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 2 3 4 m /Wb 10 10 10 VE B qE vB = = = Þ = Þ 3. l T µ 2 21 21 L I ; ll TT µ = ( ) 2 4 TT 4 LL 2 LL II 21 22 2 2 2221 21 = = Þ úû ù êë é = = = Þ 21 TT 2T T 12 1 2 = Þ = Þ 4. 3 . W 60 tan ) H ( 0 = = t 7. Mass = kg 5 8 . 949 = . 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. C 16 T 10 10 25 40 T 0 5 6 = D Þ ´ ´ = D - - 13. [ ]2 0 0 2 0 0 0 0 C 1 C 1 C 1 = úû ù êë é e m Þ = e m Þ = e m [C] = LT-1 or [C]2 = L2T-2 14. 2 2 R t M or R m 21 I µ µ = For disc X, ( ) 2 2 2 x ) R(. t r 21 ) R )( m ( 21 I p = = For disc Y, 2 2 ] R 4 [ ] 4 /t . ) R 4 ( [ 21 p = y I x y 3 yx I 64 I ) 4 ( 1 II = Þ = Þ 15. 3 21 2 21 3 2 RR TT R T ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ Þ µ 2 /3 2 /3 21 21 41 RR TT ÷ø ö çè æ = ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ Þ 8 ) 4 ( TT 2 /3 12 = = Þ hours 40 40 5 8 T 8 T 1 2 = = ´ = ´ = Þ < x 10m 10m 10m/sec V22 16. w . E . K L energy Kinetic frequency Angular 1 momentum Angular = Þ µ µ r 4L L 4 KE w wE . K LL 2 2 2 1 1 21 = Þ = ´ ÷ ÷ø ö ç çè æ = 17. RMIVUXGEg sin Decrea ¾¾ ¾ ¾ ¾ ¾ ® ¾l 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 g Þ rays has least wavelength 18. Applying the principle of conservation of linear momentum 238 u 4 v ) 238 ( ) v ( ) u ( ) 4 ( = Þ = 19. Distance between the surface of the spherical bodies = 12R -R -2R = 9R Force µ Mass, Acceleration µ Mass, Distance µ Acceleration 1 2 21 21 S 5 S 51 SS 51 SM M aa = Þ = Þ = = ÞS1 + 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) 6 2 m 2500 F ) 6 )( F ( ) 50 ( . m 21 2 ´ = Þ = Þ ) S )( F ( ) 100 ( m . 21 hr /km 100 v For 2 = = S 6 2 m 2500 ) 100 ( m 21 2 ÷ø ö çè æ ´ = Þ m 24 2 2500 2 6 100 100 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 = 66 . 8 10 2 3 10 10 g ) 30 2 ( sin ) 10 ( g 2 sin u 2 2 = ´ ´ ´ = ´ = q 24. ] MLT [ ] T ][ L ][ M [ ] momentum [ 1 1 - - = = (Planck’s Constant) = 1 2 1 2 1 T ML T ] LT ][ M [ 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 ( ) ) ( q /q dA . E 1 2 0 0 0 f - f e = Þ e = ò [ ò = f dA . E ce sin ] 27. N 10 ¯ mg N f m = W 10 . W N mg f = m Þ = m Þ = W 10 2 . 0 = ´ Þ N 2 W = \ 28. 10 6 g a = m = [ using v = u + at] 06 . 0 10 10 6 g 106 = ´ = ´ = m Þ 31. Since the displacement for both block and rope is same so, the acceleration must be same for both >P M >P >T T T = Ma ........ (i) M m P a a ) M m ( p + = Þ + = Þ M mPM a . M T + = = 33. Elastic energy = x F 21 ´ ´ F = 200 N, x = 1 mm = 10-3 m J 1 . 