NARAYANA I I T / P M T A C A D E M Y. S. No. Ans. S. No. Ans. S. No. Ans.

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NARAYANA I I T / P M T A C A D E M Y C o o n P r a c t i c e T e s t XII STD BATCHES [CF] Date: 04.07.6 ANSWER PHYSICS CHEMISTRY MATHS S. No. Ans. S. No. Ans. S. No. Ans.. (A). (C) 6. (A). (B). (D) 6.

1 NARAYANA I I T / P M T A C A D E M Y C o o n P r a c t i c e T e s t XII STD BATCHES [CF] Date: ANSWER PHYSICS CHEMISTRY MATHS S. No. Ans. S. No. Ans. S. No. Ans.. (A). (C) 6. (A). (B). (D) 6. (B). (C). (B) 6. (C) 4. (D) 4. (B) 64. (B) 5. (C) 5. (A) 65. (C) 6. (A) 6. (C) 66. (A) 7. (B) 7. (C) 67. (D) 8. (A) 8. (B) 68. (A) 9. (D) 9. (C) 69. (A) 0. (C) 40. (B) 70. (B). (B) 4. (A) 7. (C). (B) 4. (D) 7. (A). (C) 4. (C) 7. (D) 4. (A) 44. (B) 74. (A) 5. (C) 45. (B) 75. (D) 6. (C) 46. (B) 76. (A) 7. (A) 47. (D) 77. (B) 8. (B) 48. (B) 78. (A) 9. (A) 49. (C) 79. (A) 0. (C) 50. (B) 80. (B). (D) 5. (B) 8. (C). (D) 5. (C) 8. (C). (B) 5. (D) 8. (D) 4. (C) 54. (C) 84. (D) 5. (A) 55. (A) 85. (A) 6. (B) 56. (C) 86. (A) 7. (B) 57. (A) 87. (D) 8. (B) 58. (B) 88. (B) 9. (A) 59. (A) 89. (B) 0. (B) 60. (A) 90. (C) NARAYANA IIT/PMT ACADEMY ()

2 (Hint & Solution) PART A : PHYSICS. Here, Radius, r = 0 c = 0 0 Nuber of turns, N = 00 Current, I = A Area of the coil, A r Magnetic oent of the coil, M NIA NI r.4 A. Magnetic oent of the current loop M IA I a kˆ A. We know that force on a carrying wire placed a agnetic field, is given by F I l B Given that I A, l 0.5iˆ B iˆ 4 ˆj T and F 0.5iˆ iˆ 4 ˆj iˆ ˆj kˆ kn ˆ 4 0 Therefore, agnitude of force is N and its direction is along positive z-ais. v 5. As, r, Bq r q 4 Hence, p : r q p p 6. The agnetic field in between because of each will be in opposite direction in between 0 i ˆ 0 i B j ˆj d ˆj 0i d at d, Bin between 0 for d, Bin between ˆj for d, Bin between ˆj towards, net agnetic field will add up and direction will be ĵ towards ', net agnetic field will add up and direction will be ĵ NARAYANA IIT/PMT ACADEMY ()

3 8. Using F qv B, the force on the charge in the region a a ust be in the direction iˆ ˆ, j that is, in +Z-direction, which is vertically upward. And in the region a a, the force on the charge will be in Z-direction, which is vertically downward. For a a, path will be concave upwards. For a a, path will be concave downward Moreover, there ust not be any kind in the path at = a. 9. Conductor A attracts conductor C towards A Force towards A 0 IIl r N Again force towards B N 0. (C) 7 Resultant force towards B N N I I L d L I L F 0 d Where L is the overlapping length. Two long parallel conductor carrying currents in the sae direction attract each other.. or v qvbsin 90 r qbr v T r v qb. Magnetic field will not apply any force on the electron. Electric field will apply a force on the electron in the opposite direction to the otion. Also fro Lorentz equation, F e E v B ee F ee a ee vt v0 t Thus, velocity decreases. NARAYANA IIT/PMT ACADEMY ()

4 . (C) When a +ve charge of velocity v enters a unifor field of B acting in Z-direction k ˆ, the charge akes a circular trajectory. v qvb R 4. The agnetic field due to circular coil, 0i 0i B ; B r r As both the coils are perpendicular to each other, hence B is to B. 5. F B B B 0 qvb (directed radially outward) v N g sin qvb R v or N g sin qvb R Hence, at gr Na g qb gr R g qb gr Wb / 6. and 0II F l d 0 I I F ' l d F F. 7. Since, F is perpendicular to v, therefore work done per second W F v Fv cos i l B ; where r r (l being the length of the wire) 0 B i l Now, when the sae wire is bent into a circular loop of n turns n r ' l l r ' n n0i 0i Now, B ' n n B r ' l v r r v q 4 qb r v q NARAYANA IIT/PMT ACADEMY (4)

