21 AIEEE 2003 PHYSICS & CHEMISTRY SOLUTIONS 1. Force is ^r to displacement Þ the work done is zero 2. Since there is no deviation in the path of the charged particle, so net force due to presence of electric and magnetic field must be zero 3 2 4 10 Wb/m 10 10 VE ÞvB=qEÞB = = = 3. T µ l 2 21 21 ; I L ll TT = µ ( ) 4 2 TT 4 LL 2 LL II 21 22 2 2 2221 21 = = Þ úû ù êë é Þ = = = 21 TT 2T T 11 2 2 Þ = Þ = 4. t = (H) tan600 = W. 3 7. Mass = 5kg 9.8 49 = . When lift is moving downward, apparent weight = 5(9.8 -5) = 5´4.8 = 24 N 8. Potential µ R R µ length Þ Potential difference µ l 11. T 16 C 25 10 10 40 T 0 6 5 Þ D = ´ ´ D = - - 13. [ ]2 0 0 2 0 0 0 0 C 1 C 1 C 1 = úû ù êë é m e = Þ m e = Þ m e [C] = LT-1 or [C]2 = L2T-2 14. m R2 or M t R2 21 I = µ µ For disc X, 2 ( 2) 2 x r t.(R) 21 (m)(R) 21 I = = p For disc Y, [ (4R)2.t /4][4R]2 21 = p y I 3 y x yx I 64 I (4) 1 II Þ = Þ = 15. 3 21 2 22 3 1 RR TT R T ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ µ Þ 3/2 3/2 21 21 41 RR TT ÷øö çèæ = ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ Þ (4) 8 TT 3/2 1Þ 2 = = T 8 T 8 5 40 40 hours 2 1 Þ = ´ = ´ = = < x 10m 10m 10m/sec V22 16. w K.E. Kinetic energy L Angular frequency 1 Angular momentumµ µ Þ = r 4L 4 L KE w wK.E LL 2 2 2 1 1 21 = Þ = ´ ÷ ÷ø ö ç çè æ = 17. RMIVUXGE¾l¾D¾ec¾re¾asi¾n¾g® R ® Radio waves ; M ®Micro waves; I ® Infra red rays; V ® Visible rays; U® Ultraviolet rays; X ® X rays; G ® g rays; C ® Cosmic rays Þ g rays has least wavelength 18. Applying the principle of conservation of linear momentum 238 4u (4)(u)=(v) (238)Þv = 19. Distance between the surface of the spherical bodies = 12R -R -2R = 9R Force µ Mass, Acceleration µ Mass, Distance µ Acceleration 2 1 21 21 S 5S 51 SS 51 SM M aa Þ = = Þ = Þ = S1 + S2 = 9 Þ 6S1 = 9 Þ S1 = 69 = 1.5, S2 = 1.5 ´ 5 = 7.5 Note: Maximum distance will be travelled by smaller bodies due to the greater acceleration caused by the same gravitational force 21. Energy = Work done by force (F) 2 62500m m.(50) (F)(6) F 21 2 ´ Þ = Þ = .m(100) (F)(S) 21 For v = 100km/hr 2 = S 2 62500m m(100) 21 2 ÷øö çèæ ´ Þ = 24m 2500 2 100 100 6 2 S = ´ Þ = ´ ´ ´ 22. From, the question if the horizontal distance is none other than the horizontal range on the level of the roof of building Range = 8.66 2 10 10 10 3 g (10) sin(2 30) g u2 sin2 2 = ´ q = ´ = ´ ´ 24. [momentum] = [M][L][T-1]=[MLT-1] (Planck’s Constant) = 2 1 1 1 2 ML T T [M][LT ] uE - - - = = 25. According to triangle law of forces, the resultant force is zero. In presence of zero external force, there is no change in velocity23 26. According to Gauss’s Law (E.dA) q /q ( )0 0 0 2 1 ò = e Þ = e f - f [ sincef = òE.dA] 27. 10N ¯ mg f =mN f = mgÞmN = WÞm.10 = W Þ0.2´10 = W \W = 2N 28. 10 6 a = mg = [ using v = u + at] 0.06 10 10 6 10 g 6 = ´ = ´ Þm = 31. Since the displacement for both block and rope is same so, the acceleration must be same for both M >P > >P T T T = Ma ........ (i) m M P p (m M)a a + Þ = + Þ = m M PM T M.a + = = 33. Elastic energy = F x 21 ´ ´ F = 200 N, x = 1 mm = 10-3 m 200 1 10 0.1J 21 \E = ´ ´ ´ -3 = 34. Escape velocity of a body is independent of the angle of projection. Hence, changing the angle of projection is not going to effect the magnitude of escape velocity 35. KM T = 2p ....... (i) KM m 2 35T = p + ......... (ii) Dividing equation (ii) by equation (i), MM m 35 = + . Squaring both the sides 9 16 1 9 25 Mm Mm 1 MM m 9 25 = + = + Þ = - =24 36. External amount of work must be done in order to flow heat from lower temperature to higher temperature. This is according to second law of thermodynamics. 