0 10 1 200 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 2 T p = ....... (i) Km M 2 3T 5 + 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. k I m A V 2 max = w = w = mk mk 2 = w Þ = w Þ 21 21 kk or K = ww µ w Þ max V max V B A = 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 ) 2 log( T log ; gl 2 T ) g log( 21 ) l log( 21 ) 2 log( T log - + p = Þ Differentiating 100 ll 21 100 TT 0 ll 21 0 TT ´ D ´ = ´ D Þ - D ´ + = D % 10 5 . 10 21 21 » = ´ = Note: In this method, the % error obtained is an approximate value on the higher side. Exact value is less than the obtained one. 39. ÷ø ö çèæ p + - = - 3 x 2 t 600 sin 10 y 4 . Comparing it with standard equation s /m 600 v ); kx vt ( sin A y = - = 40. H 1 . 0 L 05 . 0 ) 2 ( 2 ) L ( 8 dt dI L e = Þ - - = Þ - = 41. 2 Q q = *44. ÷ø ö çèæ = Þ ÷ø ö çè æ = 1250 5000 ln 51 K NN ln f1 K 2 ln 4 . 0 2 ln 52 ) 4 ln( 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. A 5 . 1 23 I = = W 3 W 3 W 3 3V 3V W 3 3VW 6 W 2 50. ) t sin t (cos 4 x p + p = ] t sin ] t 2 [sin 4 p + ÷ø ö 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 sin 2 425 úû ù êë é ÷ø ö çè æ p + p - p = t 4 cos . 4 sin 8 úû ù êë é p - p = úû ù êë é p - p = 4 t cos 2 4 4 t cos . 2 8 Comparing it with standard equation X = A cos (wt -Kx) 2 4 A = Þ 51. Potential due to spherical shell, R 4 q v 0 1 pe = . Potential difference due to charge at the centre R 4 q R 4 Q 2 V V V ; r 4 Q 2 V 0 0 2 1 0 2 pe + pe = + = pe = 52. Work done J 10 32 10 100 2 ) 10 8 ( c q 21 32 8 2 18 2 - - - ´ = ´ ´ ´ = = 53. 2 y 2 x t 3 dt dy V , t 3 dt dx V b = = a = = 2 2 2 y x t 3 V V v 2 2 b + a = + = r 54. 3 21 21 3 TT PP T P ÷ ÷ø ö ç çè æ = ÷ ÷ø ö ç çè æ µ Comparing it with standard eq. 23 CCvp = g = 56. 273 627 ) 27 273 ( ) 273 627 ( + + - + = h 32 900 600 900300 900 = = - = work = Heat ) ( ´ h = J 10 4 . 8 J 2 . 4 10 3 32 6 6 ´ = ´ ´ ´ 57. Required work done 4 2 2 3 21 22 10 ] 5 10 [ 10 5 21 ) x x ( K 21 - ´ - ´ ´ = - = 75 . 18 10 10 75 5 21 4 3 = ´ ´ ´ ´ = - 58. m 1 l ; T l 21 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 ÷ø ö çè æ = dt dV . v . m dt dV m F constant = C kdt vdv k mC dt dV = Þ = = Þ c kt 2 V dt k dv v 2 + = Þ = Þ ò ò 2 /1 ) t ( v µ Þ 2 /1 t . c dt ds = 2 /3 2 /1 t 32 . C S dt ) t . c ( ds = Þ = Þò ò 2/3 2/3 t s 2 /3t.c S µ Þ = Þ 67. Thrust = Mass ´ Acceleration = 3.5 ´ 104 ´ 10 = 3.5 ´ 105 N 69. The force body diagram 2 3 1 2 2 3 1 1 bq q . 0 4 1 F ; aq q . 