May 2011 - Career Point

Q 1 = 1200 JEnginel = 21 cma = 6 cmSol. When tube is rotated, liquid starts to flow radiallyoutward and air in sealed arm is compressed. Let theshift of liquid be x as shown in figure.xA(l – x)Ba–xLet cross-sectional area of tube be S.Initial volume of air, V 0 = Sa and initial pressureP 0 = 10500 Nm –2Final volume, V = S(a – x)P V∴ Final pressure P = 0 0 P= 0 .a.V (a – x)Por Pressure at B, P 2 = P + xρg = 0 a+ xρg(a – x)Centripetal force required for circular motion ofvertical column of height x of liquid is provided byreaction of the tube while that to horizontal length(l – x) is provided by excess pressure at B.Force exerted by pressure difference is⎡ PF 1 = (P B – P A ) S = (P 2 – P 0 ) S = 0 x ⎤⎢ + xρg⎥S⎣(a– x) ⎦Mass of horizontal arm AB of liquid is,m = S (l – x)ρRadius of circular path traced by its centre of mass isl – x ⎛ l + x ⎞r = x + = ⎜ ⎟2 ⎝ 2 ⎠∴ Centripetal force, F 2 = mω 2 0 r⎧ l + x ⎫ 2But F 2 = F 1 or {S ρ (l – x)} ⎨ ⎬ ω 0⎩ 2 ⎭⎧ P= 0 x⎨ + xρgS⎩(a– x) ⎭ ⎬⎫or x = .01 m = 1cm∴ Length of air column in sealed arm = (a – x)= 5cm5. A reversible Carnot engine is coupled with three heatreservoirs A,B and C as shown in Figure.Temperature of these reservoirs isxABQ 2Q 3CT 1 = 400 K T 2 = 300 K T 3 = 200 KT 1 = 400 K, T 2 = 300 K and T 3 = 200K, respectively.During an integral number of complete cycles, theengine absorbs Q 1 = 1200 joule of heat energy fromreservoir A and performs W = 200 joule ofmechanical work. Calculate quantities Q 1 and Q 2 ofheat energy exchanged with the other two reservoirsB and C respectively. State whether the reservoirsabsorb or lose heat energy.Sol. Given engine can be assumed to work in two ways ;(i) engine working between a source of temperatureT 1 and sink of temperature T 2 and(ii) an engine working between source of temperatureT 1 and sink of temperature T 3 . Hence, efficiency ofthese two isT1– T21 Tη 1 = = and1 – T31η2 = = respectivelyT14 T14If an engine having efficiency η absorbs heat Q fromsource then heat energy converted into mechanicalwork by the engine will be equal to ηQ andremaining part (Q – ηQ) = (1 – η)Q should berejected to the sink.Since, heat rejected to sink by first engine, havingefficiency η 1 , is Q 2 , therefore, heat absorbed by itQ24Qfrom source is equal to = 2(1– η1)3and work done by this engine isQ2η1QW 1 = = 2(1– η1)3Similarly, theat absorbed by the other engine fromQ3source is equal to = 2Q 3(1– η2)Q3.η2and work done by this engine is W 2 = = Q 3(1– η2)But W 1 + W 2 = 200 J1∴ Q2 + Q 3 = 200 J ...(1)3Total heat extracted from source A is⎛ 4 ⎞ ⎛ 4 ⎞Q 1 = ⎜ Q2 + 2Q3⎟ or ⎜ Q2 + 2Q3⎟ = 1200 J⎝ 3 ⎠ ⎝ 3 ⎠...(2)Solving equation (1) and (2),Q 2 = 1200 J and Q 3 = – 200 JIt means reservoir B absorbs 1200 J of heat andreservoir C loses 200 J of heat.In fact engine works as an engine between reservoirsA and B and as heat pump between reservoirs B andC.XtraEdge for IIT-JEE 18 MAY 2011