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Answer solution with analysis aieee 2011 aakash
 

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    Answer solution with analysis aieee 2011 aakash Answer solution with analysis aieee 2011 aakash Document Transcript

    • 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
    • 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.
    • 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)
    • 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)
    • 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)
    • 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)
    • 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)
    • 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)
    • 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)
    • 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)
    • 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)
    • 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)
    • AIEEE-2011 (Code-R)Ans. (1) Al2O3 < MgO < Na2O Sol. Neural FeCl3 gives violet colored complex Increasing basic strength OH 3– and while descending in the group basic character   of corresponding oxides increases. +FeCl3 3H +  O Fe + 3HCl  6   6 Na2O < K2O  Increasing basic strength Violet coloured complex  Correct order is54. A vessel at 1000 K contains CO2 with a pressure of Al2O3 < MgO < Na2O < K2O 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at 57. A 5.2 molal aqueous solution of methyl alcohol, equilibrium is 0.8 atm, the value of K is CH3OH is supplied. What is the mole fraction of (1) 0.18 atm (2) 1.8 atm methyl alcohol in the solution? (3) 3 atm (4) 0.3 atm (1) 0.050 (2) 0.100Ans. (2) (3) 0.190 (4) 0.086Sol. CO2(g) + C(s) 2CO Ans. (4) No. of moles of compound at 1000 K P = 0.5 atm 0 Sol. x  Total no. of mole of solution 0.5 – p 2p For 1 kg solvent According to given condition 0.5 + p = 0.8 5.2 5.2 x   0.086 p = 0.3 55.5  5.2 60.7 (0.6)2 0.36 58. The presence or absence of hydroxy group on which So, K p    1.8 atm carbon atom of sugar differentiates RNA and DNA? (0.2) 0.255. The strongest acid amongst the following (1) 4th td .) (2) 1st compounds is : sL (3) 2nd i ce (4) 3 rd (1) ClCH CH CH COOH 2 2 2 Ans. (3) e rv lS (2) CH COOH a ti Sol. Fact. on a du c 3 (3) HCOOH E h 59. Which of the following statement is wrong? (4) CH CH CH(Cl)CO H ka s Aa 3 2 2 (1) N2O4 has two resonance structures fAns. (4) io no (2) The stability of hydrides increases from NH3 to O (D i vi s BiH3 in group 15 of the periodic table (3) Nitrogen cannot form d-p bondSol. CH3—CH2—CH—C—OH (4) Single N - N bond is weaker than the single P - P bond Cl Ans. (2) Presence of electron withdrawing group nearest to the carboxylic group increase the acidic strength to Sol. Stability of hydrides from NH3 to BiH3 decreases maximum extent due to decreasing bond strength.56. Which one of the following orders presents the 60. Which of the following statements regarding correct sequence of the increasing basic nature of sulphur is incorrect? the given oxides? (1) The oxidation state of sulphur is never less than (1) K O  Na O  Al O  MgO +4 in its compounds 2 2 2 3 (2) S2 molecule is paramagnetic (2) Al O  MgO  Na O  K O 2 3 2 2 (3) The vapour at 200°C consists mostly of S8 rings (3) MgO  K O  Al O  Na O 2 2 3 2 (4) At 600°C the gas mainly consists of S2 molecules (4) Na O  K O  MgO  Al O 2 2 2 3 Ans. (1)Ans. (2 ) Sol. The oxidation state of S may be less than +4.Sol. While moving from left to right in periodic table basic character of oxide of elements will decrease. i.e., in H2S it is –2.Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (12)
