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

  • question_answer
    A cylindrical rod of aluminum is of length 20 cm, and radius 2 cm. The two ends are maintained at temperatures of \[0{}^\circ C\] and \[50{}^\circ C\] [the coefficient of thermal conductivity is \[\frac{0.5\,cal}{cm\times sec{{\times }^{o}}C}\] ]Then the thermal resistance of the rod in \[\frac{cal}{sec{{\times }^{o}}C}\]

    A) 318      

    B)        31.8  

    C) 3.18     

    D)        0.318

    Correct Answer: D

    Solution :

    [d] Thermal resistance \[R=/lKA\] Where / =20cm. & A(cylindrical rod) \[=\pi {{r}^{2}}=40\pi c{{m}^{2}}\] So\[R=\frac{20}{0.5\times 40\pi }=0.318\frac{cal}{\sec \,\,\times {}^\circ C}\]


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