TARGET IIT-JEE-2015 PART TEST-04 XII-PASS ANSWER WITH SOLUTION Excellence Delivered by Experts Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12,Odisha Ph:-2565949,6530219. Damana Study Center. Plot No. 248, Damana Square, Near Bank of India,Sailashree Vihar Road, Chandrasekhar Pur BBSR-Ph:-0674-2745599,6444399. Cuttack Study Center PLot No.-2867 & 2868, 3rd floor, Yamaha Showroom, Infront of Arundoya Market Link Road, Cuttack-754012. Ph.:-0671-2314999, 8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 PHYSICS 1. A mA I O I O 1 A B A B mB O A IB IB O A 3 2. B mA But f mB f uA and f uB 1 7.5 uB 3 7.5 30 For minimum deviation i = e. Therefore m 2i A or A m 2 i or 3. A m sin 2 sini Now A A sin sin 2 2 (C) 2 1 2 1 u R / A m i 2 A 1 60 Sin i = sin 2 2 sin 2 2 sin30 2 where 1 1, 2 1.6, 50 cm & R = + 12 cm. / 160 cm 3 the v irtual object for the second surface so that 2R is the object distance for this surface. 1 2 1 2 u R 4. 9.5 cm D The amplitudes of the two coherent waves will be A 1 = 2A and A2 = A. therefore 2 2 Imax A 1 A 2 2A A 9 Hence the correct choice is D Imin A1 A 2 2A A 5. C White light consists of colours between violet source S 1 will decrease. The intensities of the two virtual sources will be different. Hence the correct is C 6. (B) path diff = (n1L1 – n2L2) : Phase diff = (2 / ) (n1L1 – n2L2). C We know that in case of a convexlens when object is placed at C / , the image is obtained at C. This situation is represented in the graph by the point corresponding to u = – 10 cm, v = 10 cm. 7. Therefore R 10 R 5 cm f Lens formula is 2 (for maximum error in f) u L (A) x 9. df 0.1 0.1 df 25 0.1 2 0.01 0.05 2 2 25 10 10 v L 8. 1 1 1 df d du 2 2 2 f v u f v u I v u v Here, (v / u) = m1 and (u / v) = m2 m1 – m2 = (v / u) – (u / v) = {(v2– u2) / uv} = [{(v – u) (v + u)} / uv] = [x × (1 / f )] [ f = {uv / (u + v)} and v – u = x] f = {x / (m1 – m2)} (B) The power of the lens is given by P1 = (1 / f1) = (µ – 1) [(1 / R) + (1 / R)] = (1.5 – 1) (2 / R) = (1 / R) AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -2Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 the power of each water lens (plano-concave) P2 = P3 = (1 / f2) = (µ – 1) (1 / R) = {(4 / 3) – 1}(–1 / R) = – (1 / 3R) The power of combination P = P1 + P2 + P3 = (1 / R) – [(1 / 3R) + (1 / 3R)] = (1 / 3R) = {1 / (3 × 10)} = (1 / 30 cm) = (1 / 0.3) dioptre = 3.33 dioptre 10. (A) Here (90 – r) should be just greater than critical angle. Thus cos r = (1 / 1µ2) = (µ1 / µ2) 1µ2 = [1 / (sin (90 – r)] = (1 / cos r) 2 (1 – sin r) = (µ1 / µ2) .....(1) From figure, (sin / sin r) = (µ2 / µ1) sin r = (µ1 / µ2) sin .......(2) F r o m eqs. (1) and (2), we get [1 – {(µ1 / µ2) sin ] = (µ1 / µ2) [1 – (µ1 / µ2)2 sin2 ] = (µ1 / µ2)2 Solving we get sin = [(µ2 / µ1)2 –1] 11. Ans. B Let d = the diameter of the tube. / 4 24.1 0.3 d , and 3 / 4 74.1 0.3d 3 / 4 4.1 0.3 d or 50 or 100 cm 2 2 0.3d / 4 24.1 cm 25 24.1 cm 0.9 cm or d = 3 cm 12. Ans. D V 2V 1 L L 2L Frequency = where = the velocity of transverse waves on the string. 