Narayana IIT Academy

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1 INDI Sec: LT-IIT-SPRK Jee-dvanced Date: -3-8 Time: 3. 4_P MODEL Ma.Marks:8 KEY SHEET RPT-4 PHYSICS,B,C,D,B,C,D 3 B,C 4,C,D 5,B,C 6,B,C,D 7,B,C,D 8,B,C,D 9,B,C,D,B,D CHEMISTRY,C,D,C 3,B,C,D 4,B 5,D 6,B,C 7 B,C,D 8,B,C 9,C 3,B,C,D MTHS 4 B,C 4,B,C 43 B 44 B,C 45 B,D 46 B,C 47 B,C,D 48 B,C,D 49,B,C 5 B

2 SOLUTIONS PHYSICS. t t = capacitor behave like plain conducting wire and inductor behave like infinite resistor. t t = capacitor behave like infinite resistor and inductor behave like plain conducting wire.. The induced emf in loop BHFG d db B V dt dt The induced emf in loop BCDH & DEFH volt. KVL in top left loop is, y z y y 4y z () KVL in right loop : y z 4 y z () By equation () & () it is seen that y no current in DH {This can also be seen by symmetry} This makes solution very simple now the circuit is, V V V ssume VB, VF V, then 4 4 V no current in FB, and current in GFEDCB is 5. rate of heat production = 4 / 8 watt. 3. The potential is not quadratic so the motion is not sinusoidal or simple harmonic. The motion is periodic since the particle is bounded by the potential. The energy is constant and so the particle will move back and forth between a maimum negative position and the position. The potential energy is continuous and hence K.E. = TE U is continuous. Therefore, speed is continuous. The derivative of the potential energy is not continuous at =. Since, force is equal to the negative F derivative of the potential energy, the force is not continuous at =. Therefore, acceleration m is not continuous at mv ' mv mv cos i ˆ mvsin ˆ j v v ' cos iˆ sin ˆj 3 3 Sec: LT-IIT-SPRK Page

3 sin tan cos K k k i f 3 3 mv mv m v mv 6 dv F F m (Given v = constant) dt 5. et gr F g v 3 ' 5 cos F g v g Wet F. d v W K U H et g v g Sec: LT-IIT-SPRK Page 3 v H H v d d 6. Magnetic field along -ais varies B. B B B v dt dt d e B v. pply e i dt R R B v t v v t anticlockwise R 7. v v u, i B R anticlock wise P T Pt s given,..(i) Pr mplitude of transmitted wave T / and T / T /.(ii) From equation (i) and (ii) or. (ssume )

4 5 or.4 (i) r. r r. r r. [Note : We have taken r s r r is given in the question (ii) 5..4 (iii) The wave number, K K Kr Ki K K K Kt. 5 There is no phase change on reflection K yr sin K t sin t 3 5 du du Energy density constant and y t sin K t sin K t 3 8. Energy density du vdt dv vdt Where rea of cross section of the rod then amplitude constant K density K f f V Time taken Y Y then V f V K d dt K Y Y K d dt 3/ K t 3 Y K Y K Sec: LT-IIT-SPRK Page 4

5 9. P V The cycle is ellipse with semi-major ais = and semi-minor ais = and centre is P V P maimum pressure P P 3 V 3 t maimum temperature P P cos 45 P and V V sin 45 V PV nrt 9 P V Tma, similar for T min 4 nr Total work = area of ellipse = ab P V P V gh h h g.. h h h h h h h h h h h h If the cylinder is displace down by y F hg h y h y g net hg h h g gy gy h W g hv V u gh h F a Initially, force equation for particle: ku mg mg where, k = constant mg So, k u Let at any instant, speed of ball is v: gv dv gv So, a g or g u dt u Sec: LT-IIT-SPRK Page 5

6 t dv ul n 6l n dt v or t g u g u V v f ' f f ' V v s f " f ' f " f " Hz 3. mplitude of SHM, = 5 cm, initial position of jockey is at / and moving towards etreme. T T t 6 4 v sin 4. Total time of flight T g v Time from to B t g sin Then, B to C = T t v But, T t (maimum possible time) 3 g v sin v v g g sin g 3 sin sin 3 3 sin, 6 V V V 5cos f ' f f V VS V 5cos 45 f ' f.5khz f 6. Magnification m m 3 f f Taking the direction in the right hand side V Vim V om velocity of image w.r.t mirror velocity of object w.r.t mirror V 4u in Sec: LT-IIT-SPRK Page 6 im m V mu u Vm velocity of platform (mirror) m m 3 V V V im i m V velocity of image w.r.t ground i om

