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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and cross-section A as shown in diagram. The cylinder contains one mole of an ideal diatomic    gas.    The current flows through the element such that the temperatures rises with time t as \[\Delta t=\alpha t+\frac{1}{2}\beta {{t}^{2}}\] (\[\alpha \]and\[\beta \]are constants), while pressure remains constant. The atmospheric pressure above the piston is \[{{P}_{0}}.\]Then

    A)  the rate of increase in internal energy is\[\frac{5}{2}R(\alpha +\beta t)\]

    B)  the current flowing in the element is \[\sqrt{\frac{5}{2r}R(\alpha +\beta t)}\]

    C)  the piston moves upwards with constant acceleration

    D)  the piston moves upwards with constant speed

    Correct Answer: A

    Solution :

     We know that Internal energy, \[U=\frac{nfRT}{2}=\frac{5R}{2}\left( \alpha t+\frac{1}{2}\beta {{t}^{2}} \right)\] Differentiate with respect to t \[\frac{dU}{dt}=\frac{5R}{2}[\alpha +\beta t]\] But \[dQ=n{{C}_{p}}dT\] \[\therefore \]   \[\frac{dQ}{dt}=n{{C}_{p}}\times \frac{dT}{dt}\Rightarrow {{i}^{2}}r=\frac{7}{2}R\times [\alpha +\beta t]\] \[i=\sqrt{\frac{7R}{2r}(\alpha +\beta t)}\] \[pV=nRT\] or \[V=\frac{nRT}{p}=\frac{nR}{p}\left( \alpha t+\frac{1}{2}\beta {{t}^{2}} \right)\] \[X=\frac{nR}{pA}\left( \alpha t+\frac{1}{2}\beta {{t}^{2}} \right)\] \[v=\frac{nR}{pA}(\alpha +\beta t)\] and acceleration \[=\frac{nR}{pA}\times \beta \] So, the rate of increase in internal energy is \[\frac{5}{2}R(\alpha +\beta t)\]and the piston moves upwards with constant acceleration.


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