0 4 1 F pe = pe = úû ù êë é + q pe = + q = 21 23 1 2 1 X bq sin aq 0 4q F sin F F ÷ø ö çè æ + q µ Þ 22 23 bq sin aq Fx 70. ( ) 1000 220 R or R V p 2 2 = = Power consumed = 1000 220 220 110 110 R V2 ´ ´´ = = 250 watt 73. According to Image formula 1 360 3 1 360 n - q = Þ - q = 90 360 4 360 = 4 = q Þ = q Þ 74. ( ) ( ) 1 n dt dH n 1 2 = Þ q D = q - q µ 75. ( ) ( )( ) l 2 r l r or l 2 L 2 2 2 1 p = p = 2 2 rl R ; 2 r r p r = = Þ ( ) R 4 r) ( l4) ( 2 /r ) ( l2 ) ( R 2 2 new ´ = pr = p r = R 3 R R 4 R = - = D \ % 300 100 RR 3 % 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 ) H C ( 2 ) C C ( 1 ) H H ( 1 ) C C ( 1 ) H C ( 6 ) C C ( 1 ) H H ( 1 ) H C ( 4 ) C C ( 1 H - - - - - + = = - - - - - + - + = = D . kJ 125 1175 1050 414 2 347 435 615 - = - = ´ - - + = 83. Th Th X Th 230 90 234 92 234 91 234 90 ¾® ¾ ¾® ¾ ¾® ¾ a - b - b - 84. t1/2 = 3 hrs. Initial mass (C0) = 256 g g 4 64 256 ) 2 (256 2C C 6 n0 n = = = = \ 86. z1 µ W Å 85 . 0 57 71 06 . 1 zz 2 2 12 21 = W Þ = 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 0 4 x 4 = - + Þ 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 = 23 10 02 . 6 5 . 580 . 1 ´ ´ since in NaCl type of structure 4 formula units form a cell \ unit cells = 4 5 . 58 10 02 . 6 0 . 1 23 ´´ ´ = 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 059 . 0 2 295 . 0 059 . 0 n E K log 0cell ´ = ´ = 10 10 1 K or 10 59 590 K log - ´ = = = 107. For spontaneous reaction, dS > 0 an G D 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 8 . 10 6 . 21 = 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. [ ] [ ] [ ] [ ] L/mol 10 3 10 8 . 4 10 2 . 1 O NNO K 3 2 2 2 4 2 2 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. 4 2 2 HgI K KI 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 = g 8 . 10 108 10 1 = ´29 118. [ ] [ ]22 0cell cell Zn Cu log n059 . 0 E E ++ + = V 07 . 1 0295 . 0 10 . 1 ] 1 . 0 log[ 2059 . 0 10 . 1 = - = + = 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 OH 2 CH CH 2 CH LAH COOH CH CH2 - = ¾ ¾ ® ¾ - = 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. + - - - + ® + H CrO 2 OH O Cr 24 27 2 . 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. e cos Glu D 6 O 12 H 6 nC H 2 5 10 6 O nH n ) O H C ( - ¾® ¾ + + 130. AgNO3 ® Ag + NO2 + 2 O 21 131. H O CH CH H OH CH CH 2 3 1 step 2 3 - - ¾ ¾® ¾ + + HÅ Protona ted alcohol 132. The solubility is governed by solution H D i.e. solution H D = Hydration lattice H H D - D . 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. O H 2 BaCl HCI 2 ) OH ( Ba 2 2 2 + ® + Applying Molarity equation, 2 ) HCI ( 1 ) 2 ) OH ( Ba ( 2 2 1 1 V M V M = or 25´M1 = 07 . 0 107 . 0 25 2 35 1 . 0 M 2 35 1 . 0 1 = = ´´ = \ ´ 137. Rate1 = k [A]n [B]m; Rate2 = k[2A]n [½B]m m n m n m n m n m n 12 2 2 . 2 [½] ] 2 [ ] B [ ] A [ k ] B [½ ] A 2 [ k Rate Rate - - = = = = \ 138. CH3CH2N =® C + H2O ¾® ¾ + 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. m 10 10 10 60 10 6 . 