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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    m grams of a gas of molecular weight M is flowing in an isolated tube velocity v. If the gas flow is suddenly stopped the rise in the temperature is: (y = ratio of specific heats, R = universal gas constant, J = mechanical equivalent of heat).

    A)  \[\frac{M{{v}^{2}}(\gamma -1)}{2RJ}\]                 

    B)  \[\frac{m}{M}\frac{{{v}^{2}}(\gamma -1)}{2RJ}\]

    C)   \[\frac{m{{v}^{2}}\gamma }{2RJ}\]                                       

    D)  \[\frac{M{{v}^{2}}\gamma }{2RJ}\]

    Correct Answer: A

    Solution :

                     \[\Delta U=\frac{\mu RJ\Delta \Tau }{(\gamma -1)}\] \[\Rightarrow \]               \[\Delta T=\frac{(\gamma -1)\Delta U}{\mu RJ}\] \[\Rightarrow \]               \[\Delta T=\frac{(\gamma -1)\times \frac{1}{2}m{{v}^{2}}}{(m/M)RJ}\]                 \[=\frac{M(\gamma -1){{v}^{2}}}{2RJ}=\frac{M{{v}^{2}}(\gamma -1)}{2RJ}\]


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