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JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer
    Two rods of length \[{{d}_{1}}\]and \[{{d}_{2}}\] and coefficients of thermal conductivities \[{{K}_{1}}\]and \[{{K}_{2}}\] are kept touching each other. Both have the same area of cross-section the equivalent of thermal conductivity is

    A)  \[{{K}_{1}}+{{K}_{2}}\]

    B)  \[{{K}_{1}}{{d}_{1}}+{{K}_{2}}{{d}_{2}}\]

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

    D)  \[\frac{{{d}_{1}}+{{d}_{2}}}{({{d}_{1}}/{{K}_{1}}+{{d}_{2}}/{{K}_{2}})}\]

    Correct Answer: D

    Solution :

     When two rods are connected in series \[Q=\frac{A({{T}_{1}}-{{T}_{2}})t}{\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{1}}}}=\frac{A({{T}_{1}}-{{T}_{2}})t}{({{d}_{1}}+{{d}_{2}})/K}\] \[\therefore \] \[\frac{{{d}_{1}}+{{d}_{2}}}{K}=\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}};\] \[\therefore \] \[K=\frac{({{d}_{1}}+{{d}_{2}})}{\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}}}\]


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