7 u 3 Vi 4u Vi u m / s Vi 7. Conceptual 8. From the conservation of the energy we have, Initial internal energy = dissociation energy + final internal energy T ' 3 T 8 T ' T ; T 4 9. In equilibrium, power released = power absorbed 4 4 Or 4 R T 4 R 3. Height of water column = 4cm 5 Mass of water = gm By work energy theorem W U U KE 8 Joule et piston water water CHEMISTRY. () is Cis acid (B) see the structure, use Bent s rule. (C) Open book structure bent structure (D) Bent molecule sp. 3 3 (B) and (D) dsp diamagnetic. and B are sp and sp d, unpaired e. 3. One element is in intermediate oidation state. So can act both as oidizing and reducing agent. 4. (C) is C6H5NHCOCH3, D : CH3CN cannot give NH 3 gas. 5. (B) gives more than two mono chlorodeviatives. (C) gives only one. 6.,B,C are condensation polymers. 7. () Cl Conc.does not change. NH H O NH H O. Cation concentration decreases (B) (C) Since H Conc.Inc poh >> and (D) ph > 7 M. Note the solution is acidic due to cation hydrolysis. 8. () and B are isodiaphaers. ( Z) or (n p) value is same for isodiaphers (B) D has same mass number as that of. (C) B,C,D have same mass number but different atomic number. (D) and C do not have same number of neutrons H 4 9. Svap TBP 4 () p atm BP 4K P BP S K vap (C) G H T S 4 4 ve G Sec: LT-IIT-SPRK Page 7

8 (D) H U n RT g 3 4 u u kj / mol 3. () n l 3 (B) n l (C) ngular nodes = l, spherical nodes = n l 3 (D) p orbital is angular dependent Ecept, 6, others have sp d hybridization. 3. (i), (iii), (vi), (viii), (i) do not give Cl ph pka log C H COO 6 5 ; log C H COO C H COOH C H COOH C6H5COO C H COOH Let volume of acid is V ml.3 V, V 5 ml.v Ea/ RT k e E a Ea/ RT RT e. e Ea Ea 6, Ea 8 kj / mol 5 4 G H T S H 4 6 kj / mol H H C T T p H 4 H kj / mol 36. Ecept NaCl, CsCl, RaCl, Others have formal + charge on a non metal atom or molecule. 37. OH and SH consume each one mole and COOCH5 consumes two moles of RMg to give tertiary alcohol. Total : 4 moles. 38. RCONH Br 4NaOH RNH NaCO3 NaBr HO 39. Zn 3 3 dust 3 3 CH C C CH CH C C CH 4. DDT: p, p dichlorodiphenyl trichloro ethane ( 5) CCl NO y Chloropicrin: Sec: LT-IIT-SPRK Page 8

9 MTHS 4. a O k cos, b OB k sin a b b a k cos k sin k sin k cos 4 4 k k sin sin 4 Minimum value k 4 when is 45 k k lso area of OB 4 4. n 3 lt n n, / 3 f 3 3 / 3, / 3 3 lt n n Clearly a = b = c = d a 3a 3a : eternally. f m y 44. Homogenisation gives f m y m my m m m f m m f m 45. Sec: LT-IIT-SPRK Page 9 m sin cos sin sin.sin f Lt Lt f Lt cos

10 D a 46. tan C b D ab C ab C D CD a b b b a b 4 a a a 4 a a 4 a b a a 47. y ; i,,3 i i Let i If i, then,,3,...9 For i,3,,,,3,...9 Total no.of ways of choosing the pair, is C C... a a a a... Compaining Coeff. Of a a a3 On solving n,3, 4 3, 49. sin Sec: LT-IIT-SPRK Page

11 sin n, n, n I Let f a b c d and f t f t dt F, f3 t f4 t dt F f t f3 t dt F3 and f t f4 t dt F4 f On solving 4 5. y y 5. cos 3 5 sin 3 5 Fied point(,) y 4 y X=(3,3) PR k QR h k y h k y k (Which is a circle for different values of k centers of S, S, S 3 are an PQ and none intersects with Other. Hence common tangents are either or 4) ' ' ' ' lso S, S, S 3are the circles passing through P and Q. Hence have two common tangents S k has PQ PQ as diameter so radius since ' S k is orthogonal to S i PQ Therefore PQ / PQ / therefore a 53. a 3 a a a a, roots a a a a a Sec: LT-IIT-SPRK Page

12 3 3 a a a a lt lt lt / / e e a lt a kl / t f t dt f t dt sin cos a a pply t f d f a d f t dt f t f t dt sin cos Diff. both side w.r.t, then f t dt f cos sin (i) gain diff. both side w.r.t, we get (i) Diff (ii) w.r.t then f " f ' cos sin..(iii) dding (ii) & (iii) we get f " f cos B B B B B B B B adj adjb.. B. B 3 B. B y y 4 a 5 5 f f ' sin cos (ii) Now consider parabola t t I y 6 5 Sec: LT-IIT-SPRK Page

13 4 5 5 t t t 5 t, t t.4 a t t a t t 6 g g R 57. Diff. w.r.t we get g ' g ' Put,,3 / Verify Rolle s theorem. g g Replace by We have 4 ' ' 4 g g g g 58. BC t t 7 a 5 t t 7 5a.(i) If,,, C t t D t t then t, t are roots of equation (i) and CD =a t t t 7 5a, a a a a 7 5a a 5a 3 a 6 5 or a 4 5 Minimum area is when a 4 5 a B3 P B3 E 59. P E P E 3 6. pply. M G. M on,, Sec: LT-IIT-SPRK Page 3

ALL INDIA TEST SERIES

ALL INDIA TEST SERIES 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