SLIDEREAD

Test - 4 (Code-E) (Answers)All India Aakash Test Series for Medical All India Aakash Test Series for Medical - TEST - 4 (Code-E)Test Date : 12/01 ANSWERS1.(1)37.(3)73.(2)109. (1)145. (4)2.(3)38.(3)74.(3)110. (2)146. (4)3.(4)39.(4)75.(1)111. (1)147. (4)4.(3)40.(1)76.(4)112. (3)148. (3)5.(2)41.(4)77.(3)113. (2)149. (3)6.(3)42.(1)78.(2)114. (4)150. (2)7.(4)43.(3)79.(1)115. (4)151. (2)8.(2)44.(3)80.(4)116. (4)152. (4)9.(3)45.(4)81.(2)117. (4)153. (3)10.(2)46.(2)82.(2)118. (2)154. (3)11.(2)47.(3)83.(1)119. (1)155. (3)12.(1)48.(1)84.(3)120. (3)156. (1)13.(1)49.(4)85.(2)121. (3)157. (3)14.(4)50.(2)86.(4)122. (4)158. (3)15.(3)51.(2)87.(3)123. (4)159. (1)16.(2)52.(3)88.(2)124. (4)160. (3)17.(3)53.(1)89.(1)125. (3)161. (3)18.(4)54.(1)90.(3)126. (2)162. (4)19.(2)55.(4)91.(2)127. (3)163. (3)20.(2)56.(2)92.(3)128. (1)164. (4)21.(4)57.(3)93.(2)129. (4)165. (2)22.(2)58.(2)94.(4)130. (3)166. (3)23.(1)59.(3)95.(1)131. (2)167. (2)24.(4)60.(1)96.(1)132. (1)168. (3)25.(3)61.(4)97.(4)133. (3)169. (2)26.(3)62.(2)98.(2)134. (1)170. (4)27.(1)63.(4)99.(3)135. (4)171. (4)28.(2)64.(2)100. (3)136. (3)172. (4)29.(3)65.(3)101. (1)137. (1)173. (3)30.(1)66.(3)102. (2)138. (2)174. (1)31.(2)67.(2)103. (4)139. (4)175. Delete32.(3)68.(2)104. (3)140. (4)176. (3)33.(2)69.(1)105. (4)141. (4)177. (1)34.(4)70.(1)106. (4)142. (2)178. (2)35.(1)71.(4)107. (3)143. (1)179. (3)36.(4)72.(4)108. (1)144. (3)180. (3)Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234561/16

All India Aakash Test Series for Medical-Test - 4 (Code-E) (Hints & Solutions)HINTS & SOLUTIONS[PHYSICS]1.Answer (1)Hint : Mean free path λ 330 20 n ′ 1240 330 20 12πd 2 n Sol. : PV NkBT N P kBT V P nkBTn kBT4.5.2πd 2P8.20 5 28 75RT7Answer (4)Hint & Sol. : Temperature of both the gases issame, i.e. equal to temperature of container, sointernal energy per mole will be same.Answer (3)Hint & Sol. : Kinetic energy per molecule per 1 degree of freedom is kBT it becomes zero at2 absolute zero temperature but potential energy isnon-zero.Answer (2)Hint & Sol. : Cp – Cv RAnswer (2)P constantρTSol. :PV mRTMPR Constant ρT MFor a given ρ, P1 P2 so T1 T29.Answer (3)Hint & Sol. :fR2 12 Mv rms2 1 mnv rms2 2P V 33 Vf Cp R 1 2 Answer (3)P 2E3V10. Answer (2)Hint & Sol. : In SHM v v0 Hint : Apparent frequency n ′ n . v vs Sol. : vs 72 km/h 72 ( )Hint : f Cp Cv R R R 2 6.3R ( 4T ) 2 2vM2 v′PV Cv 3RTMv Sol. : Number of moles in 20 g N2 gas 3.3RT.MSol. : T1 T, vrms vλ TAnswer (3)Hint : PV nRT Answer (4)Hint : r.m.s. speed v PkBTλ 2.7 350 1400 Hz3105 50 m/s18 F x F KxSo, graph will be straight line passing throughorigin with negative slope.Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234562/16

