AMU Medical AMU Solved Paper-2001

  • question_answer
    An object is cooled from \[75{}^\circ C\] to \[65{}^\circ C\] in 2 min in a room at \[30{}^\circ C\] the time taken to cool the same object from \[55{}^\circ C\] to \[45{}^\circ C\] in the same room (in min) is

    A)  7                                            

    B)  6

    C)  5                                            

    D)  4

    Correct Answer: D

    Solution :

                     From Newtons law of cooling Rate of cooling oc temperature difference \[\therefore \]  \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] Given,   \[{{\theta }_{1}}={{75}^{o}}C,\,\,\,{{\theta }_{2}}{{65}^{o}}C,\,\,\,t=2\] min                 \[\frac{75-65}{2}=K\left( \frac{75+65}{2}-30 \right)\] \[\Rightarrow \]               \[5=K\times 40\] \[\Rightarrow \]               \[K=\frac{1}{8}\] When body cools from \[{{55}^{o}}C\] to \[{{65}^{o}}C\], we have                 \[\frac{55-45}{{{t}_{2}}}=K\left( \frac{55+45}{2}-30 \right)\] \[\Rightarrow \]               \[\frac{10}{{{t}_{2}}}=\frac{1}{8}\times 20\] \[\Rightarrow \]               \[{{t}_{2}}=4\,\min \]


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