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

  • question_answer
    The temperature of the cold junction of thermo‑couple is \[0{}^\circ C\] and the temperature of hot junction is \[T{}^\circ C\]. The e.m.f. is \[E=16T-0.04{{T}^{2}}\mu \] volts. The temperature of inversion is                                     [EAMCET 1994]

    A)                    \[200{}^\circ C\]          

    B)                    \[400{}^\circ C\]

    C)                    \[100{}^\circ C\]          

    D)                    \[300{}^\circ C\]  

    Correct Answer: B

    Solution :

                       Comparing the given equation with \[E=\alpha \,t+\frac{1}{2}\beta \,{{t}^{2}}\] We get \[\alpha =16\] and \[\frac{1}{2}\beta =-0.04\,\Rightarrow \,\,\beta =-0.08\] Þ \[{{t}_{n}}=-\frac{\alpha }{\beta }=-\frac{16}{-0.08}=200{}^\circ C\] Also \[{{t}_{i}}=2{{t}_{n}}-{{t}_{c}}\]Þ \[{{t}_{i}}=2\times (200)-0=400{}^\circ C\]


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