6 mv h 33 3 34 - - - = ´ ´ ´ = = l31 AIEEE 2003 MATHEMATICS SOLUTION 1. dx e xx 3 dx e x3 or x e ) x ( F dx d 3 3 x sin 41 41 32 x sin x sin ò ò= = Let x3 = t, 3x2dx = dt when x = 1, t = 1 & x -4, t = 64 ò ò - = = =64 1 64 1 t sin ) 1 ( F ) 64 ( F dt ) t ( F dt t e ) t ( F K = 64. 2. n = 9 then median term = th th 5 2 1 9 = ÷øö çèæ + 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÷ø ö çè æ + ÷ø ö çè æ + ÷ø ö çè æ + ÷ø ö çè æ ¥ ® ¥ ® 43 43 4 n 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 0 dx ) x ( 4. Fundamental theorem (fact) 1 1 2 t2 t t - - = 5. 2 1 2 1 C C r r = - 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 x 2 2 = - 45 12 15 144 81 1 e , 25 81 b , 25 144 a = = + = = = Foci = (3, 0), focus of ellipse = (3, 0) Þ 43 e = 7 16 9 1 16 b2 = ÷øö çè æ - = 8. ò - = t0 dy ) y ( g ) y t ( f ) t ( F32 ò ò- - = =t0 t0 y t y t ydy e e ydy e[ ] [ ]t0 y y t t0 y y t e ye e e ye e - - - - + - = - - = [ ] ) t 1 ( e e e 1 t e 1 0 e te e t t t t t t t + - = úû ù êë é - + = - - + - = - - 9. ( ) 1 x x log ) x ( f 2 + + = ( ) 1 x x log ) x ( f 2 + + - = - ) x ( f ., e . i ), x ( f ) x ( f - = - is an odd function. 10. ac , ab , 0 c bx ax2 = ab - = b + a = + + As for given condition, 2 2 1 1 b + a = b + a 22 22 2 2 2 2 ac ac 2 ab ab - = - b a b + a - = b + aOn simplification 2a2c = ab2 + bc2 bc & , ab , ba cb ac ba 2 \ + = Þ are in H.P. 11. 3 2 2 C 2 C C 0 c c 4 1 b b 3 1 a a 2 1 - ® = 1 2 2 2 3 3 R R R , R R R 0 c c 2 1 b b 1 a 0 1 - ® - ® = 0 b c b c 2 0 a b b 0 a 0 1 = - - - 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 tan y ÷øö çèæ a + p = 4p a C Y B A X O33 ÷ø ö çèæ + p - = \ ^ 2 4 cot CA of slope OB to r CA Equation of CA ( )a - ÷øö çèæ+p -=a - cos a x 2 4 cot sin a y . a ) sin (cos x ) cos (sin y = a - a + a + 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). 1 pq 1q p 22 1q - = Þ -- = - = . 14. 1 x 3 t sin b t cos a 3 1 t sin b t cos x - = + Þ + + = y 3 t sin b t sin a 3 t cos b t sin a y = - Þ - = Squaring & adding, (3x -1)2 + (3y)2 = a2 + b2 15. K x ) x 3 log( ) x 3 log( Lim0 x = - - + ® (by L’Hospital rule) K 32 K 1 x 3 1 x 3 1 Lim0 x = \ = - - - + ® 16. q = ´ = sin rp a ; p r a r r r r ( ) ë ûq = q + q = cos 90 sin cos rp H 0 Q r G = rp sin q ......... (1) H = rp cos q .......... (2) x = rp sin( ) a + q .......... (3) From (1), (2) & (3), a + a = sin H cos a x 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) 6 12R 15 24Q 30 24P 2 2 2 - = - = + - 2 : 3 : 2 R : Q : P or 6 R 9 Q 6 P 2 2 2 2 2 2 = = =34 18. 8 n , 21 p . 21 q 2 npq 4 np = = = Þ þ ý ü= = ( ) 32 1 21 21 . 8 21 21 C 1 X p 5 8 7 1 8 = = = ÷ø ö çè æ ÷ø ö çèæ = = 19. ( ) 1 ) 1 ( f x x f n = Þ = n ) 1 ( f nx ) x ( f 1 n = ¢ Þ = ¢ - ( ) )1 n(n )1( f x 1 n n ) x( f 2 n - = ¢ Þ - = ¢ - ......................... ! n ) 1 ( f ! n ) x ( f n n = Þ = = ! n! n ) 1 ( ......... ! 