Answer solution with analysis aieee 2011 aakash

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Answer solution with analysis aieee 2011 aakash

  1. 1. AIEEE 2011 Answers by (Division of Aakash Educational Services Ltd.) CODE CODE CODEQ.N. P Q R S Q.N. P Q R S Q.N. P Q R S 01 2 1 4 1,3 31 2 2 1 3 61 4 2 2 4 02 3 1 4 1 32 3 4 1 1 62 1 1 3 1 03 1 2 2 3 33 4 2 4 2 63 4 4 3 2 04 4 3 4 1 34 3 4 4 4 64 2 3 1 4 05 3 3 1 3 35 2 1 4 2 65 3 2 1 3 06 2 1 1 3 36 3 3 2 4 66 4 2 2 4 07 1 3 3 3 37 1 4 3 4 67 3 4 3 2 08 3 4 3 1 38 1 2 2 1 68 3 4 4 4 09 4 3 3 2 39 2 4 1 3 69 3 4 4 3 10 1 1 4 1 40 2 3 1 2 70 1 4 1 4 11 1 1 1 3 41 2 2 3 2 71 3 2 1 1 12 3 1 2 2 42 2 3 2 1 72 2 4 4 2 13 4 4 2 1 43 2 2 4 4 73 4 4 2 2 14 1 3 2 2 44 1 4 4 4 74 2 2 3 4 15 3 3 2 3 45 4 4 2 3 75 2 2 4 1 16 4 3 1 3 46 2 4 2,4 3 76 1 1 1 4 17 3 1 4 3 47 4 1 2 1 77 1 3 3 4 18 3 1 4 1 48 4 3 4 1 78 1 2 3 2 19 3 1 2 2 49 2 1 1 3 79 3 4 1 4 20 1 2 4 4 50 2 1 4 4 80 1 3 1 1 21 4 4 3 2 51 3 1 2 2 81 2 3 1 3 22 2 1 3 1 52 4 1 1 1 82 3 3 3 4 23 3 2 3 4 53 2 1 1 4 83 3 1 3 2 24 4 1 3 4 54 4 2 2 1 84 2 2 3 4 25 4 1 3 1 55 2 1 4 4 85 2 2,4 1 4 26 1 3 3 1 56 3 1 2 1 86 2 3 3 2 27 2 4 3 2 57 2 1 4 4 87 4 3 4 3 28 4 1 4 2 58 4 2 3 3 88 2 3 3 2 29 1 2 2 2 59 2 2 2 4 89 1 4 3 2 30 1,3 3 1 3 60 4 3 1 3 90 2 2 3 4 Though every care has been taken to provide the answers correctly but the Institute shall not be responsible for typographical error, if any. (Division of Aakash Educational Services Ltd.) Regd. Office: Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-110075. Ph.: (011) 47623456. Fax: (011) 47623472 Toll Free: 1800-102-2727 Distance Learning Program (DLP) / Correspondence Course DivisionAakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-110075. Ph.: (011) 47623417/23/14 Fax: (011) 47623472 TOP RANKERS ALWAYS FROM AAKASH
  2. 2. DATE : 01/05/2011 Code - R Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472Time : 3 hrs. Solutions Max. Marks: 360 for AIEEE 2011 (Mathematics, Chemistry & Physics) Important Instructions : 1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited. 2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the question paper A, B, C consisting of Mathematics, Chemistry and Physics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.. 6. Candidates will be awarded marks as stated above in Instructions No. 5 for correct response of each question. ¼ (one-fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. 7. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above. 8. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination hall/room. 9. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However the candidates are allowed to take away this Test Booklet with them. 10. The CODE for this Booklet is R. Make sure that the CODE printed on Side-2 of the Answer Sheet is the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet 11. Do not fold or make any stray marks on the Answer Sheet.
  3. 3. AIEEE-2011 (Code-R) PART–A : MATHEMATICSDirections : Questions number 1-3 are based on the following  1  cos{2(x  2)} paragraph. 3. lim   x 2  x2 1. Consider 5 independent Bernoulli’s trials each with   probability of success p. If the probability of at least 1 (1) Equals (2) Does not exist 31 2 one failure is greater than or equal to , then p 32 (3) Equals 2 (4) Equals  2 lies in the interval Ans. (2)  11  1 3 (1)  , 1 (2)  ,  12  2 4   1  cos2( x  2)  Sol. lim  x2  x2    3 11   1   (3)  ,  (4)  0,   4 12   2 2|sin( x  2)|Ans. (4)  lim x2 x2Sol. Probability of atleast one failure which doesnt exist as L.H.L.   2 whereas 31  1  no failure  R.H.L.  2 32 4. Let R be the set of real numbers. 31 Statement-1 :  1  p5  32 R ) A = {(x, y)  R × td.: y – x is an integer} is an L es equivalence relation on R. rvic 1  p5  Se Statement-2 : n al 32 B tio{(x, y)  R × R : x = y for some rational u ca =  p 1 E d number } is an equivalence relation on R. 2 ka sh (1) Statement-1 is false, Statement-2 is true f Aa Also p  0 io no (2) Statement-1 is true, Statement-2 is true; (D i vi s Statement-2 is a correct explanation for Statement-1  1 Hence p  0,   2 (3) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for2. The coefficient of x 7 in the expansion of Statement-1 (1 – x – x2 + x3)6 is (4) Statement-1 is true, Statement-2 is false (1) 132 (2) 144 Ans. (4) (3) –132 (4) –144 Sol. Statement-1 is trueAns. (4) We observe that ReflexivitySol. We have (1  x  x 2  x 3 )6  (1  x )6 (1  x 2 )6 xR x as x  x  0 is an integer, x  A Symmetric coefficient of x7 in Let ( x , y ) A 1  x  x  6 2  x3  6C1 . 6C3  6C3 . 6C2  6C5 . 6C1  y – x is an integer  x – y is also an integer  6  20  20  15  6  6 Transitivity  144 Let ( x , y ) A and ( y , z ) AAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (2)
  4. 4. AIEEE-2011 (Code-R)  y – x is an integer and z – y is an integer 7. The number of values of k for which the linear equations  y – x + z – y is also an integer 4x + ky + 2z = 0  z – x is an integer kx + 4y + z = 0 2x + 2y + z = 0  (x , z ) A possess a non-zero solution is Because of the above properties A is an equivalence (1) Zero (2) 3 relation over R (3) 2 (4) 1 Statement 2 is false as B is not symmetric on  Ans. (3) We observe that Sol. For non-trivial solution of given system of linear equations 0Bx as 0  0.xx  but (x ,0) B 4 k 25. Let ,  be real and z be a complex number. If z2 + k 4 1 0 z +  = 0 has two distinct roots on the line 2 2 1 Re z = 1, then it is necessary that (1)   (1, ) (2)   (0, 1)  8  k (2  k )  2(2 k  8)  0 (3)   (–1, 0) (4) || = 1  k 2  6k  8  0Ans. (1)  k 2  6k  8  0Sol. Let the roots of the given equation be 1 + ip and  k  2,4 1 – ip, where p Clearly there exists two values of k. 8. Statement-1 :    product of roots The point A(1, 0, 7) is the mirror image of the point x .) y  1 z  2  (1  ip)(1  ip)  1  p2  1, p  B(1, 6, 3) in the line : Ltd  .  (1, ) Statement-2 : rvi ce s 1 2 3 e lS 2 d x a ti The line :on a x  y  1  z  2 bisects the line segment du c6. equals 1 2 3 dy 2 E ka sh joining A(1, 0, 7) and B(1, 6, 3). Aa 3 1  d 2 y   dy   d2y  (1) Statement-1 is false, Statement-2 is true f no (1)   2    (2)  2  i si o  dx   dx   dx  (2) Statement-1 is true, Statement-2 is true; v (D i Statement-2 is a correct explanation for 1 3 2  d 2 y   dy   d 2 y   dy  Statement-1 (3)   2    (4)  2     dx   dx   dx   dx  (3) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation forAns. (1) Statement-1Sol. We have (4) Statement-1 is true, Statement-2 is false Ans. (3) d 2 x d  dx  d  1  Sol. The mid point of A(1, 0, 7) and B(1,6,3) is (1, 3,5)    dy 2 dy  dy  dy  dy     dx  x y 1 z2 which lies on the line   1 2 3 d  1  dx 1 d2 y 1 Also the line passing through the points A and B is  .  . 2. perpendicular to the given line, hence B is the dx  dy  dy 2  dy  dx  dy     dx  mirror image of A, consequently the statement-1 is  dx   dx      true. Statement-2 is also true but it is not a correct 3  d2 y  explanation of statement-1 as there are infinitely 1 d2 y  dy   .     2 many lines passing through the midpoint of the line 3 2  dx   dx   dy  dx   segment and one of the lines is perpendicular  dx  bisector.  Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (3)
  5. 5. AIEEE-2011 (Code-R)9. Consider the following statements 11. A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent P : Suman is brilliant months his saving increases by Rs. 40 more than the saving of immediately previous month. His total Q : Suman is rich saving from the start of service will be Rs. 11040 after R : Suman is honest (1) 21 months (2) 18 months The negation of the statement "Suman is brilliant (3) 19 months (4) 20 months and dishonest if and only if Suman is rich" can be Ans. (1) expressed as Sol. Total savings = 200 + 200 + 200 + 240 + 280 + ... to (1) ~ (P  ~ R)  Q (2) ~ P  (Q  ~ R) n months = 11040 (3) ~ (Q  (P  ~ R)) (4) ~ Q  ~ P  RAns. (3)  400  n –2 2  400   n – 3 . 40  11040 Sol. Suman is brilliant and dishonest is P  ~ R  (n – 2) ( 140 + 20n) = 10640 Suman is brilliant and dishonest iff suman is rich is  20n2 + 100n – 280 = 10640 Q  (P  ~ R)  n2 + 5n – 546 = 0 Negative of statement is expressed as  (n – 21) (n + 26) = 0 ~ (Q  (P  ~ R))  n = 21 as n  – 2610. The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect 12. Equation of the ellipse whose axes are the axes of the line L3 : y + 2 = 0 at P and Q respectively. The coordinates and which passes through the point bisector of the acute angle between L 1 and L 2 intersects L3 at R. 2 .) (– 3, 1) and has eccentricity td 5 is Statements 1 : The ratio PR : RQ equals 2 2 : 5 . L Statement 2 : In any traingle, bisector of an angle i (1) 5 x 2  erv2 – 32  0 3y ce s divides the triangle into two similar triangles. n al S (2) io 2 cat 3 x  5y – 32  0 2 (1) Statement-1 is true, Statement-2 is false u (2) Statement-1 is true, Statement-2 is true; h Ed Statement-2 is the correct explanation aka of s (3) 5 x 2  3 y 2 – 48  0 Statement-1 of A i on (4) 3 x 2  5 y 2 – 15  0 (3) i vi s Statement-1 is true, Statement-2 is true; (D Ans. (2) Statement-2 is the not the correct explanation of Statement-1 x2 y2 (4) Statement-1 is false, Statement-2 is true Sol. Let the equation of the ellipse be 2  =1 a b2Ans. (4) 9 1Sol. The figure is self explanatory Which will pass through (– 3, 1) if 2  2 1 y=x a b L2 Y L1 b2 2 and eccentricity = e = 1– 2  a 5 X O y + 2x = 0 2 b2 (–2, –2) (1, – 2)   1– 2 L3 5 a P R Q y+2=0 b2 3  2  PR OP 2 2 a 5   RQ OQ 5 3 2 Statement-2 is false  b2 = a 5Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (4)
