NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • question_answer
    Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are \[{{K}_{1}}\] and \[{{K}_{2}}\]. The thermal conductivity of the composite rod will be                                                                     [NEET-2017]

    A)  \[2({{K}_{1}}+{{K}_{2}})\]    

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

    C) \[\frac{3({{K}_{1}}+{{K}_{2}})}{2}\]  

    D)       \[{{K}_{1}}+{{K}_{2}}\]

    Correct Answer: B

    Solution :

    [b] Thermal current
    \[H={{H}_{1}}+{{H}_{2}}\]
    \[=\frac{{{K}_{1}}A({{T}_{1}}-{{T}_{2}})}{d}+\frac{{{K}_{2}}A({{T}_{1}}-{{T}_{2}})}{d}\]
    \[\frac{{{K}_{EQ}}2A({{T}_{1}}-{{T}_{2}})}{d}=\frac{A({{T}_{1}}-{{T}_{2}})}{d}[{{K}_{1}}+{{K}_{2}}]\]
    \[{{K}_{EQ}}=\left[ \frac{{{K}_{1}}+{{K}_{2}}}{2} \right]\]


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