertices of a regular decagon for roots are tenth roots of unity other than r r r ; r 0 z z... z9 N 9 ; r 0 7 9 r0 i r i z 4 z Arg z j 5 i 7 (B) C 5 4 (C) C 5 z & z i j can t be adjacent vertices, Sec: Sr. IIT_IZ Page 0
4. 4 5 5 (D) zi z j cos 5 f x sec x sec x tan x gx sin nx sin x (A) tan xsec xdx tan x tan x C (B) sec x sec x tan x.4 lim en n0 sin nx (C) I n In 0 0 sin n x sin nx sin x cosn xdx sinn n,,,,, for n, 5 7 9 5 7 (D) gx 0 x or sin8x=0 and sin x 0 x,,,,,, 8 4 8 8 4 8 47. Conceptual 48. L: 49&50 For x<0, n x e f x lim e n r rr For x>0 x Sec: Sr. IIT_IZ Page
n r r f x ax lim n n r r r n r r ax lim n r n n n ax lim ax n n x e, x 0 f x b, x 0 ax, x 0 f(x) is differentiable in R a b 5. f x ', x 0 x e, x 0 ' ' ' S f log f log / f log /... 5... 4 5. Sec: Sr. IIT_IZ Page
5&54 z z z z z z z z z z z or z z z z (squaring) z z z z 0 z z or z z 0 z purely imaginary or QAmin Let z x iy for (i) x y 0 and xy 0 If y 0 then x 0 not possible. z or purely imaginary.(ii) z i denotes distance of P z from A i z i AM min Similarly, z i QA z or non positive real (from (ii) where Qz QA is minimum when Q is at QAmin Sec: Sr. IIT_IZ Page
55 & 5 57. Conceptual 58. (P) Let A be given and let P p, Q q, Rr Let K p q q r r p p q r 5 (Q) Let I sin x 4cos x 0 0 (R) Let t t t t t t 0 L lim t t 0 p q r k max p q r 5 5 sinx dx 5 t t t t e t t e lim t t t L L L... Now, 5 L e & L e L L L 4e 4 0 sin xdx 0 Sec: Sr. IIT_IZ Page 4
(S) x y 0 40 x y 4 x y 4 x y 99 499 number of solutions = 4... 99 980 59. Conceptual 0. Let F, and F, Image of F in y=x is be the foci F Here F, 0 and OF OF 4 0 F are collinear. 4 0, or, F F ac 4 4a e Sec: Sr. IIT_IZ Page 5