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JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    Two different metal rods of equal lengths & equal cross section area have their ends kept at the same temperatures \[{{\theta }_{1}}\And {{\theta }_{2}}.\]If \[{{K}_{1}}\And {{K}_{2}}\] be the thermal conductivities of rod, \[{{\rho }_{1}}\And {{\rho }_{2}}\] are their densities and \[{{s}_{1}},{{s}_{2}}\] are their specific heats, then the rate of flow of heat in the two rods will be same if

    A) \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{\rho }_{1}}{{s}_{1}}}{{{\rho }_{2}}{{s}_{2}}}\]                             

    B) \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{\rho }_{1}}{{s}_{2}}}{{{\rho }_{2}}{{s}_{1}}}\]

    C) \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{\theta }_{1}}}{{{\theta }_{2}}}\]                       

    D) \[{{K}_{1}}={{K}_{2}}\]

    Correct Answer: D

    Solution :

    Rate of heat flow is, \[H=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})}{\ell }\] For 1st rod \[{{H}_{1}}=\frac{{{K}_{1}}A({{\theta }_{1}}-{{\theta }_{2}})}{\ell }\] For 2nd rod\[{{H}_{2}}=\frac{{{K}_{2}}A({{\theta }_{1}}-{{\theta }_{2}})}{\ell }\] For \[{{H}_{1}}={{H}_{2}},{{K}_{1}}={{K}_{2}}\]


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