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1 CPT-0_XI (LJ) NARAYANA IIT ACADEMY ANSWER KEY XI STUD (LJ) IITJEE MAINS MODEL Eam Date : physics Chemistry Mathematics. (A). (A) 6. (B). (B). (D) 6. (C). (C). (A) 6. (C). (A). (A) 6. (C) 5. (C) 5. (C) 65. (C) 6. (B) 6. (B) 66. (D) 7. (A) 7. (A) 67. (B). (C). (B) 6. (C) 9. (A) 9. (C) 69. (B) 0. (D) 0. (B) 70. (C). (C). (D) 7. (C). (D). (A) 7. (C). (C). (B) 7. (D). (B). (A) 7. (A) 5. (A) 5. (B) 75. (D) 6. (C) 6. (B) 76. (A) 7. (A) 7. (D) 77. (A). (A). (A) 7. (A) 9. (C) 9. (B) 79. (B) 0. (A) 50. (C) 0. (C). (C) 5. (C). (D). (C) 5. (A). (B). (B) 5. (A). (B). (D) 5. (C). (B) 5. (B) 55. (B) 5. (D) 6. (B) 56. (A) 6. (B) 7. (C) 57. (A) 7. (B). (A) 5. (A). (A) 9. (B) 59. (B) 9. (A) 0. (C) 60. (A) 90. (C) Page No.
2 CPT-0_XI (LJ) A v u at;. B g g 5g g Sn g n ; Ratio : :... n. C vy tan and V V u U sec y U. A n y n 5. C v u 5 as, v 0; u s 6. B u gt h g 7. A Compare the equation with g y tan u cos. C u u cos iˆ u sin ˆj 9. A d dy u, u y dt dt and u u uy 0. D gt H. C V u gh. D gt Tan u. C = a 0 a t + a t HINTS & SOLUTION PHYSICS sn n d = a + a dt t Page No.
3 CPT-0_XI (LJ) d dt = a. B Horizontal component of velocity = 70 5/ = 00 m/s Let t be the time taken for a freely falling body from 90. Then y = (/) gt 90 = (/) 9. t t = 0 second Now horizontal distance = Velocity time = 00 0 = 000 m Hence the bomb missed the target by 000 m 5. A v cos, vsin y tan 0cos 6. C v v v v y y y g g v 7. A y = (/) gt (downward) 000 = (/) 0 t t =.5 sec 0 = ut =.5= 57. m A Distance = R Displacement = R. D h h 6. B u sin g u sin 90 g sin R u g g y tan u cos. A r = 00 m r. 00 average speed = 0ms time 6.0 average velocity = 0 Page No.
4 CPT-0_XI (LJ) B v v v v f i 0. C Let the total distance travelled be. Time taken to travel first half distance / t = = hr 0 0 Time taken to travel the rest half distance t = / 60 = 0 hr Average speed = Total distance Total time Hence correct answer is (). No change in the direction of = = km/hr ( /0) ( /0) CHEMISTRY. A NaCO HCl NaCl CO HO 06g.L. g L? 06 g NaCO liberated CO.. g NaCO liberated CO 06.. g NaCO liberated CO. = 0. L 06. D Zn HSO ZnSO H 65g. L 00 ml. (A) v 0. v = 5 ml 5. C Energy is released in this process; and the radiation will belong to visible region ( Balmer series) hc E RH Z n n RH Z hc n n m 6. B Wavelength of st line in Balmer series, 0 66A Page No.