  6. 6. AIEEE-2011 (Code-R) 9 1  /4 Thus  2 1  1 – tan   a 2 b = 8  log 1  1  tan   d 0   9 5  1 a 2 3a 2  /4  2   27 + 5 = 3a2 = 32 = 8  log  1  tan   d 0   32 3 32 32  a  , b    2 2  3 5 3 5 = 8  log2 . –I 4 Required equation of the ellipse is 3x2 + 5y2 = 32  2I = 2log213. If A = sin2x + cos4x, then for all real x  I = log2 3 13 3 (1) A (2)  A1 y –1 z – 3 4 16 4 15. If the angle between the line x   and 2  13 (3)  A1 (4) 1  A  2  5  16 the plane x + 2y + 3z = 4 is cos–1  14  , then   Ans. (2) equalsSol. We have, A = sin2x + cos4x = 1 – cos2x + cos4x 5 2 (1) (2) 3 3 2  1 1 = 1   cos x – 2  2  –4  3 2 (3) (4) 2 L t d. ) 5 es rvic 2 3  1 3 Ans. (2) =   cos x –   2 4  2 4 l Se Sol. The direction ratios of the given line a x ca i n tyo– 1 z – 3 3 u Clearly  A1 h E d1  2   are k as 4 fA a 1, 2,  8log 1  x  no 1  io Hence direction cosines of the line are14. The value of dx is vi s 0 1  x2 (D i 1 , 2 ,  (1) log 2 (2) log2 5 2 5  2 5  2   Also the direction cosines of the normal to the plane (3) log2 (4) log2 8 2 1 2 3 are , , .Ans. (2) 14 14 14Sol. We have, Angle between the line and the plane is , 1  4  3 8 log 1  x  1 then cos(90° –) = = sin I  0 1 x 2 dx  5  2  14 Put x = tan 3 5  3    14   5 14 2 log 1  tan   4  I=  8. sec  2 sec 2  d   9  2  45  9 2  30  25 0  30 = 20  /4 = 8  log1  tan  d  0 = 2 3Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (5)
  7. 7. AIEEE-2011 (Code-R)    25  2616. For x   0,  , define Sol. From the given data, median = a = 25.5a  2 2 Required mean deviation about median x f  x   t sint dt 2 | 0.5  1.5  2.5  ...  24.5 | = | a |  50 0 50 Then f has  |a| = 4 (1) Local maximum at  and local 2   1 1 ˆ ˆ ˆ (2) Local maximum at  and 2 19. If a  (3i  k ) and b  (2i  3 ˆ  6 k ) , then the ˆ j 10 7 (3) Local minimum at  and 2       value of (2 a  b )  [( a  b )  ( a  2 b)] is (4) Local minimum at  and local maximum at 2 (1) 3 (2) –5Ans. (1) (3) –3 (4) 5Sol. We have, Ans. (2) x Sol. We have f(x) =  t sin t dt          2a – b .  a  b  a  2b     0        f (x) = x sin x   = 2a – b . b – 2a    = –  2a – b  For maximum or minimum value of f(x), f (x) = 0 2  x = n, n Z  2  2   We observe that = –  4 | a |  | b | – 4a. b   .)  – ve = – [4 + 1] = – 5 s Lt d –ve +ve e rvic 0  2 5 Se 20. The values of p and q for which the function al  n u c a ti o  sin( p  1)x  sin x  Ed f (x) changes its sign from +ve to – ve in the , x0 neighbourhood of  and – ve to +ve in the sh  x  neighbourhood of 2 a ka f ( x )  q , x0 Hence f(x) has local maximum at x = ion local and of A   xx  x 2 i vi s  , x0 minima at x = 2  x 3/2 (D is continuous for all x in R, are 1 1 3 1 317. The domain of the function f ( x )  is (1) p  ,q (2) p  ,q | x|  x 2 2 2 2 5 1 3 1 (1) (–, ) – {0} (2) (–, ) (3) p  ,q (4) p   , q  2 2 2 2 (3) (0, ) (4) (–, 0) Ans. (4)Ans. (4) Sol. The given function f is continuous at x = 0 ifSol. The given function f is well defined only lim f (0  h )  f (0)  lim f (0  h ) h 0 h 0 when |x| – x > 0 1  x<0  p2 q  2 Required domain is (– , 0) 3 1  p ,q18. If the mean deviation about the median of the 2 2 numbers a, 2a, ......, 50a is 50, then |a| equals 21. The two circles x2 + y2 = ax and x2 + y2 = c2 (c > 0) (1) 5 (2) 2 touch each other, if (3) 3 (4) 4 (1) |a| = 2c (2) 2|a| = cAns. (4) (3) |a| = c (4) a = 2cAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (6)
  8. 8. AIEEE-2011 (Code-R)Ans. (3) Sol. We have y-axis CD  CD=C  C  P(C  D) P(C )  P     P(C ) D P(D) P(D) (c, 0) as 0  P(D)  1 (0, 0) x-axis a/2 24. Let A and B be two symmetric matrices of order 3.Sol. Statement-1 : A(BA) and (AB)A are symmetric matrices. Statement-2 : AB is symmetric matrix if matrix multiplication of A with B is commutative. The figure is self explanatory (1) Statement-1 is false, Statement-2 is true Clearly c = |a| (2) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for22. Let I be the purchase value of an equipment and Statement-1 V(t) be the value after it has been used for t years. (3) Statement-1 is true, Statement-2 is true; The value V(t) depreciates at a rate given by