Test - 4 (Code-E) (Hints & Solutions)All India Aakash Test Series for Medical-011. Answer (2) g Sol. : Elevator coming down with retardation . 4 Hint : The motion will be SHM, when ( a x ) .Sol. : x 3 sin 2πt 4 sin 4πt is superposition of1two SHM of time periods T1 1s and T2 s.2dx 6π cos 2πt 16π cos 4πtdtSo, g eff g T 2πa –4π [3 sin 2πt 16 cos 4πt] Since a is not proportional to x a / xx a cosωt , v Sol. : v So, motion will be oscillatory and periodic withperiod 1 s but not SHM.12. Answer (1)at t a – ω2A cos ωtSol. : Amplitude 2 cm, T 4 sv 2 2π 2π 2 a 4 2cos 4 3 π2cm/s2413. Answer (1)Hint & Sol. : Potential energy of oscillating bodyfrom mean position11mω2 y 2 mω2a 2 sin2 ωt2214. Answer (4)Hint : As rope has mass, so tension of ropechange at each points, hence speed of wave willchange and frequency remains same.U 2πaπsin3TT 2πlgT2 lT 2 KlSo, graph will be straight line passing throughorigin.18. Answer (4)Hint & Sol. : x A sinωtdx Aω cosωtdtd2x –ω2A sinωtdt 2T1T22g 1 8g 2λ 1m15. Answer (3)Hint : Time period of a simple penduluml.g effT6πa 3T17. Answer (3)Hint & Sol. : Time period of simple penduluma T 2πdx aω sin(ωt )dtv 4 π21 2 1620.5 λ2dxdt2π T 2π v a sin T 6 T Hint : a –ω2xv1λ1v1Sol. : n λ2 v 2 v 2nl 4l 4π5g5g16. Answer (2)Hint : Pendulum starts from extreme position, sodisplacement equation will bed2x –12π sin 2πt – 64π cos 4πtdt 2a g 5g 44a ω2A sin (ωt π)So, phase difference is π.19. Answer (2)Hint & Sol. : Total energy of oscillating bodyremains constant, so time period is infinity andfrequency is zero.20. Answer (2)Hint & Sol. : In case of forced oscillation, thefrequency of oscillation will be equal to frequencyof applied periodic force.Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234563/16

All India Aakash Test Series for Medical-Test - 4 (Code-E) (Hints & Solutions)21. Answer (4)Hint & Sol. : In stationary waves the particles ofadjacent loops of a node vibrates in oppositephase. So phase difference between them is π.22. Answer (2)Hint & Sol. : Speed of longitudinal waves in a gasis independent of pressure at constant0temperature. So v P .23. Answer (1)Hint & Sol. : For sound wave water behaves as ararer medium and air as denser medium, so soundwaves bend towards normal.24. Answer (4)Hint & Sol. : Displacement antinode is pressurenode, so there is minimum change in pressure anddensity of air.25. Answer (3)Hint : Speed of sound at room temperaturev 2n (l2 – l1)Sol. : In first casen1 – n 5n1 n 5In second casen – n2 5n2 n – 5 (ii)n1 n 5 n2 n 5l2 n 5 l1 n 521 n 5 20 n 5n 205 Hz.29. Answer (3)Hint : Because second resonance is not exactly atthree time, so there must be some end correction.Sol. : Length of first resonance, l1 17 cmSol. : l1 17 cm, l2 53 cmLength of second resonance, l2 52 cml2 e 3l1 eFrequency of source, n 1000 Hzv 2n (l2 – l1) 2 1000 (52 – 17) 10–2 700 m/s26. Answer (3)Hint & Sol. : Reflected wave has a phase changeof π, when reflected from rigid support.27. Answer (1)Hint : The amplitude of resultant wavee l 2 3 l1253 512e 1 cmEffective length of first resonance 18 cmEffective length of third resonance 18 5 φ A 2a cos for same amplitude. 2 90 cmLength of third resonance 90 – 1Sol. : y1 10 sin(ωt – kx) 89 cm2π y2 10 sin ωt kx 3 30. Answer (1)Hint : After each lower octave frequency becomeshalf, and of 20 forks, 19 pairs will form.So, phase difference φ 120 A (10 )2 (10 )2 2 10 10 cos120 A 100 100 100Sol. : Assume frequency of first fork is 2n then oflast fork will be n.2n – n 19 3A 10 units28. Answer (2)Hint : Frequency of sonometer wire1 Tn 2l µIf T and µ are constant, then n (i)n 57 Hz2n 114 HzFrequency of tenth fork 114 – 9 3 87 Hz31. Answer (2)1lHint : Third overtone of open pipe is fourthharmonic.Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234564/16