5 CPT-0_XI (LJ) Z RH R H Z B 6 6 Or B 5RH Z Wavelength of lst line in Lyman series is, Z R B L Or 7 6 Difference B L RH Z RH Z Z Or Z = Hydrogen like species is 0. B For He ion, Z R n n For hydrogen atom, Li J R R 6 R 5 H Z R n n So, Or n n i.e., n and n. D v Number of revolutions per second.(i) r.0 cm / sec n.0. 0 cm / sec n 0 r 0.59A Z cm cm Number of revolution per sec = Page No.5
6 CPT-0_XI (LJ) A. B R n n R n n n and n metre R R Or 9.0 m R We know that, c E hv h =.7 0 J 9.0. B h We know, mr n n n =. 5. (B) 6. B R H n n. 0 cm 0967 E E E E 0 5 Man of oalic acid =.5 g A Equivolent of oolic acid Eq. Wt. 6 6 Vol. Of solution = 50 ml Normality = M D KClO KCl O Page No.6
7 CPT-0_XI (LJ) g 96 g.5 g 5 g of KClO liberak oygen = g of KClO liberate oygen = 5.5 g of KClO liberate oygen = 0.96 g A M 0.M (B) Moles per litre of solution is called molarlity. 50. C BaCl Al SO BaSO AlCl 5. C 6.90 M solution of KOH contains 6.90 mole of KOH in 000 ml of solution. Wt of KOH in solution = g Wt of KOH in 000 ml solution = 6. g Since the solution is 0% by weight it means that 0 g of KOH are present in 00 g of solution g of KOH is present in 6. = g of solution 0 weight Density. g / ml volume (A) v vh z 5. A Eq. of NaCO Eq. of KCO Eq. of HCl X = 0.7 g Mass of NaCO 0.7g Mass of KCO = % NaCO 00 9.% B h. c E hv And R n n Page No.7
8 CPT-0_XI (LJ) E R. h. c. n n E h. c. R =.65 0 J For hydrogen like species, E Z Rhc n n Z R n n 7 Z Z Or Z = The species is He. 56. A The transition occurs like Balmer series as spectral line is observed in visible region. Thus, the line of lowest energy will be observed when transition occurs from rd orbit to nd orbit, i.e. n and n R 5 R 6 c E hv h J per atom 6 Energy corresponding to.0 g atom of hydrogen 9.00 Avogadro ' s number J 57. A V 5. (A) Molarity Molality = B Milli mole of HSO w or w= (A) Normality = Molarity Basicity. N = Molarity. Thus molarity =. Page No.
9 CPT-0_XI (LJ) MATHEMATICS 6. B If 0 t tan0, then tan 0 0 t t t t t t t 6 7t t 6. C If the sum is S, then 6 S.sin sin sin sin sin sin 6 0 sin sin sin sin 0 sin sin. S 6. C sin sin 6. C We know that. But cos sin cos sin cos ;sin. Both can not be true simultaneously 65. C n cot 5 cot 5 n 6 n 66. D sin cos sin cos sin cos 9 sin cos sin cos 67. B 7 Lies in IV quadrant 7 n 6. C Given equation cos sin sin 5 sin cos sin sin 5 cos cos 5 n Page No.9
10 CPT-0_XI (LJ) B cos cos cos sin which is not possible 70. C 7. C 7. C sin cos Let t 7 t t or 7 t sin cos 0 or sin cos sin sin 0 sin only There are solutions in tan A tan A tan A tan A 0, and two more in, tan A tan A tan A tan A tan A 7. D sin cos thus, nmin a cos bsin c has a solution If a b c a b 7. A cos y cos y 75. D tan y tan y tan y tan y tan tan y tan y tan cos cos cos 76. A sin Asin B sin Asin B o cosa B cosa B cosa B cos90 cosa B Maimum value of sin Asin B tan cos tan sin sin as Page No.0
11 CPT-0_XI (LJ) A sin sin cos cos cos sin cos cos sin cos sin sin cos sin sin sin cos cos cos sin sin cos sin 7. A 79. B n sin n ( ) 6 0. C cos 0 K ; cos K = cos 0 cos 0 On verification, 0 and 0 satisfies.. D sin 0, cos 0 But, b sin, cos. (B) The given equality holds for minimum of sin cos i.e., 5 5 G.S. is n. (B) 9 5 k k k K =,,,, 0,,,. (B) 5 cos sin. Minimum value of R.H.S. is and maimum value of L.H.S. is. So equality holds only when both sides are simultaneously. i.e., cos and sin 0 cos n and sin 0 n. So, solution set n. 5. (D) tan tan y tan y 6 6 Page No.
12 CPT-0_XI (LJ) n n 6 (5 n) and y options (A), (B), (C) correspond to n = 0,, 0 respectively (B) 7. (B) sin A sin B 0 ; cosa cosb 0 A B A B cos sin 0 ; A B A B sin sin 0 A B sin 0. : (A) cos Squaring and simplify. 9. A) sin5 sin sin sin.cos sin sin 0 or, satisfies C cos 6 cos cos 0 cos A B n A n B Page No.
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