ANSWERS, HINTS & SOLUTIONS PART TEST I

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1 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 IITJEE JEE(Min)-09 ANSWERS, HINTS & SLUTINS PART TEST I (Min) ALL INDIA TEST SERIES Q. No. PHYSICS Q. No. CHEMISTRY Q. No. MATHEMATICS. A. B 6. B. A. C 6. C. A. B 6. A. D. B 6. B 5. A 5. B 65. A 6. A 6. C 66. B 7. B 7. D 67. D 8. D 8. B 68. B 9. D 9. C 69. C 0. C 0. D 70. A. A. B 7. D. C. A 7. C. B. C 7. C. D. D 7. B 5. D 5. B 75. D 6. B 6. B 76. B 7. C 7. C 77. D 8. B 8. C 78. A 9. A 9. B 79. A 0. B 50. B 80. D. A 5. C 8. C. C 5. A 8. C. A 5. D 8. B. A 5. C 8. B 5. A 55. D 85. C 6. C 56. B 86. C 7. A 57. D 87. B 8. D 58. C 88. B 9. B 59. D 89. B 0. C 60. C 90. A IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

2 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 Physics. vrm vr vm vrmdt vrdt vmdt dt dt dt v v 0. rm r PART I SECTIN A. m m 5m 8m cm 0 m m 5m 8m m m 5m 8m ycm 8m 8m R 6R 8m y m(, ) m(, ) cm 5 m(, ) 8m (, ). Vector long the norml of plne is ˆ i ˆ j b Coefficient of ˆk is zero lwys lso it becomes î when b =.. I0 Kmb b I K. m. I Kmb I 0 I. 5. v v 5 Now using conservtion of mechnicl energy gh gh IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

3 AITS-PT-I-PCM (Sol.)-JEE(Min)/ h h m 0 6. v 0 v v 0 ms 7. p p V p V E m E E 8. I = k. m km. I I. 9. Bsic concept of BD nd equilibrium 0. Bsic concept of pseudo force. 5 KE mv J. 0% of this is stored in the spring. k.5 0 = m k = 5 0 N/m.. Bsic concept of projectile. n L n L X cm.. n n L n n n =. With equl initil kinetic energy, one cn write for these rotting disks tht KE = ½ I = L /I. As result L = I(KE) which mens L X < L y s object Y hs more mss. By pplying the sme force on the outside of ech disk, the torque from the centre of ech disk is the sme. rom the ngulr impulse momentum theorem, ()t = L nd since the torques nd times re equl, the chnge in ngulr momentum is the sme for ech. Consequently, L X < L Y fter the push s well. The kinetic energy depends on the totl distnce through which ech force cted. Since disk X is rotting t higher rte thn disk Y nd the pplied force will increse the ngulr speed of ech disk, there will be greter ngle swept out by disk X compred to disk Y. As result, the kinetic energy chnge for disk X is greter thn for disk Y mening tht fter the pus, K X > K Y. IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

4 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 5. CM net mi gcos 6. ˆ ˆ Bg gsin i j Accelertion of bo with respect to ground g sin ˆi gcos ˆj Accelertion of prticle Pg with respect to ground gcos ˆ PB j Accelertion of prticle with respect to Bo u ucos 5 ˆ i usin5 ˆj Initil velocity of PB prticle with respect to Bo Time of flight = usin5 u gcos gcos y gsin y gsin g gcos 7. AP 5. h 5 rom work energy theorem : g gh v A v Where vb v A cos ; cos 5 n solving : v v A A. m / sec. 5 B v A m m A A kg m v B B P B h kg 8. Bird hs to fly wy from wind direction s shown so s to rech P. V V Bird wind If t is time tken; 5 t 5cos (i) sin t (ii) Solving for t we get 8t t 0 t hrs. 6 S N Bird 5 S Wind P 5 km E IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

5 5 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 9. tn R = R tn d d R sec Rsec dt dt v = R sec dv d R sec.sec tn dt dt. R sec tn v R sec 60 R 8R R R R v 0. Let frog tkes off t ngle with v speed from ground frme mv cos v cos v P m v cos vf /P v v v cos v cos f P v cos v sin v f /p v sin R y f /P g v sin g vmin i ˆ j ˆ kˆ. ˆv i 9 i ˆ 6j ˆ kˆ ˆp 7 vˆ f vi v ˆ ˆ i.ppˆ g ˆ ˆ ˆ i ˆ j ˆ kˆ i 6 j k ˆi 7 ˆj 6 8 kˆ = 9 9. Time of flight = 0 s; 0 Rnge = 5 = 0 m u v. Are = = 6 Ns Chnge in momentum = 6 mv = 5.6 v = m/s v H 9.8 m g Due togrvittion = 6 0. = 5.6 Ns IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

6 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 6 v0. r = v 0 R, nd z = R cot ˆ ˆ Zk r ( i) v ˆ ˆ A r A Rcot (i) R(k) v0 v0 = ˆ ˆ k i R cot ˆi Rkˆ Rcot R v A (v0 v ˆ 0 )i v ˆ 0i v A = v 0 Similrly, v B r B Where, r ˆ ˆ B Rcot i Rj v 0 v ˆ ˆ ˆ B v0k i v0 j cot z r A A C R v0 B 0 0 v v v cot = 0 v tn sint sint R sint m m m m 0 Here R m v 0 cos t & v 0 sin t m m 0 v m, nd v cos t cost R cost Rt sin t & y R cos t 6. No eternl force is cting long horizontl direction 7. E = U + K, for given E, k will be mimum where U will be minimum. 8. Tking torque bout A mg cos fs sin N cos A BD of Rod- N N fs N f s B mg N BD of Rod- mg tn IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

