NDA Sec: Sr. T_Z Jee-Advanced Date: 9-01-18 Time: 0:00 PM to 05:00 PM 011_P Model Max.Marks: 40 KEY SHEET CHEMSTRY 1 C D A 4 B 5 B 6 C 7 A 8 A 9 ABC 10 ABCD 11 ABC 1 ABC 1 8 14 9 15 1 16 7 17 6 18 19 A-PQS, B-PS, C-PR, D-PQRS 0 A-QS, B-PQRS, C-PQR, D-PQR PHYSCS 1 D A C 4 D 5 C 6 B 7 B 8 A 9 ABCD 0 AB 1 CD BC 4 9 5 6 0 7 6 8 1 9 A-R B-PQ C-S D-PQ 40 A-PS B-QS C-PR D-QS MATHS 41 D 4 D 4 C 44 C 45 C 46 A 47 D 48 B 49 BC 50 AB 51 ACD 5 AD 5 4 54 55 5 56 9 57 7 58 9 59 A-T, B-S, C-Q, D-P 60 A-PQRST B-PRST C-PQRST D-S
9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s CHEMSTRY 1. Calamine - ZnCO ; Calcite CaCO, Argentite Ag S, Cuprite CuO, Chalcocite CuS, Cinnabar HgS, Tinstone SnO, Sylvine KCl. A B A B. N O cannot oxidize FeSO 4. t is in +1 oxidation state lower than NO. 4. 19 kcal 4.1810 J 1eV 1mol H 0.8eV 19 mol kcal 1.6010 J 6.010 atoms K(g) F(g) F (g) K (g) H 19kcal 0.8eV K(g) K (g) e E = 4. ev F(g) e F (g) EA H E 4. 0.8.5eV 5. l 4; number of degenerate orbitals = l 1 9; maximum total spins = 9X 1 maximum multiplicity = S 1 minimum total spins =0 9 X 1 10 6. minimum multiplicity = 0 1 1 CH NaNH CH NH Cl Cl AgNO Cl AgCl 7. Pinacole-pinacolone reaction 1) One of two groups leaves which results in formation of more stable carbon cation. n given compound Ome is better stabilizing group ) Relatively electron rich ring mignation to form ketone Sec: Sr.T_Z Page
9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s NH 8. CH5 B H C C CH C C CH equ. O 1 Equi H O 5 5 C H C C CH C H C C CH O 09. Since the same number of moles of gas appears on both sides of the equation, the equilibrium constant expression can be stated as a ratio of moles instead of concentrations 46.0g 54g 0.181 mol initially present 1.00g H.00g H 1.9g 54g 0.500mol H 0.0075mol at equilibrium nitial Produced Used up Equilibrium [ ] 0.181 0.174 0.0075 [H ] 0.500 0.174 0.6 [H] 0 (0.174) 0.48 [H] (0.48) K 50 [ ][H ] (0.6)(0.0075) 10. All four statements are correct 11. FeO does not give O on heating 1. carbohydrates which differ at 1 and carbons only form same osazone. 1. 14. p 1 0.77 0.7atm 0.7 V 0.1 0.08 00 V 9L Ne Sec: Sr.T_Z Page
9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s 15. Normality = 10 1.78 5.6 Milii equivalents of H O = milliequivalents of KMnO 4 1.78x 11 = 0.04 x 5 x volume in ml 16. ii- is not disproportionation reaction i) Na Cl NaCl NaOCl Cold& dil ii) Na F NaF OF Cold& dil iii) Na P4 PH NaH PO iv) HPO HPO4 PH v) NO HNO NO Hot HO vi) Na ClO NaClO NaClO vii) HClO ¾¾ Auto ¾ ¾ HCl + HClO Oxidation viii) HNO HNO NO 17. 18. CH CH, HO, HO 0. A. OCH OCH H CH OCH Sec: Sr.T_Z Page 4
B. OCH OCH H CH 9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s H C. Oph Oph H ph Oph D. CH CH CH H CH CH CH CH CH CH H CH CH CH CH CH CH H CH CH CH PHYSCS 1. let separation between particles become infinite. Applying energy conservation and momentum conservation So, for the system to be bound. Up thrust acting on the object with the glue intact is 5000-100h N, where h is in meter. Other forces are weight = 500 N downwards And force due to glue = 000 N in limiting condition. Hence 5000-100h=500+000 Or h=5 meter Sec: Sr.T_Z Page 5
