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

  • question_answer
    On a linear temperature scale Y, water freezes at \[160{}^\circ  Y\]and boils at \[- 50{}^\circ  Y\]. On this Y scale, a temperature of 340 K would be read as : (water freezes at 273 K and boils at 373 K)

    A) \[-73.7{}^\circ Y\]

    B) \[-\,233.7{}^\circ Y\]

    C) \[-86.3{}^\circ Y\]

    D) \[-106.3{}^\circ Y\]

    Correct Answer: C

    Solution :

    [c] \[\frac{Reading\,on\,any\,scale-LFP}{UFP-LFP}\] =constant for all scales \[\frac{340-273}{373-273}=\frac{{}^\circ y-\left( -160 \right)}{-50-\left( -160 \right)}\Rightarrow \frac{67}{100}=\frac{y\,+160}{110}\]\[\therefore y=-86.3{}^\circ Y\]


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