EAMCET Medical EAMCET Medical Solved Paper-2010

  • question_answer
    For a thermocouple the temperature of cold junction (Tc), neutral temperature (Tn) and temperature of inversion (Ti) are \[0{}^\circ C\], \[285{}^\circ C\] and \[585{}^\circ C\] respectively. If the temperature of cold junction is raised to \[10{}^\circ C\], then

    A)                 \[{{T}_{n}}=\text{ }275{}^\circ C\text{ }and\text{ }{{T}_{i}}=\text{ }570{}^\circ C\]

    B)                 \[{{T}_{n}}=\text{ }275{}^\circ C\text{ }and\text{ }{{T}_{i}}=\text{ }560{}^\circ C\]

    C)                 \[{{T}_{n}}=\text{ }285{}^\circ C\text{ }and\text{ }{{T}_{i}}=\text{ }560{}^\circ C\]

    D)                 \[{{T}_{n}}=\text{ }295C\text{ }and\text{ }{{T}_{i}}=\text{ }580{}^\circ C\]

    Correct Answer: C

    Solution :

                     No change in neutral temperature when the temperature of cold junction is raised to \[10{{\,}^{o}}C\]                 \[{{T}_{n}}=\frac{{{T}_{i}}+{{T}_{c}}}{2}\]                 \[{{285}^{o}}=\frac{{{T}_{i}}+(0+10)}{2}\]                 \[285\times 2={{T}_{i}}+{{10}^{o}}\]                 \[570-10={{T}_{i}}\]                 \[{{T}_{i}}=560{{\,}^{o}}C\]                 So, the \[{{T}_{n}}=285{{\,}^{o}}C\]and \[{{T}_{i}}=560{{\,}^{o}}C\]


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