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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    The cold junction of a thermocouple is at \[0{}^\circ C\]. The thermo-emf \[\varepsilon ,\] in volt, generated in this thermocouple varies with temperature \[t\,\,{}^\circ C\] of the hot junction as \[\varepsilon =6+4t-\frac{{{t}^{2}}}{32}\] The neutral temperature of the thermocouple is

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

    B) \[76{}^\circ C\]

    C) \[64{}^\circ C\]                 

    D)        \[50{}^\circ C\]

    Correct Answer: C

    Solution :

    Given,\[\varepsilon =6+4t-\frac{{{t}^{2}}}{32}\] For neutral of thermocouple                 \[\frac{d\varepsilon }{dt}=0\] Here,     \[\frac{d\varepsilon }{dt}=4-\frac{2t}{32}\]                 \[4-\frac{2t}{32}=0\]                 \[t=\frac{4\times 32}{2}\]                 \[t={{64}^{o}}C\]


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