NARAYANA. C o m m o n P r a c t i c e T e s t 9 XII STD BATCHES [CF] Date: PHYSICS CHEMISTRY MATHEMATICS 16. (D) 31. (A) 46. (D) 61.
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![NAAYANA I I T / N E E T A C A D E M Y. (C). (B). (A). (D) 5. (B) 6. (D) 7. (B) 8. (B) 9. (B) 0. (D). (D). (C). (B). (D) 5. (C). (C) C o m m o n P r a c t i c e T e s t 9 XII STD BATCHES [CF] Date: 9.](/docs-images/77/74538831/images/1-0.jpg "06.7 ANSWE PHYSICS CHEMISTY MATHEMATICS 6. (D). (A) 6. (D) 6. (C) 7. (A). (C) 7. (A) 6. (A) 8. (D). (B) 8. (A) 6. (A) 9. (A). (B) 9. (A) 6. (A) 0. (B) 5. (B) 50. (B) 65. (A). (D) 6. (D) 5. (A) 66.")
1 NAAYANA I I T / N E E T A C A D E M Y. (C). (B). (A). (D) 5. (B) 6. (D) 7. (B) 8. (B) 9. (B) 0. (D). (D). (C). (B). (D) 5. (C). (C) C o m m o n P r a c t i c e T e s t 9 XII STD BATCHES [CF] Date: ANSWE PHYSICS CHEMISTY MATHEMATICS 6. (D). (A) 6. (D) 6. (C) 7. (A). (C) 7. (A) 6. (A) 8. (D). (B) 8. (A) 6. (A) 9. (A). (B) 9. (A) 6. (A) 0. (B) 5. (B) 50. (B) 65. (A). (D) 6. (D) 5. (A) 66. (B). (C) 7. (A) 5. (B) 67. (B). (C) 8. (B) 5. (A) 68. (B). (B) 9. (A) 5. (A) 69. (D) 5. (C) 0. (A) 55. (D) 70. (D) 6. (A). (C) 56. (D) 7. (D) 7. (D). (D) 57. (A) 7. (D) 8. (A). (B) 58. (C) 7. (C) 9. (D). (B) 59. (D) 7. (D) 0. (D) 5. (B) 60. (A) 75. (B) (Hint & Solution) PAT A : PHYSICS 76. (B) 77. (B) 78. (C) 79. (B) 80. (B) 8. (D) 8. (D) 8. (A) 8. (A) 85. (C) 86. (B) 87. (C) 88. (B) 89. (C) 90. (C) Sunt resistance is given by I G S G I IG 0 Here, I G I 0. J and G I 0. S I 0. I 0.9. (B) Let be resistance connected in series wit te galvanometer to read V. Ten V V Ig G or G I V A g NAAYANA IIT/NEET ACADEMY ()
2 . Here, G.5, Ig 50mA 0.05A, I 5A. (D) IgG Now, S I I g Net resistance of te circuit is V Current in te circuit, I 0.A Here, G 60, r 0.0 Wen te galvanometer is sunted by a resistance r, is effective resistance is Gr P 0.0 r G Total resistance of te circuit P Current in te circuit, I A; 0.0 5; Here, G 0, I.0mA 0 A, V 0.V g V I G g g V 0.V G 0 70 I 0 A 7. A galvanometer can be converted into a voltmeter by connecting a ig resistance in series wit it. 8. (B) In case of a galvanometer, I IG I IG I I I S I G; I I S I G G G G G G or G 0 were G is te resistance of te galvanometer 9. Here, I g A, I 0A G I I 0A A 9 For ammeter, g IgG I Ig S; S I g A 0. Te resistance of an ideal voltmeter is infinity.. Wen tese two bulbs are connected in parallel, te total power consumed by tem is P P P P p NAAYANA IIT/NEET ACADEMY ()
3 But P 0W given p 0W 0W P or P 55W V V As P, 0V 880 P 55W V. As, P For and value of V is same. P 0 P P 0 P : P :. Internal resistance l 50 r l. e E l r L V P 0 Q 80 P 5 0 P 5 Q 60 Q Q Q Q Q esistance, P Effective voltage on carging..6v V.6 Total resistance 8 i 0.5A 8 l 7. Let lengt of eac wire be V r For ma. power r Maimum power watt. NAAYANA IIT/NEET ACADEMY ()
4 8. P V / ; V / P P V P ; 9. m s law V = I V I (for ammeter) V I V I 0 0 = 0. Switc S is open so capacitor is not in circuit Current troug resistor A Let potential of point sown in fig. is V 0 ten using om s law V 0 V a = = V (i) Now current troug 5 resistor A 5 So V 0 V b = = V From equation (i) and (ii) V b V a = = 8V.. Total current troug te circuit 0 i A Now voltmeter reading iv V 500 V. 50. (C) Let emf of te cell VA VB VA VD VB VD. 5 NAAYANA IIT/NEET ACADEMY ()
