JEE MAIN 2021 TEST-6 SOLUTIONS PDF

T E S T-6 S O L U T IO N S STANDARD ANSWER KEY Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 4 1 1 1 2 1 3 2 1 3 4 3 1 1 1 Q 16 17 18 19 20 21 22 23 24 25 26 27 2...

T E S T-6 S O L U T IO N S STANDARD ANSWER KEY Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 4 1 1 1 2 1 3 2 1 3 4 3 1 1 1 Q 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A 1 3 3 2 1 3 3 8 5 3 0 7 8 7 4 Q 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 A 3 2 2 4 2 3 3 2 2 3 1 4 2 2 4 Q 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 A 2 3 3 1 4 2 1 6 8 2 2 1 2 5 10 Q 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 A 2 4 2 2 1 2 2 4 2 4 2 3 1 2 2 Q 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 A 1 2 1 1 3 2 8 2 4 2 6 3 2 8 3 (1) (4). N2O4 + H2O HNO2 + HNO3 Van arkel process is used for Zr. Cl2O6 + H2O HClO3 + HClO4 (12) (3). Assertion is correct as Haemetite ore is used for extraction of Fe. (2) (1). NH3 is strongest ligand amongs the Haemetite is an oxide ore so reason is following Co+3 complexes. incorrect (3) (1). Only ionic chloride gives chrome chloride (13) (1). test. (1) During leaching when bauxite is treated (4) (1). (NH4)2Cr2O7 Cr2O3 + 4H2O + N2 with concentrated NaOH, then sodium (5) (2). b, c, d complex shown spin only magnetic aluminate and sodium silicate is formed in moment equivalent to 5 unpaired the soluble form, whereas Fe2O3 is electrons. precipitated. (a) K4[Mn(CN)6]; (2) The blistered appearance of copper during the metallurgical process is due to the Mn+2 = 3d5 4S0 (in octahedral complex) evolution of SO2. CN– ; SFL t2g221 eg00 (due to pairing of e–) (3) Cast iron is obtained from pig iron. (b) [Fe(H2O)6]Cl3; Fe+3 = 3d5 4S0 (4) Hall-Heroult process is used for H2O; WFL So t2g111 eg11 production of only aluminium. (c) K3[FeF6]; Fe+3 = 3d5 4S0 F– ; WFL, So t2g111 eg11 (14) (1). Phen = N N is a strong field (d) K4[MnF6]; Mn+2 = 3d5 4S0 symmetrical bidentate ligand. F– ; WFL, So t2g111 eg11 (1) [Fe(phen)3]2+ [Fe(phen)3]3+ (6) (1). [Ma3b3] exhibits fac & mer isomerism. Fe2+ : 3d6 Fe3+ : 3d5 (7) (3). CCl4 is used as firer extinguisher, CHI3, as antiseptic, DDT used as insecticide and + 0.6  0 + 0.6  0 CHCl3 (Chloroform) also as anesthetic. –0.4  0 –0.4  0 (8) (2). Siderite : FeCO3 C.F.S.E. = –2.4  0 C.F.S.E. = –2.0  0 (9) (1). HOCl < HClO2 < HClO3 < HClO4 By oxidation of Fe2+ into Fe3+, the CFSE value decrease. (+1) (+3) (+5) (+7) Acidic nature EN  O.S. For compounds of an element. (2) [Zn(phen)3]2+ [Zn(phen)3]3+ (10) (3). A Zn2+ : 3d10 Zn3+ : 3d9 A C.F.S.E = 0 C.F.S.E = –0.60 M en By oxidation of Zn2+ into Zn3+, the CFSE B value increase. B (3) [Ni(phen)3]2+ [Ni(phen)3]3+ This structure does not contain plane of symmetry hence it is optically