    • AIEEE-2011 (Code-R) PART–C : PHYSICS61. A carnot engine operation between temperatrues T1 63. Three perfect gases at absolute temperatures T1, T2 1 and T3 are mixed. The masses of molecules are m1, and T 2 has efficiency . When T2 is lowered by m2 and m3 and the number of molecules are n1, n2 6 1 and n3 respectively. Assuming no loss of energy, the 62 K, its efficiency increases to . Then T1 and T2 final temperature of the mixture is 3 are respectively n1 T12  n2 T22  n3 T32 2 2 2 T1  T2  T3 (1) 310 K and 248 K (2) 372 K and 310 K (1) n T  n T  n T (2) 1 1 2 2 3 3 3 (3) 372 K and 330 K (4) 330 K and 268 K n1T1  n2T2  n3T3 n1T12  n2T22  n3T32Ans. (2) (3) n1  n2  n3 (4) n1T1  n2T2  n3T3 1 T T2 5 Ans. (3)Sol.  1 2   6 T1 T1 6 n1 n n T1  2 T2  3 T3 T  62 2 Also, 2  Sol. U = N N N T1 3 n1 n2 n3   T2 5 N N N  T1 6 n1T1  n2 T2  n3T3 = T2  62 4 n1  n2  n3   T2 5 64. A boat is moving due east in a region where the or T2 = 62 × 5 earths magnetic field is 5.0 × 10–5 NA–1 due north ) and horizontal. The boat.carries a vertical aerial 2 m or T2 = 310 K s Lt d long. If the speed of the boat is 1.50 ms –1 , the  T1 = 372 K magnitude of rvice the induced emf in the wire of aerial62. A pulley of radius 2 m is rotated about its axis by is na l Se a force F = (20t – 5t2) newton (where t is measured (1) atio mV 0.15 (2) 1 mV uc in seconds) applied tangentially. If the moment of Ed 0.75 mV (3) (4) 0.50 mV inertia of the pulley about its axis of rotation is as h 10 kg m2 , the number of rotations made by theak Ans. (1) pulley before its direction of motion if reversed of A n is Sol.  = Bvl v i si o (D i (1) More than 9 = 5.0 × 10–5 × 2 × 1.50 (2) Less than 3 = 15 × 10–5 (3) More than 3 but less than 6 = 0.15 mV (4) More than 6 but less than 9 65. A thin horizontal circular disc is rotating about aAns. (3) vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insectSol. T = F × r = 40t – 10t2 now moves along a diameter of the disc to reach its  = 4t – t2 other end. During the journey of the insect the angular speed of the disc t3 2t 2  0 (1) First increase and then decrease 3  t=6s (2) Remains unchanged (3) Continuously decreases 2t 3 t 4 Also,  =  (4) Continuously increases 3 12 Ans. (1) 2666 =  108 Sol. L = I = Constant 3  I first decreases, then increases. n  5.73 2   first increases and then decreasesAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (13)
    • AIEEE-2011 (Code-R)66. Two identical charged spheres suspended from a Ans. (4) common point by two massless strings of length l are initially a distance d(d << l) apart because of N Sol. At t  t2 , N 2  (Left) their mutual repulsion. The charge begins to leak 3 from both the spheres at a constant rate. As a result 2N the charges approach each other with a velocity v. t  t1 , N 1  3 Then as a function of distance x between them  t2 – t1 = half life = 20 min 1  69. Energy required for the electron excitation in Li++ (1) v  x (2) v  x 2 from the first to the third Bohr orbit is 1 (1) 122.4 eV (2) 12.1 eV (3) v  x–1 (4) v  x 2 (3) 36.3 eV (4) 108.8 eVAns. (2) Ans. (4) FSol. tan   1 1 mg Sol. E = 13.6 (3)2  2  2  l  3 1  8 x kq 2 T = 13.6  9  or  9 2l mgx 2 F = 108.8 eV x3 kq 2 x  70. The electrostatic potential inside a charged spherical 2l mg mg ball is given by  = ar2 + b where r is the distance from the centre ; a, b are constants. Then the charge dx dq 3x 2 2 kq density inside the ball is dt  dt (1) –6 a0 (2) –24  a0r 2l mg Also, q  x3/2 (3) –6 a0r td .) (4) –24  a0 sL vice Ans. (1) dx x 3/2  2 , i.e., v  x 1/2 Sol.  = ar2 +Ser b  dt x al d i on at  2 ar  E  2 ar67. 