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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are\[{{\operatorname{T}}_{2}}\] and \[{{T}_{1}}\]\[({{\operatorname{T}}_{2}}>{{T}_{1}})\]. The rate of heat transfer through the slab, in a steady state is \[\left( \frac{A\left( {{\operatorname{T}}_{2}}>{{T}_{1}} \right)K}{x} \right)f\], with f equal to             

    A) \[\frac{2}{3}\]

    B) \[\frac{1}{2}\]

    C) 1

    D) \[\frac{1}{3}\]

    Correct Answer: D

    Solution :

    [d] the thermal resistance is given by \[\frac{x}{KA}+\frac{4x}{2KA}=\frac{x}{KA}+\frac{2x}{KA}=\frac{3x}{KA}\] \[\therefore \frac{dQ}{dt}=\frac{\Delta T}{\frac{3x}{KA}}=\frac{\left( {{T}_{2}}-{{T}_{1}} \right)KA}{3x}\] \[=\frac{1}{3}\left\{ \frac{A\left( {{T}_{2}}-{{T}_{1}} \right)K}{x} \right\}\therefore f=\frac{1}{3}\]


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