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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Sample Paper Topic Test - Thermometry, Calorimetry & Thermal Expansion

  • question_answer
    At \[15{}^\circ C,\] a wheel has diameter of 30 cm and inside diameter of a steel rim is 29.9 cm. Coefficient of thermal expansion of steel is \[1.2\times {{10}^{-5}}{{/}^{o}}C\]. To what temperature, the rim must be heated so that it can slip over wheel?

    A) \[2787{}^\circ C\]                    

    B) \[2772{}^\circ C\]

    C) \[2529{}^\circ C\]                    

    D) \[2802{}^\circ C\]

    Correct Answer: D

    Solution :

    Rim will slip if its diameter increases to 30 cm. Hence \[\Delta l=0.1\,cm\]
    \[\Delta T=\frac{\Delta l}{l\times a}=\frac{0.1}{29.9\times 1.2\times {{10}^{-5}}}={{2787}^{o}}C\]
    Rim to be heated to temperature
    \[2787+15={{2802}^{o}}C\]


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