100 g of water is heated from 30°C to 50°C. Ignoring E ducdr the slight expansion of the water, the change in sh its internal enetgy is (specific heat of water ak a E  4r 2  q f A is 0 4184 J/kg/K) no io vi s  q = –80ar3 (D i (1) 2.1 kJ (2) 4.2 kJ dq (3) 8.4 kJ (4) 84 kJ  = dVAns. (3) 240 r 2 drSol. Q = U + W = 4r 2 dr 100 = –60a As W = 0  U = Q = 4184   20 1000 71. Work done in increasing the size of a soap bubble = 8368 from a radius of 3 cm to 5 cm is nearly (Surface = 8.368 kJ tension of soap solution = 0.03 Nm–1)  8.4 kJ (1) 0.4 m J (2) 4 m J68. The half life of a radioactive substance is 20 minutes. (3) 0.2 m J (4) 2 m J The approximate time interval (t2 – t1) between the Ans. (1) 2 time t2 when of its has decayed and time t1 when Sol. W = SE 3 1 = 2 × T × 4(52 – 32) × 10–4 of its had decayed is = 2 × 0.03 × 4(16) × 10–4 3 (1) 28 min (2) 7 min = 3.84 × 10–4 (3) 14 min (4) 20 min = 0.4 mJAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (14)
    • AIEEE-2011 (Code-R)72. A resistor R and 2F capacitor in series is connected 74. An object, moving with a speed of 6.25 m/s, is through a switch to 200 V direct supply. Across the decelerated at a rate given by capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up dv  2.5 v 5 s after the switch has been closed. (log102.5 = 0.4) dt (1) 3.3 × 107  (2) 1.3 × 104  where v is the instantaneous speed. The time taken (3) 1.7 × 105  (4) 2.7 × 106  by the object, to come to rest, would beAns. (4) (1) 8 s (2) 1 sSol. Charge on capacitor q = q0(1 – e–t/RC) (3) 2 s (4) 4 s q0 Ans. (3) or V  (1  e t /RC ) C 0 t V = 200(1 – e–t/RC) dv Sol.  v  — 2.5 dt 120 = 200(1 – e–t/RC) v 0 e–t/RC = 0.4 2 v  2.5 t t  ln (0.4) RC 2 6.25  2.5t t  10  t=2s  ln    2.303  0.4 RC  4  75. Direction : The question has a paragraph followed by 5 two statements, Statement-1 and Statement-2. Of the R = 6 given four alternatives after the statements, choose the 2  10  2.303  0.4 .) one that describes the statements. = 2.7 × 106  Lt d A thin air film cesformed by putting the covex73. A current I flows in an infinitely long wire with surface of a erv i is plane-convex lens over a plane glass lS ona pattern due to light reflected from the corss section in the form of a semi-circular ring of plate. With monochromatic light, this film gives an raidus R. The magnitude of the magnetic induction a ti interference along its axis is E duc (coNvex) surface and the bottom (glass plate) top ka sh surface of the film. (1) 0 I (2) 0 I o f Aa Statement-1 : When light reflects from the air-glass 4 R 2 R n v i si o plate interface, the reflected wave 0 I 0 I (D i suffers a phase change of . (3) (4) 22 R 2 R Statement-2 : The centre of the interference patternAns. (2) is dark. dl (1) Statement-1 is false, Statement-2 is trueSol. (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of B   dB sin  Statement-1 0 di B sin  (4) Statement-1 is true, Statement-2 is true and 2 R Statement-2 is not the correct explanation of Statement-1 0  i   2 R  R      R sin  d 0 Ans. (4) Sol. The centre of interference pattern is dark, showing 0 i  that the phase difference between two interfering 2 R waves is  . But this does not imply statement-1.Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (15)
    • AIEEE-2011 (Code-R)76. Two bodies of masses m and 4 m are placed at a Ans. (3) distance r. The gravitational potential at a point on Sol. q  q0 cos t the line joining them where the gravitational field is zero is q0 energy is halved when q  9Gm 2 (1)  (2) Zero r   t  4Gm 6Gm 4 (3)  (4)  r r 1   tAns. (1) Lc 4Sol. Field is zero at O.   t  Lc r/3 2r/3 4 79. This question has Statement-1 and Statement-2. Of m o 4m the four choices given after the statements, choose the one that best describes the two statements. GM 4GM Statement-1 : A metallic surface is irradiated by a V – – monochromatic light of frequency r /3 2r /3 v > v0 (the threshold frequency). The 9GM maximum kinetic energy and the – r stopping potential are Kmax and V077. This question has Statement-1 and Statement-2. Of respectively. If the frequency incident the four choices given after the statements, choose on the surface is doubled, both the the one that best describes the two