NEET Physics Transmission of Heat NEET PYQ-Transmission Of Heat

  • 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:                                                                              [AIPMT 2001]

    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         \[{{H}_{2}}=2{{H}_{1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner