SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION-2014 VERSION A1
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1 SOLUTIONS & ANSWERS OR KERALA ENGINEERING ENTRANCE EXAMINATION-04 VERSION A [PHYSICS & CHEMISTRY]. Ans: 8% g l T g l T % % 8%. Ans: mm and 50. Ans: 78.4 m t LC mm s v r 00 4 s 5 fall h gt m 4. Ans: : : 5 5. Ans: m s S : S : S : : 5 :. S ut at u a 9 4 u 6 a 6 Solving a m s u m s 6. Ans: 7 m s p mv 0 0 v 0.54 v m s 8. Ans: zero zero at maximum height 9. Ans: 0t v x 8t v y 6t v v x vy 0t 0. Ans: Provide required centripetal force. Ans: 40 N 0 a 0 6 ma 0 40 N. Ans: 5.5 J KE tota 7 0 mv J. Ans: 6.66 cm v v a 0 40a v 4 v a(0 x)..v 0 60 x 6.66 cm v ( x) 4. Ans: Both momentum and kinetic energy are conserved 5. Ans: kg m L L L Ι ω Ι ω L x 5 0 (x 0 ) 4 x Ans: î ĵ kˆ L r p ( î ĵ kˆ ) ( î ĵ kˆ ) î ĵ kˆ 6. Ans: ω ω 0 αt
2 7. Ans: ω ω α 0 t rad s L L mvr mv m r r L L rad s 8. Ans: Gravitational potential energy gravitation potential mass of the body 9. Ans: 0. Ans:. Ans: mgr GMm GMm GMm R R R GM m mgr.r. R R mv Total energy KE mv r r p V p V pv T 4 T. r. r r r r r. Ans: Bulk modulus. Ans:.6 cm 4. Ans: J 4 T r. r 4 hρg h wρ wg h ρ h w ρ w m.6 cm kx J r ( ) k 0 k 0 6 k Ans: m / T / ( ) 4 k Additional work 4 J kt / v T m m 6. Ans: 50 K, 00 K T η T 0. T T ( T 50) 0.4 T T T T T T 50 K T 00 K 7. Ans: R 8. Ans: 5 6 degrees R per mole per degree R 9. Ans: s Diatomic 5 degrees of freedom d y y d y ω y ω T s ω 0. Ans: The effective spring constant K of springs in parallel is given by... K K K. Ans: 50 J TE mω A 00 KE mω ( A x )
3 mω A A mω 50 J A 9. Ans: 0 V qe, E q 500 N C V Ed V. Ans: k 40. Ans: 6 V Equivalent circuit is Combined form A sinkx cos kct or antinode sinkx kx x k Ω Ω 4 4 V 6 V Ω Ω 4 V. Ans: 4 and 4 N and A 4. Ans: In a stretched string, the first overtone is the same as the second harmonic 5. Ans: 44.7 µc σ σ qe e. E ε 0 ε0 σ ε Aσ µc 6. Ans: QV 7. Ans: Product of the charges 8. Ans: N 0 6ε q q A A A N N N 4. Ans: 5 Ω (Assuming 0 V instead of 0 Ω) 0R 500 R 4. Ans: 4. Ans: V R 5 Ω 5E E R 5r r R 5 R 5r 5R r 4r 4R R r R r P VΙ, 44. Ans: 6 kj Ι V P Loss in transmission Ι r P.r V V mv kq q r r v T k q q. m r r r v kq q 0 4 4ε0 6ε q q q q Ι Rt J 6 kj 45. Ans: Introducing a medium of higher permeability. 46. Ans: 4 Am T T Ι mb 4 Ι mb
4 Ι m 4. BT ( 0.75) 4 Am C sin 58. Ans: or rear view concave mirror 47. Ans: Sodium paramagnetic (contains unpaired electron in s orbital) 48. Ans: mυ r KE mυ r 49. Ans: MB MB cos0 MB cos80 MB 50. Ans: Lenz s law 5. Ans: The current and voltage are in phase 59. Ans: The wavelength of the source is increased β Dλ d 60. Ans: sin 5 sinθ λ b θ sin (0.4) sin Ans: V 5. Ans: 55 φ r B B 0.0 dφ BAω cosωt E max BAω volt 6. Ans: λ mp me h p λe λp mp me h me m Current step up Voltage step down n s 6 n p Ans:. 0 J m U ε 0 E Average U J m 55. Ans: The energy contribution of both electric and magnetic fields are equal. 56. Ans: Refraction 57. Ans: sin tan 60 n sinc n 6. Ans: 40 minutes 6. Ans: or 50% decay half life or % decay time is half life Interval half life 40 minutes 5 9 R A / / RTe 5 RAl 7 A 5 A Ans: more than twice its initial value KE hυ φ KE hυ φ more than double 65. Ans: whole of the positive charge is concentrated at the centre of atom. 66. Ans:.5 and MeV