37. V A m 2I k max = w = w =mk mk Þw2 = Þw = 21 21 kk K or = wÞwµ w V max V max A B = 12 21 2 2 1 1 kk AA (A ) mk (A ) mk = Þ ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ Þ 38. ÷ ÷ø ö ç çè æ = p = p + gl log 21 ; logT log(2 ) gl T 2 log(g) 21 log(l) 21 ÞlogT = log(2p)+ - Differentiating 100 ll 21 100 TT 0 ll 21 0 TT D = + ´ D - Þ D ´ = ´ D ´ 21 10.5 10% 2= 1 ´ = » Note: In this method, the % error obtained is an approximate value on the higher side. Exact value is less than the obtained one. 39. ÷øç ö è= - æ - + p3 y 10 4 sin 600t 2x . Comparing it with standard equation y = Asin (vt - kx); v = 600m/s 40. L 0.1H 0.05 2 ( 2) 8 (L) dt dI e = -L Þ = - - Þ = 41. 2 Q q = *44. ÷øç ö èæ = Þ ÷øö çè= æ 1250 5000 ln 51 K NN ln f1 K ln2 0.4 ln2 52 ln(4) 51 = = 45. No. of a particles emitted = 8, No. of b- particles emitted = 4, No.of b+ particles emitted = 2 z = 92 -2 ´ 8 + 4 -2 = 78 48. 1.5A 23 I = = 3W 3W 3W 3V 3V 3W 3V6W 2W 50. x = 4(cospt + sin pt) t ] sin t] 2 [sin 4 p + ÷øö çè= æ p - p ú ú ú ú ûù ê ê ê ê ëé ÷ ÷ ÷ ÷ øö ç ç ç ç èæ p - p + p ÷ ÷ ÷ ÷ øö ç ç ç ç èæ p - p - p = ´ 2 t 2 t cos 2 t 2 t 4 2 sin25 úû ù êë é ÷øö çè= p æ- p + pt 4 .cos 4 8 sin úûù êëé p - p = úûù êë= ép - p 4 4 2 cos t 4 .cos t 2 8 Comparing it with standard equation X = A cos (wt -Kx) ÞA =4 2 51. Potential due to spherical shell, 4 R q v 0 1 pe = . Potential difference due to charge at the centre 4 R q 4 R 2Q ; V V V 4 r 2Q V 0 0 1 2 0 2 pe + pe = + = pe = 52. Work done 32 10 J 2 100 10(8 10 ) c q 21 32 8 2 18 2 - - - = ´ ´ ´ = = ´ 53. 2 y 2 x 3 t dt dy 3 t , V dt dx V = = a = = b2 2 2 x y v V V 3t 2 2 = + = a +b r 54. 3 21 23 1 TT PP T P ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ µ Comparing it with standard eq. 23 CCvp = g = 56. 627 273 (627 273) (273 27) + h = + - + 32 900 600 900900 300 = - = = work = (h) ´ Heat = 3 10 4.2J 8.4 10 J 32 ´ ´ 6 ´ = ´ 6 57. Required work done 2 3 2 2 4 1 22 5 10 [10 5 ] 10 21 K(x x ) 2= 1 - = ´ ´ - ´ - 5 75 10 10 18.75 2= 1 ´ ´ ´ 3 ´ -4 = 58. ; l 1m T 2l 1 n = m = T = 10 Kg wt. = 10 ´ 10 = 100 N m = 9.8 g/m = 9.8 ´10-3 kg/m n = 50 hz26 66. Power = F. V = Þ ÷øö çè= æ dt dV m . v . dt dV F m constant = C k vdv kdt mC dt dV Þ = = Þ = kt c 2 V vdv k dt 2 Þò = ò Þ = + Þv µ(t)1/2 c.t1/2 dt ds = 1/2 t3/2 32 Þòds=ò(c.t )dt Þ S = C. 3/2 3/2 s t 3/2 t.c ÞS= Þ µ 67. Thrust = Mass ´ Acceleration = 3.5 ´ 104 ´ 10 = 3.5 ´ 105 N 69. The force body diagram 2 1 3 2 2 1 3 1 bq q . 4 0 1 ; F aq q . 4 0 1 F pe = pe = úû ù êë é q+ pe = q+ = 21 21 3 X 1 2 bq sin aq 4 0 q F sin F F ÷øö çèÞ µæ q+ 22 23 bq sin aq Fx 70. ( ) 1000 220 or R R V p 2 2 = = Power consumed = 1000 220 220 110 110 R V2 ´ ´= ´ = 250 watt 73. According to Image formula 1 360 1 3 360 n - q - Þ = q = 90 360 4 360 = 4 = Þq= q Þ 74. ( ) ( ) n 1 dt dH n 2 1 µ q - q = Dq Þ = 75. L 2l or ( r l) ( r2 )(2l) 2 2 1 = p = p 2 2 rl ; R 2 r r p Þ = =r ( ) 4 R ( )r()4l ( )r/2 2l R () new 2 2 = ´ p= r p = r \DR = 4R - R = 3R 100 300% R3R % RR D = ´ = < < ˙ ˙ ˙ q q a -q3 (-q1) F2 F1 a +q227 76. Liquid hydrogen and liquid oxygen are used as excellent fuel for rockets. H2(l) has low mass and high enthalpy of combustion whereas oxygen is a strong supporter of combustion 77. CH3 NH2 + CHCl3 + 3KOH® CH3 N=C+ 3KCl + 3H2O 78. Nylon is a polyamide polyer 79. More is the no . of + I groups attached to N atom greater is the basic character. 80. C6H5I will not respond to silver nitrate test because C-I bond has a partial double bond character. 