3 ) 2 n ( ) 1 n ( n ! 2 ) 1 n ( n ! 1n 1 n - + + - - - - + - 0 C ) 1 ( .......... C C C C n n n 3 n 2 n 1 n 0 n = - + + - + - = 20. Since n r is perpendicular v u n , v and u r r r r r ´ = k ˆ2k ˆ2 2 2 0 1 1 0 1 1 k j i nˆ - = - = ´ - = 3 3 ) k ( . ) k 3 j 2 i ( nˆ . = - = - + + = w r 21. k 4 j 2 i 7 F F F 2 1 - + = + + r r r k 2 j 2 i 4 A of V . P B of V . P d - + = - = r r r unit 40 8 4 28 d . F W = + + = = r r 22. AD C B j 4 i 3 r r+ k 2 j 2 i 5 r r r - + P.V of 2 k ) 4 4 ( j ) 2 0 ( i ) 5 3 ( AD + + - + + = 33 1 16 16 AD or k 4 j i 4 = + + = + - = 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 dx ) 1 x ( ) x 3 ( dx ) 1 x ( ) x 3 ( dx ) 1 x ( ) x 3 ( Aò ò ò - - + + + = 01 21 10 dx ) x 2 4 ( dx 2 dx ) x 2 2 ( [ ] [ ] [ ]21 2 10 01 2 x x4 x2 x x2 - + + - = - = 0 -(-2+1) + (2 -0) + (8 -4) -(4 -1) = 1 + 2 + 4 -3 = 4 sq. units 24. Shortest distance = perpendicular distance = 26 16 9 144 327 3 3 1 4 12 2 = + + - ´ + ´ + ´ - \ Shortest distance ] r 26 [ 13 13 26 9 15 1 4 26 - = - = + + + - = Q 25. ' c ' d 3 1y ' a ' b x ; c d 3 1y a b x - = = - - = = - For perpendicular 0 c c 1 a a = ¢ + + ¢ 26. 0 n m l n m l z z y y x x 2 2 2 1 1 1 1 2 1 2 1 2 = - - - 0 1 1 2 k k k 1 2 1 0 0 0 1 2 k k 1 1 1 1 1 = + - + - Þ = -- - k2 + 3k2 = 0 Þ k(k + 3) = 0 or k = 0 or -3 27. ò ò - + - + = = ba ba dx ) x b a ( f ) x b a ( dx ) x ( xf I ò ò - + - - + + = ba ba dx ) x b a ( xf dx ) x b a ( f ) b a ( ò ò- - + + = ba ba dx ) x ( xf dx ) x b a ( f ) b a ( ò + = ba dx ) x ( f ) b a ( I 2 ò ò - + + ( = + = ba ba dx ) x b a ( f 2 ) b a I ; dx ) x ( f 2 ) b a ( I 28. Portion OA, OB corresponds to motion with acceleration ‘f’ and retardation ‘r’ respectively. Area of , t OL Let . t OB and S OAB 1 = = = D36 tS 2 v ; v . t 21 AL . OB 21 S , v AL and t LB 2 = = = = = Also, trs 2 tfs 2 t t t ; trs 2 rv t , tv r and tfs 2 fv t , tv f 2 1 2 2 1 1 + = + = = = = = = = ÷ø ö çè æ + = Þ ÷ø ö çè æ + = r1 f1 s 2 t ts 2 r1 f1 t 29. t ) cos u ( gh 2 u R ´ q = = gh 2 cos 1 t q = ....... (1) Now, 2 gt 21 t ) sin u ( h + q - = Substituting ‘t’ from (1), úû ù êë é q + qq - = 2 cos g h 2 g 21 gh 2 cos sin u h q + q - = 2 sec h tan gh 2 u h h tan h tan gh 2 u h 2 + q + q - = hg 2 u tan ; 0 tan hg 2 u tan2 = q \ = q - q 30. Applying R1®R1 + R2 + R3 As, 0 ; 0 1 n 2 n = D \ = w + w + 31. R 2a n sin ; r 2a n tan = ÷øö çèæ p = ÷øö çèæ p ÷ø ö çèæ p = + Þ úû ù êë é p + p = + n 2 cot . 2a R r n ec cos n cot 2a R r 32. Taking co-ordinates as ( ) ( ) yr , xr & y , x ; ry , rx ÷øö çè æ . Above coordinates satisfy the relation y = mx Therefore lies on the straight line. 33. 