active, rest Ni2+ : 3d8 Ni3+ : 3d7 of all options has plane of symmetry and C.F.S.E = –1.20 C.F.S.E = –1.80 they are optically inactive. By oxidation of Ni2+ into Ni3+, the CFSE (11) (4). Liquation is used for Sn. value increase. Zone refining is used for Ga. (4) [Co(phen)3]2+ [Co(phen)3]3+ Mond’s process is used for Ni. 2  Co2+ : 3d7 Co3+ : 3d6 [CoIII(H2O)4(SCN)2]3[CoIII(ox)3] C.F.S.E = –1.80 C.F.S.E = –2.40 Linkage isomerism is due to presence of By oxidation of Co2+ into Co3+, the CFSE SCN– value increase. Optical isomerism is exhibited due to (15) (1). Ti+2 = 1s2 2s2 2p6 3s2 3p6 3d2 presence [Co(ox)3]3– complex ion, which Unpaired electrons = 2. has asymmetric structure. Spin only magnetic moment Geometrical isomerism is exhibited due to (µ) = cationic part: + Ti+3 = 1s2 2s2 2p6 3s2 3p6 3d1 SC N SC N SC N NH3 H3N H3N Unpaired electrons = 1 III III Co Co H3N NH3 H3N NH3 µ= NH3 SC N cis trans V+2 = 1s2 2s2 2p6 3s2 3p6 3d3 (21) 3. Paramagnetic unpaired electrons µ= KO2 K+O2– Sc+3 = 1s2 2s2 2p6 3s2 3p6 according to MOT paramagnetic µ=0 Na2O2 2Na+O2– diamagnetic (16) (1). The total number of isomers for a square CaO Ca+O2– diamagnetic planar complex [M(F)(Cl)(SCN)(NO2)] is F SCN F N CS KF K+F– diamagnetic. M M :N Cl N O 2 /S C N /C l Cl N O 2 /N C S /C l NO2 O O odd electron species (3) (3) Paramagnetic F SCN F N CS NO odd electron species M M Cl O N O /S C N /C l Cl O N O /N C S /C l Paramagnetic (22) 3. (3) (3) (17) (3). en and C2O42– are bidentate ligand. So PCl5 + 3H2O H3PO4 + 3HCl coordination number of [Co(Cl)(en)2]Cl is (basicity-3) 5 and K3[Al(C2O4)3] is 6. (18) (3). E.N  + ive charge (basicity-2) t + 2 /5 (basicity-2) – 3 /5  t (basicity-2) (23) 8. Hybridisation Geometry = –0.4 t (a) Ni(dmg)2 dsp2 Square planar (19) (2). Pt Cl4.5NH3 [Pt(NH3)5Cl]Cl3  (b) [Ni(CN)4]2– dsp2 Square planar [Pt(NH3)5Cl]+3 + 3Cl– (c) [Ni(CN)4]4– sp3 Tetrahedral (20) (1). Formula of the complex : (d) [Ni(CO)4] sp3 Tetrahedral 3  (e) [Ni(Cl)4]2– sp3 Tetrahedral -N-irridate (III). (26) 0. The chromium is in zero oxidation state (f) K2Cr2O7 d3s Tetrahedral having configuration [Ar]18 3d5 4s1. The (g) K2CrO4 d3s Tetrahedral CO is a strong field ligand so compels for (h) KMnO4 d3s Tetrahedral the pairing of electrons. Thus the complex (i) K2MnO4 d3s Tetrahedral has d2sp3 hybridisation & is diamagnetic. [Cr(CO)6] - (j) CrO2Cl2 d3s Tetrahedral 3d 4s 4p (24) 5. All are correct. (i) CO bond order R earrangem ent d 2 sp 3  hybridization Correct sequence of CO Bond order: H3B  CO >CO>Fe(CO)5 > [Mn(CO)5]– as there is no unpaired electrons. (BO > 3.0) (BO = 3.0) (BO < 3.0) (BO < 3.0) (27) 7. (B) is incorrect. (ii) Mn(CO)6 can act as reducing agent (28) 8. (NH4)2S and Na2S are water soluble ZnS because the metal carbonyl complex