81. For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path. 82. CH2 = CH2 (g) + H2(g) ® CH3 -CH3 DH = 1(C = C) + 4(C -H) +1(H-H) -1(C- C) - 6(C-H) = 1(C = C) +1(H-H) -1(C -C) - 2(C-H) = 615+ 435- 347 - 2´ 414 = 1050 -1175 = -125kJ. 83. Th X Th 230Th 90 234 92 234 91 234 90 ¾¾-b® ¾¾-b® ¾¾-a® 84. t1/2 = 3 hrs. Initial mass (C0) = 256 g 4g 64 256 (2) 256 2C C n 6 0 n \ = = = = 86. z1 W µ 0.85Å 57 1.06 71 zz 2 1 2 2 21 = Þ W = W = Þ WW 88. Co(NH3)5 Cl3[Co(NH3)5Cl]+2 + 2Cl-\ Structure is [Co(NH3)5 Cl] Cl2. 89. 4 (+1) + x + (-1) ´ 4 = 0 Þ 4 + x - 4 = 0 x = 0 91. An acidic solution cannot have a pH > 7. 92. In neopentane all the H atoms are same (10). CH3 -C -CH3 CH3 CH3 94. PH3 + 4Cl2 ® PCl5 + 3HCI 95. Fe+2 = 3d6 . 4s0 96. 4HCI + O2 ® 2Cl2 + 2H2O Cloud of white fumes 99. The properties of elements change with a change in atomic number. 100. Ammonia can dissolve ppt. of Agcl only due to formation of complex as given below: AgCl + 2NH3 ® [Ag(NH3)2] Cl28 101. Glass is a transparent or translucent super cooled liquid. 102. For s-electron, l = 0 \ angular momentum = zero 103. Number of formulas in cube shaped crystals = 6.02 1023 58.5 1.0 ´ ´ since in NaCl type of structure 4 formula units form a cell \ unit cells = 58.5 41.0 6.02 1023 ´´ ´ = 2.57 ´ 1021 unit cells. 104. H -C -O || O Q « H -C = O || O 105. As adsorption is an exothermic process. \ Rise in temperature will decrease adsorption. 106. The equilibrium constant is related to the standard emf of cell by the expression 0.059 2 0.295 0.059 n logK E0cell = ´ = ´ 10 or K 1 10 10 59 590 logK = = = ´ - 107. For spontaneous reaction, dS > 0 an DG and dG should be negative i.e. < 0 108. [A] = 1.0 ´10-5, [B] = [1.0 ´ 10-5] Ksp = [2.B]2 [A] = [2´10-5]2 [1.0´10-5] = 4´10-15 109. No. of moles of boron = 2 10.821.6 = for BCl3 \ 1 mole of Boron = 3 mole of Cl \ 2 mole of Boron = 6 mole of Cl H2 + Cl2 ®2HCI Þ 3 moles of Hydrogen is required = 3 ´ 22.4 = 67.2 Litre 110. [ ] [ ] [ ] [ ] 3 10 mol/L 4.8 10 1.2 10 N O NO K 3 2 2 2 2 42 2 C - - - = ´ ´ = = ´ 111. Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles. \ High pressure will be required. 113. 2 2 4 HgI +KI®K HgI (insoluble) (soluble) On heating HgI2 decomposes as HgI2 Hg + I2 117. No. of moles of silver = moles 10 1 96500 9650 = \ Mass of silver deposited = 108 10.8g 10 1 ´ =29 118. [ ] [ 2]2 0cell cell Zn Cu log n0.059 E E ++ = + log[0.1] 1.10 0.0295 1.07V 20.059 = 1.10 + = - = 120. f-block elements show a regular decrease in atomic size due to lanthanide/actinide contraction. 122. LiAlH4 can reduce COOH group and not the double bond CH2 CH CH2OH CH CH COOH LAH 2 = -= - ¾¾¾® 123. According to kinetic theory the gas molecules travel in a straight line path but show haphazard motion due to collisions. 125. A chiral object or structure has four different groups attached to the carbocation. 126. Cr O - +OH- ®2CrO2- +H+ 4 22 7 . The above equilibrium shifts to L.H.S. on addition of acid. 127. It is because mercury exists as liquid at room temperature. 128. Gypsum is CaSO4.2H2O 129. D Glu cos e 6 O 12 H 6 nC H 6 10 5 2(C H O )n nH O - + ¾¾® + 130. AgNO3 ® Ag + NO2 + 2 O 21 131. CH CH OH H CH CH O H 3 2 step 1 3 2 + + ¾¾¾® - - HÅ Protona ted alcohol 132. The solubility is governed by solution DH i.e. solution DH = lattice Hydration DH -DH . Due to increase in size the magnitude of hydration energy decresaes and hence the solubility. 133. The rain water after thunderstorm contains dissolved acid and therefore the pH is less than rain water without thunderstorm. 