Manipal Medical Manipal Medical Solved Paper-2004

  • question_answer
    A beaker of hot water cools from\[75{}^\circ C\] to\[70{}^\circ C\] in \[{{t}_{1}}\] minutes, from \[70{}^\circ C\] to \[65{}^\circ C\] in \[{{t}_{2}}\] minutes and from\[65{}^\circ C\]to\[60{}^\circ C\]in\[{{t}_{3}}\] minutes. Then:

    A)  \[{{t}_{1}}={{t}_{2}}={{t}_{3}}\]

    B)  \[{{t}_{1}}<{{t}_{2}}<{{t}_{3}}\]

    C)  \[{{t}_{1}}>{{t}_{2}}>{{t}_{3}}\]

    D)  \[{{t}_{1}}<{{t}_{2}}<{{t}_{3}}\]

    Correct Answer: B

    Solution :

     The rate of cooling is given as \[\frac{dQ}{dt}=K({{T}_{1}}-{{T}_{2}})\]         ...(i) where\[({{T}_{1}}-{{T}_{2}})\]is temperature difference between the temperature of body and surroundings. Since, the temperature difference between \[75{}^\circ C\]and surrounding temperature is greater than the temperature difference between   \[70{}^\circ C\]and surrounding temperature Hence,       \[{{t}_{1}}<{{t}_{2}}\]                       ...(2) Similarly, the temperature difference between   \[70{}^\circ C\]and surrounding temperature is greater than temperature difference between\[65{}^\circ C\] and surrounding temperature. Hence,       \[{{t}_{2}}<{{t}_{3}}\]                ...(3) Thus, from eqs. (2) and (3), \[{{t}_{1}}<{{t}_{2}}<{{t}_{3}}\]


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