5 . Let = resistance of te bulb V0 V0 P0 or P V V V P P0. V0 / P0 V0. Wen switc S is open total current troug ammeter. 0 i A. 0 Wen switc is closed i 5 A. t / 5. Q Q e 0.Q or e t / or t / e 0 / 9 6. Te resistance in te middle plays no part in te carging process of C, as it does not alter eiter te potential difference across te C combination or te current troug it. i G 7. By using i S g i S 000 S 9 V P i P P P l P l V P so V V and P 00 V Now W., W. W : W : W 5: 5 : 6 or W W W and W Current I is given by: 0I 0I.5 or I amp 0 V V.5 I 0.5volt A B But V V V V V V.5 A D D B A B Since, tere is no current troug.5 volt cell, so, V V.5volt D Hence, V V 0volt A B D NAAYANA IIT/NEET ACADEMY (5)
6 PAT B : CHEMISTY. Products as & S configuration and diastereomers. THS does not react wit G.. wile can react. Grignard reagent prefer, addition. Compount X is P C CH5 P C C H 5 NH H H + P H 5 C N H PCl 5 C NH C H 5 P 5. (B) CH H, Zn CH CH H Aldol CH 6. It is cannizaro reaction 7. (A) CH AC /AC H CH H CH CH H + 8. Intramolecular cannizaro reaction 9. In cannizaro reaction H transfer is.d.s 0. Wold Kiscner reagent does not cange H group. -N grouop increases more + on C. Attacking site in NaHS is S. (B) HC H C CH C CH CH,, (, M, P). P CH = form given Acetal 5. P CH = + AC AC P CH = CH CH P HC CH CH Br NAAYANA IIT/NEET ACADEMY (6)
7 6. (D) CH CH C CH CH H C CH C CH CCH CH CH CCH CH C CH 7. (A) CH CH Bayers H HI H H CH 8. It is a bayer villager oidation 9. Clemension reagent ydrolysis acetal in Con. H S is used 50. NaH will give olfmans elimination 5. For D ecange H must be present 5. (B) CH CH C C et et et CH CH C C Cet et Cet 5. (A) - + CH MgX 5. (A) Br CH CH CH 55. for C N bond formation N H group must be present 56. it is a retro aldol condensation 57. N increases electropilicity of carbonyls 58. Cu / Feling Solution Propanol- Propanal ed ppt. 59. diastereomers are formed 60. actual formation NAAYANA IIT/NEET ACADEMY (7)
8 , 0 f e 0, 0 lim f lim 0 0 e lim 0 e lim f lim 0 0 e lim 0 e 0 e lim f lim f f We ave, Hence, 0 0 f 6. We ave, f is discontinuous. PAT C : MATHEMATICS log a log b, if 0 k, if 0 f is continuous at =. lim f f 0 0 log a log b lim 0 k f 0 k log a.a log b. b lim lim k 0 a 0 b a b k log lim 0 f,, f() is continuous at =. lim f f 6. We ave, 0 lim 0 6. We ave, f [ [] + [ ] = 0, integer and [] + [ ] =, integer] = cos, 0 f a, 0, 0 6 NAAYANA IIT/NEET ACADEMY (8)
9 For f() is continuous at = 0, lim f lim f f lim f Now, and 0 0 lim f 0 0 cos lim cos lim 0 sin lim 8 0 lim lim lim a 65. Given, lim 6 8 lim f lim f f 0 f 0 sin cos, for f a sin b, for cos, for LHL at, lim f lim sin cos 0 lim cos sin 0 HL at lim f lim a sin b 0 0 lim a cos b a b 0 0 f() is continuous at. a b.(i) Now LHL at, lim f lim a sin b a b 0 a NAAYANA IIT/NEET ACADEMY (9)
10 HL at, lim f lim cos a b 0 n solving Eqs. (i) and (ii), we get a, b 66. f Grap of f() is as sown below..(ii) f as points of non-differentiable in set {, 0,,, } 67. Given, f g and Now, f g g g Clearly, g() is not defined at = 0. g() is discontinuous at = 0 Also f9) is not defined at =, F[g()] is also not defined at =, Wen g() =, ten Hence, te point of discontinuity of f g are,, 0, / / e e, if 0 / / f e e 0, if 0 f 0 f 0 Lf ' 0 lim 0 / / e e e / e / lim 0 / lim e 0 e / = f 0 f 0 f ' 0 lim 0 / / e e e / e / lim 0 / e lim 0 / e Lf ' 0 f ' We ave, Now, NAAYANA IIT/NEET ACADEMY (0)