135. Ba(OH) 2HCI BaCl 2H O 2 2 2 + ® + Applying Molarity equation, 2 (HCI) 1 (Ba(OH)2 ) 1 1 2 2 M V M V = or 25´M1 = 0.07 100.7 2 25 0.1 35 M 2 0.1 35 1 = = ´´ \ = ´ 137. Rate1 = k [A]n [B]m; Rate2 = k[2A]n [½B]m n m n m n m n m n m 12 [2] [½] 2 .2 2 k[A] [B] k[2A] [½B] Rate \Rate = = = - = - 138. CH3CH2N =® C + H2O ¾¾® H+ CH3CH2NH2 + HCOOH. Therefore it gives only one mono chloroalkane. 140. On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the b.p. of water increases. 141. r = k[O2] [NO]2. When the volume is reduced to 1/2, The conc. will double. \ New rate = k[2O2] [2NO]2 = 8 k [O2] [NO]2. The new rate increases by eight times. 142. Magnesium provides cathodic protection and prevent rusting or corrosion.30 143. Both NO2 and O3 have angular shape and hence will have net dipole moment. 144. N3-, F-and Na+ contain 10 electrons each. 145. Permanent hardness of water is due to chlorides and sulphates of calciumand magnesium. 146. In H2S, due to low electronegativity of sulphur the L.P. -L.P. repulsion is more than B.P. -B.P. repulsion and hence the bond angle is 920. 147. Both XeF2 and and CO2 have a linear structure. 148. Electronic configuration of Cr is 3d 4s So due to half filled orbital I.P. is high of Cr. 149. The lines falling in the visible region comprise Balmer series. Hence the third line would be n1 = 2, n2 = 5 i.e. 5 ® 2. 150. 10 m 60 10 10 6.6 10 mv h 33 3 34 - - - = ´ ´ l = = ´31 AIEEE 2003 MATHEMATICS SOLUTION 1. e dx x3x e dx x3 or x e F(x) dx d 3 sinx3 41 41 32 sinx sinx = ò =ò Let x3 = t, 3x2dx = dt when x = 1, t = 1 & x -4, t = 64 =ò =ò = - 64 1 64 1 sin t dt F(t) dt F(64) F(1) t e F(t) K = 64. 2. n = 9 then median term = th th 5 2 9 1 ÷ = øç ö èæ + term. Last four observations are increased by 2. The median is 5th observation which is remaining unchanged. \ There will be no change in median. 3. þ ý ü î í ì - + + ïþ ïý ü ïî ïí ì ÷øö çèæ + ÷øö çèæ + ÷øö çèæ + ÷øö çèæ ®¥ ®¥ 43 43 n 4 4 4 4 4 n nn ........ n2 n1 n1 Lim nn ............ n3 n2 n1 Lim ò = úû ù êë é - = 10 10 5 4 51 5 x (x) dx 0 4. Fundamental theorem (fact) 1 2 1 t2 t = -t - 5. 1 2 1 2 r - r = C C for intersection Þ r -3 < 5 Þ r < 8 ......... (1) and r1 + r2 > C1C2, r+3 > 5 Þ r = 2 ..........(2) From (1) and (2), 2 < r < 8. 6. y2 = 4a(x -h), 2yy1 = 4a Þ yy1 = 2a Þ y12 + yy1 = 0 Degree = 1, order = 2, 7. 25 1 81 y 144 x2 2 - = 45 12 15 144 81 , e 1 25 81 , b 25 144 a = = = + = = Foci = (3, 0), focus of ellipse = (3, 0) Þ 43 e = 7 16 9 b2 16 1 ÷ = øö çè= æ - 8. = ò - t0 F(t) f(t y)g(y)dy32 =ò- =ò- t0 t0 etyydy et eyydy [ ] [ ]t0 t t y y 0 = et - ye-y - e-y = - e ye- + e- [ ] e (1 t) e t 1 e e te e 0 1 e t t t t t t t + - = úû ù êë = - - + - - - = é + - 9. f(x) = log(x + x2 +1) f(-x)= - log(x + x2 +1) f(-x)= - f (x), i.e., f(x) is an odd function. 10. ac , ab ax2 + bx + c = 0, a + b = - ab = As for given condition, 2 2 1 1 b + a a + b = 22 22 2 2 2 2 ac a2c ab ab - - = a ba +b = - a + b On simplification 2a2c = ab2 + bc2 bc , & ab , ba cb ac b2a Þ = + \ are in H.P. 11. 2 2 3 0 C C 2C 1 4c c1 3b b1 2a a = ® - 3 3 2 2 2 1 0 R R R , R R R 1 2c c1 b b1 0 a = ® - ® - 0 0 2c b c b0 b b a 1 0 a = - -- b (c -b) -(b -a) (2c -b) = 0 On simplification, c1 a1 b2 = + \ a, b, c are in Harmonic Progression. 12. Co-ordinates of A = (acosa, a sin a) Equation of OB, x 4 y tan ÷øç ö è= æ p + a 4p a C Y B A O X33 ÷øç ö è^ \ = - æ p + 2 4 CA r to OB slope of CA cot Equation of CA ÷(- a) øç ö è- a=- æp+2 x a cos 4 y a sin cot . y(sina + cosa) + x(cosa - sina) = a 13. Equation of bisector of both pair of straight lines, px2 + 2xy -py2 = 0 ... (1) qx2 + 2xy -qy2 = 0 .... (2) From (1) and (2). pq 1 1q 2p 2 1q Þ = - -= - - = . 