1 z = w ..... (1) As, i z therefore 2 z Arg = w p = ÷øö çèæ w1 z = w \ ....... (2) From (1) & (2), 0 z z ; 0 z z and 1 z = w + w = w + w = w = i i z ; . . z z z 2 - = w - = w w w w - = w - = w37 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 ) e x ( ) y 1 ( y tan 2 1 = - + + - ) y 1 (e ) y 1 ( x dy dx e x dy dx ) y 1 ( 2y tan 2 y tan 2 1 1 + = + + Þ = + + - - I.F. = y tan e e 1 dy ) y 1 ( 1 2 - ò + = ydy tan y tan y tan 1 1 1 e y 1 e ) e ( x - - - ò + = k e xe 2 C 2 e ) e ( x y tan 2 y tan y tan 2 y tan 1 1 1 1 + = \ + = - - - - 36. Let f(x) = ex ò ò - = \10 10 x 2 x dx ) e x ( e dx ) x ( g ) x ( fò ò- =10 10 x2 x 2 dx e dx e x [ ] [ ] [ ]10 x 2 10 x x 10 x 2 e 21 e xe 2 e x - - - = [ ] 23 2e e 1 e e2 21 2e e 2 2 - - = + - -úû ù êë é - - = 37. 7 r 154 r2 = Þ = p 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 x 2 1 ) C ( P , 4x 1 ) B ( P , 3 1 x 3 ) A ( P - = - = + = These are mutually exclusive 1 2 x 2 1 0 and 1 4x 1 0 , 1 3 1 x 3 0 £ - £ £ - £ £ + £ 1 x 2 1 and 1 x 3 , 2 x 3 1 £ £ - £ £ - £ £ -38 21 x 21 and , 1 x 3 , 32 x 31 £ £ - £ £ - £ £ - Also 1 2 x 2 1 4x 1 3 x 3 1 0 £ - + - + + £ 3 13 x 31 13 x 3 1 12 x 3 13 0 £ £ Þ £ £ Þ £ - £ þ ý üî í ì£ £ þ ý üî í ì- - - 3 13 , 21 , 1 , 32 min x 31 , 21 , 3 , 31 max úû ù êë é Î Þ £ £ 21 , 31 x 21 x 31 39. n(S) = 5C2; n(E) = 2C1 + 2C152 C C C ) S ( n ) E ( n ) E ( p 2 5 1 2 1 2 = + = = 40. 3 a 5 a 2 2 & 3 a 5 a a 3 1 3 2 2 2 + - = a + -- = a 3 a 5 a 2 ) 3 a 5 a ( ) a 3 1 ( 91 2 2 2 2 2 + - = úû ù êë é + -- 1 a 6 a 9 or 9 ) 3 a 5 a ( ) a 3 1 ( 2 2 2 + - = + -- 32 a or 26 a 39 or 27 a 45 a 9 2 = = + - = 41. r 1 r ) x ( ! r ) 1 r n ( .......... ) 2 n ( ) 1 n ( n T + - - - = + For first negative term, n -r + 1 < 0 or 5 32 r > \ r = 7. Therefore, first negative term is T8. 42. 8 /r 2 r 256 r 256 r 8 r 256 r 256 1 r ) 5 ( ) 3 ( C ) 5 ( ) 3 ( C T - - + = = Terms will be integral if 8r & 2 r 256- both are +ve integer. As 256 . .......... , 24 , 16 , 8 , 0 r 256 r 0 = \ £ £ For above values of ÷ø ö çèæ - 2 r 256 , r is also an integer. 43. After t; velocity = t f ´ a - + = - + = cos ut f 2 u t f ) u ( t f V 2 2 2 BA r r For max. and min.39 ( ) f cos u t or 0 cos fu 2 t f 2 V 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 1 n 1 r n r n r n 1 r n 1 r n C C C C C C C + + = + + + = + + - + 4 4 3 4 4 2 1 1 r 2 n r 1 n 1 r 1 n C C C + + + + + Þ + 45. a b q a - q = b b + a = q or 4h h 43 - 160 h . 40 h 1 160 h 40 h 53 or tan . tan 1 tan tan tan + - = a q + a - q = b h2 -200 h + 6400 = 0, h = 40 or 160 metre Therefore possible height = 40 metre 46. A E B D C P4/3 0 60 x 0 90 0 30 8/3 3 38 x or x3 /8 60 tan 0 = = Area of 3 316 3 38 4 21 ABD = ´ ´ = D \ Area of 3 332 3 316 2 ABC = ´ = D 47. If 2b 3 2A cos c 2C cos a 2 2 = ÷øö çèæ + ÷øö çèæ 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. 