7 7 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 fs min fsm N mg min 5N tn mg K K K L cos mg mg L r L mg L L L r L L L L L r L r L L r L L r L L L L C L K L m mg 0. mv mgcos R () R M mv mg cosr Mg () Mg M mgcos mgcos M mcos m cos M Chemistry PART II SECTIN A. o S 96 o sp but not idel tetrhedrl. 09 o 8' S Idel tetrhedrl IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

8 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 8 S Trigonl plner S See-sw. * p z LUM * * p py, HM p p p Z * s p, py s s s * s s s s 5. S g S g 8 C 0 C C C V.D V.D initil eqm moles moles eqm initil 8 / C C C 96 C 9 % decomposition 00.% 9 6. Hf form interstitil hydride, e.g. HfH.98 KH is n ionic hydride SiH is covlent hydride Co don t form hydride (hydride gp) 7. Actul boiling point order H S < H Se < H Te < H IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

9 9 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 8. n heting N N N it is dimgnetic in nture. N C NC (Suitble for use in ir purifiction) N is good oidizing gent. 6. S S e idtion hlf rection 6 Cr e Cr Re duction hlf rection NCr is oidizing gent Eq. wt. of N Cr (oidizing gent) = M/. CH g H g C g 8H g Cg H g C g H g (i) (ii) Eqution (i) Eqution (ii) CH g H g C g 6H g (iii) Equilibrium constnt for Eqution (iii) By multiplying Eq. (iii) by we get K K CH g H g C g H g K Equilibrium constnt K. Possible vlue of quntum number of d orbitl. n =,, m = -, -, 0, +, + s = +/, - / / K K. H P H H n pk 9 7 n pk 9 7 H H n 0 pk H S n pk 9 7 H IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

10 AITS-PT-I-PCM (Sol.)-JEE(Min)/ n P D H 0 = y = 0 y = Mn M M Mn n-ctor of Mn = 5 n-ctor of M = Number of equivlent of Mn = Number of M + 5 mole of Mn = 6 Mole of Mn. 5 CH CN H CH CH N meq. 0 meq meq. 0 meq. 0 CH CH 0.0 M 500 ph pk logc.75 log log 5.75 log Uncertinty in velocity v 0. ms 00 h. p h m v k.0 A 0 log t A t log A t A0 k.t IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

11 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 = = 0.88 A 0 = Anti log (.88) A t =.9 A t rction remin A.9 rction decomposed NH SH s NH g H S g At eqm P P Ptotl P K P = P When H S is dded NH SH s NH g H S g P NH P H S According to question P HS Ptotl P P K P.P P HS NH P P NH.P P P NH P P NH : P P NH N C H NHC N m.eq. initil m.eq. 0 m.eq. m.eq. fter miing 0 m.eq. 0 0 m.eq. 0 m.eq. Now, Miture contins N C nd NHC so it will cts s buffer. NC ph pk log NHC = log0/0 = Moles of CS in 00 ml solution Molrity = mol/lit IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

12 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 K sp C S.5 0 = A t/rd t/rd log log K / A 0 K A0.0 log K min hlf life K Time tken for completion of 75% t min / A 5. Cr.87 n n.87 n (unpired e ) Cr y Mn n n.87 n (unpired e ) Mn y y = Rte of disppernce of A We know A B C t t t Rte of ppernce of C C A t t = 0 - mol L - s - A 0 0 mol L s t Here n ve By dding inert gs bckwrd rection fvoured By dding D(g) bckwrd rection tkes plce. IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

13 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 57. H P 0 0 H P 0 P Mss of H = Mss of H + mss of neutron = =.06 mu Actul mss =.006 mu mss defect = = 0.0 mu. Bond Energy = Mev = 9.5 Mev. 59. Dimond Grphite Resistivity Stndrd molr entropy Density C 08 o C o S 9 o S Br 96 o Br IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

14 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 Mthemtics PART III SECTIN A 6. If lim f nd lim g m 6. Volume = = 5 lim f g m 6. (p q) (r q) is equivlent to [~(p q) (r q)] [~(r q) (p q)] 6. y = f( + ) + + = f() + f( ) + f() y = f( + ) + + = f() + f( ) + f( ) f() = f( ) d d /6 / /6 5/6 = c c 5 5 / 6 0 P() = h h f e f h e he e f f lim h0 h f() = f() + e f() = e 67. ne of them hs to be mim nd other minim 68. There re only points where f() hs locl minim 69. Are = tn d tn d Circle is circumcircle of the tringle 7. n + = n + n lim lim n n Let lim n n n n n n f () = g() h() = n 7. yd dy dy yd y y y d y d y y y IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

15 5 AITS-PT-I-PCM (Sol.)-JEE(Min)/9 7. Apply chin rule 75. f() is constnt function 76. cot > dy d y + = 0, y = 0 d y d 79. Substitute = r sec. y = r tn rdr cos sin cos r sec d n r sinsin nc y n y sin nc n m / / /n lim f f d m n /6 n / m n n = lim m n 6 = n = 6 8. h() is constnt function nd is lwys periodic 8. 0 sin 9 lim n 9 = 8. e 5 d = e d = 8. or lim f to eist, lim f must be 0 e c r lim 08 r r 09 r 09 IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659

16 AITS-PT-I-PCM (Sol.)-JEE(Min)/ = 00, re points of non-differentibility 87. is point of globl minim e b = b e 88. Let I 0 Let t tn d I 0 cot t dt t I d If f() is differentible, then f() should hve unique vlue t tht point 90. If n(p M C) = 00 n(m C) 0 n(m C) = 0 IITJEE Ltd., IITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , 65699, 659