14. a cos t bsin t 3x 1 3 cos t bsin t 1 x = + + Þ + = - a sin t bsin t 3y 3 a sin t b cos t y = - Þ - = Squaring & adding, (3x -1)2 + (3y)2 = a2 + b2 15. K x log(3 x) log(3 x) Limx 0 + - - = ® (by L’Hospital rule) K 32 K 1 3 x 1 3 x 1 Limx 0 - = \ = - - + ® 16. a = r ´p; a = rpsinq r r r r H = rp cosqësin(900 + q)= cosqû Q r G = rp sin q ......... (1) H = rp cos q .......... (2) x = rp sin(q+a) .......... (3) From (1), (2) & (3), x = a cosa +Hsina r r 17. R2 = P2 + Q2 + 2PQ cos q ......... (1) 4R2 = P2 + 4Q2 + 4PQ cos q ......... (2) 4R2 = P2 + Q2 -2PQ cosq ......... (3) On (1) + (2), 5R2 = 2P2 + 2Q2 ......... (4) On (3)´2+(2), 12R2 = 3P2 + 6Q2 ......... (5) 2P2 + 2Q2 -5R2 = 0 ......... (6) 3P2 + 6Q2 -12R2 = 0 ......... (7) 12 6 R 24 15 Q 24 30 P2 2 2 - = - = - + or P :Q :R 2 : 3: 2 6 R 9 Q 6 P 2 2 2 2 2 2 = = =34 18. ,n 8 21 p . 21 q npq 2 np 4 Þ = = = þ ý ü = = ( ) 32 1 21 21 8. 21 21 p X 1 C 8 5 7 1 8 = = = ÷øö çèæ ÷øç ö è= = æ 19. f(x)=xnÞf(1)=1 f¢(x)=nxn-1Þf¢(1) =n f¢(x)=n(n-1)xn-2Þf¢(1)=n(n-1) ......................... f n (x) = n!Þf n (1)= n! = n!n! ......... ( 1) 3! n(n 1)(n 2) 2! n(n 1) 1! n 1- + - - - - + + - n C C C C .......... ( 1) C 0 n n n 3 n 2 n 1 n 0 =n - + - + + - = 20. Since n r is perpendicular u and v, n u vr r r r r = ´ k ˆ 2k ˆ 2 2 21 1 01 1 0i j k nˆ = - = - ´ - = 3 3 ) k ( . ) k 3 j 2 i ( nˆ . = - = - + + = w r 21. F F F 7i 2j 4k 1 2 + + = + - r r r d = P.V of B- P.V of A = 4i + 2j- 2k r r r W = F.d = 28+ 4 + 8 = 40 unit r r 22. AB D C 3i 4 j r r + 5i 2 j 2kr r r + - P.V of 2 (3 5)i (0 2) j (4 4)k AD = + + - + + = 4i - j+ 4k or AD = 16+16+1 = 33 23. y = -x+1 (-1, 2) y = 3+x y = 3-x (1, 0) (2, 1)y = x-1 (0, 3)35 ò{ } ò{ } ò { } - = + - - + + - - - + + - - - - 01 10 21 A (3 x) ( x 1) dx (3 x) ( x 1) dx (3 x) ( x 1) dx ò ò ò - = + + + - 01 21 10 (2 2x)dx 2dx (4 2x)dx [ ] [ ] [ ]21 1 2 0 01 =2x-x2 +2x +4x-x - = 0 -(-2+1) + (2 -0) + (8 -4) -(4 -1) = 1 + 2 + 4 -3 = 4 sq. units 24. Shortest distance = perpendicular distance = 26 144 9 16 2 12 4 1 3 3 327 = + + - ´ + ´ + ´ - \ Shortest distance = 26 - 4 + 1+ 15 + 9 = 26 -13 = 13 [Q26 - r] 25. c' 3 d' 1y a' x b' ; c 3 d 1y a x - b = = - - = = - For perpendicular aa¢ +1+ cc¢ = 0 26. 0 l m nl m n x x y y z z 2 2 21 1 1 2 1 2 1 2 1 = - - - 0 k 2 1 12 1 k k0 0 1 0 k 2 11 1 k1 1 1 = + + -- - = Þ - - k2 + 3k2 = 0 Þ k(k + 3) = 0 or k = 0 or -3 27. = ò = ò + - + - ba ba I xf(x)dx (a b x) f (a b x)dx = + ò + - -ò + - ba ba (a b) f(a b x)dx xf (a b x)dx = + ò + - -ò ba ba (a b) f(a b x)dx xf (x)dx = + òba 2I (a b) f(x)dx = + ò = ( + ò + - ba ba f(a b x)dx 2 a b) f(x)dx; I 2 (a b) I 28. Portion OA, OB corresponds to motion with acceleration ‘f’ and retardation ‘r’ respectively. Area of OAB S and OB t. Let OL t , 1 D = = =36 t 2S t.v; v 21 OB.AL 21 LB t and AL v, S 2 = = = = = Also, tr2s tf2s ; t t t tr2s rv , t tv and r tf2s fv , t tv f 2 1 2 2 1 1 = = = = = = = + = + ÷øö çèæ + = Þ ÷øö çè= æ + r1 f1 t 2s t2s r1 f1 t 29. (u cos ) t g2h R = u = q ´ g2h cos 1 t q = ....... (1) Now, gt2 21 h = (-usinq)t + Substituting ‘t’ from (1), úû ù êë é q + q= - q g cos2 2h g 21 g2h cos usin h = - tan q + h sec 2 q g2h h u tan h tan h g2h h = -u q + 2 q + hg 2 tan 0; tan u hg 2 tan2 q- u q = \ q = 30. Applying R1®R1 + R2 + R3 As, 1+ wn + w2n = 0; \D = 0 31. 