11 f() is not differentiable at = f cos f ' 5 ; cos ( ) and cos are differentiable for all, but is not differentiable at = Hence, f() is non-differentiable at =. f a a f lim a a f a lim a 0 af a a f ' 0 a f ' f 5 f 5 lim 0 f 5.f f 5 lim 0 f f 5.lim 0 0 form 0 [using L` Hospital rule] 7. We ave, f() is differentiable and f ' lim f ' 5 0 f lim 0 is 5, wen f() = f() = [n + psin] = n + [psin] F() is not differentiable at tose points, were psin is an integer. psin is an integer of sin =, and p r r i.e.,, sin, sin p p were, 0 r p but, 0 r r Function is not differentiable at, sin, sin, were 0 r p p p So, te required number of points are + (p ) = p 77. Let be te semi-vertical angle of te cone C/AB wose eigt C is m and radius B = m, ten Let V be te volume of water in te cone i.e. te volume of te cone CA 'B'. After time t min, be te eigt of te water, ten V 'B' C ' 'B' 'B' V tan tan 'B' tan C' NAAYANA IIT/NEET ACADEMY ()
12 V dv d d Now, dt dt dt d 770. dt d 77 0 dt d cm / min dt L 0 cm 77L 770 cm dv i.e. 77L 770 cm dt 78. Given, and y are te sides of squares. Ten, te area of te squares are and y. y y We ave to obtain, d (i) d d Given curve is y = Tus, d (ii) d y y d [from Eqs. (i) and (ii)] d y d d y d d y d 79. Given, y.(i) Let, y be te point on te curve (i), ten Also, point y..(ii), y lies on te line + y = c. y c.(iii) From Eq. (i), we ave d d, y Now, slope of tangent at (, y ) = Slope of te line NAAYANA IIT/NEET ACADEMY ()
13 8 y c From Eq. (ii), we ave c c c 8 8 c c 8 c Given y y =..(i) y y y 0 d d d 6y y y 0 y y d 6y d, 6 Equation of te tangent is y y y 0.(ii) From Eqs. (i) and (ii), we ave , 5 NAAYANA IIT/NEET ACADEMY ()
14 Wen 6, ten 5 y 0 ter point on te curve were tangent meet again is 8. We ave y n differentiating w.r.t., we get 8 d Since, te tangent is parallel to -ais, terefore 0 d 8 and y 8. We ave, y f d f ' d,, f ' f ' d It is given tat, slope of normal tan 6, 5 0 tan f ' f ' 8. Te equation of te curve is y d Let m and m be te slopes of tangents at (, 0) and (, 0) to tat given curve. Ten, m 5 d and, 0, 0 m 5 d Clearly, m.m So te required angle is. 8. Given, f() = + a + b, a b (i) From Eq. (i), we ave f ' a f ' a a a f ' b b a NAAYANA IIT/NEET ACADEMY ()
15 According to te question, f ' a f ' b a a b a a b a b a b Now, a b f a b b b a b f() = 85. Equation of te curve is y y e 0 y e y 0 d d y y e y.e d d y d Clearly, 0 So, te required point is (, 0). 87. We ave, subnormal, y d y and subtangent d sunnormal Subtangent 88. Given, y 8a a 0 8a a y 8a d d y y a Lengt of subnormal y. y. a d y y d y / d Slope of te tangents d 89. Let f(, y) = ( + y ) /6 Taking =, = 0.08 and y =, y = 0. Differentiating () w.r.t., treating as constant. f 5/6 y 6 8 5/ and differentiating () w.r.t. y treating as constant 9 f 5/6 8 5/6 5 y y 6 6 y 6 6 f f df.d y /6.9. f, fd NAAYANA IIT/NEET ACADEMY (5)
16 90. We ave, pv. = k(constant) log p.log v log k dp. dp.p 0 p dv v dv v dp Now, p v dv.p p v p v. v p v p v p v p v given p v TPIC/SUB TPIC Q. NS. PHYSICS Electric Instrument,,,, 5, 6, 8, 9, 0,,, 5, 6, 9,,,, 7 Heating Effect,, 8, 8, 9 Maimum Power Teorem 7 -C Circuit 0,, 6, 0 Power in Circuit CHEMISTY Nucleopillic Addition,, 55 G, 5, 5, 57, &, addition Beckmans rearrangement Aldol 5, 7, 56 Cannizaro 6, 8, 9 Perkin 7, 5 eduction 0, 9 NaHS addition imes idation 8, 58 Wolf Kiscner 50 MATHEMATICS Continuity 6, 6, 6, 6, 65, 67 Differentiability 66, 68, 75 ate measure 76, 77, 78 Tangent and normal 79, 80, 8, 8, 8, 8, 85, 86, 87, 88 Percentage error 89, 90 NAAYANA IIT/NEET ACADEMY (6)
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