PROGRESS TEST-2 RB , RBK-1806 RBS-1803 (JEE ADVANCED PATTERN)

Transcription

1 PROGRESS TEST- RB-83-84, RBK-806 RBS-803 (JEE ADVANCED PATTERN) Test Date:

2 [ ] PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ (A). (C) v 3v 3 m / s in downward direction B 3. (A) g 5 A PHYSICS 4. (B) 5. (A) 6. (A) 7. (C) 8. (A) 9. (C) 0. (C) Using v u R v = 30 cm v 5 0 Now, using V im = v V u om v (Vi V m) (V o V m) u ( 30) V i () [( 0) ( )] ( 5) V i = 45 cm/s So the image will move with velocity 45 cm/s.. (B, C). (B, C) 3. (B, C) 4. (A, B) 5. (A, B, C) 6. () 7. (3) 8. (6) 9. (8) V = (velocity of B w.r.t ground) V 4 v pulley = = m/s ; V is velocity of B w.r.t ground Solving we get v = 8 m/s

3 PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ [ 3 ] 0. (5) The velocity of the sphere is same as that of the cube, which is given as v 5ti ˆ ˆj Hence, acceleration of the sphere: or a (5i ˆ 0j)ms ˆ dv a dt y mg N a y a Hence, a 5ms and ay 0ms From FBD of sphere, N y N ma 5 0N Ny mg may Ny 0 0 0N Total force N N (0) (0) 0 5 N y. (D) Mass of S 8 in sample = 60 g; CHEMISTRY 60 Mole of S Number of moles of O required = Volume of O required =.4 5. (C) Vol. of air required L 3. (B) Order of R effect NO CN CHO Ph COOH Priority order : COOH > OH (B) CH OH CH OH COOH acts as principal f.g. (suffi : oic acid) acts as substituent (hydroy)

4 [ 4 ] PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ (A) L.R. 4Ca (PO ) F8SiO 30C 3P CaF 8CaSiO 30CO (D) 7. (C) 8 mole SiO gives 3 mole P mole SiO will give mole 8 CO ehaled in 5 minutes = = 3.33 g moles of CO = mole moles of KO consumed = mole mass of KO consumed = 5.38 g Let mass of Ag O = g Ag O Ag O 3 g 3 6gO g g %of AgO %.6 8. (A) CH CHO Br 9. (B) 30. (C) Be has stable configuration S 5 C H 5 3. (A,B) 3 C CH CH 3 3-Cyclopropyl--pentene 3. (A), (B), (D)

5 PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ [ 5 ] 33. (A,B) W (A) 46g of 70% HCOOH(d Solution.4g / ml) V ml solution. W 70% HCOOH 70g HCOOH in 00 V Mass of solution =.4 00=40g So, in 40g solution, mass of HCOOH = 70g in 46g, mass 70 of HCOOH 46 3g 40 (B) 0M HCOOH 0mole HCOOH in 000 ml solution mass of solution=000g Mass of HCOOH = 0 46= 460g So in 50g solution mass of 460 HCOOH 50 3g 000 (C) 34. (B), (D) 35. (A,C,D) W 5% HCOOH 5gHCOOH in 00g solution. W So in 50g solution, mass of HCOOH =.5 g V strength = 8; 8 M.5. L contain.5 moles of H O or.5 34 = 85 g H O Mass of litre solution = 65 g d 65g / L HO w 80g of moles of H O = 0.5 XH O w.5 34 % v m

6 [ 6 ] PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ (8) 37. (5) 38. (3) 39. (4) 0 34 Number of moles of Ba (OH) 0.4mole 00 7 or moles of OH = 0.4 Number of moles of HNO 3 =. =.4 mole Hence, the final solution is acidic due to presence of ecess H H 0.888M O CH 3, N CH 3 Zn HCl ZnCl H ; moles of H evovled = moles of HCl required = 4 CH 3 O, CH=CH, O C R V ; V mL 36.5, N=O CH 3 N N, H O, H CH 3, N H 40. (4) CH6 3.5O CO 3HO; CH4 3O CO HO Let volume of ethane is litre,.4 4 = (8 ) at constant T and P, V n; Mole fraction of C H 6 in miture =. litre.litre 0.4 8