is dissolves in dilute HCl others are insoluble stable when EAN is equal to nearest noble in dilute HCl as well as water. gas configuration (29) 7. K4 [Fe(CN)6] + K2CO3 + 5S [Mn(CO)6] [Mn(CO)6]+ Fe + CO2 + KCNO + 5 KCNS EAN = 37 EAN = 36 (30) 4. [Fe(H2O)6]2+ has four unpaired electrons. (Less stable) (More stable) 2– 3 (31) (3). Induced emf = (iii) [CuCl5] ; sp d Hybridisation (Outer orbital A I = 5 B complex) (Magnetic moment = B.M.) 1 15V 5V (iv) [Pd(Cl) (Br)(gly)] is square planer VA – 5 – 15 + 5 = VB complex hence it can exhibits geometrical isomerism. VA – VB = 15 volt (v) trans-[Co(gly)3] is optically active (32) (2). complex, hence equimolar mixture of 'd'  3 and  form on mixing gives racemic O mixture. A C (25) 3. (i), (ii) and (iii) are correct (i) NaCl (s) + K2Cr2O7 (s) + H2SO4 (few drops) + H2O A C 1 B 2 9 B 2 2 2 CrO2Cl2 + OH– (aq)  + Cl– + 2H2O | VA – VO | = (ii) | VO – VC | = (iv) Correct I.U.P.A.C. name of complex compound is | VA – VC | = = [Cr (NH3)5(CN)] [Ir (NO2)6] - (33) (2). Pentaamminecyanidochromium(III)hexanitrito 4  L R a b C R em f start em f co n tin u e d istan ce trav elled in th is d u ratio n But, ; N o em f till th is situ atio n ; t = RC ln 2 ag ain em f start (34) (4). So total distance travelled up to which emf induces = 2a (35) (2).  = NBA = LI  time t = 2a/v N µ0 nIR2 = LI ; N µ0 IR2 = LI (39) (2). N & R constant. Self inductance (L) = 25 × 2 ×2 × 10–4 = 10–2C = 0.01 C (36) (3). LI dI = I2R (40) (3). The direction of propagation of an EM wave is direction of e–t/2 = 1 – e–t/2 ; t = 2 ln 2 (37) (3). sin [t + (6y – 8z) B = 2 × 10–8 T along z direction. where and (41) (1). X L = L Z  60° Direction of propagation, R (38) (2).....(i) R 30° Z' 1 Xc  C.....(ii) From (i) and (ii) 5  (46) (2). Induced current in loop will be in clockwise direction. I ind  or 2 = 108 or  = 104 (47) (3). Rnet = 40 + 40 = 80  2f = 104  XL = 100  (42) (4). Maximum Electric field E = (B) (c) = (3 × 10–5) c XC = 40  ; Z = 100  = (2 × 10–6) c Power factor = cos  = Maximum force (48) (3). VL = VC  IXL = IXC  XL = XC Voltage across LC combination = 0.9 N = I (XL – XC) = 0  XL – XC = 0 Frms = = 0.6 N (approx)  Voltage across LC combination is always (43) (2). outer = (µ0nKte–t) 4R2 0. (49) (1). iinduced = At t = 0, iinduced = –ve. (44) (2). (50) (4). As XL = XC so it is condition of resonance [For transparent medium µr µ0] Hence | Z | = R and VL = –VC  VL + VC = 0 so V = 0  and = Intensity = (51) 2. XL = L = 2000 × 5 × 10–3 = 10  E02 = kE2v  So it is condition of resonance Total resistance of the circuit Similarly, R = RL + Re = 4 + 6 = 10 (45) (4). Induced emf = Bv  = 0.3 × 10–4 × 5 × 10 = 1.5 × 10–3 V 6  and  (52) 1. (56) 2. 