5 . Radius of the circular path of a charged particle in a agnetic field is given by: v K r Bq Bq ( kinetic energy, For the sae value of K and B, r q p d rp : rd : a : : q q q p d K v ) 4 : : : : or r rp rd e e e. Considering the aperian loop (shown by dotted line), no current is enclosed by this loop. Therefore, the agnetic field will be zero inside the tube because according to Apere s law, B. dl 0 current enclosed by aperian loop. 6. Magnetic induction at the centre of a circular loop of radius R carrying current I is 0 I 0I B 4 R R RB I Area of the loop, 0 A R RB Magnetic oent of the loop, M IA R 7. vr = l..(i) M = NIA (ii) BR Electric field can deviate the path of particle as shown in fig, when it is along negative Y- ais. This is not any option. The given path can be possible only if electric force iparts velocity along X-ais and agnetic force is along negative Y-ais. This is possible only in option (B). In this case F ae qv B qaiˆ qviˆ ckˆ aj ˆ qaiˆ qvcj ˆ. 0. If r is ore than l then the particle enters region V III where r (in region II) qb qbr i.e., V qbl in region III () is correct and () is wrong qbl Clearly path length in region II will be aiu when V () is correct Tie T, which is independent of V so tie spent in region II is sae for any velocity qb V as long as particle coes back to region I. NARAYANA IIT/PMT ACADEMY (5)

6 PART B : CHEMISTRY. Glycerol + oalic acid 0C HCH C. + NaH dil.hcl H/H CH CCl CH CNa CH CH CH CN (X) (Y) (Z) Acetonitrile. The order of increasing boiling point aong the acid derivatives is CH CCl < CHC CH < (CH C) < CH CNH. Absence of Higher Etensive Hydrogen olecular ass interolecular Bonding and ore surface area hydrogen bonding 4. The rate deterining step for esterification is the addition of alcohol. The rate deterining step involves change of hybridization of carbonyl carbon fro sp² to sp³ and consequently steric retardation increases as the size of alcohol increases and hence the rate of esterification decreases. +CHC 5. H CCH ne H group on acylation bring about a change in H olecular ass = 4, Molecular ass of C4H0 =4+0+ 6=06 Actual increases in olecular ass = 84 Nuber of H groups = 84/4 =. 7. Acetic acid is soluble in cold water while benzoic acid is not. 8. Larger the nuber of electron releasing alkyl groups at carbon, less the susceptibility of the acid to the nucleophilic attack of alcohol. Thus the order of esterification is I > II > III. 9. CH = C CCH CH CH Q= C = C H H H /H + CH CH CCH CH = C + H /H CHCH H Tautoetrises CH CH C = C Acetone and propanal can be distinguished by Fehling solution H H CH C CH Acetone H Tautoetrises CH CH CH Propanal 40. NH is stronger electron releasing than CH group. Therefore broination will take place at p-position w.r.t NH group. 4. Epect (A) other do not provide iodofor test 4. The electron- Releasing CH group facilitates the release of hydride ion. 44. Fehling s reagent. Tollen s reagent and Schiff s reagent react with acetaldehyde only. Grignard s reagent reacts with both acetaldehyde and acetone. 45. It is an eaple of nucleophilic addition. NARAYANA IIT/PMT ACADEMY (6)

7 46. Since the copound A reduces Fehling solution is ust be an aldelyde. CCl CH (A) chloral chloral could be obtained by the action of Cl on ethyl alcohol. ( ) Cl () C H H +Cl CH CH CCl CH CCl CH 5 HCl (A) (B) 47. C6H5CH H C CH CN NaH+Ca (A) H CHCCl Anhyd.AlCl (B) C CH H C CH H CH HCN (C) (D) 48. H AlCl CH C Cl HCl CH H C C H CH (trans diol) C CH C CH C CH + Mg.Hg/THF 49. This is perkin s reaction 50. CH CH CH although has no hydrogen ato, will not give aldol condensation. Cl Cnc.KH CH + HCH Cl CH H HCK 5. In Cannizzaro reaction of aroatic aldehyde and HCH. It is always foraldehyde which is oidized. H H H Δ 5. CH C + CH C CH C CH H H CH C CH CH CH CH C CH C CH Mesityl oide CH 54. C H C CH 6 5 H C H CH CH 6 5 Cl NaH(aq) Cl C 6 H 5 C CH H H H C H C CH + NaH/,H 6 5 Internal Cannizzaro NARAYANA IIT/PMT ACADEMY (7)

