JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Critical Thinking

  • question_answer
    The thermo e.m.f. 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 \[{{700}^{o}}C.\] If the cold junction is kept at \[{{0}^{o}}C,\] then the neutral temperature is        [AIEEE 2004]

    A)            \[{{700}^{o}}C\]

    B)            \[{{350}^{o}}C\]

    C)            \[{{1400}^{o}}C\]

    D)            No neutral temperature is possible for this thermocouple

    Correct Answer: D

    Solution :

                       Comparing with standard equation \[E=\alpha t+\frac{1}{2}\beta {{t}^{2}}\]  a = a and b = 2b Þ  \[{{t}_{n}}=-\frac{a}{2b}=-\frac{1}{2}\times 700=-{{350}^{o}}C\] This is not possible.


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