Statement-2 is not a correct explanation for dV ( t ) Statement-1 differential equation   k(T  t ) , where k > 0 dt (4) Statement-1 is true, Statement-2 is false is a constant and T is the total life in years of the Ans. (3) equipment. Then the scrap value V(T) of the equipment is Sol. Clearly both statements are true but statement-2 is not a correct explanation )of statement-1. .td (1) e–kT (2) T 2  I 25. If ( 1) is a cubece s Lof unity, and (1 + )7 = A + B. k Then (A, B)Ser vi root equals al (1) (–1,n1) kT 2 k(T  t )2 ti o (2) (0, 1) uca(1, 1) (3) I  (4) I  Ed 2 2 (3) (4) (1, 0)Ans. (3) k a sh (3) Ans. a of A Sol. (1 + )7 = A + B i on V (T ) T vi s  (–2)7 = A + B  dV (t )   k(T  t )dt (D iSol. I t 0  –2 = A + B T  1 +  = A + B  (T  t )2   V (T )  I  k  2   A = 1, B = 1   0   (A, B) = (1, 1) T 2  26. Statement-1 : The number of ways of distributing  V (T )  I   k   10 identical balls in 4 distinct boxes such that no  2  box is empty is 9C3. kT 2 Statement-2 : The number of ways of choosing any  V (T )  I  2 3 places from 9 different places is 9C3.23. If C and D are two events such that C  D and (1) Statement-1 is false, Statement-2 is true P(D)  0, then the correct statement among the (2) Statement-1 is true, Statement-2 is true; following is Statement-2 is the correct explanation for Statement-1 P( D) (1) P(C|D)  (2) P(C|D) = P(C) (3) Statement-1 is true, Statement-2 is true; P(C ) Statement-2 is not a correct explanation for (3) P(C|D)  P(C) (4) P(C|D) < P(C) Statement-1Ans. (3) (4) Statement-1 is true, Statement-2 is falseAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (7)
  9. 9. AIEEE-2011 (Code-R)Ans. (3) 1 e =  [ln x ]1Sol. Number of ways to distribute 10 identical balls in 2 four distinct boxes such that no box remains empty = 10–1C4–1 = 9C3 1 3 =  1  0  sq. units 2 2 Number of ways to select 3 different places from 9 places = 9C3 dy 29. If  y  3  0 and y(0) = 2, then y(ln 2) is equal Clearly statement-2 is not a correct explanation of dx statement-1 to27. The shortest distance between line y – x = 1 and (1) –2 (2) 7 curve x = y2 is (3) 5 (4) 13 4 3 Ans. (2) (1) (2) 3 4 Sol. We have 8 dy 3 2 y3 (3) (4) dx 8 3 2Ans. (3) 1  dy  dxSol. The equation of the tangent to x = y2 having slope y3 1 1 is y  x   ln |(y + 3)| = x + k, where k is a constant of 4 integration 1  (y + 3) = c ex 1 Hence shortest distance = 4 Initially when x = 0, y.) 2 = 2 L td  c=5 rvi ce s 3 3 2 Finally l Serequired solution is y + 3 = 5ex the = 4 2  8 units na  tio (ln2) = 5e ln2 – 3 = 10 – 3 = 7 a y uc d28. The area of the region enclosed by the curves y = x, s hE   a ka 30. The vectors a and b are not perpendicular and 1 x = e, y  and the positive x-axis is o fA       c and d are two vectors satisfying b × c  b × d i on x v is    (1) 5 square units (2) 1 (D i square units and a  d  0 . Then the vector d is equal to     2 2   a c     b c   3 (1) c      b  ab  (2) b      c  ab  (3) 1 square units (4) square units     2       ac     bc Ans. (4) (3) c      b  ab  (4) b      c  ab      Ans. (1) 1 y=x y=x Sol. We have x=e     (1,1) bc  b d        a  (b  c )  a  (b  d )Sol.            ( a  c )b  ( a  b )c  ( a  d )b  ( a  b )d         ( a  b )d  ( a  c )b  ( a  b)c Required area e   1 1 =  1  1   dx  (a  c )   2 x  d     bc 1 (a  b)Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (8)