2R a n ; sin 2r a n tan ÷ = øç ö è÷ = æ p øç ö èæ p ÷øç ö èæ p = + Þ úûù êë+ = é p + p 2n .cot 2a r R n cosec n cot 2a r R 32. Taking co-ordinates as ;(x,y)&(xr, yr) ry , rx ÷øö çèæ . Above coordinates satisfy the relation y = mx Therefore lies on the straight line. 33. zw = 1 ..... (1) As, i z therefore 2 z Arg = w ÷ = p øç ö èæ w1 z = w \ ....... (2) From (1) & (2), 0;z z 0 z z z 1 and = w + w = w + w = w = . . ; z i i z z z 2 w w w = - w = - w w= - w = -37 34. z2 + az + b = 0; z1 + z2 = -a & z1z2 = b 0, z, z2 form an equilateral D \ 02 + z12 + z22 = 0.z1 + z1.z2 + z2.0 (for equation D , z12 + z22 + z32 = z1z2 + z2z3 + z3z1) z12 + z22 = z1z2 or (z1 + z2)2 = 3z1z2 \ a2 = 3b. 35. 0 dx dy (1 y2 ) (x etan 1 y ) + + - = - (1 y ) e (1 y ) x dy dx x e dy dx (1 y ) 2tan y 2 2 tan y 1 1 + = + + + = Þ + - - I.F. = e e tan 1 y dy (1 y ) 1 2 - ò + = tan ydy tan y tan y 1 1 1 e 1 y e x(e ) - - - ò + = C 2xe e k 2 e x(e ) tan y 2tan y 2tan y tan y 1 1 1 1 = + \ = + - - - - 36. Let f(x) = ex \ò =ò - 10 10 f(x)g(x)dx ex (x2 ex )dx =ò -ò 10 10 x2exdx e2xdx [ ] [ ] [ ]10 1 2x 0 1 x x 0 2 x e 21 = x e -2 xe -e - [ ] 23 2e 2e e 1 e 21 2e e 2 2- - = + - -úûù êëé = - - 37. pr2 = 154Þr = 7 For centre on solving equation 2x -3y = 5 & 3x -4y = 7 or x = 1, y = 1 centre = (1, -1) Equation of circle, (x -1)2 + (y + 1)2 = 72 x2 + y2 -2x + 2y = 47 38. 2 1 2x , P(C) 41 x , P(B) 3 3x 1 P(A) = + = - = - These are mutually exclusive 1 2 1 2x 1 and0 41 x 1, 0 3 3x 1 0 £ + £ £ - £ £ - £ -1 £3x £ 2, - 3 £ x £ 1 and -1£2x£ 138 21 x 21 , 3 x 1, and 32 x 31 - £ £ - £ £ - £ £ Also 1 2 1 2x 41 x 3 1 3x 0 £ + + - + - £ 3 13 x 31 0 £ 13- 3x £ 12Þ1£ 3x £ 13Þ £ £ þ ý ü î í ì £ £ þ ý ü î í ì - - - 3 13 , 21 ,1, 32 x min 31 , 21 , 3, 31 max úûù êë£ £ Þ Îé 21 , 31 x 21 x 31 39. n(S) = 5C2; n(E) = 2C1 + 2C152 C C C n(S)n(E) p(E) 2 5 1 2 1 2 = = + = 40. a 5a 3 2 &2 a 5a 3 1 3a 3 2 2 2 - + a = - + a = - a 5a 3 2 (a 5a 3) (1 3a) 91 2 2 2 2 2 - + = úû ù êë é - + - 9 or 9a 6a 1 (a 5a 3) (1 3a) 2 2 2 = - + - + - 32 = 9a2 - 45a + 27 or 39a =26 or a = 41. r r 1 (x) r! n(n 1)(n 2)..........(n r 1) T = - - - + + For first negative term, n -r + 1 < 0 or 5 32 r > \ r = 7. Therefore, first negative term is T8. 42. 2 r /8 256 r r 256 r 8 r 256 r 256 r 1 T C ( 3) ( 5) C (3) (5) - - + = = Terms will be integral if 8r & 2 256- r both are +ve integer. As 0 £ r £ 256 \r = 0,8,16,24,...........256 For above values of ÷øç ö èæ - 2 256 r r, is also an integer. 43. After t; velocity = f ´t V = f t + (-u)= f 2 t2 + u2 - 2f ut cosa BA r r For max. and min.39 ( ) f u cos V 2f t 2fu cos 0 or t dt d 2 2 BA = - a = = a Therefore, total no. of values of r = 33. 44. Using nCr + nCr-1 = n+1Cr r n r n 1 r 1 n r n r n r 1 n r 1 =n C + C + C + C = C + + C + C + 14-42443 + r 1 n 2 r n 1 r 1 n 1C C C+ + + + + + Þ 45. a b q q=a+b or b=q-a 4h h 43 - 160 h . 40 h 1 160 h 40 h 53 or 1 tan . tan tan tan tan + - = + q a b = q - a h2 -200 h + 6400 = 0, h = 40 or 160 metre Therefore possible height = 40 metre 46. A E B D C P4/3 600 x 900 300 8/3 3 3 8 or x x8 /3 tan 600 = = Area of 3 3 16 3 3 8 4 21 DABD = ´ ´ = \ Area of 3 3 32 3 3 16 DABC=2´ = 47. If 23b 2A c cos 2C a cos2 2 ÷ = øç ö è÷ + æ øç ö èæ a[cos C + 1] + c[cos A + 1] = 3b (a + c) + (a cos C + c cos B) = 3b a + c + b = 3b or a + c = 2b or a, b, c are in A.P. 48. a + b + c = 0Þ(a + b + c).(a + b + c) = 0 r r r r r r r r ra b c 2(a.b b.c c.a) 0 2 2 2 + + + + + = r r r r r r r r r7 2 1 4 9 a.b + b.c + c.a = - - - = - r r r r r r 49. = ò - 10 I x(1 x)n dx - = ò- - = ò - - - 10 n 10 I x(1 x)n dx (1 x 1)(1 x) dx40 =ò- + -ò- 10 10 (1 x)n 1 dx (1 x)n dx n 1 1 n 2 1 (n 1) (1 x) (n 2) (1 x) 10 1 n 1 0 n 2 + - + = úû ù êë é - + - - úû ù êë é - + = - + + n 2 1 n 1 1 I + - + = 50. sinx xcosx sec x .2x Lim (xsinx) dx d sec t dt dx d Lim 2 2 x 0 x0 2 x 0 2 + = ® ® ò (by L’ Hospital rule) 1 1 12 1 cosx xsinx 2sec x Lim 2 2 x 0 = += ´ ÷øö çèæ + ® 51. OA C centre of sphere = (-1, 1, 2) Radius of sphere 1+1+ 4 +19 = 5 Perpendicular distance from centre to the plane 4. 