Test - 4 (Code-E) (Hints & Solutions)All India Aakash Test Series for Medical-Sol. : Assume length of open pipe is lo and ofclosed pipe is lc.Frequency of third overtone of open pipe 420 4v2lol 1800 2.14 m84034. Answer (4)5v4l cFrequency of fifth harmonics of closed pipe 6 3002lHint : When tube is dipped in water it will behaveas closed organ pipe.4v5v 2lo 4lcSol. : Frequency of 2nd overtone of open pipe frequency of fundamental tone of closed pipelo 16 lc 10v3v 2l 4l ′lc 5 lo 8l′ 32. Answer (3)Hint : Pressure variation is maximum atdisplacement nodes.Sol. : In third harmonic, 2 nodes and 2 antinodeswill be produced.l6Length of tube inside water l l 5l 6 635. Answer (1)Hint : Speed of transverse wave in a string is givenas v TµSol. : Displacement equation of wave is.y 0.021 sin (0.1x 30 t)l 3λ4speed of wave v λ l 60 20 cm4 33300 First node will be formed at 20 cm from open end.TµT 300 300 1.3 10–4 11.7 N33. Answer (2)Hint : Frequency of stretched string n 36. Answer (4)p T2l µHint & Sol. : All the points between two nodesvibrates in same phase, so they reach their meanposition simultaneouslyp 1, 2, 3 Sol. : Frequency of pth mode np p T2l µFrequency of (p 1)th mode np 1 p 1 T2lµ (i)37. Answer (3)Hint : (vp)max aωWave velocity (ii)ωkDivide (i) by (ii)Sol. : (vp)max 3vwp420 490 p 1aω 3p 62π a 3λPut in equation (i)420 ω 30 300 m/sk 0.16 1.8 1032l0.02 λωk2π4π (2)m33Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234565/16

All India Aakash Test Series for Medical-Test - 4 (Code-E) (Hints & Solutions)38. Answer (3)Hint : Speed of sound in gas at temperature Tv Sol. : y 8 sin 2π (0.1x – 2t)2π 2π 0.1λλ 10 cm2π2π2π φ x2radλ10542. Answer (1)Cp2Hint : γ 1 and γ Cvf3RT.MSol. : According to question(v0 )t C 21 (vH )0 C22γR ( 273 t )32 1 γR 27322( Q )p140 7 cal/mol-K n T2 10Cv Cp – R 5 cal/mol-KCp 7 γ Cv 5273 t 2734t 273 3 819 C39. Answer (4)Sol. : CpIHint : L2 L1 10 log10 2I1(in dB)γ 1 ISol. : L2 L1 10 log10 2I1(in dB)90 – 40 10 log1050 10 log105 log1072 1 f52 2 5 ff 543. Answer (3)Hint : As the observer moves away from sourceapparent frequency goes on decreasing.I2I1I2I1I2I1 v v0 Sol. : Apparent frequency n ′ n v I2 105I1v n′ 0 n 1 v v n 0nvv8 0100v40. Answer (1)Hint : Equation of stationary wave from free end2πx2πt cosλTSol. : Given equation of stationary waveπxy 5 sin cos 40 πt3On comparing2πx πx λ3λ 6 cm y 2a sinDistance between adjacent node and antinode 2f8100 28 m/s.v2 u2 2as28 28 s 196 m2 244. Answer (3)Hint & Sol. : When observer moves away fromsource.v0 350 λ4 1.5 cm.41. Answer (4)Hint : Phase difference between two particles at2π xdistance x, φλ v n ′ n 1 0 v nn′ v0 nvAakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234566/16

