PHASE TEST-1 RB-1804 TO 1806, RBK-1802 & 1803 RBS-1801 & 1802 (JEE ADVANCED PATTERN)

Transcription

1 PHASE TEST- RB-804 TO 806, RBK-80 & 80 RBS-80 & 80 (JEE ADVANCED PATTERN) Test Date:

2 [ ] PHASE TEST- (ADV) RB , RBK & RBS-80-80_ (B) En = ; Ea = y IP EA IP y EN I.P. y. (A) CHEMISTRY NH (g) N (g) H (g) Before sparking After sparking 76 at equilibrium Increase in pressure = 8 ; = 9 cm Hg Partial pressure of H = = 9 cm Hg. (A) Moles of BaSO4... moles of M (SO 4 ) wt. of M (SO 4 ) (B) 5. (B) 6. (A).74 0 (M 96 ) M = 6.9 When the compound, on heating dissociate into its at least one compound then these solubility of that compound is decided by % ionic chr. higher the ionic chr. great will be thermal stability. (A) LiCO < Na CO <K CO (B) Li O > Na O > K O order of there stability (C) HF > HCl > H Br (D) BeSO 4 < MgSo 4 < CaSO 4 Indicated bond is a double bond maimum delocalisation (by Resonance & Hyperconjugation) maimum single bond character hence minimum rotational energy barrier. Rotational energy barrier bond strength. Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

3 PHASE TEST- (ADV) RB , RBK & RBS-80-80_ [ ] N O * High delocalisation * Higher single bond character O * Moderate delocalisation * 7 Hyper conjugating structure 7. (B) * Hyper conjugating structure * Less single bond character When phenolphthalein is used as an indicator : 0.05 = 0 0. ; = 40 ml. when methyl orange is used as an indicator y = 0 0. y = 0 ml y = 80 ml 8. (A,B,C) 9. (A), (B), (D) (A) H + N N Sp H H Sp H (C) Sp S Chr = % P Chr = 66% (B) OH OH Al OH Sp OH Tetrahedral (D) Hybridized orbitals always from6(sigma) bond due to bond form by pure orbital only. 0. (A, B, C) 7 7g / L H O mol L H O 4 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

4 [ 4 ] PHASE TEST- (ADV) RB , RBK & RBS-80-80_ Molarity of HO V M V volume H O means ml will give 5.6 ml at 7 K and atm P V P V 5.6 V V.8 ml. (A, B, C, D). (C). (C) 4. (A) 5. (B) For Boyle s law T = constant PV = K = constant Ortho effect causes less resonance. Two group showing resonance in similar direction so resonance energy is higher. Q = Alkali metal Q 6. (A) 7. (9) F QF F = Fluorine Q 0 Q O st Hion of Q and R is very low. 6 Pf atm. Final height = 9 cm P i = atm, initial length = h i cm Boyle s law Pi Vi Pf Vf hia 9A 76 hi 8 cm The length by which the Hg column shifts down = hi h = 8 9 = 9 f Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

5 PHASE TEST- (ADV) RB , RBK & RBS-80-80_ [ 5 ] 8. () 9. (9) V, n constant. 0. (6) P P T 09 P P P T T T 00 i f f f i i i f f % Pressure increases P P f i i X% = 9% X = 9 Moles of HO 9P 00 00% 00P i 9 Pincrease P P P P 00 f i i Fe 4HO FeO4 4H Consumed moles of Fe 4. (4) Mass of Fe 56 4, 4 w M O (Metallic oide) H M(s) H O( ) 0.0 g g POAC on O M M 0.7 M E 6. (4). () E 4 9 At constant temperature PA da MB 4 P d M.5 B B A Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

6 [ 6 ] PHASE TEST- (Adv) RB , RBK & RBS-80-80_ (A) 5. (B) 6. (B) 7. (D) Let tension be T then. T Common acc ms. n Frictional force on lower block, f 0 0N sn 50N f 0N u, v, R 0, 0 0cm fk 8. (D) mg sin During up the incline motion, u v 0, a g sin v 0 O v gsin S 0 v 0 S 4g sin ma. For block M, PHYSICS F ma F T MA A M When the object is placed at the focus of the lens, the refracted rays will be incident normally on the silvered surface. So, they will retrace their path. Hence, the image will be formed at the location of the object. In this way, the combination behaves as a concave mirror of radius of O I 0 cm Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

