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

  • question_answer
    The emf of a thermocouple, one junction of which is kept at \[{{0}^{o}}C,\] is given by \[e=at+b{{t}^{2}}\] the Peltier co-efficient will be

    A)            \[(t+273)\,\,(a+2bt)\]      

    B)            \[(t+273)\,\,(a-2bt)\]

    C)            \[(t-273)\,\,(a-2bt)\]        

    D)            \[(t-273)\,\,(a-2bt)\]

    Correct Answer: A

    Solution :

                       \[\because \] Peltier coefficient \[\pi =T\frac{de}{dT}\] and \[{{t}^{o}}C=T-273\]                    \ \[e=a(T-273)+b{{(T-273)}^{2}}\]                    Differentiating w.r.t. T  \[\frac{de}{dT}=a+2b(T-273)\]            \[\pi =T\frac{de}{dT}=T[a+2b(T-273)]\]Þ\[\pi =(t+273)\,(a+2bt)\]


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