At t 9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s. 0, the waves reaching at A are emitted at t = -1s and t 1 S Respectively by S 1 and S So, Phase difference between the waves At A is Hence resulting amplitude is 5 mm 4. X 1, X 5 L C Hence Z 5. X L R 7 X C Z V Z So. V X X sin 10 t 7 AB L C 5. Apply superposition Principle. 6. Considere an elementary circulare section of radius r and width dr. Emf = r Current db r dt r di r 1mm dr Which upon integration gives i 4A 7. Kinetic energy = m Ta Potential energy = m Ta So energy= m Ta Sec: Sr.T_Z Page 6
9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s 0.4rL 8. Force due to absorption of light is F1. C df Considering energy of each photon is E. charge in momentum due to reflection is ZE C cos Number of photos stirrings the elementary area is 0.8 R cos d h E So 1.6Rh df cos. d C Force due to reflection is / / df cos 4 1.6Rh C 4 rh Fnet 1.6 0.4 C 8 rh 15 C 9. Area under the graph 0. 1. av s N 0 VdN a Vavg and dn vdv N v V dn a Vrms and dn vdv N v 0 0 Sec: Sr.T_Z Page 7
T sin f 9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s T cos mg N.. nitially steady state heat flow rate 4. 4 4 1 A 4 4 B AT i AT AT AT Also And 4 4 4 T A T1 T B 4 4 4 T B T A T 1 4 4 So i AT1 T 5. When N is maximum dn dt 0 or N N 1 1 6. Phase difference between i 1 and i is 90 So when i 1 is max imum i is zero MATHEMATCS 41. The required angle is the angle between OA OB and AB AC 4. BCD 1,, or 0 or 6 6 6 60 60 90 510 1 4. The lines x y x y 4 0 are concurrent at 1 A, in a variable line be B h,k then 1 1 AP PB h k P,. The image of hence locus, of P is x y x 4y 4 0 Sec: Sr.T_Z Page 8
44. f,,, be root, of f x then 9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s f x x x x x f i 1 1 1 1 1 0 4 f x x a b c d 0 45. f ' x x 6ax a 1 The root, of f ' x 0 should lie in, 4 a 1, 4 and a 1, 4 a, and a 1, 5 a 1, 46. 0 xf x dx, put x t xdx dt 9 1 f xdt 0 47. xdy y dy ydx ydx xdy x y dy y c y y x y c y 48. 1 1 1 5 sin x cos y sec z t x 1,y 1,z 1 49. z1 z z1 z z1 z1z z1z 0 evg OPQ is right angle triangle with QP z hypoteneous 50. Let y=m x be the chord. x m x m 1 4 4 0 Sec: Sr.T_Z Page 9
9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s 4m 4 x1 x,x 1x 1 m 1 m Since 0 0, divides the join of points 4m x 4x,then x1 1 m 4 4x1 9 9m 9 16m 4m 1 m 4 m 0, the lines are y 0, 7y 4x 0 7 x,y, x,y in the ratio 1:4 1 1 51. x n y n n so S n denotes number of divisions of n S 6 9 f P is prime then S P 5. tan x tan x tan x tan x 5. Let Then 8 15 8 7 tan x 8tan x 7 0 18, 4 18 (rejected) tan x 8tan x 4 0 tan x 4 1 1 r1 1 r 1 r r 54. f x is discontinuous when cot x is integer. in 0 1,, cot x x f is not continuous at 1-1 x cot,cot, 4 55. Sum of Eigen values of matrix A is equal to trace of A 56. When e x e log x 1 log x e x e Sec: Sr.T_Z Page 10
8 10 1dx dx 9 8 57. Let N be 6, 7, 7 9-01-18_Sr.T_Z_JEE-Adv_(011_P)_GTA-10_Key & Sol s such PN is per pendicular to the line. Then 1 N, 5, 9 PN 7 7 8 1 58. Required probability = 64 C 18 59. A. circle touch each other S S 0 is a common tangent B. S1 S C. y 1 m x 4 is tangent if m m 1 4 1 16 9 D. Center is point of intersection of y 4x and x y 0 and other asymptotes also passes through centre 1, 5 5 and perpendicular x y 0. 60. A. 6 6 c c 1 540 B. a 1,a,a each has ways and a 4,a 5,a 6 has each ways C. required number of way = 6P 10 16 D. 6 6 79 10 609 P Sec: Sr.T_Z Page 11