3 12 1 4 4 1 2 4 7 OC d = = + + = = - + + + AC2 = AO2 -OC2 = 52 -42 = 9 Þ AC = 3 52. Vector perpendicular to the face OAB5i j 3k 2 1 31 2 1i j k = OA´OB = = - - Vector perpendicular to the face ABCi 5j 3k 2 1 11 1 2i j k AB AC = - - - -= ´ = - Angle between the faces = Angle between their normals ÷øö çèq = + + = q = - æ 35 19 or cos 35 19 35 35 5 5 9 cos 141 53. 4 9(k) f(x) k9(x) kf(x) lim x a = -- ® (By L’Hospital rule) 4 or k 4. 9' (x) f' (x)9' (x) f' (x) lim k x a = = -- ® 54. 3 lim 2 x ( 2x) . (1 sinx) 2x 4 tan p - - ÷øö çèæ p - ®p Let y; y 0 2 x = p + ® .8 8 y ( 8). 2y 2sin 2y tan ( 2y) . (1 cosy) 2y tan 3 2 lim 3 y 0 lim y 0 - - = - - ÷øö çè- æ- = ® ® 32 1 y /2 siny /2 . 2y 2y tan 32 1 2 lim y 0 = úû ù êë é ÷øö çèæ = ® 55. 2 2 2 2 2 1 2 1 (h - a ) + (k - b ) = (h - a ) + (k - b ) (a b a b ) 0 21 (a a )x (b b )y 21 21 22 21 2 1 2 2 - + - + + - - = (a b a b ) 21 C 21 21 22 22 = + - - 56. 0 c c cb b ba a a c c 1b b 1a a 1 0 c c 1 cb b 1 ba a 1 a 2 32 32 3 222 2 32 32 3 = Þ + = +++ (a -b) (b -c) (c -a) + abc(a -b) (b -c) (c -a) = 0 (abc + 1) [(a -b) (b -c) (c -a)] = 0 As 0 1 c c1 b b1 a a222 ¹ (given condition) \abc = -1 57. x2 - 3 x +2=0 or(x-2)(x-1)=0 x =1,2 or x = ±1,± 2 or \No. of solution = 4 58. f(x) = 2x3 -9ax2 + 12a2x + 1 f¢(x) = 6x2 -18ax + 12a2; f¢(x) = 12x -18a For max. or min. 6x2 -18ax + 12a2 = 0 Þ x2 -3ax + 2a2 = 0 x = a or x = 2a, at x = a max.and at x = 2a min. p2 = q a2 = 2a Þ a = 2 or a = 0 but a > 0, therefore, a = 2.42 59. ÷ ÷ø ö ç çè æ - + = = x1 x1 f(0) 0; f(x) xe R.H.L. 0 eh (0 h) e 2 /h lim h 0 lim 2 /h h 0 + = = ® - ® L.H.L Lim(0 h)e h 0 1 h1 h 0 - = ÷øö çè-æ - ® Therefore, f(x) is continuous R.H.D. 0 h (0 h)e he h1 h1 h1 h1 lim h 0 + - = ÷øö çèæ + - ÷øö çè-æ + ® L.H.D. 1 h (0 h)e he Lim h1 h1 h1 h1 h 0 = - - - ÷øö çèæ + - ÷øö çè-æ - ® Therefore, L.H.D. ¹ R.H.D. f(x) is not differentiable at x = 0. 60. -+ -+ -1 0 1 log (x x) 4 x 3 f(x) 3 2 10 + - - = 4 - x2 ¹ 0;x3 - x > 0;x ¹ ± 4 \D = (-1,0)U(1,¥)-{ 4} D = (-1,0)U(1,2)U(2,¥) . 61. f(x + y) = f(x) + f(y). Let f(a) = ma f(1) = 7; \ m = 7, f(x) = 7x å å = = + = n1 nr 1 2 7n(n 1) f(r) 7 r 62. x2 1 1 dx dy or x1 y = x + = - For max. or min., 0 x 1 x1 1 2 - = Þ = ± 2( ve minima) dxd y x2 dxd y x 2 2 2 2 3 2 + = ÷ ÷ø ö ç çè æ = Þ = Therefore x = 1 63. Let b be the inclination of the plane to the horizontal and u be the velocity of projection of the projectile g(1 sin ) u andR g(1 sin ) u R 2 2 2 1 - b = + b = úû ù êë é + = + = = g u R R2 R1 R1 or u2g R1 R1 2 1 2 2 1 2 Q Therefore, R1, R, R2 are in H.P.43 64. Sx = 170, Sx2 = 2830 increase in Sx =10 , then Sx'= 170+10 = 180 Increase in Sx2 =900 - 400 = 500 then Sx'2 =2830+ 500 = 3330 Variance = 2 2 x' n1 x' n1 ÷øö çèS - æ S 180 222 144 78. 15 1 3330 15 1 2 ÷ = - = øö çè= ´ - æ ´ 65. As for given question two cases are possible. (i) Selecting 4 out of first five question and 6 out of remaining 8 question = 5C4´8C6 = 140 choices. (ii) Selecting 5 out of first five question and 5 out of remaining 8 questions = 5C5´8C5 = 56 choices. Therefore, total number of choices = 140 + 56 = 196. 66. úû ù êë é úû ù êë é = úû ù êë é b aa b = b aa b b aa b A2 a = a2 + b2; b = 2ab 67. No.of ways in which 6mm can be arranged at a round table = (6 -1)! Now women can be arranged in 6! ways. Total number of ways = 6!´5! 