Manipal Medical Manipal Medical Solved Paper-2003

  • question_answer
    The dimensional formula for thermal resistance is:

    A)  \[{{M}^{-1}}{{L}^{-2}}{{T}^{3}}\theta \]

    B)  \[{{M}^{-1}}{{L}^{2}}{{T}^{-3}}\theta \]

    C)  \[M{{L}^{2}}{{T}^{-2}}\theta \]

    D)  \[M{{L}^{2}}{{T}^{-2}}{{\theta }^{-1}}\]

    Correct Answer: A

    Solution :

     When thermal conductivity is K, thermal resistivity is\[\frac{1}{K},\]then thermal resistance Now heat conducted is given by \[\frac{{{Q}_{1}}}{t}={{\theta }_{1}}\frac{kA({{\theta }_{1}}-{{\theta }_{2}})}{L}=\frac{{{\theta }_{1}}-{{\theta }_{2}}}{R}\] Therefore, dimensions of R = dimensions of\[\frac{({{\theta }_{1}}-{{\theta }_{2}})t}{Q}\] \[=\frac{\theta T}{M\times ({{L}^{2}}{{T}^{-2}}{{\theta }^{-1}})\theta }={{M}^{-1}}{{L}^{-2}}{{T}^{3}}\theta \]


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