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

  • question_answer
    \[e=\alpha \,t-\frac{1}{2}\beta \,{{t}^{2}}\], If temperature of cold junction is \[{{0}^{o}}C\] then temperature of inversion is \[(\text{if }\alpha =500.0\mu V/{{}^{o}}C,\,\beta =5.0\mu V/\text{Squar}{{\text{e}}^{o}}C\])            [DCE 2001]

    A)            100                                           

    B)            200

    C)            300

    D)            400

    Correct Answer: B

    Solution :

               \[{{t}_{n}}=\frac{\alpha }{\beta }=\left( \frac{500}{5} \right)=100{}^\circ C\] Also \[{{t}_{n}}=\frac{{{t}_{i}}+{{t}_{c}}}{2}\Rightarrow 100=\frac{{{t}_{i}}+0}{2}\]Þ \[{{t}_{i}}=200{}^\circ C\]


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