  10. 10. AIEEE-2011 (Code-R) PART–B : CHEMISTRY31. In context of the lanthanoids, which of the following Sol. Cr +3 in octahedral geometry always form inner statements is not correct? orbital complex. (1) Availability of 4f electrons results in the 35. The rate of a chemical reaction doubles for every formation of compounds in +4 state for all the 10°C rise of temperature. If the temperature is raised members of the series by 50°C, the rate of the reaction increases by about (2) There is a gradual decrease in the radii of the (1) 64 times (2) 10 times members with increasing atomic number in the (3) 24 times (4) 32 times series Ans. (4) (3) All the members exhibit +3 oxidation state (4) Because of similar properties the separation of rate at (t + 10)°C Sol. Temperature coefficient () = lanthanoids is not easy rate at t°CAns. (1)  = 2 here, so increase in rate of reaction = ()nSol. Lanthanoid generally show the oxidation state of +3. When n is number of times by which temperature is raised by 10°C.32. In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face 36. a and b are van der Waals constants for gases. centre positions. If one atom of B is missing from one Chlorine is more easily liquefied than ethane of the face centred points, the formula of the because compound is (1) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6 (1) A2B5 (2) A2B (2) a and b for Cl2 > a and b for C2H6 (3) AB2 (4) A2B3 (3) a and b for Cl2 < a and b for C2H6Ans. (1) (4) a for Cl2 < a for C2tH.6) but b for Cl2 > b for C2H6 d L esSol. ZA  8 Ans. (2) e rvic 8 Sol. Both a and S for Cl is more than C2H6. al b 5 ti o n u ca ZB  37. The hybridisation of orbitals of N atom in 2 EdNO3 – , NO2 and NH4 are respectively So formula of compound is AB5/2 ka sh i.e., A2B5 o f Aa (1) sp2, sp3,sp (2) sp, sp2,sp333. The magnetic moment (spin only) of [NiCls]onis (3) sp2, sp,sp3 (4) sp, sp3,sp2 4i 2– (1) 1.41 BM (2) 1.82 BM(D i vi Ans. (3) Sol. NO3– – sp2 ; NO2+ – sp ; NH4+ – sp3 (3) 5.46 BM (4) 2.82 BM 38. Ethylene glycol is used as an antifreeze in a coldAns. (4) climate. Mass of ethylene glycol which should beSol. Hybridisation of Ni is sp3 added to 4 kg of water to prevent it from freezing at –6°C will be: (Kf for water = 1.86 K kgmol–1 and Unpaired e– in 3d8 is 2 molar mass of ethylene glycol = 62gmol–1) So   n(n  2) BM (1) 304.60 g (2) 804.32 g = 2  4  8  2.82 BM (3) 204.30 g (4) 400.00 g34. Which of the following facts about the complex Ans. (2) [Cr(NH3)6]Cl3 is wrong? Sol. Tf = Kf.m for ethylenglycol in aq. solution (1) The complex gives white precipitate with silver w 1000 nitrate solution Tf  K f  mol. wt. wt. of solvent (2) The complex involves d 2 sp3 hybridisation and is octahedral in shape 1.86  w  1000 6 (3) The complex is paramagnetic 62  4000 (4) The complex is an outer orbital complex  w = 800 gAns. (4) So, weight of solute should be more than 800 g.Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (9)
  11. 11. AIEEE-2011 (Code-R)39. The outer electron configuration of Gd (Atomic No.: 43. A gas absorbs a photon of 355 nm and emits at two 64) is wavelengths. If one of the emissions is at 680 nm, (1) 4f7 5d1 6s2 (2) 4f3 5d5 6s2 the other is at : (1) 518 nm (2) 1035 nm (3) 4f8 5d0 6s2 (4) 4f4 5d4 6s2 (3) 325 nm (4) 743 nmAns. (1) Ans. (4)Sol. Gd = 4f 75d 16s 240. The structure of IF7 is  1 1 Sol.   (1) Pentagonal bipyramid  1  2 (2) Square pyramid  1 1   355 680  2 (3) Trigonal bipyramid (4) Octahedral 1 680  355 Ans. (1)  2 355  680Sol. Hybridisation of iodine is sp 3d 3  2 = 743 nm So, structure is pentagonal bipyramid. 