May 2011 - Career Point

21. A current carrying loop is placed in a uniformmagnetic field pointing in negative z direction.Branch PQRS is a three quarter circle, whilebranch PS is straight. If force on branch PS is2 F. Force on branch PQR is given a × F. Thenvalue of a is.yPQR22. One of the sides of the frame shown in the figurecan move. The frame is in uniform magnetic fieldof B 0 . Which is perpendicular to the plane of theframe. At the time t= 0. the magnetic field beginsB 0to decrease as B(t) = . Side of the frame is1+ktmoving with velocity v in order not to induceelectric current in it. If the value of k is 3 sec –1 ,then v = ........... × a.at = 0bBSxBv23. Two circular rings of identical radii and resistanceof 36 Ω each are placed in such a way that theycross each other's centre C 1 and C 2 a shown infigure. Conducting joints are made at intersectionpoints A and B of the rings. An ideal cell of emf 20V is connected across AB. The power delivered bycell is ........ × 25 watt.AC 1BC 2CHEMISTRYQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. For a general nth order process A → P with initialconcentration of reactant "a" and rate constant k,the expression for time for 75% completion ofreaction is -(A)1 ⎛⎜2n – 1⎝ an–1– 2 ⎞⎟⎠n–11k(B)1 ⎛⎜2n – 1⎝2n–2an–1– 2 ⎞⎟⎠1k(C)1 ⎛⎜2n – 1⎝ a2n–2n–1– 1⎞⎟⎠1k(D)1 ⎛⎜2n – 2⎝ a2n–22n–1– 1⎞⎟⎠2. An important application of radioactive decay is inthe dating of rocks, fossils, and ancient objects.Naturally occurring radioactive uranium -23892decays in a series of 238 U steps that finally lead tothe stable isotope 206 Pb.A sample of rock is found to contain 23.2 mg ofuranium -238 and 1.42 mg of lead -206. If the halflifeof 238 U 92 is 4.51 × 10 9 years, what is the age ofrock ?(A) 1.8 × 10 10 years (B) 4.4 × 10 8 years(C) 1.7 × 10 9 years (D) 2.8 × 10 7 years3. For a reaction A → Product, half-life measured fortwo different values of initial concentrations 5 ×10 –3 M and 25 × 10 –4 M are 1.0 and 8.0 hrsrespectively. If initial concentration is adjusted to1.25 × 10 –3 M, the new half-life would be :(A) 16 hrs(B) 32 hrs(C) 64 hrs(D) 256 hrs4. Consider the following standard reductionpotentials :Half reaction(V)Ni 2+ (aq) + 2e – Ni (s) Eº = – 0.23 VFe 2+ (aq) + 2e – Fe (s) Eº = – 0.41 VMn 2+ (aq) + 2e – Mn (s) Eº = – 1.03 VCo 2+ (aq) + 2e – Co (s) Eº = – 0.28 VCr 2+ (aq) + 2e – Cr(s) Eº = – 0.74 VWhich of the following metals could be usedsuccessfully to galvanize steel?(A) Ni only (B) Ni and Co(C) Fe only (D) Mn and Cr5. A beaker contains a small amount of gold dust(Au(s)). Which of the following aqueous solution,when added to the beaker, would dissolve the golddust (ie, convert Au (s) to + Au 3+ (aq))?Half reactionEº (at 25ºC)Zn 2+ + 2e – ⎯→ Zn – 0.76Al 3+ + 3e – ⎯→ Al –1.66O 2 + 2H + + 2e – ⎯→ H 2 O 2 0.702–Cr 2 O 7 + 6e – + H – ⎯→ 2Cr 3+ 1.23Au 3+ + 3e – ⎯→ Au 1.50O 2 + 4H + + 4e – ⎯→ 2Η 2 Ο 1.30(A) Cr 2 O 7 2– (acidic solution)(B) H 2 O 2 (acidic solution)(C) Al 3+(D) Zn 2+1kXtraEdge for IIT-JEE 49 MAY 2011