Narayana IIT Academy INDIA Sec: Sr. IIT_IZ Jee-Advanced Date: Time: 02:00 PM to 05:00 PM 2014_P2 MODEL Max.Marks:180 KEY SHEET

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1 INDIA Sec: Sr. IIT_IZ Jee-Advnced Dte: -9-8 Time: : PM to : PM 4_P MDEL M.Mrks:8 KEY SEET PYSICS A B C 4 A B 6 D 7 A 8 B 9 A D D C A 4 B D 6 B 7 A 8 D 9 A D CEMISTY B B C 4 D A 6 B 7 A 8 C 9 D C D D C 4 D B 6 D 7 D 8 C 9 A 4 D MATS 4 B 4 B 4 B 44 A 4 D 46 C 47 A 48 C 49 B B C D C 4 A C 6 A 7 B 8 C 9 C 6 D

2 . Field t A: SLUTINS PYSICS Due to is E prllel to C C Due to is E 4 prllel to C C 4. 8 Due to is E prllel to C A E E + E + E A 6. I bttery is connected cross points C nd D, points A nd B will be t sme potentil. Similrly, E nd F re t sme potentil. There will be no current in the resistnce directly connecting A to B nd E to F. The circuit cn be redrwn s AC CB EC FC AD BD 6. Conceptul 4. Consider long wire element s shown. The igure shows the cross section o the wire. I I Current through the wire element is di d d Sec: Sr. IIT_IZ Pge

3 Mgnetic ield due to the current di t is di db I d I d component o the ield is db I sind / I I B db sind From symmetry B y B I esultnt B t is. When plte is moving to right: B B + B Molecules bounce bck with speed V- u Chnge in momentum o one molecule P m V u mv mv mu Number o molecule hitting per unit time n A(V-u)n When plte is moving to let: y F n P An V u m V u mna V u Molecule bounce bck with speed V + u Chnge in momentum o one molecule P m V + u mv mv + mu And number o molecule hitting per unit time n A V u n +. F n P An V + u m V + u man V + u F F mna V + u V u 8mnAV.u dn 6. Let the eqution o prbol be ( u u ) du dn u u du N N du Sec: Sr. IIT_IZ Pge

4 N 6N 6 dn 6N u u du v rms u dn N dn u dn u du u u du du 4 But ( ) 6N N 4 v rms unit Since the speed is shows speed in m/s vrms m / s 7. q q 4 q + + 6q 9 q 7 C mgnitude o chrge lown through the bttery is q 7 C 8. The electric potentil o shell B is zero V B Sec: Sr. IIT_IZ Pge 4

5 kq kq + 4 q Q 9. Given C C V + T T dv CV + CV + T V dt T dv dt + V n k T n kv kv e T Ve T constnt. When r B r r j rdr r Br 4 r B, r When r > B r j rdr Br r rdr Br B, r r Sec: Sr. IIT_IZ Pge

6 t r, B 8 t r, B B B 8 Sec: Sr. IIT_IZ Pge 6 4. The right chmber hs mole N. ut o this Thereore, the chmber hs miture o tomic gs 4 mole N dissocites into toms. mole o ditomic gs nd + n C V V CV n n n+ n n C CV + C + V CP CV + CP C. For the dibtic compression in this prt PV PV AL P P AL P 4 P 8P 4 Work done by nitrogen during the process V PV 8 PV PV P W 4 Ad 4PV 4P AL Work done on nitrogen by the gs in the other chmber is W 4P AL Let the inl pressure o e chmber be P. For equilibrium L P A k + PA 4 kl P + 8P 4 A Chnge in temperture o e is PV P V T n. n V kl AL + 8 P AL P AL 4 A 4 4 kl 9P AL + 6 mole o mono. The torque o electric orce on the rod is initilly lrger thn the grvittionl orce torque. This provides n ngulr ccelertion to the rod nd it speeds up. Beyond

7 certin point, the grvittionl orce s torque becomes dominnt nd slows it down to eventully bring the rod t rest when it becomes horizontl. E cn be ound by pplying lw conservtion o energy. Let potentil t line AB be V Electrosttic potentil energy o the chrge on the rod when it is verticl is U i L.V When rod becomes horizontl, we cn write its Electrosttic potentil energy s L U V d [V potentil t distnce rom end A V E ] L V E d V L E Loss in Electrosttic potentil energy i L U U E L This is equl to gin in grvittionl potentil energy L L Mg E Mg E 4. The electric orce on the rod is F e LE. Force cn be considered t centre or writing the torque. In equilibrium, L Fe [torque bout A] mg L L LE cos Mg sin Mg Putting E we get 4 o L The rod hs mimum KE t this position. KE (loss in Electrosttic PE)- (gin in grvittionl PE) Sec: Sr. IIT_IZ Pge 7