0 ) c b a ( . ) c b a ( 0 c b a = + + + + Þ = + + r r r r r r r r r 0 ) a . c c . b b . a ( 2 c b a 2 2 2 = + + + + + r r r r r r r r r 7 2 9 4 1 a . c c . b b . a - = - - - = + + r r r r r r 49. ò - = 10 n dx ) x 1 ( x I ò ò - - - = - - = - 10 n 10 n dx ) x 1 ( ) 1 x 1 ( dx ) x 1 ( x I40 ò ò- - - = + 10 10 n 1 n dx ) x 1 ( dx ) x 1 ( 1 n 1 2 n 1 ) 1 n ( ) x 1 ( ) 2 n ( ) x 1 ( 10 1 n 10 2 n + - + = úû ù êë é + - - - úû ù êë é + - - = + + 2 n 1 1 n 1 I + - + = 50. x cos x x sin x 2 . x sec Lim ) x sin x ( dx d dt t sec dx d Lim 2 2 0 x x0 2 0 x 2 + = ® ® ò (by L’ Hospital rule) 1 1 1 1 2 x cos xx sin x sec 2 Lim 2 2 0 x = +´ = ÷ø ö çè æ + ® 51. OC A centre of sphere = (-1, 1, 2) Radius of sphere 5 19 4 1 1 = + + + Perpendicular distance from centre to the plane . 4 3 12 4 4 1 7 4 2 1 d OC = = + + + + + - = = AC2 = AO2 -OC2 = 52 -42 = 9 Þ AC = 3 52. Vector perpendicular to the face OABk 3 j i 5 3 1 2 1 2 1 k j i OB OA - - = = ´ =Vector perpendicular to the face ABCk 3 j 5 i 1 1 2 2 1 1 k j i AC AB - - = - - - = ´ = Angle between the faces = Angle between their normals ÷ø ö çè æ = q = + + = q - 35 19 cos or 35 19 35 35 9 5 5 cos 141 53. 4 ) x ( f ) k ( 9 ) x ( kf ) x ( 9 k lim a x = -- ® (By L’Hospital rule) . 4 k or 4 ) x ( ' f ) x ( ' 9 ) x ( ' f ) x ( ' 9 k lim a x = = -- ® 54. 3 lim 2 x ) x 2 ( ) x sin 1 ( . 2x 4 tan - p - ÷ø ö çè æ - p p ® Let 0 y ; y 2 x ® + p = 8 . 8 y . ) 8 ( 2y sin 2 2y tan ) y 2 ( ) y cos 1 ( . 2y tan 3 2 lim 0 y 3 lim 0 y - - = - - ÷ø ö çè æ- - = ® ® 32 1 2 /y 2 /y sin . 2y 2y tan 32 1 2 lim 0 y = úû ù êë é ÷ø ö çè æ = ® 55. 2 2 2 2 2 1 2 1 ) b k ( ) a h ( ) b k ( ) a h ( - + - = - + - 0 ) b a b a ( 21 y ) b b ( x ) a a ( 21 21 22 22 2 1 2 1 = - - + + - + - ) b a b a ( 21 C 21 21 22 22 - - + = 56. 0 c c c b b b a a a 1 c c 1 b b 1 a a 0 c 1 c c b 1 b b a 1 a a 3 2 3 2 3 2 222 3 2 3 2 3 2 = + 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 c c 1 b b 1 a a 1 222 ¹ (given condition) \abc = -1 57. ( )( ) 0 1 x 2 x or 0 2 x 3 x2 = - - = + - 4 solution of . No or 2 , 1 x or 2 , 1 x = \ ± ± = = 58. f(x) = 2x3 -9ax2 + 12a2x + 1 ) x ( f¢ = 6x2 -18ax + 12a2; ) x ( f ¢ = 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 xe ) x ( f ; 0 ) 0 ( f R.H.L. 0 eh e ) h 0 ( h /2 lim 0 h h /2 lim 0 h = = + ® - ® L.H.L 0 e ) h 0 ( Lim h1 h1 0 h = - ÷ø ö çè æ - - ® Therefore, f(x) is continuous R.H.D. 0 h he e ) h 0 ( h1 h1 h1 h1 lim 0 h = - + ÷ø ö çè æ + - ÷ø ö çè æ + - ® L.H.D. 1 h he e ) h 0 ( Lim h1 h1 h1 h1 0 h = - - - ÷ø ö çè æ + - ÷ø ö çè æ - - ® Therefore, L.H.D. ¹ R.H.D. f(x) is not differentiable at x = 0. 