Test - 4 (Code-E) (Hints & Solutions)All India Aakash Test Series for Medical-It is an equation of straight line with negative slope.1n , If T and µ are constantslSol. : n1 : n2 : n3 1 : 2 : 3l1 : l2 : l3 45. Answer (4)Hint : Frequency of stretched stringn 12l1 1 1::n1 n2 n3 1:Tµ1 1:2 3 6:3:2[CHEMISTRY]46. Answer (2)Sol. :Hint : IUPAC naming of an ether is Alkoxy alkane.Sol. :It has total 8 α-H atoms which will participate inhyperconjugation.52. Answer (3)2-Methoxypropane.Hint : Basic strength of a base depends onavailability of lone pair of electrons on N-atom.47. Answer (3)Hint :Sol. : In compounds,and48. Answer (1)Hint : –I effect decreases in the following order, lone pair is delocalizedhence, weak bases.–NO2 –F –OH – C CHSo,49. Answer (4)Hint:Cyclicplanarstructureshaving(4n 2)π delocalized electrons are aromatic.is most powerful base.53. Answer (1)Hint : % of N Sol. :1.4 meq. of NH3Mass of compoundSol. : Meq. of NH3 25 1 % N in the compound 1.4 25 1 28%1.2554. Answer (1)Hint : For keto-enol tautomerism molecule musthave enolizable H-atom.50. Answer (2)Hint : Electronelectrophile.Sol. :deficientspeciesiscalledSol. :– nucleophile, AlCl3 – electrophile Nucleophile ,NH3– Nucleophile51. Answer (2)2Hint : α-H atoms on carbon next to sp hybridisedcarbon take part in hyperconjugation.Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234567/16

All India Aakash Test Series for Medical-Test - 4 (Code-E) (Hints & Solutions)55. Answer (4)Sol. :Hint : Acidic strength increases with increase instability of conjugate base.isomerism.62. Answer (2)Sol. :Hint :p-nitrophenol is strongest acid as phenoxide ion isstabilized by strong –R effect of –NO2 group.56. Answer (2)63. Answer (4)Hint : Resonance stabilized carbanion is highlystable.Hint :Sol. :Sol. :Negative charge is delocalized over oxygen atom,hence, highly stable anion.57. Answer (3)Hint : Aromatic compounds in which strongelectron withdrawing groups are bonded tobenzene ring don’t undergo Friedel Crafts reactioneasily.Sol. :does not show geometricalshows geometrical isomerism.This molecule has a chiral carbon, hence, showsoptical and geometrical isomers.64. Answer (2)Hint : R — COONa NaOHRH Na2CO3Sol. :doesn’t undergo Friedel Craftsreaction easily.65. Answer (3)58. Answer (2)Hint : Reductive ozonolysis of an alkene producesaldehyde or ketone.Hint : Chiral compound is optically active.Sol. : Butan-2-ol has one chiral carbon hence,optically active.Sol. :59. Answer (3)Hint : Alkanes can exhibit chain and positionisomerism.Sol. : Hexane (C6H14) has 5 structural isomers:n-hexane, 2-methylpentane, 2,2-dimethylbutane,2,3-dimethylbutane, 3-methylpentane.66. Answer (3)60. Answer (1)Hint : Such liquids are made to boil at atemperature lower than their normal boiling pointsby reducing the pressure on their surface.61. Answer (4)Hint :Hint :67. Answer (2)type molecules do not exhibitHint : Acetylene contains acidic hydrogens.δ δ δ Sol. : CH3 — MgBr H— C CH CH4 (g)geometrical isomerism.Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-476234568/16

Test - 4 (Code-E) (Hints & Solutions)All India Aakash Test Series for Medical-Sol. : Rate of nitration follows the order68. Answer (2)Hint : Addition of HX to alkynes finally producesgemdihalides.Sol. :75. Answer (1)Hint : CaC2 2H2O HC CH Ca(OH)2(P)Sol. :69. Answer (1)76. Answer (4)Hint : Alkenes give electrophilic addition reactions.Hint : HBr will show Markovnikov addition.70. Answer (1)Sol. :Hint : For isomeric alkanes, as the surface areadecreases, boiling point of alkane decreases.Sol. : B.P. order :n-pentane iso-pentane neo-pentane.71. Answer (4)Hint : Ozonolysis followed by hydrolysis doescleavage of C C.Sol. : Aldehydes have general formula CnH2nO,So, 12n 2n 16 4477. Answer (3)14n 16 4414n 28 n 2Hint :So, the aldehyde is CH3 — CHOSol. :Hence, alk