'Immediately' after pressing the switch S, the current in the coil L, due to its self- induction will be zero, that is i2 = 0. The current will only be found in the =  × 10–7 S.I. units resistance R1 and this will be the total current in the circuit. Q ( 2 ,2 ,0 ) A B  i = i1 = = 2.0 ampere.  /4 P C X I = 2A ( 2 ,0 ,0 ) R (57) 1. e=–L = – 10 × 10–3 = –1Volt. (53) 6.  | e | = 1 volt. (58) 2. In EMW, electric field and magnetic field have same energy density and same intensities hence ratio is 1 : 1 = [0.1 × 5 + 0.2 × 10 + 0.4 × 1.25] (59) 5. = 200 × 40 × 10–6 × 2 × 50 = 2.5 A (54) 8. (60) 10. L = 40m H C = 100µ F 9V ienc = 2Avr I R V = 10 sin (314t) x r XL = L = 314 × 40 × 10–3 = 12.56 dx Using ampere’s law,  B (2r) = µ0 (2Avr) B = µ0 Av Phasor - IX L (55) 2. From KVL, I FBD of slider, ma = mg – iB – kx kx IX C ma = mg – (k + B22) x iB Vm = Im (XC – XL) Now, mg 10 = Im (31.84 – 12.56) 7 (61) (2). (x) = f (x) + f (2 – x)  '(x) = f '(x) – f '(2 –x)......(1) Since f ''(x) > 0  f '(x) is increasing (0, 2) h = 10 Case-I : When x > 2 – x  x > 1    '(x) > 0 (1, 2)  f (x) is increasing on (1, 2) Case-II : When x < 2 – x x < 1   '(x) < 0 (0, 1)  (x) is decreasing on (0, 1) (62) (4). (65) (1). h = 2r sin  a = 2r cos  v =  (r cos )2 (2r sin ) v = 2r3cos2sin  = (p + 1) + 1 = p + 2 for continuity LHL = RHL = f (0)  (p, q) = (–3/2, 1/2) (63) (2). f (x) = x or h For f ' (x) 0 (66) (2). y = f (f ( f (x))) + ( f (x))2 a kx – x2  0 = f '(f (f (x))) f '(f (x)) f '(x) + 2f (x) f '(x) x2 – kx 0 x (x – k)  0, so x [0, 3] = f '(1) f '(1) f '(1) + 2f (1) f '(1) + ve [x  3] = 3 × 5 × 3 + 2 × 1 × 3 = 27 + 6 = 33 3kx – 4x2  0 (67) (2). Using LMVT f ' (c) = 4x2 – 3kx  0 4x (x – )0 ; 3– 0 ; k4 Minimum value of k is m = 4 (68) (4). y = ax3 + bx2 y' = 3ax2 + 2bx y'' = 6ax + 2b For point of inflection y'' = 0 (64) (2). x = –b / 3a 3a + b = 0.....(i) (as x = 1) Point satisfy the curve also, so 3=a+b.....(ii) 8  from (i) & (ii), a = –3/2, b = 9/2 4 a 2b = × 3ct2 = (0,2) 6 0 4 (73) (1). f (x) = {1 + cos x} sin x (69) (2). –2 + – 0 / 3  f (x) = f '(x) =  f (x) = = maximum value So f (x) is continuous only. f (0) = f () = minimum value (70) (4). Let f (x) = 5x, x = 2, x = 0.01 (74) (2). x2(1 + y) = y2(1 + x) f (x + x) = f (x) +.x  x2 – y2 + xy(x – y) = 0  (x – y) (x + y + xy) = 0 f (2 + 0.01) = f (2) + (5x) x x  y [does not satisfy the given equation] = 52+ 5x ln 5 x = 25 + (25) (1.6095) (.01) = 25.402375 = 25.4024  x + y + xy = 0  y = (71) (2).   