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JIPMER Jipmer Medical Solved Paper-2003

  • question_answer
    A cup of tea cools from \[80{}^\circ C\] to \[60{}^\circ C\] in one minute. The ambient temperature is\[30{}^\circ C\]. In cooling from \[60{}^\circ C\] to \[50{}^\circ C\]. It will take:

    A)  50 sec                  

    B)         90 sec                  

    C)  60 sec                  

    D)         48 sec

    Correct Answer: D

    Solution :

    Rate of cooling \[\propto \] excess of temperature \[\frac{80{}^\circ -60{}^\circ }{60{}^\circ }=K\left( \frac{80{}^\circ +60{}^\circ }{2}-30{}^\circ  \right)\] \[\frac{1}{3}=K(40)\]                      ?(1) \[\frac{60{}^\circ -50{}^\circ }{t}=K\left( \frac{60{}^\circ +50{}^\circ }{2}-30{}^\circ  \right)\] \[\frac{10}{t}=K(25)\]                     ...(2) Dividing equation (1) by (2), we get                 \[\frac{t}{30}=\frac{40}{25}\] or \[t=\frac{40}{25}\times 30=48\,\sec \]


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