AIEEE 2008 Solved Question Papers

This booklet contains 27 printed pages. PAPER – 1 : MATHEMATICS, CHEMISTRY & PHYSICS Test Booklet Code Read carefully the Instructions given in the booklet. IMPORTANT INSTRUCTIONS: D5 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 105 questions of 3 marks each. The maximum marks are 315. 5. There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. Part A – Mathematics (105 marks) –35 Questions Part B – Chemistry (105 marks) –35 Questions Part C – Physics (105 marks) –35 Questions 6. Candidates will be awarded three marks each for indicated correct response of each question. One mark will be deducted for indicated 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. Use Blue /Black Ball Point Pen only for writing particulars /marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil of strictly prohibited. 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. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in 4 page (Pages 20–23) at the end of the booklet. 10. 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. 11. The CODE for this Booklet is D5. 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. 12. Do not fold or make any stray marks on the Answer Sheet. AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 2 PART-A MATHEMATICS Solutions of SET-D5 Note: Questions with (*) mark are from syllabus of class XI. *1. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 806. Then which one of the following gives possible values of a and b? (1) 6,1ba (2) 4,3ba (3) 7,0ba (4) 2,5ba Sol.: Variance is 8.622nxnx and 6nx (given)  8.6365100256422ba  0.34922ba  2522ba Correct choice: (2) 2. The vector kjiaˆˆ2ˆ lies in the plane of the vectors jibˆˆ and kjcˆˆ and bisects the angle between b and c. Then which one of the following gives possible values of  and ? (1) 1,2 (2) 1,1 (3)2,2 (4) 2,1 Sol.:  ba, and c are coplanar.  ][cba = 0  2 …(i) Also a bisects the angle between b and c.  cbaˆˆ  2ˆˆ2ˆkjia …(ii) Comparing (ii) with kjiaˆˆ2ˆ, we get ,2  = 1 and 1, which also satisfies (i). Correct choice: (2) 3. The non-zero vectors ba,and care related by ba8 and bc7. Then the angle between a and c is (1) 2 (2)  (3) 0 (4) 4 Sol.: Clearly a and c are anti parallel.  Angle between a and c is . Correct choice: (2) *4. The line passing through the points a,1,5 and 1,,3b crosses the yz-plane at the point 213,217,0. Then (1) 4,6ba (2) 2,8ba (3) 8,2ba (4) 6,4ba Sol.: Equation of given line in symmetric form is aazbyx11125 …(i)  Any point on (i) can be aab1,11,25 …(ii)  213,217,0 lies on (i)  25 …(iii) Using (iii) in (ii) and comparing with given point we get 4,6ba Correct choice: (1) 5. If the straight lines 33221zykx and 21332zkyx intersect at a point, then the integer k is equal to (1) 2 (2) –2 (3) –5 (4) 5 Sol.:  Two given lines are intersecting.  02332211kk  5k is the required integral value. Correct choice: (3) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 3 6. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is (1) 2222252yyy (2) 2222252yyx (3) 222252yyx (4) 222252yyy Sol.: Equation of circle can be 25222yax …(i)  yyxa2 …(ii) Using (ii) in (i), we get    2 2 2 2 25 2      y y y Correct choice: (1) 7. Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that cxazybzcyx, and aybxz. Then abccba2222 is equal to (1) 0 (2) 1 (3) 2 (4) –1 Sol.: According to given condition 0111abacbc  12222abccba Correct choice: (2) 8. Let A be a square matrix all of whose entries are integers. Then which one of the following is true? (1) If det 1A, then 1A exists and all its entries are integers (2) If det 1A, then 1A need not exist (3) If det 1A, then 1A exists but all its entries are not necessarily integers (4) If det 1A, then 1  A exists and all its entries are non-integers Sol.: Obviously (1) is the correct answer. Correct choice: (1) *9. The quadratic equations 062axx and 062cxx have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is (1) 3 (2) 2 (3) 1 (4) 4 Sol.: Let the roots of 062axx be , 4 and the roots of 62cxx = 0 be , 3  64 …(i) a4 …(ii) c3 …(iii) and 63 …(iv) (ii) and (iv)  8a  1st equation reduces to 0862xx Clearly 2 and 1  Common root is 2. Correct choice: (2) *10. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent? (1) 47.8.6C (2) 4846..7CC (3) 4746..8CC (4) 48.7.6C Sol.: 1M, 4I‟s and 2P‟s can be arranged by !2!4!7 and in the 8 gaps 4 S can arranged with 48C ways, so total ways are 4846..7CC Correct choice: (2) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 4 11. Let dxxxI10sin and dxxxJ10cos. Then which one of the following is true? (1) 32I and 2J (2) 32I and 2J (3) 32I and 2J (4) 32I and 2J Sol.: We know that 1sinxx, when 1,0x  xxxsin  32sin10xx Again xxx1cos when 1,0x  2cos10xx Correct choice: (4) 12. The area of the plane region bounded by the curves 022yx and 132yx is equal to (1) 32 (2) 34 (3) 35 (4) 31 Sol.: 0 2 2   y x  22xy parabola 1 3 2   y x  1312xy parabola Solving equation of two parabolas simultaneously, we get x = –2; 1y Area of the region ABCA A D (–2, –1) B (–2, 1) 22xy (–2, 0) C 1312xy 32131313121031021022yydyydyyy Hence area of region bounded by given curves is equal to 34. Correct choice: (2) 13. The value of 4sinsin2xdxx is (1) cxx4sinlog (2) cxx4coslog (3) cxx4coslog (4) cxx4sinlog Sol.: Let tx4  tx4 dtdx  cttdttdtttdxxxesinlogcot1sin4sin24sinsin2 cxxcxxee4sinlog4sinlog4 Correct choice: (1) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 5 *14. The statement pqp is equivalent to (1) qpp (2) qpp (3) qpp (4) qpp Sol.: p q qp pq pqp qpp T T T T T T T F T T T T F T T F T T F F F T T T Correct choice: (4) 15. The value of 32tan35coseccot11 is (1) 174 (2) 175 (3) 176 (4) 173 Sol.: 32.4313243tancot32tan43tancot32tan35coseccot11111 176176cotcot61289tancot11 Correct choice: (3) Directions: Questions number 16 to 20 are Assertion-Reason type questions. Each of these questions contains two statements: Statement-I (Assertion) and Statement-2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. 16. Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr(A), the sum of diagonal entries of A. Assume that IA2. Statement-1: If IA and IA, then 1detA. Statement-2: If IA and IA, then 0trA. (1) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1. (2) Statement-1 is true, Statement-2 is false. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Sol.: Let dcbaA IA2  12bca, 12dbc, 0,0cdabda Out of all possible matrices if we consider1001A, then Atr = 0.  Statement-2 is wrong. Again if ,IA then 1A  Statement-1 is correct. Correct choice: (2) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 6 *17. Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “x is a rational number iff y is a transcendental number”. Statement-1: r is equivalent to either q or p. Statement-2: r is equivalent to qp~~. (1) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1. (2) Statement-1 is true, Statement-2 is false. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Sol.: p : x is an irrational number q : y is a transcendental number r : x is a rational number iff y is a transcendental number  qpr~: pqsor:1 qps~~:2 p q p~ q~ qpr~ pqsor1 qp~ qps~~2 T T F F F T F T T F F T T T T F F T T F T T T F F F T T F F F T Clearly 1s and r are not equivalent  Statement-1 is false. Also 2s and r are not equivalent  Statement-2 is also false. Hence none of the option is correct. *18. In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement-1: The number of different ways the child can buy the six ice-creams is 510C. Statement-2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6 A‟s and 4 B‟s in a row. (1) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1. (2) Statement-1 is true, Statement-2 is false. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Sol.: Statement-1: Number of ways = number of non negative integral solutions of the equation 654321TTTTT =41015156CC  Statement-1 is wrong. Statement-2: Number of different ways of arranging 6A‟s and 4 B‟s in a row = 410!4!6!10C  Statement-2 is correct. Correct choice: (3) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 7 *19. Statement-1: 10221nnrrnnCr. Statement-2: 10111nnnrrrnxnxxxCr. (1) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1. (2) Statement-1 is true, Statement-2 is false. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Sol.: nrrrnrnrrnrnrrnxCxCrxCr000..1nrnrrrnrrnxCxCnx10111nnxxnx111 …(i) Statement-2 is true. Putting 1xin (i), we get 102.2.1nnrrnnCr. Statement-1 is also true. Correct choice: (4) *20. Statement-1: For every natural number 2n, nn1......2111. Statement-2: For every natural number 11,2nnnn. (1) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1. (2) Statement-1 is true, Statement-2 is false. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Sol.: Statement-2: 1nn is true for 2n. Statement-1: 1nn  n......432 Now n2  n121 n3  n131 : : nn  nn11 Also n111. So nnnn1......312111 Correct choice: (4) *21. The conjugate of a complex number is 11i. Then that complex number is (1) 11i (2) 11i (3) 11i (4) 11i Sol.: 11iz  21iz  21iz =1211iii  11i Correct choice: (1) 22. Let R be the real line. Consider the following subsets of the plane RR: AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 8 20and1:,xxyyxS integeranis:,yxyxT. Which one of the following is true? (1) S is an equivalence relation on R but T is not (2) T is an equivalence relation on R but S is not (3) Neither S nor T is an equivalence relation on R (4) Both S and T are equivalence relations on R Sol.: For S, 1xy for reflexive 1xx  0 = 1  S is not reflexive. So S can not be equivalence. For T, Iyx, then Ixx0  T is reflexive. Iyx, then Ixy  T is symmetric also. Now Iyx and Izy  Izx  T is transitive also. Hence T is an equivalence relation. Correct choice: (2) 23. Let YNf: be a function defined as 34xxf, where NxxyNyYsomefor34:. Show that f is invertible and its inverse is (1) 43yyg (2) 43yyg (3)343yyg (4) 434yyg Sol.: Clearly fis bijective function so it is invertible. 