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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Thermometry

  • question_answer
    Recently, the phenomenon of superconductivity has been observed at 95 K. This temperature is nearly equal to [CPMT 1990]

    A)            ? 288°F                                   

    B)            ? 146°F

    C)            ? 368°F                                   

    D)            +178°F

    Correct Answer: A

    Solution :

               \[\frac{F-32}{9}=\frac{K-273}{5}\] Þ\[\frac{F-32}{9}=\frac{95-273}{5}\] Þ\[F=-288{}^\circ F\]


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