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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    Two unequal bars in length and cross-section with thermal conductivities \[{{k}_{1}}\] and \[{{k}_{2}}\] are connected in series. One end is maintained at \[{{\theta }_{1}}^{o}C\] and other end at \[{{\theta }_{2}}^{o}C\]. If the ratio of length of two bars is \[{{n}_{1}}\] and ratio of cross sectional area of two bars is \[{{n}_{2}}\], then find the value of \[n{{  }_{1}}/{{n}_{2}}\] for which temperature of the junction point is \[\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]

    A) \[\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}}\]     

    B)  \[\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{2}}}\]

    C) \[\frac{{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]                                 

    D) \[\frac{{{k}_{1}}}{{{k}_{2}}}\]

    Correct Answer: D

    Solution :

    \[\theta =\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] \[2\theta ={{\theta }_{1}}+{{\theta }_{2}}\] \[{{\theta }_{1}}-\theta =\theta -{{\theta }_{2}}\]                             ?.(1) Bars are in series Heat current\[=H=\] \[\frac{{{\theta }_{1}}-\theta }{{{R}_{1}}}=\frac{\theta -{{\theta }_{2}}}{{{R}_{2}}}\]                         ?(2) From eq. (1) and (2) \[{{R}_{1}}={{R}_{2}}\] \[\frac{{{l}_{1}}}{{{K}_{1}}{{A}_{1}}}=\frac{{{l}_{2}}}{{{K}_{2}}{{A}_{2}}}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{K}_{1}}}{{{K}_{2}}}\times \frac{{{A}_{1}}}{{{A}_{2}}}\] \[{{n}_{1}}=\frac{{{K}_{1}}}{{{K}_{2}}}.{{n}_{2}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{K}_{1}}}{{{K}_{2}}}\]


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