8 L L E sind Mg cos E L L MgL L Mg Mg ML g MgL L. Consider the origin t the strting point nd the nd y es s shown. Let the velocity o the prticle t ny time be V V iˆ+ V ˆj Force on the prticle is: y ence, nd ˆ F mgj ˆ + q V i ˆ + V ˆ j Bk ˆ ( y) mg qbv j + qbv iˆ dvy m mg qbv..() dt dv m dt qbv...() Dierentiting eqution () with respect to time y m dvy dt dv qb dt using () dvy qb Vy dt m Solution to this dierentil eqution is (s lernt in chpter o SM) V y nd re constnts. sin qb Vy Vy t + m It is known tht V y t t.. () Sec: Sr. IIT_IZ Pge 8

9 sin qb Vy Vy t m Dierentiting wrt time dvy qb qb y Vy cos t dt m m t time t ; y g Put in () g V y VyqB m mg qb mg qb Vy sin t qb m dvy qb gcos t dt m V mg qb cos t qb m..(4) Time ter which the prticle strts climbing up is ter V y becomes zero (or the irst time ter relese). V y qb m t t m qb mg At this time V is V qb This is the required speed. 6. Mgnetic orce does not perorm ny work during the motion o the chrge prticle. mgh mv m g h qb 7. Conceptul 8. Conceptul 6Kq 9. (P) (Q) U U + U sel int erction Kq Kq Kq U + + q 4 KQ Kq Kq U + 6Kq U 6 Sec: Sr. IIT_IZ Pge 9

10 () Kq Kq Kq U + + ( + + ) Kq 7 Kq 4 Kq 4 6 Kq ( 9 ) 4 (S) W U U et i Kq Kq 4. Due to, it will behve s solenoid nd due to v, it will behve s cylindricl wire crrying current long its length. B t P will be il (due to solenoid nd zero to wire. Net B t P is il. B nd B y depends on. At Q B By B z (due to solenoid) i B By (or wire) nd Bz where I depends on v r B z depends on v nd inversely proportionl to distnce rom pipe. CEMISTY NN nc 4 9 Br(eq) C C C C N C C CN N C C C C 7 ecess / lindlrs Ctlyst ( C ) 8 C 4 9 q.n4cl N ( C ) 8 C 4 9. Me Me Me Me Me Me Me CCl mole + AlCl MeCCl + AlCl leum S. Cl Cl Sec: Sr. IIT_IZ Pge

11 CC. C Its distereomers re CC CC & 4.. C So ns is C C N idises D-glucose to Scchric cid nd D option orm Scchric cid + Me C Cl Me C C EtN (X) N Me C + Ph C N (Y) Li (i) C CuLi Ts C Ts NGP o Ph- is observed in such cses, so retention o conigurtion is observed so nswer is A C Ts Tovoidsteric C groupcomes romoppositeside o Ts C Ts Sec: Sr. IIT_IZ Pge

12 - CN Sec: Sr. IIT_IZ Pge 8. Thereore ns is C 9. [Y] is n ldehyde thereore it will give Tollen s nd Fehling test but not the iodoorm test..&. Solution: ) Fct ) It is rection so P is ormed s mjor product SN 4. Conceptul 4. Let Put / I e d y... nd dding, we get I e d put / MATS e d e d I e e I 4. we hve, e e t - d d

13 on integrting, we get c; As c Are d 6( squre units) And M cos is strictly decresing on, m M m M, I, N-, C M N IC 4 4. :, n is lso continuous on is strictly incresing on so is one-one unction Also lim nd lim rnge o is is onto lso 46. () is discontinuous in [,7] t two points Viz.,7 Sec: Sr. IIT_IZ Pge

14 Using this in given question, lim lim lim b n n n n n n 49. n n r n C n r r n d r lim C.. r r n n n r n lim d lim. d e. d 6 e n n. y g y g y g g d c y y y (-) nd g(-) g c y d put t g y c tdt t d tdt 7 &.we hve, g... k & g... k & g k Sec: Sr. IIT_IZ Pge 4

15 As () & g() k 4 & ; g. & 4.we hve, ()(-) +b (i) since, olle s theorem is pplicble to g() t. So, g()g(b) nd g. b b b b nd 6 b b b b... ii nd 6,... iii rom (i), /. From nd lso, rom (i) nd (ii) b b d 64 Fig, n d n d n. n. (using Leibniz ule). So n 4 9 Sec: Sr. IIT_IZ Pge

16 6. gives / /. chnges sign rom positive to negtive while pssing through so it is point o locl mimum. Then put 7. we hve, n solving, we get e Now, veriy it. 8. (P) Put Limit n d. Use integrtion by prts tking h (Q) y ) 96lim h y n n d d e tn sin tn 96 n e 7 cos cos cos cos or Also, sin sin sin or sin ( rejected ) Common vlue o., 6 6 (S) g 4 So, b 8 b sin sin sin... sin A) lim tn. tn 4 6 b -+ hs integrl roots s irst unction, to get. Sec: Sr. IIT_IZ Pge 6

17 At, roots re +,+,-,- Q,S,T 4 B) b b but So,,, b,,,, 6,, 4, 6, 4,,,,, P.Q,T C) b ; b, b,,,, nly,,,, hs integrl roots X,, ST D) 4,b -, T 6. 6,6 6 P) number o integers in domin is4,,,4, Q) number o integers where discontinuous is ; 4, ) number o integers where not dierentible is ;,4, S) 4, 9 Sec: Sr. IIT_IZ Pge 7