7 PHASE TEST- (Adv) RB , RBK & RBS-80-80_ [ 7 ] curvature (R) = 0 cm f = R = 0 cm 9. (A) Let retardation be f and initial velocity be u. For AB, a up fp (i) A u p B q C a b For AC, a b u(p q) f(p q) (ii) From (i) and (ii) (a fp ) a b (p q) f(p q) p a b a fp fp fq, p q p a a b ap aq ap bp fq p p q p(p q) aq bp fq, p(p q) (aq bp) f pq(p q) 0. (A) F (A) 4 0 q Q q r df For maimum repulsion force Q q 0 dq Q q. (B, C) v = m/s a =.0 m/s Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

8 [ 8 ] PHASE TEST- (Adv) RB , RBK & RBS-80-80_ v u a or 0 or = 4.5 m. Also v u a t 0.0t or t = s y u t a t y y and v a t 0.5 y y v v iˆ v y.5 m/s ˆj 0.5 ˆj and r iˆ yj ˆ 4.5ˆ i.5 ˆj. (A, C, D) T sin mg (i) T cos mg (ii) T cos -.5 m.5 ĵm/s m T cos mg T cos T sin T sin T cos T sin C mg T sin B T sin mg mg T cos. (A) (B) (C) F a > 0 when 50 4 F a 0 when 00 4 F = 00 N, F 00, F 400 F / / 4 50 a 5 m/s 5 5 a = 0 If F = 500 N a = 5 m/s, a =.5 m/s F/ F F/ F/4 F/4 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

9 PHASE TEST- (Adv) RB , RBK & RBS-80-80_ [ 9 ] 4. (A, B) For microwaves, c f = 00 m. = d sin = I d sin = (50sin ) 00 I 0 sin cos = sin Solution for [5 To 7] At a general point P on the screen = ( SP t) t ( d sin SP) P S d t D y S = yd S P SP) t( ) sin = t( ) d sin D ( d D 5. For central maima = 0, y = ( d sin ( ) t) y = 0. d Screen (B) 6. If slab is removed, d sin d sin sin 7 (A) Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

10 [ 0 ] PHASE TEST- (Adv) RB , RBK & RBS-80-80_ D d 5 4 mm. 8. (A) 9. (B) (B) N N sin N a B a A m (A) B mg mg sin N' mg sin a B a A mg N = ma A (i) mg sin Nsin mab.(ii) a a sin... (iii) A B from (I), (II) & (III), A mgsin mg ma A )sin m sin a 40. (0) m aa m sin mg sin mg sin sin m m g sin m m g sin aa and a B m m sin m m sin m m mg sin mmg m g sin m gsin N mg m m sin m m sin mmg mmg sin m mg cos mgcos N m m sin m msin sin a = retardation of block = m/s u = initial velocity = m/s By v u at u t (time after which its velocity becomes zero) a = s So block will come to rest after s. After that the applied force is 0 N towards left which is less than the maimum limiting friction. So it will remain at rest after that. 5N R 50N 0N 0N Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

11 PHASE TEST- (Adv) RB , RBK & RBS-80-80_ [ ] 4. (4) Path difference of the two rays (one coming directly to antenna and other after reflection B from water surface AB AC = h h cos cos cos h A C h cos = cos or h cos For constructive interference h cos n or 4. (0) Tension in string is zero. n n h =,,... etc cos 4cos 4 cos Middle pulley T T T T = T T = 0 4. () The light escape is confined within a cone of ape angle ' c ' where c is the critical angle. Imagine a sphere with source of light as its centre and the surface area abc is A. c here A R sin d R ( cosc ) 0 b a c c R P c c c Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

12 [ ] PHASE TEST- (Adv) RB , RBK & RBS-80-80_ R c sin 60º 44. (7) A Power transfer P 4R 4 W P 0 0 W P 0 40 W Optical Path Difference, 7 t 0 m Pnet P P P P cos 7 0 Watt 45. (4) 46. (8) Common acceleration, 8 a ms 6 f f 8N Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

13 PHASE TEST- (ADV) RB , RBK & RBS-80-80_ [ ] 47. (C) 48. (A) 49. (C) MATHEMATICS min(a).(n ) (n )... (n ). AX n n r (n r) C r f() [], I f(f ()) f () f () Consider e, > 0 e Let f() = e e ( ) f '() 4 f() is in (0,); and in (, ). Y Min (f()) = e 4 No solution for > 0 e has real solution from graph Again g() = 4 and g'() with g(0) < 0 (0, -4) X g() = 0 has real positive root. m + n = 50. (C) lim cos( n ) = n 5. (A) lim cos( n n ) = n lim cos n n n = n l lim n ln(ncosec n) = lim 0 sin ln = sin lim ln 0 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