68. No option satisfied wrong. A = (7, -4, 7), B = (1, -6, 10), C = (-1, -3, 4) and D = (5, -1, 5) AB = (7 -1)2 + (-4 + 6)2 + (7 -10)2 = 36 + 4 + 9 = 7 Similarly BC = 7, CD = 41 , DA = 17 69. (uvw).(uvuwvvvw)r r r r r r r r r r r+- ´-´ -´+´ ( ) 0 u.(uv) u v w .(u v u w v w) r r r r + r - r r ´ r - r ´ r + r ´ r = ´ v.(u w) 0 v.(u v) u.(v w) 0 u.(u w) r r r r r r r r r r r r - ´ + ´ + ´ - ´0 w.(u w) 0 w.(u w) w.(u v) 0 v.(v w) r r r r r r r r r r r r + ´ - ´ + ´ - ´ u.(v w) v.(u w) w.(u v) r r r r r r r r r = ´ - ´ - ´ [uvw] [vwu] [wuv] u.(v w) rr r r r r r rr r r r = + - = ´ 70. sin-1 x = 2sin-1 a 2 2sin a 2 ; 2 sin x 2 1 1 - p £ - £ p \- p £ - £ p 2 1 a 21 or 4 sin a 4 - p £ -1 £ p - £ £ ) 21 2 1 (As 2 1 \ a £ > . Out of given four option no one is absolutely correct but (c) could be taken into consideration. 2 1 ® a £ is correct, if 2 1 a < is taken as correct then it domain satisfy for 3 1 a = but equation is satisfied. 21 3 1 2 1 > >44 71. Eq. of planes be 1 cz by ax 1 & cz by ax 1 1 1 + + = + + = (^ r distance on plane from origin is same.) 21 21 21 2 2 2 c1 b1 a1 1 c1 b1 a1 1 + + = - + + - 0 a1 a1 21 2 \S -S = 72. 1 1 i (1 i) 1 1 i1 i x 2 x = úûù êëé -÷ = Þ + øç ö èæ -+ + Î = \ = Þ = ÷ ÷ øö ç ç èæ ++ + 1 (i) 1; x 4n; n 1 1 1 1 i 2i x 2 x 73. f : N ®1 f(1) = 0, f(2) = -1, f(3) = -1, f(4) = -2, f(5) = 2, and f(6) = -3 so on. >>>>>> 123456 0-1 1-2 23 In this type of function every element of set A has unique image in set B and there is no element left in set B. Hence f is one-one and onto function. 74. f(x) = ax2 + bx + c f(1) = f(-1) Þa + b + c = a -b + c or b = 0 \ f(x) = ax2 + c or f¢(x) = 2ax Now f¢(a) ; f¢(b) ; and f¢(c) are 2a(a); 2a(b); 2a(c). If a, b, c are in A.P. then f¢(a) ; f¢(b) and f¢(c) are also in A.P. 75. - + .......... .......... .....¥ 3.4 1 2.3 1 1.2 1 Let ( ) ÷øç ö èæ + = - + = n 1 1 n1 nn 1 1 Tn S = T1 -T2 + T3 -T4 + T5 .............. ¥ .......... 51 41 41 31 31 21 21 11 ÷øç ö è÷ - æ - øç ö è÷ + æ - øç ö è÷ - æ - øç ö è= æ - úûù êë= - é - + - ............¥ 51 41 31 21 1 2 ÷øç ö è= - - + + = - = æ e4 1 2[ log(1 1) 1] 2log2 1 log38. D 39. A 40. B 41. D 42. C 43. C 44. A 45. A 46. NONE 47. B 48. C 49. D 50. D 51. D 52. B 53. B 54. D 55. B 56. C 57. C 58. D 59. C 60. A 61. A 62. C 63. A 64. B 65. C 66. C 67. A 68. C 69. C 70. C 71. A 72. B 73. D 74. B 75. A A I E E E 2 0 0 3 K E Y P h y s i c s A n d C h e m i s t r y 1. B 2. A 3. B 4. A 5. D 6. A 7. A 8. D 9. C 10. B 11. A 12. C 13. C 14. D 15. C 16. A 17. A 18. A 19. C 20. B 21. C 22. D 23. D 24. B 25. D 26. A 27. D 28. NONE 29. C 30. B 31. D 32. A 33. D 34. C 35. C 36. A 37. C 38. D 39. B 40. D 41. C 42. A 43. D 44. A 45. B 46. A 47. A 48. B 49. B 50. C 51. C 52. D 53. B 54. D 55. C 56. B 57. B 58. A 59. A 60. A 61. D 62. B 63. C 64. C 65. B 66. B 67. A 68. D 69. B 70. C 71. C 72. D 73. B 74. D 75. D 76. B 77. C 78. B 79. B 80. D 81. C 82. C 83. B 84. D 85. D 86. C 87. A 88. D 89. A 90. A 91. B 92. C 93. C 94. B 95. C 96. A 97. C 98. C 99. B 100. A 101. A 102. A 103. D 104. B 105. A 106. D 107. A 108. D 109. A 110. B 111. B 112. D 113. A 114. d 115. B 116. D 117. A 118. B 119. B 120. B 121. C 122. A 123. B 124. A 125. A 126. A 127. D 128. A 129. B 130. C 131. D 132. A 133. D 134. B 135. D 136. C 137. C 138. D 139. B 140. A 141. B 142. B 143. B 144. A 145. B 146. B 147. A 148. A 149. A 150. D 1. A 2. A 3. A 4. B 5. B 6. D 7. D 8. C 9. C 10. D 11. D 12. A 13. A 14. C 15. D 16. B 17. C 18. B 19. D 20. A 21. D 22. D 23. D 24. D 25. A 26. D 27. A 28. A 29. A 30. B 31. D 32. B 33. A 34. D 35. C 36.. D 37. D Mathematics 45