34xy  xy43  43yyg Correct choice: (2) *24. AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is 45°. Then the height of the pole is (1) 13237 m (2) 131237 m (3) 131237m (4) 13237m Sol.: BCh60tan …(i) and BCh745tan …(ii)  hBC7  7hBC From (i)  73hh  hh373  373hh A h B D 7 m 60° 45° C 1337h  21337 m Correct choice: (4) *25. A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then BAP is (1) 1 (2) 52 (3) 53 (4) 0 Sol.: BAn6,5,4,3,2,1  1BAP Correct choice: (1) 26. It is given that the events A and B are such that 41AP, 21|BAP and 32|ABP. Then BP is (1) 32 (2) 21 (3) 61 (4) 31 AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 9 Sol.: 41AP BPBAPBAP  BPBAP21 …(i) APBAPABP  4132BAP  61BAP. Putting the value of BAP in (i)  31612BP Correct choice: (4) *27. A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 21. Then the length of the semi-major axis is (1) 34 (2) 35 (3) 38 (4) 32 Sol.: Perpendicular distance from focus on directrix = aeea140  224aa  234a  a = 38 Correct choice: (3) *28. A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at (1) (0, 1) (2) (2, 0) (3) (0, 2) (4) (1, 0) Sol.: Vertex will be mid-point of F and M. So, (1, 0) Correct choice: (4) F (0, 0) V x y x = 2 M (2, 0) *29. The point diametrically opposite to the point P(1, 0) on the circle 034222yxyx is (1) 4,3 (2) 4,3 (3) 4,3 (4) 4,3 Sol.: Given 0 3 4 2 2 2      y x y x ; Centre 2,1 C is the mid-point of P and Q so 31x and 41y  4,3Q Correct choice: (1) (1, 0) P (x1, y1) Q C (–1, –2) *30. The perpendicular bisector of the line segment joining 4,1P and 3,kQ has y-intercept –4. Then a possible value of k is (1) –2 (2) –4 (3) 1 (4) 2 Sol.: Equation of perpendicular bisector of PQ is 21127kxky Y-intercept is, 4282k  4k Correct choice: (2) *31. The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is (1) 12 (2) 4 (3) –4 (4) –12 Sol.: Let G.P., 32,,,ararara. Given 12ara and 4832arar  4812rar. So 42r  2r 2r, then 12a Correct choice: (4) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 10 32. Suppose the cubic qpxx3 has three distinct real roots where p > 0 and q > 0. Then which one of the following holds? (1) The cubic has minima at both 3p and 3p (2) The cubic has maxima at both 3p and 3p (3) The cubic has minima at 3p and maxima at 3p (4) The cubic has minima at 3p and maxima at 3p Sol.: Let qpxxxf3 pxxf23 For maxima or minima 0xf  3px xxf6  0xf for 3px and 0xf for 3px Correct choice: (3) 33. How many real solutions does the equation 0560301614357xxxx have? (1) 3 (2) 5 (3) 7 (4) 1 Sol.: 560301614357xxxxxf 03048707246xxxxf  f is increasing also xfxlim;xfxlim xf Clearly 0xf have exactly one real root. Correct choice: (4) 34. Let 1if01if11sin1xxxxxf. Then which one of the following is true? (1) fis differentiable at x = 0 but not at x = 1 (2) fis differentiable at x = 1 but not at x = 0 (3) f is neither differentiable at x = 0 nor at x = 1 (4) f is differentiable at x = 0 and at x = 1 Sol.: 1if01if11sin1xxxxxf; hhhhfRhh1sinlim01sinlim100, which does not exist. 0211cos1111sin0xxxxxf = 1cos1sin. So xf is differentiable at x = 0 but not at x = 1. Correct choice: (1) 35. The solution of the differential equation xyxdxdy satisfying the condition 11y is (1) 1xxey (2) xxxyln (3) xxyln (4) 2lnxxxy Sol.: Given xydxdy1  1xydxdy dxxeIF1= x1 cdxxxy1.11. cxxyln  11y, so c = 1  xxxyln Correct choice: (2) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 11 PART-B CHEMISTRY *36. Which one of the following is the correct statement? (1) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase. (2) B2H6.2NH3 is known as „inorganic benzene‟. (3) Boric acid is a protonic acid. (4) Beryllium exhibits coordination number of six. Sol.: Inorganic benzene is B3N3H6, Boric acid is a Lewis acid and beryllium exhibits co–ordination number of 4 only. BeCl2 and AlCl3 both exhibit bridged structures in the solid state. Correct choice: (1) *37. The treatment of CH3MgX with CH3CC–H produces (1) H CH3–C=C–CH3 H (2) CH4 (3) CH3–CH=CH2 (4) CH3CC–CH3 Sol.: CH3MgX + CH3–CC–H  CH4 + CH3–CC MgX Correct choice: (2) *38. The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (1) –CHO, –COOH, –SO3H, –CONH2 (2) –CONH2, –CHO, –SO3H, –COOH (3) –COOH, –SO3H, –CONH2, –CHO (4) –SO3H, –COOH, –CONH2, –CHO Sol.: The correct decreasing order of priority for the functional groups according to IUPAC nomenclature is –CO2H > –SO3H > –CONH2 > –CHO Correct choice: (3) *39. The pKa of a weak acid, HA is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be (1) 7.01 (2) 9.22 (3) 9.58 (4) 4.79 Sol.: B+ + A– + H2O BOH + HA pH = 21pKw + 21pKa – 21pKb = 21 (14 + 4.80 – 4.78) = 7.01 Correct choice: (1) *40. The hydrocarbon which can react with sodium in liquid ammonia is (1) CH3CH=CHCH3 (2) CH3CH2CCCH2CH3 (3) CH3CH2CH2CCCH2CH2CH3 (4) CH3CH2CCH Sol.: CH3CH2CCH 3NHliquidinNa CH3CH2CC–Na+ + 2H21 Correct choice: (4) 41. Given oCr/Cr3E= –0.72 V, oFe/Fe2E= –0.42 V. The potential for the cell Cr | Cr3+ (0.1 M) || Fe2+ (0.01 M) | Fe is (1) –0.339 V (2) –0.26 V (3) 0.26 V (4) 0.339 V Sol.: 2Cr(s) + 3Fe2+(aq)  2Cr3+(aq) + 3Fe(s) Ecell = oFe|Fe2E– oCr|Cr3E– 3223]Fe[]Cr[log60059.0= –0.42 – (–0.72) – 32)01.0()1.0(log6059.0 = 0.26 V Correct choice: (3) 42. Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl (1) reduces permanganate to Mn2+. (2) oxidises oxalic acid to carbon dioxide and water. (3) gets oxidised by oxalic acid to chlorine. (4) furnishes H+ ions in addition to those from oxalic acid. Sol.: KMnO4 can oxidise HCl also (along with H2C2O4) into Cl2 and itself gets reduced to Mn2+. Correct choice: (1) *43. Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 12 (1) R2SiCl2 (2) R3SiCl (3) R4Si (4) RSiCl3 Sol.: n RSiCl3 3n H2O OH –3n HCl n R–Si–OH OH O R–Si–O–Si–R O O R–Si–O–Si–R O O O Cross linked silicone polymer Correct choice: (4) 44. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: 21Cl2(g) 1 Cl(g) dissH 2 Cl–(g) egH 1 hydH 2 Cl–(aq) The energy involved in the conversion of 21Cl2(g) to Cl– (aq) (using the data, 2CldissH= 240 kJ mol–1, ClegH= –349 kJ mol–1, ClhydH= –381 kJ mol–1) will be (1) –850 kJ mol–1 (2) +120 kJ mol–1 (3) +152 kJ mol–1 (4) –610 kJ mol–1 Sol.: H = 24021+ (–349) + (–381) = –610kJ mol–1. Correct choice: (4) 45. Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? (1) Metal sulphides are less stable than the corresponding oxides. (2) CO2 is more volatile than CS2. (3) Metal sulphides are thermodynamically more stable than CS2. (4) CO2 is thermodynamically more stable than CS2. Sol.: 2MS + C  2M + CS2 ; G1 = positive 2MO + C  2M + CO2 ; G2 = negative This suggests that CO2 is thermodynamically more stable than CS2. Metal sulphides are thermodynamically more stable than CS2 while metal sulphides are more stable than the corresponding oxides. Correct choice: (1) *46. Four species are listed below: (i) 3HCO (ii) H3O+ (3) 4HSO (4) HSO3F Which one of the following is the correct sequence of their acid strength? (1) (i) < (iii) < (ii) < (iv) (2) (iii) < (i) < (iv) < (ii) (3) (iv) < (ii) < (iii) < (i) (4) (ii) < (iii) < (i) < (iv) Sol.: The decreasing order of acidic strength is HSO3F > H3O+ > 4HSO > 3HCO Correct choice: (1) *47. Which one of the following constitutes a group of the isoelectronic species? (1) CN–, N2,2222C,O (2) N2, ,O2NO+, CO (3),O,C222CO, NO (4) NO+, ,C22CN–, N2 Sol.: Isoelectronic species possess same number of electrons. NO+, ,C22CN– and N2, each have 14 electrons and thus are isoelectronic. Correct choice: (4) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 13 48. Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives (1) p–nitrophenol (2) nitrobenzene (3) 2,4,6–trinitrobenzene (4) o–nitrophenol Sol.: OH Conc. H2SO4 SO3H OH HNO3 –SO3 OH NO2 The temperature is not mentioned, so it can be assumed to be room temperature at which ortho is the stable product. Correct choice: (4) *49. The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is (1) 7.56 × 105 J mol–1 (2) 9.84 × 105 J mol–1 (3) 8.51 × 105 J mol–1 (4) 6.56 × 105 J mol–1 Sol.: E2 = 226)2()1(10312.1= –3.28 × 105 J mol–1 ; E1 = –1.312 × 106 J mol–1 E = E2 – E1 = (–3.28 × 105 + 1.312 × 106) J mol–1 = 9.84 × 105 J mol–1. Correct choice: (2) 50. The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction, is (1) (CH3)2CHCl (2) CH3Cl (3) (C2H5)2CHCl (4) (CH3)3CCl Sol.: SN2 reaction is shown by primary halides more than secondary halides and secondary halides more than tertiary halides. Correct choice: (2) 51. Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotized and then heated with cuprous bromide. The reaction mixture so formed contains (1) mixture of o– and p–bromoanilines (2) mixture of o– and m–bromotoluenes (3) mixture of o– and p–bromotoluenes (4) mixture of o– and p–dibromobenzenes Sol.: Conc. H2SO4 Sn/HCl CH3 NO2 + CH3 NH2 NaNO2 /HCl CH3 NH2 + CH3 NO2 0–5°C Cu2Br2.  CH3 N2Cl– + CH3 + N2Cl– + CH3 Br + CH3 Br o– and p–bromotoluenes CH3 Conc. HNO3 Correct choice: (3) *52. In the following sequence of reactions, the alkene affords the compound „B‟ CH3CH=CHCH33OA ZnOH2 B. The compound B is (1) CH3CH2COCH3 (2) CH3CHO (3) CH3CH2CHO (4) CH3COCH3 Sol.: CH3–CH=CH–CH33O CH3–CH CH–CH3 O O O (A) OH/Zn2 2CH3CHO + H2O + ZnO (B) Correct choice: (2) *53. Which one of the following pairs of species have the same bond order? (1) 2O and CN– (2) NO+ and CN+ (3) CN– and NO+ (4) CN– and CN+ Sol.: Same bond order would be for the species which have same number of total electrons. CN– and NO+ both have 14 elelctrons and will have a bond order of 3. Correct choice: (3) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 14 54. At 80°C, the vapour pressure of pure liquid „A‟ is 520 mm Hg and that of pure liquid „B‟ is 1000 mm Hg. If a mixture solution of „A‟ and „B‟ boils at 80°C and 1 atm pressure, the amount of „A‟ in the mixture is (1 atm = 760 mm Hg) (1) 48 mol percent (2) 