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JEE Main & Advanced Physics Transmission of Heat Question Bank Radiation Newton's Law of Cooling

  • question_answer
    A body takes 4 minutes to cool from \[{{100}^{o}}C\] to \[{{70}^{o}}C\]. To cool from \[{{70}^{o}}C\] to \[{{40}^{o}}C\] it will take (room temperature is \[{{15}^{o}}C\])                                              [MP PET 1995]

    A)            7 minutes                              

    B)            6 minutes

    C)            5 minutes                              

    D)            4 minutes

    Correct Answer: B

    Solution :

                       \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\,\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\]                    \[\therefore \]\[\frac{100-70}{4}=K\,\left( \frac{100+70}{2}-15 \right)\] = 60K Þ K = \[\frac{1}{8}\]            Again \[\frac{70-40}{t}=\frac{1}{8}\left( \frac{70+40}{2}-15 \right)\]= 5 Þ t = 6 min.


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