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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    Heat is flowing through two cylindrical rods A and B of same material having the same temperature difference between their ends. The diameters of rods A and B are in the ratio 1 : 2 and their lengths in the ratio 2:1. The ratio of the rate of flow of heat in rod A to that in rod B is:

    A) \[2:1\]                                  

    B)  \[2:3\]                 

    C)  \[1:1\]                                 

    D)  \[1:8\]

    E)  4: 1

    Correct Answer: D

    Solution :

    \[H=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})}{l}\Rightarrow H\propto \frac{{{r}^{2}}}{l}\] \[\therefore \]  \[\frac{{{H}_{A}}}{{{H}_{B}}}={{\left( \frac{{{r}_{A}}}{{{r}_{B}}} \right)}^{2}}\times \left( \frac{{{l}_{B}}}{{{l}_{A}}} \right)={{\left( \frac{1}{2} \right)}^{2}}\times \frac{1}{2}=\frac{1}{8}\]


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