50 mol percent (3) 52 mol percent (4) 34 mol percent Sol.: PT = 760 = AoAXP+ BoBXP= 520 XA + 1000 (1 – XA) XA = 21 or 50 mol percent. Correct choice: (2) 55. For a reaction 21A  2B, rate of disappearance of „A‟ is related to the rate of appearance of „B‟ by the expression (1) dt]B[ddt]A[d (2) dt]B[d4dt]A[d (3) dt]B[d21dt]A[d (4) dt]B[d41dt]A[d Sol.: dt]A[d2 = Rate of reaction with respect to A. dt]B[d21 = Rate of reaction with respect to B. dt ] A [ d 2  = dt]B[d21, dt ] B [ d 41 dt ] A [ d   Correct choice: (4) *56. The equilibrium constants 1pKand 2pKfor the reactions X 2Y and Z P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressures at these equilibria is (1) 1 : 3 (2) 1 : 9 (3) 1 : 36 (4) 1 : 1 Sol.: Let initial moles of X and Z taken are „a‟ and „b‟ respectively. X 2y Z P + Q Moles at equilibrium a(1– ) 2a Moles at equilibrium b(1– ) b b 1 p K = )1(a)1(aP)a2(1T2 2pK= )1(b)1(bP)b(2T2 21ppKK= 21TTPP4 = 91 ; 21TTPP = 361. Correct choice: (3) *57. In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the following is the correct statement? (1) H2 is removed through occlusion with Pd. (2) CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in alkali. (3) CO and H2 are fractionally separated using differences in their densities. (4) CO is removed by absorption in aqueous Cu2Cl2 solution. Sol.: CO + H2 + H2O Catalyst CO2 + 2H2 2NaOH (alkali) Na2CO3 + H2O Water gas Correct choice: (2) 58. In which of the following octahedral complexes of Co (atomic number 27), will the magnitude of o be the highest? (1) [Co(H2O)6]3+ (2) [Co(NH3)6]3+ (3) [Co(CN)6]3– (4) [Co(C2O4)3]3– Sol.: Magnitude of o will be highest with the strongest ligand. Since, CN– is the strongest ligand of all, thus would lead to a greater separation between t2g and eg orbitals. Correct choice: (3) 59. The coordination number and the oxidation state of the element „E‟ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively, (1) 4 and 3 (2) 6 and 3 (3) 6 and 2 (4) 4 and 2 Sol.: (en) and oxalate ion are both bidentate ligands. Co–ordination number of E in the complex = (2 × 2) + (1 × 2) = 6. Oxidation state of E in the complex = [x + (–2) = +1] = +3. Correct choice: (2) *60. Identify the wrong statement in the following: AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 15 (1) Ozone layer does not permit infrared radiation from the sun to reach the earth. (2) Acid rain is mostly because of oxides of nitrogen and sulphur. (3) Chlorofluorocarbons are responsible for ozone layer depletion. (4) Greenhouse effect is responsible for global warming. Sol.: Ozone layer prevents the ultra violet radiations and not the infrared radiations from the sun to reach the earth. Correct choice: (1) 61. Larger number of oxidation states are exhibited by the actinoids than those by lanthanoids, the main reason being (1) more energy difference between 5f and 6d than between 4f and 5d orbitals. (2) more reactive nature of the actinoids than the lanthanoids. (3) 4f orbitals more diffused than the 5f orbitals. (4) lesser energy difference between 5f and 6d than between 4f and 5d orbitals. Sol.: The energy difference between 5f and 6d is lesser than that between 4f and 5d orbitals. Thus, in actinoids, the electrons can be removed from 5f as well as 6d, so more number of oxidation states are exhibited by them. Correct choice: (4) 62. In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be (1) X2Y (2) X3Y4 (3) X4Y3 (4) X2Y3 Sol.: Number of effective Y in a unit cell = 4. Number of effective X in a unit cell = 8 × 32 = 316. So, formula of the compound = 43/16YX= 4/13/1YX= 34YX Correct choice: (3) 63. Gold numbers of protective colloids (A), (B), (C) and (D) are 0.50, 0.01, 0.10 and 0.005, respectively. The correct order of their protective powers is (1) (A) < (C) < (B) < (D) (2) (B) < (D) < (A) < (C) (3) (D) < (A) < (C) < (B) (4) (C) < (B) < (D) < (A) Sol.: Lesser the value of gold number of a protective colloid, better is its protective power.  (A) < (C) < (B) < (D) Correct choice: (1) 64. The vapour pressure of water at 20°C is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20°C, the vapour pressure of the resulting solution will be (1) 16.500 mm Hg (2) 17.325 mm Hg (3) 17.675 mm Hg (4) 15.750 mm Hg Sol.: Moles of glucose = 18018= 0.1, moles of H2O =182.178= 9.9 SSoPPP= waterofmolesecosgluofmoles ; 17.5 – PS = 9.9P1.0S PS = 17.325 mm Hg. Correct choice: (2) 65. Bakelite is obtained from phenol by reacting with (1) CH3COCH3 (2) HCHO (3) (CH2OH)2 (4) CH3CHO Sol.: Phenol + HCHOacido– and p–hydroxybenzylalcohol  Bakelite Correct choice: (2) 66. The absolute configuration of HO2C CO2H H HO OH H is (1) R, S (2) S, R (3) S, S (4) R, R AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 16 Sol.: CO2H OH H HO H R R CO2H Correct choice: (4) *67. For the following three reactions a, b, c, equilibrium constants are given: a. CO(g) + H2O(g) CO2(g) + H2(g) ; K1 b. CH4(g) + H2O(g) CO(g) + 3H2(g) ; K2 c. CH4(g) + 2H2O(g) CO2(g) + 4H2(g) ; K3 Which of the following relations is correct? (1) K3 = K1K2 (2) K32132KK (3) 21KK= K3 (4) K2K3 = K1 Sol.: CO(g) + H2O(g) CO2(g) + H2(g) ; K1 CH4(g) + H2O(g) CO(g) + 3H2(g) ; K2 CH4(g) + 2H2O(g) CO2(g) + 4H2(g) ; K3 = K1 × K2 Correct choice: (1) 68. Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1, respectively. For the reaction, 21X2 + 23Y2  XY3, H = –30 kJ, to be at equilibrium, the temperature will be (1) 750 K (2) 1000 K (3) 1250 K (4) 500 K Sol.: 2X21 + 2Y23  3XY ; S = 50 – 23402160 = –40 JK–1 ; H = –30 kJ ; G = H – TS At equilibrium, G = 0 ;  H = TS ; T = SH= 1JK40J30000= 750 K. Correct choice: (1) *69. The electrophile, E attacks the benzene ring to generate the intermediate –complex. Of the following, which –complex is of lowest energy? (1) NO2 + H E (2) NO2 + H E (3) NO2 + H E (4) + H E Sol.: Arenium ion (–complex) formed by the attack of electrophile on nitrobenzene at any one of the three positions is less stable than that formed by the attack of electrophile on benzene. Correct choice: (4) 70. –D–(+)–glucose and –D–(+)–glucose are (1) anomers (2) enantiomers (3) conformers (4) epimers Sol.: –D–(+)–glucose and –D–(+)–glucose are those diastereomers that differ in configuration at C–1 atom. Such isomers are referred as anomers. Correct choice: (1) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 17 PART-C PHYSICS 71. This question contains Statement-1 and Statement -2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement –1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. and Statement –2: For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z. (1) Statement –1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (2) Statement –1 is true, Statement-2 is false (3) Statement –1 is false, Statement-2 is true (4) Statement –1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 Sol.: Correct choice: (2) *72. This question contains Statement-1 and Statement -2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement –1: For a mass M kept at the centre of a cube of side „a‟ the flux of gravitational field passing through its sides is 4  GM. and Statement –2: If the direction of a field due to a point source is radial and its dependence on the distance „r‟ from the source is given as 2r1 , its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface. (1) Statement –1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (2) Statement –1 is true, Statement-2 is false (3) Statement –1 is false, Statement-2 is true (4) Statement –1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 Sol.: sd.Eg– 4GMenclosed = – 4GM Correct choice: (4) *73. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on circular scale is 50. Further, it is found that screw gauge has a zero error of – 0.03mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3mm and the number of circular scale divisions in line with the main scale as 35. The diameter of wire is (1) 3.67 mm (2) 3.38 mm (3) 3.32 mm (4) 3.73 mm Sol.: Least count mm01.050mm5.0 Zero error = – 0.03 mm Measured diameter = 3 mm + 35  0.01 mm = 3.35 mm Corrected diameter = 3.35 mm – (– 0.03mm) = 3.38 mm Correct choice: (2) *74. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be (1) 2211122211VPVPTVPTVP (2) 222111221121TVPTVPVPVPTT (3) 122211221121TVPTVPVPVPTT (4) 2211222111VPVPTVPTVP AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 18 Sol.: Internal energy of the system will remain conserved. 2V21V1v21T.CnTCnTCnn RVPRVPTRTVPRTVP2211222111; 122211221121TVPTVPVPVPTTT Correct choice: (3) 75. A student measures the focal length of a convex lens by putting an object pin at a distance „u‟ from the lens and measuring the distance „v‟ of the image pin. The graph between „u‟ and „v‟ plotted by the student should look like O v(cm) u(cm) (1) O v(cm) u(cm) (2) O v(cm) u(cm) (3) O v(cm) u(cm) (4) Sol.: f1u1v1 Correct choice: (1) Directions: Questions No. 76 and 77 are based on the following paragraph Consider a block of conducting material of resistivity „‟ shown in the figure. Current „I‟ enters at „A‟ and leaves from „D‟. We apply superposition principal to find voltage „V‟ developed between „B‟ and „C‟. The calculation is done in the following steps: (i) Take current „I‟ entering from „A‟ and assume it to spread over a hemispherical surface in the block. (ii) Calculate field E(r) at distance „r‟ from A by using Ohm‟s law E = j, where „j‟ is the current per unit area at „r‟. (iii) From the „r‟ dependence of E(r), obtain the potential V(r) at „r‟. (iv) Repeat (i), (ii) and (iii) for current „I‟ leaving „D‟ and superpose results for „A‟ and „D‟. a b a A B C D V I I 76. V measured between B and C is (1) ba2Ia2I (2) ba2I (3) baIaI (4) baIaI Sol.: 2r2IjE Potential difference due to current at A dr.r2Id.EVV2abaBCCBl; ba2Ia2Ir12I'Vaba By principle of superposition, baIaI'V2V Correct choice: (3) 77. For current entering at A, the electric field at a distance „r‟ from A is AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 19 (1) 2r2I (2) 2r4I (3) 2r8I (4) 2rI Sol.: Correct choice: (1) *78. Consider a uniform square plate of side „a‟ and mass „m‟. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is (1) 2ma127 (2) 2ma32 (3) 2ma65 (4) 2ma121 Sol.: 2222cmma322am6mamdII Correct choice: (2) 79. An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by (1) a meter scale provided on the microscope (2) a screw gauge provided on the microscope (3) a vernier scale provided on the microscope (4) a standard laboratory scale. Sol.: Correct choice: (3) 80. A horizontal overhead powerline is at a height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (0 = 4  10–7 T m A–1) (1) 5  10–6 T southward (2) 2.5  10–7 T northward (3) 2.5  10–7 T southward (4) 5  10–6 T northward Sol.: T105ri24B60 Correct choice: (1) *81. The speed of sound in oxygen (O2) at a certain temperature is 460 ms–1. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal) (1) 650 ms–1 (2) 330 ms–1 (3) 460 ms–1 (4) 500 ms–1 Sol.: 0MRTV 325RT7460 … (i); 43RT5V … (ii)  46008.3V = 1419 ms–1 Correct choice: (none of the answers is correct) 82. A 5V battery with internal resistance 2 and a 2V battery with internal resistance 1  are connected to a 10  resistor as shown in the figure. The current in the 10  resistor is (1) 0.03 A P2 to P1 (2) 0.27 A P1 to P2 (3) 0.27 A P2 to P1 (4) 0.03 A P1 to P2 P2 P1 10 1 2V 2 5V Sol.: 21211221RrRrrrrri 1102101222150.03 A Correct choice: (1) *83. A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms–1. The magnitude of its momentum is recorded as AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 20 (1) 17.56 kg ms–1 (2) 17.57 kg ms–1 (3) 17. 6 kg ms–1 (4) 17.565 kg ms–1 Sol.: 565.1700.5513.3mvp kg ms–1 Since result should have only 3 significant digits  p = 17.6 kg ms–1 Correct choice: (3) 84. A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor? (1) It is a pnp transistor with R as emitter (2) It is an npn transistor with R as collector (3) It is an npn transistor with R as base (4) It is a pnp transistor with R as collector Sol.: Correct choice: (3) *85. A block of mass 0.50 kg is moving with a speed of 2.00 ms–1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is (1) 0.67 J (2) 0.34 J (3) 0.16 J (4) 1.00 J Sol.: Using momentum conservation, v5.125.0  1ms32v Loss of energy 311325.12125.02122 0.67 J Correct choice: (1) *86. A wave travelling along the x-axis is described by the equation txcos005.0t,xy. If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then  and  in appropriate units are (1) 0.1,04.0 (2) 0.2,50.12 (3) ,00.25 (4) 0.2,08.0 Sol.: 2508.022; 22T2 Correct choice: (3) 87. Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A = 10 cm2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (0 = 4  10–7 T m A–1) (1) 4.8   10–5 H (2) 2.4   10–4 H (3) 2.4   10–5 H (4) 4.8   10–4 H Sol.: lANNM210 Correct choice: (2) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 21 *88. A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes? A (1) B A (2) B A (3) B A (4) B Sol.: Surface tension of soap solution is less than surface tension of water. Correct choice: (1) *89. A jar is filled with two non-mixing liquids 1 and 2 having densities 1 and 2, respectively. A solid ball, made of a material of density 3, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for 1, 2 and 3? (1) 321 (2) 231 (3) 213 (4) 231 Liquid 1 Liquid 2 3 1 2 Sol.: Heavier liquid settles down at the bottom So, 21 ,23 otherwise, ball will sink 13, otherwise, ball will float in liquid 1  231 Correct choice: (2) 90. Suppose an electron is attracted towards the origin by a force rk where „k‟ is a constant and „r‟ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be „rn‟ and the kinetic energy of the electron to be „Tn‟. Then which of the following is true? (1) nr,n1Tnn (2) 2nnnr,n1T (3) 2n2nnr,n1T (4) Tn independent of nr,nn Sol.: 2nhL  2nhmvrn Also nn2rkrmv  kmv2  k21mv21T2n, which is independent of n. km2nhmv2nhrn  nrn Correct choice: (4) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 22 Directions : Questions No. 91, 92 and 93 are based on the following paragraph. Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure). Crystal plane Outgoing Electrons Incoming Electrons i d 91. Electrons accelerated by potential V are diffracted from a crystal. If d = 1Å and i = 300, V should be about (h = 6.6  10–34 Js, me = 9.1  10–31 kg, e = 1.6  10–19 C) (1) 500 V (2) 1000 V (3) 2000 V (4) 50 V Sol.: For constructive interference, path difference = n From the given figure, path difference = MP+PN = 2d cos i  nicosd2  Ån3 Also ÅV150  V150n32  2n50V For n = 1, V = 50 volt O i P N M Correct choice: (4) 92. If a strong diffraction peak is observed when electrons are incident at an angle „i‟ from the normal to the crystal planes with distance „d‟ between them (see figure), de Broglie wavelength dB of electrons