f (0) = 8 (75) (2). Length of subnormal = y So, f (x) is continuous at x = 0 when a = 8. y2 = 8ax 2y = 8a y = 4a (72) (3). Let any point on the curve is + – + (76) (1). f '(x) = ex (x – 1) (x – 2) 1 2 f (x) is increasing in (–, –2) & (–2, –1) & (2, ) (77) (2). f (x) = Equation of tangent is y – ct2 = – 2t3 For x intercept f ' (x) =  f ' (1) = 0 – ct2 = – 2t3  =a– (78) (1). f '(x) = For y intercept, b – ct2 = – 2t3 Now at x = 0  b – ct2 = 2t2c  b = 3ct2 8 (0) – 2 = 2 (a) (0) – b  b = 2 and a  R 9  Also, f (x) is continuous in (–1/2, 1/2). [f2(x) + g2(x)] = 2f (x) f '(x) + 2g(x) g'(x) (79) (1). = 2f (x) g (x) – 2g (x) f (x) = 0  f 2(x) + g2 (x) = constant f 2(5) + g2 (5) = 4 + 4 = 8  f 2(10) + g2 (10) = 8 (83) 2. f (x) = x3 + x2 f ' (1) + x f '' (2) + f '''(3) f '(x) = 3x2 + 2xf '(1) + f ''(x)..... (1) f ''(x) = 6x + 2f '(1).....(2) f '''(x) = 6.....(3) (80) (3). y – x3/2 = 7, (x  0) Put x = 1 in equation (1) : B (x , y ) f '(1) = 3 + 2f '(1) + f ''(2).....(4) A 1 ,7 2 Put x = 2 in equation (2) : f ''(2) = 12 + 2f '(1).....(5) ; From equation (4) & (5) : –3 – f '(1) = 12 + 2f '(1) 3f '(1) = –15 ; 3x2 = 1 – 2x f '(1) = –5  f '' (2) = 2 Put x = 3 in equation (3) : f '''(3) = 6 3x2 + 2x – 1 = 0 ; 3x2 + 3x – x – 1 = 0  3 2 f (x) = x – 5x + 2x + 6 (x + 1) (3x – 1) = 0 f (2) = 8 – 20 + 4 + 6 = –2  x = –1 (rejected) x = 1/3 (84) 4. y=7+ x3/2 =7+ = 2 · e + e = 3e T : y – e = 3e (x – 1) y = 3ex – 3e + e = (3e) x – 2e (4/3, 2e) lies on it (81) 2. Rationalize (85) 2. S = {x  R, x2 + 30 – 11x  0} = {x  R, 5  x  6} Now f (x) = 3x3 – 18x2 + 27x – 40  f '(x) = 9 (x – 1) (x – 3), which is positive in [5, 6]  f (x) increasing in [5, 6] Hence maximum value = f (6) = 122. (86) 6. x3 – 3x2 – 9x + 27 = (x – 3)2 (x + 3) form using L' hospital     In order that f (x) is continuous for all (82) 8. f '(x) = g (x) and g' (x) = –f (x) x R 10  =6 and......... (2) (87) 3. (a – 1) x2 = x (2b + 3) The above equation is satisfied by three As, f (x) = x + cos x + 2 distinct values of x therefore it is an  f ' (x) = 1 – sin x identity  f ' (0) = 1  2 – 2a = 0  a = 1 and 2b + 3 = 0 and f " (x) = – cos x f '(1) = – 1  b = –3/2  From (1) and (2), we get Now, f (x) = 2x + 1, Let g (x) = px + q  g ' (x) = p g '(3) = and g"(3) = f (g (x)) = 6x – 7  2 (px + q) + 1 = 6x – 7 Hence, (g' (3) + g"(3)) = 1 + 1 = 2  2px + 2q + 1 = 6x – 7  2p = 6  p = 3 and q = – 4  g ' (2012) = 3 (88) 2. f (x) = x + cos x + 2 f ' (x) = 1 – sin x and f ' (0) = 1 f " (x) = – cos x and f " (0) = – 1 f f (0 ) = 3 g (3 ) = 0 When y = 3 then x = 0  g ' (3) = =1............ (1) Again,  or g '' (3) = 1............ (2) Hence, g ' (3) + g '' (3) = 1 + 1 = 2 Alternatively : Clearly, f is a bijective mapping. Also, f (0) = 3 ......... (1) 11 (89) 8. Points of maxima are and  points of minima are –, and global minimum value occurs at which is zero. (90) 3. As y = f (x) is polynomial hence it is differentiable.  f '(x) = k (x – 1) (x – 2) (x – 3) f ' (0) = – 6k 6 = – 6k  k = – 1 Hence f '(x) = – (x – 1) (x – 2) (x – 3) = – (x3 – 6x2 + 11x – 6) I D I D – 1 2 3  As, f (0) = 1, hence C = 1 So, f (x) = 1 – Hence, f (2) = 1 – 4 + 16 – 22 + 12 = 3 12