statements. Kmax and V0 are also doubled. Statement-1 : Sky wave signals are used for long Statement-2 : The maximum kinetic energy and the .) Ltd from a surface are linearly distance radio communication. These stopping potential of photoelectrons signals are in general, less stable es emitted than ground wave signals. rvicdependent on the frequency of Se incident light. Statement-2 : The state of ionosphere varies from n al ti o hour to hour, day to day and season d uca Statement-1 is false, Statement-2 is true (1) E to season. sh a ka (2) Statement-1 is true, Statement-2 is false of A (1) Statement-1 is false, Statement-2 is true n (3) Statement-1 is true, Statement-2 is true and i si o (2) Statement-1 is true, Statement-2 is false Div Statement-2 is a correct explanation for ( (3) Statement-1 is true, Statement-2 is true and Statement-1 Statement-2 is the correct explanation of (4) Statement-1 is true, Statement-2 is true and Statement-1 Statement-2 is not the correct explanation of (4) Statement-1 is true, Statement-2 is true and Statement-1 Statement-2 is not the correct explanation of Ans. (1) Statement-1 Sol. kmax = hf – Ans. (3) kmax = 2hf – Sol. Because of variation in composition of ionosphere, the signals are unstable. kmax = 2kmax + 78. A fully charged capacitor C with initial charge q0 is kmax > 2kmax connected to a coil of self inductance L at t = 0. The 80. Water is flowing continuously from a tap having an time at which the energy is stored equally between internal diameter 8 × 10–3 m. The water velocity as the electric and the magnetic fields is it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance 2 × 10–1 m below the tap (1) LC (2)  LC is close to  (1) 3.6 × 10–3 m (2) 5.0 × 10–3 m (3) LC (4) 2  LC 4 (3) 7.5 × 10–3 m (4) 9.6 × 10–3 mAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (16)
    • AIEEE-2011 (Code-R)Ans. (1) 82. Two particles are executing simple harmonic motionSol. a1v1 = a2v2 of the same amplitude A and frequency  along the x-axis. Their mean position is separated by distance a1 v22 = v12 + 2gh v1 X0(X0 > A). If the maximum separation between v22 = (0.4)2 + 2 × 10 × 0.2 them is (X0 + A), the phase difference between their motion is = (0.4)2 + 4 a2   v22 = 4.16 v2 (1) (2) 6 2 v2  4.16   (3) (4) 3 4 v2 = 2.04 m/s Ans. (3) Also, a1V1 = a2V2 Sol. x1 = A sin t d V1 = d V2 1 2 2 2 x2 = x0 + A sin(t + ) V1 Separation = x0 + A sin (t + ) – A sin t d2 = d1 V 2  = x0  2 A sin   cos t 3 0.4 2 = 8  10 2.04  Maximum separation = x0  2 A sin  x0  A = 3.54 × 10 m –3 281. A mass M, attached to a horizontal spring, executes    S.H.M. with amplitude A1. When the mass M passes 3 through its mean position then a smaller mass m is 83. If a wire is stretched to make it 0.1% longer, its placed over it and both of them move together with resistance will ) A  td. (2) amplitude A2. The ratio of  1  is sL (1) Decrease by 0.05% Increase by 0.05%  A2  i ce (3) Increase erv0.2% by (4) Decrease by 0.2%  M  m (1)  1/2 M (2) M  m Ans. (3) iona lS  t u a  M  1/2 Sol.E dR c l2 as h Mm  M  (3) (4)    M  m a k R 2 l of A M  i on R lAns. (1) isSol. Dv At mean position, initial velocity = A( i 1 1 84. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the MA1 1 fountain is v, the total area around the fountain New velocity = Mm that gets wet is MA1 1  A2 2  v2 v2 M m (1)  (2)  g2 g A2  M  1   A1  M  m  2 v4  v4 (3)  (4) g2 2 g2 k 1  Ans. (3) M 2 k Sol. Area  Rmax 2  M m v2 sin 90 v2 A2 M Rmax     g g A1 m M A1 m M v 4   Area  or g2 A2 MAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (17)
