May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
16. Half a mole of photon is used to break the O 3molecule completely according to I st reaction. O 2produced in this reaction dissociates according to2 nd reaction, then what is the total energy requiredin KJ to carry out the dissociation of O 2 .(A) 384 KJ(B) 280 KJ(C) 480 KJ(D) 300 KJThis section contains 7 questions (Q.17 to 23).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W012345678901234567890123456789012345678921. Find number of 120º bond angles in O = CF 2 .22. Give number of H-atoms in H 2 C = SF4 which arein plane of axial F-atoms.23. For an element Z eff for its outermost electron is3.55. If atomic radius of element is 0.75 Å, thenfind out its electro negativity.MATHEMATICSQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1– sin 2θ + cos 2θ1. If f (θ) =, then value of2cos 2θf (11º) f (34º) equals -13(A) (B) 2 4(C) 41(D) None of these2. If 3 ≤ a < 4 then value ofsin –1 (sin [a]) + tan –1 (tan [a]) + sec –1 (sec[a])(where [.] denotes greatest integer function) is -(A) 2π (B) 2π – 3(C) 3 (D) 2π + 317. A compound exists in the gaseous phase both asmonomer (A) and dimer (A 2 ). The molecularweight of A is 60. In an experiment 240 g of thecompound was confined in a vessel of volume32.84 litre and heated to 127º C. Calculate thepressure (in atm) developed if the compound existsas dimer to the extent of 50% by weight underthese conditions.18. Li 2+ ion in its ground state absorbs a photon ofenergy 183 electron volt. Electron of Li 2+ ionstrikes the He + ion. (in Å) after being struck byelectron of Li 2+ ion.3. If sum of all solutions of equation3cot 2 kπθ + 10 cotθ + 3 = 0 in [0, 2π] is 2k ∈ I then k equals -(A) 3 (B) 6(C) 10 (D) 154. Range of k for whichk cos 2 x – k cos x + 1 ≥ 0 ∀ x ∈ R is -1(A) k < – (B) k > 42where19. First ionization energy of Li is 13.6 eV. If it isassumed that outermost electron of Li revolvesunder the influence of nucleus which is shielded byinner two electrons in first orbit of Li, then find outby what amount of charge inner two electronsshield the nucleus?20. A photon of 50 eV energy strikes a metal surface(having work function 1.64 eV). Photoelectronejected from metal with maximum kinetic energystrikes He + ion (in ground state).Find out de-Broglie wavelength of electron of He +ion finally in Å.(C) – 21 ≤ k ≤ 4(D) None of these5. If sin (sin x + cos x) = cos (cos x – sin x ) thenlargest possible value of sin x is -1(A)(B) 12(C)16 – π42(D) 4πXtraEdge for IIT-JEE 58 MAY 2011
6. If ∆ is area of ∆ABC and length of two sides are 3& 5 respectively, if third side is c, then22c + 16c + 64 c + 16c + 54(A) ∆ ≤(B) ∆ =12 38(C) ∆ >c2+ 16c + 7443(D) None of these7. A point Q is selected at random inside theequilateral triangle. If sum of lengths ofperpendicular dropped on sides from Q is P. Thenaltitude of triangle is -PP(A) (B) 2 3(C) P(D) None of theseQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. If cotθ + tanθ = x and secθ – cosθ = y, then -(A) x sinθ . cosθ = 1(B) sin 2 θ = y cosθ(C) (x 2 y) 1/3 + (xy 2 ) 1/3 = 1(D) (x 2 y) 2/3 – (xy 2 ) 2/3 = 1θ9. If 0 ≤ θ ≤ π and sin = 1 + sin θ – 1 – sin θ2then possible values of tanθ is/are -43(A) (B) – 3 4(C) – 34(D) 0⎛ 2 1 ⎞10. If ⎜cos x + ⎟ (1 + tan 2 2y) (3 + sin 3z) = 42⎝ cos x ⎠then -(A) x is a multiple of π(B) x is a multiple of 2π(C) y is a multiple of π/2(D) None of these11. In a ∆ABC, if r = 1, R = 3, s = 5 which of thefollowing is/are correct (where a, b, c representsides of triangle) -(A) ar ∆ABC = 5(B) Product of sides of ∆ABC is 60(C) a 2 + b 2 + c 2 = 24(D) Sum of exradii of ∆ABC is 13This section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)Let α ± β is not an odd multiple of π.