8 55. Cl Cl C CH although has no hydrogen ato, still it does not give Cannizzaro reaction. Cl This is because with H, it gives halofor reaction. 56. Benzaldehyde has not Hydrogen 57. (A) CH CH H C CH CH CH 60. (A) Cl CN CH CH CH CH CH CH CH CH CH CH CH CH CH + KCN H/H PART C : MATHEMATICS 6. dy y y 0 d y dy y y d y Curves are orthogonal 0 6. f ' / / / [0,]] / / f ' 0 f is decreasing on [0, ]. Greatest value of f is f 0 6. f ' 0, R. f f ' 64. Signs of f ' Miniu occurs at = f ' 6 8a a 65. f ' 0 a, a f " 8a f " 6a 0 and f " a 6a 0 p a, q a p q a a a 0 0, a a Let ' f a b c a b f c d f 0 df a b 6 0 By Roll s theore, at least one root of f =0 lies in (0, ) ' 67. y² = 8 (i) Differentiating w.r.t t dy d y 8 dt dt dy d y. 8 dt dt 9 y NARAYANA IIT/PMT ACADEMY (8)

9 Fro equation (i) is 9 9, a cos, ya sin dy a cos d a sin Equation of noral at point a cos, a sin is 9 Required point 8 sin y a sin a cos cos y cos a sin It is clear that noral passes through fied point (a, 0) 69. y 5 5 dy 5 d dy d,0 dy d,0 Product of slopes tangents is angle between Increasing in, Let f4 f ' 0 Increasing in, 7. V 4 0 r,0 r 5 dv 50. dt 4 r dr 50 dt dr dt 8 (where r = 5) 7. Any point on ellipse is P, y a cos bsin Area of rectangle = 4ab cos sin absin Maiu area = ab. b a cos b sin a 70. Let f f Increasing in Let, 4 f f ' Not increasing in, Let / ( ) 6 f f By LMVT f 6 f f ',,6 6 f 6 f And f ' 6 f 6 f dy d a cos cos sin a sin sin cos dy sin d cos Equation of noral is y a sin cos cos a cos cos sin cos y sin a Distance fro origin = a (constant NARAYANA IIT/PMT ACADEMY (9)

10 cos Slope of noral = tan sin Angle ade with ais is 75. f ' 76. Signs of f ' f ' changes sign as crosses. f has inia at =. f f f ' c, c loge 0 c c log e log e f ' cos sin sin cos Graph of y p q cuts -ais at three distinct pints dy p 0 d Maiu at p, inia at 79. Condition for shortest distance is slope of tangent to = y² ust be sae as slope of line y = + y, y 4,, y 0 4 p Shortest distance = p a b c p ' 4 and ' 0 0 c 0 p ' 4 a b D 9a b 0 9a b 0( p ' = 0 has only one root = 0) P P a 0 p has only one change of sign. = 0 ' is a point of inia. P a b d P 0 d P a b d, 0, 0 P P P P P P P is not iniu but P() is aiu 4 8. y 8 y ' y (, ) is point of contact thus y = is tangent 8. f li f k li f k f has a local iniu at = - f f f k k k Possible value of k is NARAYANA IIT/PMT ACADEMY (0)

11 8. e e AM GM e e f 0 so stateent - is correct As f() is continuous and / belongs to range 0, or f, f c for soe C. Hence correction option is (d) 84. dy dy y 6y 0 d d dy y d y ] Slope of noral at (, ) = - y Equation of noral y Solving we get point as (, ) and, So in forth quadrant 85. f is constant is [, 5] 86. V 4500 r dv dr 7 7 dt dt r r r 9 dr dt 9 f ' b a 0 b a 0; 4b a 0 a b ; b 4 a f 89. ' 6 so f() is increasing have distinct roots. equation cannot f ' 4 sin cos 4 cos sin sin 4 for increasing sin f ' 0 e 0 f ' 0 NARAYANA IIT/PMT ACADEMY ()

12 TPIC/SUB TPIC Q. NS. PHYSICS Magnetic oent,, 6, 7, 5 Force a current carrying a wire, 9, 6 Apere law 4, Motion of charge in agnetic field 5, 8,,,, 5, 7, 9,,, 4, 8, 0 Biot-Savart law 6, 4, 8, 9 Magnetic force on a oving charge 7, 0, Magnetic force of a current carrying wire 0 CHEMISTRY Foration of Acid,, 57, 47, 46 Esterification 4, 8 Variation of B.P. rder of hydrolysis 58 Cannizzaro 4, 5, 55, 54, 5 HVZ Reaction 60 Aldol 5, 50, 56 Miscellaneous 5, 6, 7, 9, 40, 4, Perkin Reaction 49 Pinacol Pincacolone 48 Nucleophilic addition 44, 45 MATHEMATICS Maia-Minia 6, 64, 65, 7, 75, 79, 8, 87 Increasing-Decreaseing 6, 70, 7, 89, 90 Tangent Noral 6, 68, 69, 74, 8, 84 Rolle s Theore 66, 76, 78, 80, 88 Mean Value 67, 7, 7, 8, 85 NARAYANA IIT/PMT ACADEMY ()

CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS

CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS PROBLEMS 5. SSM REASONING According to Equation 21.1, the agnitude of the agnetic force on a oving charge is F q 0 vb sinθ. Since the agnetic field points