    • AIEEE-2011 (Code-R)85. A thermally insulated vessel contains an ideal gas 88. The transverse displacement y(x, t) of a wave on a of molecular mass M and ratio of specific heats . It string is given by is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, y(x , t )  e  – ax 2  bt 2  2 ab xt  its temperature increases by This represents a (  – 1) (  – 1) 2 (1) Mv 2 K (2) 2(   1)R Mv K 1 2R (1) Standing wave of frequency b (  – 1) Mv 2 (3) Mv 2 K (4) K a 2 R 2R (2) Wave moving in +x direction with speed bAns. (1) b 1 2 n  RT (3) Wave moving in –x direction with speedSol. mv  a 2 (   1) (4) Standing wave of frequency b MV 2 (   1)  T  Ans. (3) 2R86. A screw gauge gives the following reading when ( a x  b t )2 Sol. y(x , t )  e used to measure the diameter of a wire. This is a wave travelling in –x direction with speed Main scale reading : 0 mm Circular scale reading : 52 divisions b Given that 1 mm on main scale corresponds to 100 a divisions of the circular scale. 89. A car is fitted with d.)convex side-view mirror of a Lt The diameter of wire from the above data is es focal length 20 cm. A second car 2.5 m behind the rvic first car is overtaking the first car at a relative speed (1) 0.005 cm (2) 0.52 cm of 15 m/s. e l S The speed of the image of the second car (3) 0.052 cm (4) 0.026 cm i on a as tseen in the mirror of the first one is d u ca hEAns. (3) 1 m/s Diameter = main scale reading + l.c. × circular scale as (1) 15 m/s (2)Sol. k 10 f Aa no reading 1 i si o (3) m/s (4) 10 m/s v 15  0  52  1 100 mm (D i Ans. (3) Sol. For mirror  0.52mm  0.052cm87. A mass m hangs with the help of a string wrapped 1 1 1   around a pulley on a frictionless bearing. The pulley v u f has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the 1 dv 1 du mass m, if the string does not slip on the pulley, is  0 v 2 dt u2 dt g 3 (1) (2) g 2 3 2 dv  v2 du  f  du  2    dt u dt  f  u  dt 2 (3) g (4) g 3 2  20  duAns. (4)     20  280  dt mg mg 2gSol. a    2 I m 3  1  du 1 m 2 m    m/s R 2  15  dt 15Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (18)
    • AIEEE-2011 (Code-R)90. Let the x-z plane be the boundary between two Sol. The data is inconsistant transparent media. Medium 1 in z  0 has a Taking boundary as x-y plane instead of x-z plane, refractive index of 2 and medium 2 with z < 0 the angle of incidence is given as has a refractive index of 3 . A ray of light in  i j  ( k ).(6 3   8 3   10 k )  cos   ˆ medium 1 given by the vector A  6 3 i  8 3 ˆ – 10 k ˆ j 108  192  100 is incident on the plane of separation. The angle of 1     60 refraction in medium 2 is 2 (1) 75° (2) 30° Now, 2 sin 60  3 sin r (3) 45° (4) 60° 1  sin r  2Ans. (3)  r  45  td .) sL i ce lS e rv n a u c a ti o d s hE ka f Aa io no (D i vi sAakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (19)
    • (Division of Aakash Educational Services Lt d. ) ANALYSIS OF PHYSICS PORTION OF AIEEE 2011 XII XI XII XI XII XII XI XI Heat & Modern Unit and Electricity Magnetism Mechanics Optics Waves Total Thermodynamics Physics MeasurementsEasy 0 3 1 7 3 0 2 1 17Medium 1 1 2 2 0 2 0 1 9Tough 2 0 0 1 0 1 0 0 4 Total 3 4 3 10 3 3 2 2 30 XI syllabus 18 XII syllabus 12 Distribution of Level of Questions in Physics Electricity Topic wise distribution in Physics Heat & Thermodynamics 7% 7% 13% Magnetism 10% 10% 13% Mechanics 30% 57% 33% Modern Physics 10% 10% Optics Unit and Measurements Waves Easy Medium Tough Percentage Portion asked from Syllabus of Class XI & XII 40% 60% XI syllabus XII syllabus
    • (Division of Aakash Educational Services Lt d. ) ANALYSIS OF CHEMISTRY PORTION OF AIEEE 2011 Organic Chemistry Inorganic Chemistry Physical Chemistry TotalEasy 0 3 3 6Medium 8 6 8 22Tough 1 0 1 2 Total 9 9 12 30 XI syllabus 9 XII syllabus 21 Distribution of Level of Question in Chemistry Topic wise distribution in Chemistry 40% 30% 7% 20% 73% 30% Easy Medium Tough Organic Chemistry Inorganic Chemistry Physical Chemistry Percentage Portion asked from Syllabus of Class XI & XII 30% 70% XI syllabus XII syllabus
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