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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    In a room, where the temperature is \[30{}^\circ C\], a body cools from \[61{}^\circ C\] to \[59{}^\circ C\] is 4 min. The time taken by the body to cool from \[51{}^\circ C\] to \[49{}^\circ C\] will be

    A) 4 min             

    B)  6 min             

    C)  5 min                         

    D)  8 min

    Correct Answer: B

    Solution :

     From the Newton?s law of cooling. Rate of cooling \[\propto \] difference in temperature. \[\frac{dT}{dt}\propto \Delta \theta \] \[\Rightarrow \]            \[\frac{dT}{dt}=K\Delta \theta \] In first case, \[dT=61-59=2\] \[\Delta \theta =50-30=20\] \[dt=4\,\min ,\] \[\therefore \]    \[K=\frac{1}{\Delta \theta }\left( \frac{dT}{dt} \right)=\frac{2}{30\times 4}=\frac{1}{60}\] For second case, \[dT=2\] \[\Delta \theta =50-30=20\] \[\therefore \]    \[dt=\frac{1}{K}\,\left( \frac{dT}{d\theta } \right)=\frac{2}{\frac{1}{60}\times 20}=6\,\min \]


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