7 PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ [ 7 ] MATHEMATICS 4. (A) Circum-centre O, ortho-centre H centroid G A O G H 8 G, 9 AG : GD : B G D C 4. (B) For f() to be defined, 6 0, 0, ( 3)( ) 0,, [,) 3 3,,,, 43. (B) 3,3 (3, ). tan 0 4sin0 sin0 4sin0 cos 0 sin0 sin 40 cos0 cos0 sin0 sin 40 sin40 sin80 sin 40 cos 0 cos0 sin 60 cos 0 3 cos0 44. (B) Positive integral solutions = {3, 4} i.e. two 45. (B) From 3 tana + 4 = 0, we get tana = 4/3, so that

8 [ 8 ] PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ sin A = tana 4 / / 9 5 tan A sin A 0 and tan A 0 in quad.ii and cosa = 3 tan A 5 cosa is negativein quad.ii Hence cot A 5 cosa + sin A = (C) According to property f f, then f (A) Option (C) is correct. hence 007 = Q() ( 5 + 6) + a + b 007 = Q() ( )( 3) + a b...() R() now R(0) = b =? Put = in () a + b = () put = 3 in () 3a + b = (3) (3) () gives a = nowb = 007 a = 007 ( ) = ) = = 6[ ] = 3[ ] ab(a c b c ) hence a = ; b = 3; c = 006 a + b + c = = 0 Ans.

9 PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ [ 9 ] 48. (B) P c,0 a, Q c 0, b OPQ = (OP)(OQ) = c ab clearly OPQ will not depend upon a, b and c if c = ab, 49. (C) i.e. a, c, b are in G.P. We have, B 6, 7, C 6, 7 BC = 7, AD = 3 Y A (0, 5) 0, y+0 = 0 C D O B 7 + y-0=0 y = 0, 0 7 ABC = = 8 7 sq. units 50. (A) This will formed right angled triangle with vertices (0,0), (0,0) & (0,4), right angle at (0,0) so orthocentre is (0,0) 5. (A) We have, log 3(log ) 9 log (log )

10 [ 0 ] PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ log 3(log ) 3 log (log ) log 3(log ) 3 log (log ) (log ) log (log ) (log ) log 0 ( log )(log ) 0 / log, log, 5. (A,C,D) When y = 4 y y = 4 (i) (ii) P = 0 then it has infinte solution if 4 < P < 0 or 0 < P < 4 then it intersect at points (iii) P 4 or P 4 then it has only one solution (0, 6) O 53. (A, C, D) As 3 3 log As, log Also, log log log log log5 log7 log3 = log = 0. As, 54. (B, D) log = log7 3 Dividing by cos(0 ), we get 3 = log Ans.] tan 0 tan tan0 tan tan(0 45 ) tan 057 Hence = k (80 0 ) Put k = 0 = 057 o 800 o = 57 o If k = = 77 o

11 PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ [ ] 55. (A,B,C) 56. (4) f & S = cos + cos cos = sin cos sin cos cos sin cos = sin cos = 57. (3) (sin cos cos sin ) sin cos = sin3 4 sin4 = 4 Ans. if = 7 n {} 3 [] 5[] (n 3){} 5[] {} n () Also, 0 {} 5[] n [], {n } n 3 5 It has 5 solutions, [] = 0,,, 3, 4 only, if 3 n 4 n,3,4 are 3 values of n. n Domain of is 0...(i) domain of 6 is [ 4, 4]...(ii) domain of log ( ) is (, 0) (, )...(iii) intersect of (i), (ii) and (iii) is (, 4] domain of sin is such that...(iv) [4n, 4n + ], n I...(v) intersection of (iv) and (v) is = 4 [n, (n + )], n I hence domain of f () is {4}. Hence range of f is {5} ]

12 [ ] PT-II (ADV) RB-83-84, RBK-806 & RBS-803_ () If sin sin cos cos cos( ) (since (0, ) ) tan 4 tan 4 tan (5) 4 4 tan tan 7 Put 0, then a b c d a b c d a b c d 5 5