α + βIf cos α + cos β = b, sin α + sin β = a, θ =2na(a – b)and sin 2θ + cos 2θ = 1 + where n ∈ I2 2a + bthen12. Value of n is -(A) 0 (B) 1(C) – 2(D) None of these13. If cosec n x = A then sin 3A. sin A is polynomial inx, whose degree is equal to -(A) 5 (B) 4 (C) 2 (D) 314. If degree of polynomial obtained in above questionis p then max. value of (p + 1) sin x + (p + 2) cos xis -(A) p + 2 (B) p + 3(C) p + 1(D) None of thesePassage : II (Q. No. 15 to 16)Let sides of a triangle are a = n + 1, b = n + 2 &c = n with sin C = 54 , then answer the following.15. Area of ∆ABC is -(A) 84 (B) 72(C) 60(D) None o these16. Largest exradius of circle escribing ∆ABC -(A) 12 (B) 14(C) 16(D) None of theseThis section contains 7 questions (Q.17 to 25).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W0123456789012345678901234567890123456789XtraEdge for IIT-JEE 59 MAY 2011
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6. If ∆ is area of ∆ABC and length of two sides are 3& 5 respectively, if third side is c, then22c + 16c + 64 c + 16c + 54(A) ∆ ≤(B) ∆ =12 38(C) ∆ >c2+ 16c + 7443(D) None of these7. A point Q is selected at random inside theequilateral triangle. If sum of lengths ofperpendicular dropped on sides from Q is P. Thenaltitude of triangle is -PP(A) (B) 2 3(C) P(D) None of theseQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. If cotθ + tanθ = x and secθ – cosθ = y, then -(A) x sinθ . cosθ = 1(B) sin 2 θ = y cosθ(C) (x 2 y) 1/3 + (xy 2 ) 1/3 = 1(D) (x 2 y) 2/3 – (xy 2 ) 2/3 = 1θ9. If 0 ≤ θ ≤ π and sin = 1 + sin θ – 1 – sin θ2then possible values of tanθ is/are -43(A) (B) – 3 4(C) – 34(D) 0⎛ 2 1 ⎞10. If ⎜cos x + ⎟ (1 + tan 2 2y) (3 + sin 3z) = 42⎝ cos x ⎠then -(A) x is a multiple of π(B) x is a multiple of 2π(C) y is a multiple of π/2(D) None of these11. In a ∆ABC, if r = 1, R = 3, s = 5 which of thefollowing is/are correct (where a, b, c representsides of triangle) -(A) ar ∆ABC = 5(B) Product of sides of ∆ABC is 60(C) a 2 + b 2 + c 2 = 24(D) Sum of exradii of ∆ABC is 13This section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)Let α ± β is not an odd multiple of π.α + βIf cos α + cos β = b, sin α + sin β = a, θ =2na(a – b)and sin 2θ + cos 2θ = 1 + where n ∈ I2 2a + bthen12. Value of n is -(A) 0 (B) 1(C) – 2(D) None of these13. If cosec n x = A then sin 3A. sin A is polynomial inx, whose degree is equal to -(A) 5 (B) 4 (C) 2 (D) 314. If degree of polynomial obtained in above questionis p then max. value of (p + 1) sin x + (p + 2) cos xis -(A) p + 2 (B) p + 3(C) p + 1(D) None of thesePassage : II (Q. No. 15 to 16)Let sides of a triangle are a = n + 1, b = n + 2 &c = n with sin C = 54 , then answer the following.15. Area of ∆ABC is -(A) 84 (B) 72(C) 60(D) None o these16. Largest exradius of circle escribing ∆ABC -(A) 12 (B) 14(C) 16(D) None of theseThis section contains 7 questions (Q.17 to 25).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W0123456789012345678901234567890123456789XtraEdge for IIT-JEE 59 MAY <strong>2011</strong>