5 67. Ans: 0, 0, 68. Ans: 45 R β. R C B 69. Ans: a heavily doped p n junction with forward bias 70. Ans: ax transmission (all others are broadcasting) 7. Ans: 8 Rh Rh 7. Ans: to 0 MHz 7. Ans: Ans: 9.7 o A 8Rh 4 hc E λ J No. of photons υ R H n n λ cm RH o A 75. Ans: J Ans: Triclinic 80. Ans: atm P 8. Ans: 80 M P S K Cr O 7 belongs to triclinic system nrt atm V 4 5 M 80 g mol 8. Ans: hydroformylation Hydroformylation of olefins yields aldehydes 8. Ans: (i), (iii), (iv) No. of electrons in (i) NH 0 (ii) CH 8 (iii) (iv) 84. Ans: K SO 4 NH 0 NH 4 0 KE hυ hυ 0. ev J 76. Ans: to separate one mole of solid NaCl into one mole of Na (g) and one mole of Cl (g) to infinite distance 85. Ans: The substance which give lilac colour in flame test is potassium K SO 4 BaCl BaSO 4 KCl white ppt NH 4 NaCl (s) lattice H Na (g) Cl (g) 77. Ans: He < Li < C < O Bond order of C, Li, and He 0.5 Stability is of the order He < Li < C < O 78. Ans: S 4 Tetrahedral O.5 The central atom N in hybridised 86. Ans: They are all paramagnetic in nature NH 4 is sp Interhalogen compounds are diamagnetic in nature 87. Ans: The two oxygen-oxygen bond lengths in O are different The two oxygen-oxygen bond lengths in the ozone molecule are identical (8 pm) S 4 see saw
6 88. Ans: MnO 89. Ans: 5,, 90. Ans: 4.90 BM 4 > CrO7 > VO The correct decreasing order is due to the increasing stability of the lower species to which they are reduced 7 Mn O4 Mn 5e s 7 6 Mn O4 Mn O 4 e 7 4 Mn O4 Mn O e s M (Z 4) [Ar] d 4 4s 0 µ n (n ) BM 9. Ans: 00 J K 4 (4 ) 4.9 BM A (l ) A (g) S 58 J K mol or A (g) A (l ) 00 S 58 J K Ans: J K Λ α c Λ0 400 K a Cα Ans:.6 0 No. of moles of C nrt Partial pressure of C V K p p C.6 0 atm 94. Ans: Ans: 400 [OH ] 0 mol / L [X ] 5 0 mol / L K SP mol L 96. Ans: 5 D.O concentration 5 mg / kg water ppm Ans: decreases by 60 mv Cu e Cu 0.06 Eel E o el log[cu ] Eel Eel 98. Ans: E > E > E [Cu ] 0.0 log [Cu ] V E E 0.0 E E E E 0.0 E > E > E 99. Ans: 0.0 mol L min d[a] 0.0 mol L min d[c] mol L min 00. Ans: rate k [NH4 ] [NO ] rom expt and rate [ NO ] rom expt and rate [ NH4 ] rate k [NH4 ][NO ] 0. Ans: reduction of gold (III) chloride with formalin AuCl HCHO H O Au(sol) HCOOH 6HCl reduction 0. Ans: Diaminetetraaquacobalt(III) chloride According to IUPAC rules 0. Ans: methanal and -pentanone (i) O CH CH C CH (ii) Zn / HO CH CH O HCHO CH CH C CH CH W W 400g
7 04. Ans: trans--butene is formed CH C CCH Na liq NH 05. Ans: RI > RBr > RCl > R H H C Order of reactivity of alkyl halides : RI > RBr > RCl > R 06. Ans:,-dimethylpropanal CH CH C CHO C C CH H CH CH CH CHO OH. Ans: IV < I < II < III CH CHCHCHO Presence of electron withdrawing group in benzoic acid increases the acid strength while presence of electron donating group decreases the acid strength. Ans: -methylbenzaldehyde CHO CH [O ] COOH COOH CH (,-Dimethylpropanal) Since it does not contain αh, it does not undergo aldol condensation 07. Ans: mixture of -methylpentane and -methylpentane 08. Ans: 4 n-hexane undergoes isomerisation when heated with AlCl / HCl(g) to a give a mixture of -methylpentane and -methyl pentane The isomers are ClCH CH CH CH, CH CH CCl CH CH, CH CH CH CHCl CH and CH CH CH CH CH Cl CH 09. Ans: carbonyl chloride CHCl O h ν 0. Ans: Diphenyl methanol COCl HCl Alcohols of the type CH CHOHR (where RH or alkyl or aryl group) undergo iodoform reaction. Diphenyl methanol (C 6H 5CHOHC 6H 5) does not undergo iodoform reaction 4. Ans: Carbylamine reaction C 6H 5NH CHCl KOH C 6H 5NC KCl H O 5. Ans: (iv) > (ii) > (iii) > (i) Aliphatic amines are more basic than and amines. Aryl amines are less basic than NH 6. Ans: ethanamine Gabriel s phthalimide synthesis is used for the preparation of aliphatic amines 7. Ans: β-d-ribose Sugar present in RNA is β-d-ribose 8. Ans: initial product obtained in the condensation of phenol and formaldehyde in the presence of acid catalyst Phenol and formaldehyde undergo polymerization to form novolac and bakelite 9. Ans: glycogen Glycogen is known as animal starch 0. Ans: utilization of existing knowledge base for reducing the chemical hazards along with developmental activities It is the definition of green chemistry. Ans: two moles of ethanal dil.naoh CH CHO
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