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EAMCET Medical EAMCET Medical Solved Paper-1997

  • question_answer
    Two cylindrical conductors A and B of same metallic material have their diameters in the ratio 1:2 and lengths in the ratio 2:1. If the temperature difference between their ends is same, the ratio of heats conducted respectively by A and B per second is:

    A)  1 : 2                                      

    B)  1 : 4

    C)  1 : 16                                    

    D)  1 : 8

    Correct Answer: D

    Solution :

                     \[H=\frac{Q}{t}=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})}{l}\] Here      \[{{K}_{1}}={{K}_{2}}\]                                 \[{{l}_{A}}:{{l}_{B}}=2:1\] \[{{r}_{A}};{{r}_{B}}=1:2\] From Eq.(i) \[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{r_{A}^{2}}{{{l}_{A}}}\times \frac{{{l}_{B}}}{r_{B}^{2}}={{\left( \frac{{{r}_{A}}}{{{r}_{B}}} \right)}^{2}}\times \left( \frac{{{l}_{B}}}{{{l}_{A}}} \right)\] \[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}\] Hence,     \[{{Q}_{A}}:{{Q}_{B}}=1:8\]


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