60. --+ + -1 0 1 ) x x ( log x 4 3 ) x ( f 3 10 2 - + - = 4 x ; 0 x x ; 0 x 4 3 2 ± ¹ > - ¹ - { } 4 ) , 1 ( ) 0 , 1 ( D - ¥ - = \ U ) , 2 ( ) 2 , 1 ( ) 0 , 1 ( D ¥ - = U U . 61. f(x + y) = f(x) + f(y). a = a m ) ( f Let f(1) = 7; \ m = 7, f(x) = 7x å å + = = = n1 n1 r 2 ) 1 n ( n 7 r 7 ) r ( f 62. 2 x1 1 dx dy or x1 x y - = + = For max. or min., 1 x 0 x1 1 2 ± = Þ = - ) ima min ve ( 2 dxy d x2 dxy d 2 x 2 2 3 2 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 ) sin 1 ( g u R and ) sin 1 ( g u R 2 2 2 1 b - = b + = úû ù êë é = = + = + g u R R2 R1 R1 or ug 2 R1 R1 2 2 1 2 2 1 Q Therefore, R1, R, R2 are in H.P.43 64. 2830 x , 170 x 2 = S = S increase in 10 x = S , then 180 10 170 ' x = + = SIncrease in 500 400 900 x2 = - = S then 3330 500 2830 ' x 2 = + = SVariance = 2 2 ' x n1 ' x n1 ÷øö çè æ S - S . 78 144 222 180 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 a b b a = a b b a a b b a A2 ab 2 ; b a 2 2 = b + = a 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) 7 9 4 36 ) 10 7 ( ) 6 4 ( ) 1 7 ( AB 2 2 2 = + + = - + + - + - = Similarly BC = 7, CD = 41 , DA = 17 69. ( ) )w v v v w u v u(. w v u r r r r r r r r r r r ´ + ´ - ´ - ´ - + ( ) 0 )v u.( u ) w v w u v u ( . w v u r r r r r r r r r r r r ´ = ´ + ´ - ´ - + ) w u .( v 0 ) v u .( v ) w v .( u 0 ) w u .( u r r r r r r r r r r r r ´ - ´ + ´ + ´ - 0 ) w u .( w 0 ) w u .( w ) v u .( w 0 ) w v .( v r r r r r r r r r r r r ´ - ´ + ´ - ´ + ) v u .( w ) w u ( . v ) w v .( u r r r r r r r r r ´ - ´ - ´ = ) w v .( u ] v u w [ ] u w v [ ] w v u [ r r r r r r r r r r r r ´ = - + = 70. sin-1 x = 2sin-1 a 2 a sin 2 2 ; 2 x sin 2 1 1 p £ £ p - \ p £ £ p - - - 2 1 a 21 or 4 a sin 4 1 £ £ - p £ £ 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 i 1 ) i 1 ( 1 i 1 i 1 x 2 x = úû ù êë é -+ Þ = ÷øö çèæ -+ + Î = \ = Þ = ÷ ÷øö ç çèæ ++ + 1 n ; n 4 x ; 1 ) i ( 1 1 1 i 2 i 1 x x 2 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 ) x ( f¢ = 2ax Now ) a ( f¢ ; ) b ( f¢ ; and ) c ( f¢ are 2a(a); 2a(b); 2a(c). If a, b, c are in A.P. then ) a ( f¢ ; ) b ( f¢ and ) c ( f¢ are also in A.P. 75. ¥ + - ..... .......... .......... 4 . 31 3 . 21 2 . 11 Let ( ) ÷ø ö çèæ + - = + = 1 n1 n1 1 n n 1 Tn S = T1 -T2 + T3 -T4 + T5 .............. ¥ .......... 51 41 41 31 31 21 21 11 ÷øö çèæ - - ÷øö çèæ - + ÷øö çèæ - - ÷øö çèæ - = úû ù êë é ¥ - + - - = .. .......... 51 41 31 21 2 1 ÷ø ö çèæ = - = + + - - = e4 log 1 2 log 2 ] 1 ) 1 1 log( [ 2 138. 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