2 13. Ans. A In a wave equaiton, x and t must be related in the form (x–t). We rewrite the given equation as y 1 1 1 x t . For t = 0, this becomes y = 1 x 2 , as given 2 1 2 for t = 2 this becoemes y = 1 x 2 1 1 x 12 2 1 or 0.5 m / s y = a coskx cos t sinkx sin t is given 14. Ans. C As y =0 ato x = 0 for all values of t(node), the cos kx term must vanish in the sum of the two waves 15. Ans. D Initially, 260 1 2 T1 , T1 50.7 g 507 N . m = (0.0075 m³) (10³ kg/m³) (10 m/s²) New tension = T 2 = (507 – 75) N = 432 N or 16. Ans. C I1 40 10 = log10 I 0 When the mass is submerged, upthrust = 75 N n= 1 T2 2 m n o r 260 T2 T1 432 144 12 507 169 13 n = 240 Hz. I1 10 4 I0 ........(i) I2 Also, 20 10log10 I 0 AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -3Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 I2 102 I0 .............(ii) I2 r2 10 2 12 I1 r2 r2 100r12 r2 10m r1 1m 17. Ans. A There will be no effect of magnetic force on time period because the magnetic force will be perpendicular to the inclined plane. q, m 18. C 2 T 2 2 m m 2 m 4 m 2 T2 or k k eff. 2k or 2k T2 2 22 2 12 2 7 22 k 6 59.15 Nm 1 4 7 19. B As here two masses are connected by two springs, this problem is equivalent to the oscillation of a reduced mass of a spring effective spring constant. T 2 k eff. Here, 1 k 1 2k 2 m 2 20. D m1m2 m m1 m2 2 k eff k1 k 2 2k n 2 k eff. 1 2k 2 2 2 m 1 k 1 0.1 1 Hz m 0.1 Here, m = 4kg, k = 800 Nm1 ; E = 4 J Is SHM, total energy is E = 4 1 800 A 2 2 1 2 kA , where A is the amplitude of oscillation. 2 2 Maximum acceleration, amax. A k A m k m 800 Nm 1 0.1 m 20 ms 2 4 kg 21. B Let m be the mass of each particle, then 1 h / m1 and 2 gh / m2 , where 1 and 2 m1 m 2 1 2 CM 2m 2 2 1c 1 CM 1 2 Velcoity of A w.r.t. C frame is, | 1c | 1 2 h h 2 m | 1c | m h 1 1 2 2 m 1 2 2 | 2c | So, required wavelength is 21 2 12 22 AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -4Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 22. (c) 4 2 l T2 T 2 l/g T 2 4 2 l/g g Here % error in l = 1mm 0 .1 100 100 0 .1 % 100 cm 100 0.1 and % error in T = 2 100 100 0 .05 % % error in g = % error in l + 2(% error in T) 0 .1 2 0 .05 = 0.2 % 23. (b) 24. (b) V 4 3 r 3 % error in volume 3 % error in radius. 25. 3 0.1 100 5.3 (c) Volume of cylinder V r 2 l Percentage error in volume V 2 r l 100 100 100 V r l 0.01 0. 1 2 100 100 (1 2)% 2 . 0 5 . 0 = 3% 26. A N2 3 7x e N1 100 We have N1 N0 e7 & N2 N0 ex x 7 ln 100 100 2.3log10 3 3 0.693 0.693 But T 12.5 1/2 100 x 7 ; e 3 = 2.3 [ 2 – 0.4771]= 2.3 × 1.5229 x–7 = 12.5 12.5 2.3 1.5229 = 1.52 63 0.693 3 x= (63 + 7) years = 70 years 27. C We have K.E. = k T = 1.38 × 10 –23 × 300 J = 1.38 1023 300 1.38 3 eV 10 2 eV 19 1.6 1.6 10 = 0.026 eV 28. C Binding energy = [ZMP A Z MN M]c 2 2 = 8MP 17 8 MN M c = 8MP 9MN M c 2 = 8MP 9MN M0 c 2 3 RhC V e.v. 4 RhC 29. C Enery released = (B.E. of product – BE of reactant) (80 × 7 + 120 ×8 – 200 × 6.5) = 220 MeV 30. C 1 1 First excitation energy = RhC 2 2 2 1 3 RhC 4 4V e.v. 3 CHEMISTRY 31. A f1 z12 n32 n1 3 n2 3 22 23 32 , Solution: 3 2 f2 n1 z 2 z1 2 z 2 1 33 1 27 AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -5Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 32. C Solution: Maximum number of electron having same spin quantum number is equal to the number of orbitals 2 1 33. C Solution: Magnetic moment = n(n 2) 3.873 25Mn [Ar]3d5 4s2 therfore Mn should be in + 4 4s 3d2 34. 