NEET AIPMT SOLVED PAPER 2001

  • question_answer
                    Rate of heat flow through a cylindrical rod is \[{{H}_{1}}\]. Temperatures of ends of rod are \[{{T}_{1}}\] and \[{{T}_{2}}\]. If all the dimensions of rod become double and temperature difference remains same and rate of heat flow becomes \[{{H}_{2}}\]. Then:                                                                                                                                                          

    A)                 \[{{H}_{2}}=2{{H}_{1}}\]               

    B)                 \[{{H}_{2}}=\frac{{{H}_{1}}}{2}\]

    C)                 \[{{H}_{2}}=\frac{{{H}_{1}}}{4}\]              

    D)                 \[{{H}_{2}}=4{{H}_{1}}\]

    Correct Answer: A

    Solution :

                    Key idea: Rate of heat flow is the ratio of heat transferred to the time interval.                 Rate of heat flow                 \[H=\frac{Q}{t}=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})}{l}\]                 or            \[H\propto \frac{A}{l}\]                 Since, dimensions of area \[(A)=[{{L}^{2}}]\]                 Dimensions of length (l) = [L] \[\therefore \]  \[H\propto l\] or            \[\frac{{{H}_{2}}}{{{H}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{2{{l}_{1}}}{{{l}_{1}}}\]                            \[(\because {{l}_{2}}=2{{l}_{1}})\]                 or            \[{{H}_{2}}=2{{H}_{1}}\]


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