14 [ 4 ] PHASE TEST- (ADV) RB , RBK & RBS-80-80_ sin lim sin 0 lim e 0 5. (A) 5. (C) sin sin l lim = lim = lim Let L k y 0, L y 0, L (k ) y 0 and L4 0, then the equation of the circumcircle is L L L L 0, 4 cos 6 Provided coefficient of = coefficient of y and coefficient of y = 0. i.e. (k y )( (k ) y ) y 0, where k = k + k = and k(k ) 0 7 i.e. 4 4y 4 y 6 0 sin cosy and cos siny (a ) and a a and a 0 Hence a = 54. (A, C, D) Y D B O L =0 4 A L =0 L =0 L =0 C X e 0 e tan Case I : > 0 e e 0 and 0 e tan e tan (e )(tan ) 0 and 0 (e tan ) tan and e tan 0, n,n,n I 4 4 ( 0 e, but tan e tan ) Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

15 PHASE TEST- (ADV) RB , RBK & RBS-80-80_ [ 5 ] Case II : < 0 tan and e tan n,n,ni 4 ( 0 e, but tan e tan ) 55. (A, B) BC = h + ; AD = h + y = - B -y+=0 P(h,k) A D C So, (h + ) = a. 56. (A, B) For y y 0, 0 and 0 are one pair of possible lines 4 y y, are the other pair of possible lines. 4 y y y ; Also (A, B, C) cos + cos y = a cos + cos y + cos cos y = a...() cos + cosy = b b cos cos y...() a b From () and (), cos cos y 4 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

16 [ 6 ] PHASE TEST- (ADV) RB , RBK & RBS-80-80_ Now, cos + cos y = c 4cos cos + 4cos y cosy = c 4(cos + cos y) (cos + cos y cos cos y) (cos + cos y) = c a + c = a( + b). Solution for 58 to 60 Centre (8, 6) is mid point of O and P Length of tangents form O and P to the circle are same and equal to 50 r = CT = 50 T T o COT T 'PC 45 TCT' 90 o O (0, 0) Q C (8, 6) P(6, ) shortest path from O to P is along tangent OT, then arc TT ' then T'P 58. (B) Length of arc TT ' r (C) Area 60. (D) r CT OT 4 6. (D) Required length = arc QTT' T'P y 8 (0,) y (-4,4) (4,4) (-6,0) (-,0) (,0) (6,0) Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

17 PHASE TEST- (ADV) RB , RBK & RBS-80-80_ [ 7 ] 6. (B) y = m touches y = ln () at p(,ln( )), then y ' and tangent is y ln( ) which passes through origin y = ln() y = m (, 0) ln( ) 0 e Hence, slope of tangent through origin is e Hence, all the lines of positive slope less than e intersect y = ln () at distinct points. 6. () Suppose n(a) = n, then number of refleive relations on A is n n 64. (5) and number of symmetric relations on A is n n 0 n. Graph of y = g() Y n n 0 X There are five points where y = g() is not differentiable. 65. () A B A B C D C D sin cos sin cos 4 Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

18 [ 8 ] PHASE TEST- (ADV) RB , RBK & RBS-80-80_ () sin A sinb sin C sind 4 A B C D 90 A B cos cos 6 6 k Let icot, k {,,...,n }, then n k k k cos cos i sin n n n k k cos isin k k k n n isin cos isin n n n n n n cos k i sin k ( ) ( ) 0 n n n n C C... 0 roots of the above equation are k icot,k {,,...,n } n n n n k k p q C i cot icot icot i cot 0 n k n k n p qn n n C n(n )(n ) (n )(n ) f(n) f() 6n 67. () f(n) lim n n The point B is (, ). Image of A(, ) in the line y + = 0 is given by y 4 5 Co-ordinates of the image is 9, 5 5 (, ) and Equation of BC is y 5 = 0 a + b =. 9, 5 5 lie on BC Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7

19 PHASE TEST- (ADV) RB , RBK & RBS-80-80_ [ 9 ] 68. (5) 69. (4) Both circles cut each other orthogonally. C and D will be centres of two circles also CD will be diameter of circumcircle of quadrilateral ACBD. Diameter The given equation is sincos sin cos 4 sin cos sincos sin 4 6 sin 6 sin 4 6 Hence sin sin 6 6 n ; n 6 6 Least positive value of is / Mentors Eduserv: Parus Lok Comple, Boring Road Crossing, Patna- Helpline No. : /4/6/7