35. 36. 37. 38. number of unpaired electron n 3 A A B D B Solution: 2NO(g) + moles 2 2 – 2 × 0.5 1 nf = 1 + 1 O2 (g) 0.5 0 0 2NO2 (g) 0 2 × 0.5 1 ni = 2 + 0.5 n (2.5 2) 0.5 moles changein pressure p nRT 1 300 0.5 2 atm. V 12 6.25 39. C Solution: c S1: Na2O2 MgO ZnO P4O10 : as non-metallic character increases the acidic character increases. MgO, Na2O are basic, ZnO amphoteric and P4O10 acidic. S2: Na Si Al Mg : Mg has higher than Na due to small size and higher nuclear charge. Mg has higher than Al because of ns2 configuration (has extra stability and high electron penetration power of s-subshell electrons) and Si has higher than Al because of higher nuclear charge and small size. IE1 : Na = 496, Al = 577, Mg = 737 and Si = 786 kJ/mole S3: There is more interelectronic repulsion in 2p-subshell of fluorine than chlorine (3p). So extra electron will be added easily in 3p-subshell of chlorine as compared to 2p-subshell of fluorine. Down the group electron affinity values generally decreases with increasing atomic number due to increase in atomic size. So Cl > F > Br S4: Isoelectronic series of ion; all have the xenon electron configuration . Ionic radius 1 nuclear charge Atomic number : Te = 52; I = 53; Cs = 55; Ba = 56. 40. A Solution: As non-metallic character of element attached to oxygen atom increases, the difference between the electronegativity values of element and oxygen decreases and the acid character of oxides increases and vice-versa. 41. C Solution: AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -6Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 42. A Solution: C O 1.43Å; C O 1.23 Å ; C O 1.09Å CO C O; partial triple bond character; bond length is inter mediate of C = O and C O bonds CO2 O C O; double bond length 43. D Solution: (d) (a) Bond order 1 bond length Bond order O 2 2.5, O 2 2 O2 1.5 So correct of bond length is O 2 O 2 O 2 . (c) 22 0. So He2 does not exist. 2 In all these molecules all electrons are paired in molecular orbitals. So they are all diamagnetic (d) F2 MOT configuration: 1s2 *1s2 2s2 * 2s2 2pz2 2pz2 * 2py2 * 2pz2 * 2pz2 (b) Bond order for He2 44. D Solution: (d) Adjacent bonds, thereby distoring the structure as doubly bonded oxygen atom require more space than a single bonding pair and repel. AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -7Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 45. C Solution: c 46. B Solution: Cyclohexane (C6H12) and carbontetrachloride (CCl4) both are non-polar compounds so there exist London dispersion force between the constituent molecules. 47. B Solution: b (2) KO2 2H2O KOH H2O2 1/ 2 O 2 (3) 4KO2 2CO2 2K 2CO3 3O2 A Solution: S1: In concentrated solution unpaired electrons pair up together and become dimagnetic. S2 and S3 are correct statements 48. S4: Be(OH)2 2OH [Be(OH)4 ]2 beryllate ion Be(OH)2 2HCl 2H2O [Be(OH2 )4 ]Cl2 49. C Solution: S1: True statement - salt of weak base and strong acid. S2: As we move down the group from Li to Cs the strength of metallic bonding decreases as size of atom increases. Due to this three occurs a decrease in close packing of atoms in crystal lattice from Li to Cs and thus the softness increases