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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2012

  • question_answer
    A hot body is allowed to cool. The surrounding temperature is constant at\[{{30}^{o}}C\]. The body takes time t, to cool from \[{{90}^{o}}C\] to \[{{89}^{o}}C\] and time \[{{t}_{2}}\] to cool from \[{{60}^{o}}C\] to \[{{59.5}^{o}}C\]. Then,

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

    B)  \[{{t}_{2}}=\frac{{{t}_{1}}}{2}\]

    C)  \[{{t}_{2}}=4{{t}_{1}}\]             

    D)  \[{{t}_{2}}={{t}_{1}}\]

    Correct Answer: D

    Solution :

    : According to Newtons law of cooling, \[\frac{{{T}_{1}}+{{T}_{2}}}{t}=K\left( \frac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{s}} \right)\] where \[{{T}_{s}}\] is the surrounding temperature. As per question, \[\frac{(90-89)}{{{t}_{1}}}=K\left( \frac{90+89}{2}-30 \right)\] \[\frac{1}{{{l}_{1}}}=K(59.5)\]                  ??.(i) and \[\frac{(60-59.5)}{{{t}_{2}}}=K\left( \frac{60+59.5}{2}-30 \right)\] \[\frac{0.5}{{{t}_{2}}}=K(29.75)\]            ??..(ii) Divide (i) by (ii), we get \[\frac{{{t}_{2}}}{0.5{{t}_{1}}}=\frac{59.5}{29.75}\] \[\frac{{{t}_{2}}}{{{t}_{1}}}=1\] or \[{{t}_{2}}={{t}_{1}}\]


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