JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

  • question_answer
    The thermo-emf of a thermocouple varies with the temperature\[\theta \]of the hot junction as\[E=a\theta +b{{\theta }^{2}}\]in volts where the ratio a/b is\[700{}^\circ C\]. If the cold junction is kept at\[0{}^\circ C,\]then the neutral temperature is           [AIEEE 2004]

    A) \[700{}^\circ C\]          

    B)      \[350{}^\circ C\]

    C) \[1400{}^\circ C\]

    D) no neutral temperature is possible for this thermocouple

    Correct Answer: D

    Solution :

    [d] The thermo emf is given by
    \[E=a\theta +b{{\theta }^{2}}\]                                       (given)
    For neutral temperature \[({{\theta }_{n}}),\frac{dE}{d\theta }=0\]
    \[\Rightarrow \]   \[a+2b{{\theta }_{n}}=0\]
    \[\Rightarrow \]   \[{{\theta }_{n}}=-\frac{a}{2b}\]
    \[\therefore \]      \[{{\theta }_{n}}=-\frac{700}{2}\]              \[\left( \because \frac{a}{b}={{700}^{o}}C \right)\]
    \[=-{{350}^{\text{o}}}C<{{0}^{\text{o}}}C\]
    But neutral temperature can never be negative (less than zero) i.e., \[{{\theta }_{n}}<|{{0}^{o}}C\].
    Hence, no neutral temperature is possible for this thermocouple.


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