can be calculated by the relationship (n is an integer) (1) dBnλisind2 (2) dBnλicosd (3) dBnλisind (4) dBnλicosd2 Sol.: For strong peak, path difference = ndB  2d cos i = ndB Correct choice: (4) 93. In an experiment, electrons are made to pass through a narrow slit of width „d‟ comparable to their de Broglie wavelength. They are detected on a screen at a distance „D‟ from the slit (see figure). Which of the following graphs can be expected to represent the number of electrons „N‟ detected as a function of the detector position „y‟ (y = 0 corresponds to the middle of the slit)? d D y = 0 N y (1) d N y (2) d N y (3) d N y (4) d Sol.: After diffraction electron beam will spread. Correct choice: (2) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 23 94. In the circuit shown, A and B represent two inputs and C represents the output. The circuit represents (1) NAND gate (2) OR gate (3) NOR gate (4) AND gate A B C Sol.: Correct choice: (2) 95. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range r0, where r is the distance from the centre of the shell? O (1) r E(r) R O (2) r E(r) R O (3) r E(r) R O (4) r E(r) R Sol.: For given situation, electric field inside the shell is zero and is inversely proportional to r2 for a point outside the shell. Correct choice: (3) *96. A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x direction with a constant speed. The position of the first body is given by x1(t) after time t and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time t? O (1) t (x1–x2) O (2) t (x1–x2) O (3) t (x1–x2) O (4) t (x1–x2) Sol.: 21at21x and vtx2  vtat21xx221  vtat21x212 At t = 0, x 12 = 0 and at any time 0dtxd2122 Correct choice: (4) 97. Relative permittivity and permeability of a material are r and r, respectively. Which of the following values of these quantifies are allowed for a diamagnetic material? (1) 5.0,5.0rr (2) 5.1,5.1rr (3) 5.1,5.0rr (4) 5.0,5.1rr Sol.: For diamagnetic material, 10r and for any material 1r Correct choice: (4) *98. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km s–1, the escape velocity from the surface of the planet would be (1) 110 km s–1 (2) 0.11 km s–1 (3) 1.1 km s–1 (4) 11 km s–1 Sol.: 122121RRMMVV 122121RRMMVV s/km1101010s/km11 Correct choice: (1) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 24 *99. A thin rod of length „L‟ is lying along the x-axis with its ends at x = 0 and x = L. Its linear density (mass/length) varies with x as nLxk, where „n‟ can be zero or any positive number. If the position xCM of the centre of mass of the rod is plotted against „n‟, which of the following graphs best approximates the dependence of xCM on n? O (1) n xCM L/2 L O (2) n xCM L/2 L O (3) n xCM L/2 L O (4) n xCM L/2 Sol.: 2nL1ndxLxkdxLxkxdmdmxxL0nL0ncm If n = 0, 2Lxcm and if n   xcm = L Correct choice: (3) *100. The dimension of magnetic field in M, L, T and C (Coulomb) is given as (1) 11–CMT (2) 12–CMT (3) 11–CMLT (4) 22CMT Sol.: Use liBF LBTCMLT2  11CMTB Correct choice: (1) 101. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is „d‟. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant 1 = 3 and thickness 3d while the other one has dielectric constant 2 = 6 and thickness3d2. Capacitance of the capacitor is now (1) 40.5 pF (2) 20.25 pF (3) 1.8 pF (4) 45 pF Sol.: dApF9C00; 2010A3/d2A3/dC1 6,321  pF929dA29C040.5 pF Correct choice: (1) *102. An athlete in the olympic games covers a distance of 100m in 10s. His kinetic energy can be estimated to be in the range (1) 20,000 J – 50,000 J (2) 2,000 J – 5,000 J (3) 200 J – 500 J (4) 2  105 J – 3  105 J Sol.: Assuming average speed of athlete to be v 1ms10sec10m100v  estimated kinetic energy 2mv21K AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 25 Let mass of athlete lies between 40 kg and 100 kg  J5000KJ2000 Correct choice: (2) *103. A spherical solid ball of volume V is made of a material of density 1. It is falling through a liquid of density 2 (2 < 1). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e. Fviscous = –kv2(k >0). The terminal speed of the ball is (1) kVg1 (2) kVg21 (3) kVg21 (4) kVg1 Sol.: The ball will acquire terminal speed in the state of equilibrium  0gVkgV122v kVg21v Correct choice: (3) 104. Shown in the figure is a meter-bridge set up with null deflection in the galvanometer. The value of the unknown resistor R is (1) 110  (2) 55  (3) 13.75  (4) 220  G 20cm R 55 Sol.: For balanced meter bridge (null deflection) 8020R55 R = 220  Correct choice: (4) *105. While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance. Then (1) 54 > x > 36 (2) 36 > x > 18 (3) 18 > x (4) x > 54 Sol.: x4'V34V V'Vcm54V'V3x As velocity of sound increases with temperature, V'V  cm54x Correct choice: (4) AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 26 AIEEE 2008 Q & S (SETD5) Brilliant Tutorials Pvt. Ltd. Head Office : 12, Masilamani Street, T. Nagar, Chennai-600 017 Phone: 044-24342099, 24343308 Fax: 24343829 27