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The thermo emf in lead-iron thermocouple with one junction at $0^{\circ}C$ , is given by $e=1784t-bt^2$ in volts, where $t^{\circ}C$ is the temperature of the other junction. The neutral temperature is $371.7^{\circ}C$. Then the value of b in $V/(^{\circ}C)^2$ is

$\begin {array} {1 1} (1)\;-2.4 & \quad (2)\;4.799 \\ (3)\;-9.6 & \quad (4)\;+9.6 \end {array}$

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(1) -2.4
answered Nov 7, 2013 by pady_1

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