1. Xtra Edge February 2012 - Career Point

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1. For the process A → B 1 3 1 U ∝ ⇒ RT ∝ ρ 2 M V or T ∝ V So AB is isobaric process (pressure in constant) 3. TV 8U 0 ⎡ 3 ⎤ T A = ⎢As 4U 0 = RTA 3R ⎥ ⎣ 2 ⎦ M VA = 2ρ 0 16U so 0ρ 0 M PA = and VB = 3M 4ρ 0 so WAB = P[VB - VA] 2. 16U 0ρ 0 ⎡ M M ⎤ 16U 0ρ 0 M ⇒ ⎢ − ⎥ = − × 3M ⎣4ρ 0 2ρ 0 ⎦ 3M 4ρ0 Option [D] is correct P The slope of the graph is 0 m = − 2V0 The equation of this graph is given by P0V 3P0 P = − + 2V0 2 nRT −P0 V 3P0 ⇒ = + V 2V0 2 −P0 V 2 3P0 ⇒ T = − V + V 2nRV0 2nR dT −2P0V0 3P0 = + = 0 dV 2nRV0 2nR P0V ⇒ nRV0 2P0 3V0 = ⇒ V = 2nR 2 3V At 0 V = , temp will be maximum [n = 1] 2 2 P0 9V0 3P0 3V0 Tmax = − × + × 2RV0 4 2R 2 9 P0V0 9 P0V0 ⇒ − + 8 R 4 R 9 P0V0 ⇒ 8 R Option [C] is correct -3 PV = k − 3 ⇒ V = k nR for this polytrophic process, PV -2 = constant so x = -2 nR( T2 − T1 ) nR( T2 − T1 ) 2 w = = = nRT0 1− x 3 3 ∆U = nCV∆T = n 3/2 R(3T0 – T0) = 3nRT0 ⎛ 2 ⎞ 11 ∆ Q = ∆U = W = ⎜ + 3⎟nRT0 = nRT0 ⎝ 3 ⎠ 3 ∆Q 11 nRT 11 ∆Q = nC∆T 0 ⇒ C = = = R n∆T 3 n( 2T0 ) 6 Option [A,C,D] is correct y 4. R ωt F Fx Fy x F = mω 2 R Fy = F sin ωt Fy = mω 2 R sin (ωt) x * In figure as cos ωt = ⇒ x = R cos ωt R * For one complete circle, θ = ωt y * R ωt vx x vx = – ωR sin ωt v = ωR Option [A,C] is correct 5. From 1 st 8 Solution Set # 9 Physics Challenging Problems Questions were Published in January Issue law, ∆Q = ∆U + W nCαT + nCvαT + pαV ⇒ nCαT = nCVαT + nRT/V αV α V αV nRT αV ⇒ nαT0e αV = C VαT0 e αV + T0e αV V ⇒ αC = αCV + R/V ⇒ C = CV + R/αV Option [B] is correct 6. Option [C] is correct 7. Option [B] is correct 8. Option [D] is correct XtraEdge for IIT-JEE 16 FEBRUARY 2012
PHYSICS 1. AB and CD are two ideal springs having force constant K1 and K2 respectively. Lower ends of these springs are attached to the ground so that the springs remain vertical. A light rod or length 3a is attached with upper ends B and C of springs. A particle of mass m is fixed with the rod at a distance a from end B and in equilibrium, the rod is horizontal. Calculate period of small vertical oscillations of the system. B a m 2a C K1 A D Sol. Let, in equilibrium, compressive forces in left and right springs be F1 and F2, respectively. Considering free body diagram of rod BC, (figure) B a 2a C F1 mg F1 + F2 = mg ...(i) Taking moments about B, mg × a = F2 × 3a 1 or F2 = mg 3 Substituting this value in equation (i), 2 F1 = mg 3 Let the particle be pressed from its equilibrium position by applying a force 'F'. Let left and right springs be further compressed through y1 and y2, respectively. Increase in compressive forces in the spring will be K1y1 and K2y2 respectively. In other words, total compressive forces in two springs will be (F1 + K1y1) and (F2 + K2y2) respectively. Now considering new free body diagram (figure) of the rod BC, Students Forum Expert’s Solution for Question asked by IIT-JEE Aspirants K2 F2 (F1 + K1y1) + (F2 + K2y2) = F + mg Taking moments about B, (figure) F ...(ii) XtraEdge for IIT-JEE 17 FEBRUARY 2012 B a (F1 + K1y1) mg (F2 + K2y2) (F + mg) × a = (F2 + K2y2) × 3a Solving above equations, ...(iii) or y1 = 2a 2 F F K1y1 = and K2y2 = . 3 3 2F 3K1 and y2 = F 3K 2 Since, distance of the particle from left spring is 'a' and that from right spring is '2a', therefore, downward displacement of the particle from equilibrium position will be 2y1 y2 y = + = 3 3 4F 9K l + F = 9K 2 C ( K1 + 4K 2 ) F 9K K 9K1K 2 or F = ...(iv) ( K1 + 4K 2) Now, if the particle be released, it starts accelerating upwards due to excess compressive force in springs. Hence, the restoring force is (K1y1 + K2y2), which is numerically equal to F. or Restoring force = F = 1 9K1K 2 . y ( K + 4K ) F 9K1K 2 or Restoring acceleration, a = = .y m m( K1 + 4K 2 ) Since, acceleration is restoring and is directly proportional to displacement y, therefore, the particle performs SHM. Its period of oscillations is y T = 2π a 2 π m( K1 + 4K 2) or T = 3 K K 1 2 1 2 Ans. 2
Page 3 and 4: Volume - 7 Issue - 8 February, 2012
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(a) temperature and (b) Pressure ?
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should be bought to meet the requir
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This work done is stored inside the
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Use : It can be used to accelerate
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NH3 NH3 NH3 Co +3 NH3 NH3 NH3 No un
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7. f(A) = A 2 - 5A + 7 I2 ⎡ 3 1
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y 2 = - (2 - x) 2 + 9 This is the e
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(0, 4) ( 2 3, 2) (4, 0) ⎡ 2 4 2
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