CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    A bucket full of hot water is kept in a room. It cools from \[{{75}^{o}}C\] to \[{{70}^{o}}C\] in \[{{t}_{1}}\] minutes, from \[{{70}^{o}}C\] to \[{{65}^{o}}C\] in \[{{t}_{2}}\] minutes and from \[{{65}^{o}}C\]to \[{{60}^{o}}C\] 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: A

    Solution :

    According to Newtons law of cooling we have, rate of cooling \[\propto \] temperature difference between the liquid and surrounding. As temperature difference decreases gradually, time taken to cool increases \[i.e.,\] \[{{T}_{1}}<{{T}_{2}}<{{T}_{3}}\]


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