44. Identify the compound that exhibits tautomerism.41. Ozonolysis of an organic compound gives (1) Phenol (2) 2–Butene formaldehyde as one of the products. This confirms (3) Lactic acid (4) 2–Pentanone the presence of : Ans. (4) (1) An acetylenic triple bond O (2) Two ethylenic double bonds (3) A vinyl group Sol. (4) An isopropyl group 2-Pentanone t d. ) has -hydrogen & hence it will exhibit tautomerism. sLAns. (3) ce 45. The entropy ichange involved in the isothermal  Ozonolysis lS e rv reversible expansion of 2 moles of an ideal gas from on aSol.  C = CH2 HCHO a volume of 10 dm3 at 27°C is to a volume of 100 dm3  a ti Vinylic group E duc 42.3 J mol—1 K—1 (1) (2) 38.3 J mol—1 K—142. The degree of dissociation () of a weak electrolyte, as h A xB y is related to vant Hoff factor (i) by A ak (3) 35.8 J mol—1 K—1 f the Ans. (2) (4) 32.3 J mol—1 K—1 expression: io no vi s x y1 (Di 1 i Sol. S  nR ln V2 (1)   (2)    x  y  1  V1 i1 100 i1 x y1 S  2.303  2  8.314 log (3)   (4)   10 x y1 i 1 S = 38.3 J/mole/KAns. (2) 46. Silver Mirror test is given by which one of theSol. Van’t Hoff factor (i) following compounds? Observed colligative property (1) Benzophenone (2) Acetaldehyde = Normal colligative property (3) Acetone (4) Formaldehyde A x B y  xA  y  yB  x  1  x y Ans. (2, 4 ) Total moles = 1 –  + x + y Sol. Both formaldehyde and acetaldehyde will give this test = 1 + (x + y – 1) [Ag(NH ) ] 1  (x  y  1) HCHO  Ag   Organic compound 3 2  i Silver mirror 1 [Ag (NH3 )2 ] i 1 CH 3 – CHO  Ag      Silver mirror x y1 Organic CompoundAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (10)
  12. 12. AIEEE-2011 (Code-R)47. Trichloroacetaldehyde was subject to Cannizzaros Sol. KBr  KBrO 3  Br2  .....  reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and Br another compound. The other compound is : OH OH H2O (1) Chloroform + Br2 Br Br (2) 2, 2, 2-Trichloroethanol (3) Trichloromethanol 2, 4, 6-Tribomophenol 50. Among the following the maximum covalent (4) 2, 2, 2-Trichloropropanol character is shown by the compoundAns. (2) (1) MgCl2 (2) FeCl2 (3) SnCl2 (4) AlCl3 Cl Cl Ans. (4) NaOH Sol. According to Fajans rule, cation with greater chargeSol. Cl–C–CHO Cl–C–COONa + Cannizaros reaction and smaller size favours covalency. Cl Cl 51. Boron cannot form which one of the following anions? Cl (1) BO  (2) BF3 2 6 Cl–C–CH2–OH 2 1 (3) BH (4) B(OH) 4 4 Cl Ans. (2) 2, 2, 2-trichloroethanol Sol. Due to absence of low lying vacant d orbital in B. sp3d2 hybridization is not possible hence BF63– will48. The reduction potential of hydrogen half-cell will be not formed. negative if : 52. Sodium ethoxide has reacted with ethanoyl chloride. The compound that is) produced in the above (1) p(H2) = 2 atm and [H+] = 2.0 M . reaction is Lt d (2) p(H2) = 1 atm and [H+] = 2.0 M es e rvic (1) Ethyl ethanoate (2) Diethyl ether al S (3) p(H2) = 1 atm and [H+] = 1.0 M (3) 2–Butanone (4) Ethyl chloride Ans. (1) tion u ca (4) p(H2) = 2 atm and [H+] = 1.0 M dAns. (4) s hE ka Aa O 1 f no  Sol. H  e  H 2  – io Sol. CH3—CH2ONa + Cl—C—CH3 i vi s 2 Sodium ethoxide Apply Nernst equation (D 1 0.059 PH 2 O E  0 log 2 1 [H ] CH3 —CH2—O—C—CH3 0.059 2 1/2 Cl E log – 1 1 –Cl Therefore E is negative. O49. Phenol is heated with a solution of mixture of KBr CH3—CH2—O—C—CH3 and KBrO3 . The major product obtained in the Ethylacetate above reaction is : (1) 2, 4, 6-Tribromophenol 53. Which of the following reagents may be used to (2) 2-Bromophenol distinguish between phenol and benzoic acid? (3) 3-Bromophenol (1) Neutral FeCl (2) Aqueous NaOH 3 (4) 4-Bromophenol (3) Tollens reagent (4) Molisch reagentAns. (1)Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (11)

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