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BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
    Thermal conductivity of a material in CGS  system is 0.4. In steady state, the rate of flow of heat is\[\text{1}0\text{ cal}/\text{s}-\text{c}{{\text{m}}^{\text{2}}}\].The thermal gradient will be

    A) \[\text{5}{}^\circ \text{C}/\text{cm}\]                  

    B) \[\text{15}{}^\circ \text{C}/\text{cm}\]

    C) \[\text{25}{}^\circ \text{C}/\text{cm}\]                

    D) \[\text{5}0{}^\circ \text{C}/\text{cm}\]

    Correct Answer: C

    Solution :

    \[\frac{\Delta Q}{\Delta t}=KA\frac{\Delta \theta }{\Delta x}\] Thermal gradient \[=\frac{\left( \frac{\Delta Q}{\Delta t} \right)}{KA}=\frac{10}{0.4}={{25}^{o}}C/cm\]


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