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Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    The thermo emf of a thermocouple is given by\[e=2164r-6.2{{t}^{2}},\] the neutral temperature and a temperature of inversion are

    A)                  349, 174.5                          

    B)  174.5, 349

    C)                  349, 698                              

    D)  698, 349

    Correct Answer: B

    Solution :

                    At neutral temperature, dE /dt = 0; so                 \[\frac{dE}{dt}=2164-6.2\times 2\times {{t}_{n}}=0\]or\[{{t}_{n}}={{174.5}^{o}}C\]                 At temperature of inversion                 \[e=0=2164{{t}_{i}}-6.2_{i}^{2}\]                 Hence         \[{{t}_{i}}~=349{}^\circ C\]


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