from top to bottom. S3: It is hemihydrate of CaSO4(2CaSO4.H2O) S4: KHCO3 is soluble in water and cannot be precipitated like NaHCO3 50. D Solution: d AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -8Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 (a) P4 3NaOH 3H2O PH3 3NaH2PO 2 (b) 4S 6NaOH Na 2S2O3 2Na 2S 3 H2O (c) 3 Cl2 6 NaOH 5 NaCl NaClO3 3 H2O (d) 2B 6 NaOH 2 Na3BO3 3 H2 51. C Solution: c (a) It is a weak monobasic acid soluble in water and in aqueous solution the boron atom completes its octet by (b) (c) 52. (d) A accepting OH– from water molecules: B(OH)3 2H2O [B(OH)4 ] H3O Only TI3+ acts as oxidising agent on account of inert pair effect In the solid state, the B(OH)3 units are hydrogen bonded together into two dimensional sheets with almost hexagonal symmetry B(OEt)3 imparts green colour to burner flame. Solution: NH4ClO 4 HNO3 HClO 4 NH4NO3 2NH4NO3 2N2O 4H2O 53. D Solution: d (1) (2) (3) (4) 54. A Solution: a 55. NH3 has higher boiling point than AsH3 and PH3 on account of H-bonding. SbH3 has highest boiling point on account of highest molecular weight. As the size of element increases, the strength of M–H bond decreases and thus stability decreases. Basicity decreases down the group. (Drago rule). HEH angle (°) : NH3 = 107.8 ; PH3 = 93.6, AsH3 = 91.8 ; SbH3 = 91.3 Solution: d (a) (b) (c) (d) Disproportionation is slow at the ordinary temperature for OCl , it is fast for OBr and very fast for OI– E° for the 1/2 F2 | F– electrode = + 2.9 V HI stronger acid than HBr because of its low bond dissociation enthalpy Because of low bond dissociation energy and high hydration energy of F2, F2 acts as strong oxidising agent and oxidising power decrease down the group. E° 1/2 F2 | F = + 2.9 V AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. -9Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020 Amogh Classes–PART TEST-04-TARGET IIT-JEE-2015 56. C Solution: 57. A Solution: Sulphite gives SO 2 with H2SO 4 which turns acidified K 2Cr2O7 green according to following reaction K 2Cr2O7 H2SO4 3SO2 K 2SO 4 Cr2 (SO 4 )3 (green) H2O.H2S also gives green coloured solution with acidi- 58. fied K 2Cr2O7 solution but a slightly coloured precipitate of sulphur is also obtained. The H2S gas has rotten egg smell. D 2 HCl NaOH MnCl2 MnO2 (Brown / black) Solution: Mn H2S MnS (buff / light pink coloured) [O] 59. fusion MnO2 2KNO3 Na2CO3 Na 2MnO 4 (green compound) KNO 2 2NaOH CO2 A [O] Solution: MnO 4 / H [O] : Fe2 Fe3 [Fe(CN)6 ]4 Fe 4 [Fe(CN)6 ]3 60. D Solution: d (b) Cu2 4NH4 OH [Cu(NH3 )4 ]2 (deep blue) 4H2O Bi3 3NH4OH Bi(OH)3 (white) 3NH4 (c) Cu2 [Fe(CN)6 ]4 Cu2 [Fe(CN)6 ] (chocolate brown) Bi3 [Fe(CN)6 ]4 no change observed MATHEMATICS 61. 66. 71. B D A 62. 67. 72. B A C 63. 68. 73. C A C 64. 69. 74. C A B 65. 70. 75. B A C 76. B 77. D 78. C 79. C 80. C 81. 86. B A 82. 87. C A 83. 88. C C 84. 89. C B 85. 90. B D AMOGH CLASSES:- Head Office: Plot No.-1189, Nilakantha Nagar, Besides New Passport Office, Devray College Road, Nayapalli, BBSR-12, Odisha Ph:-2565949,6530219. Damana Study Center. Ph:-0674-2745599,6444399. - 10 Cuttack Study Center Ph.:-0671-6544599,2314999,8456887020
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