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R l=sin 45º sin 30ºR sin 45º = = 2l sin 30º9.[A,B] As a T = a Nvdv − v 2∴ = which can alsods Rdv v 2be written as = – dt RIntegrating the above equations answer isobtained.10.[A,C] (A) ρ B LAg = ρ × 4L Ag + 2ρ × 4L Ag11.[B,D]12.[A,B,C]L 3L(B) F B = ρ × Ag + 2ρ × Ag4 47 FF B = ρLAg; a = B − mg4 m8V3Ω13Va•b6V← eq. ckt.1011Using Kirchoff's Law Solve the circuit.CVV–CVIf battery is disconnected and plate are pulledapart, then charge will remain constantQE =2A ∈ × 2 = Q0 Α ∈0∴ E remain same (A) is correct work is doneagainst attractive force+ −Felc. Fext←⎯⎯⎯ ⎯⎯⎯→by Fext. (B) is correct.U = 21 CV2V = constant [as battery is connected]∈ AC = 0dcas d increaseC decrease ∴ U decreaseoption (C) is correct.13.[C] As springs are in parallelFnet(k1+ k 2 )x∴ a = =mass (m + m )14.[A] Frictional force on m 2 will act in direction ofdisplacement ifk 2 x > m 2 a15.[A] as k 2 A – µm 2 g = m 2 a max⎡ k1+ k 2∴ k 2 A – µm 2 g = m 2 ⎢⎣ m1+ mSolve to get answer. ]16.[C] Induced emf across OP =17.[B](i) current =1Bωl28R2l / 22⎤⎥ A⎦21 ⎛ ⎞Bω⎜l ⎟ =2 ⎝ 2 ⎠Torque on the rod = 2∫Bi x dx =2204Bil24Bωl28… (i)…(ii)Ml dωB ωl∴ × = − [substituting τ = I α]12 dt 32Rω2 2 tdω3B l–∫=ω 8RM ∫dtω00Solving this eq. & eq. (i)2Bω i = 0 le –αt8R∞θ =∫idt018.[A] Heat generated =1I ω2Column Matching19. [A] → p; [B] → p; [C] → r; [D] → qAs cube is floating ρ s ALg = ρ L Axg⎛ ρ ⎞∴ x =⎜ s⎟ L⎝ ρL⎠20. [A] → r; [B] → p; [C] → s; [D] → q20S = 21 × 2 × 16 = 16 m| Wg | = mg S =W N = m(g + a) cos 2 θ. SW f = m(g + a) sin 2 θ . SXtraEdge for IIT-JEE 93APRIL 2010