Haryana PMT Haryana PMT Solved Paper-2004

  • question_answer The dimensional formula for thermal resistance is :

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

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

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

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

    Correct Answer: A

    Solution :

                                             When thermal conductivity is K, thermal resistivity is \[\frac{1}{K}\], then thermal resistance                                 \[R=\frac{1\times L}{KA}\] 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 \[\frac{({{\theta }_{1}}-{{\theta }_{2}})t}{Q}\] \[=\frac{\theta T}{M\times ({{L}^{2}}{{T}^{-2}}{{\theta }^{-1}})}={{M}^{-1}}{{L}^{-2}}{{T}^{3}}\theta \]

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