A thermocouple of resistance 1.6,Omega is con | vedclass

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(A) Let the temperature of the molten metal be $t\,^{\circ}C$.The thermo-emf generated is $e = 10 	imes 10^{-6} 	imes t\,	ext{V} = 10^{-5}t\,	ext{

Vedclass · Accepted Answer

$960$

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(A) Let the temperature of the molten metal be $t\,^{\circ}C$.The thermo-emf generated is $e = 10 	imes 10^{-6} 	imes t\,	ext{V} = 10^{-5}t\,	ext{V}$.The total resistance of the circuit is $R_{total} = R_{thermocouple} + R_{galvanometer} = 1.6\,\Omega + 8\,\Omega = 9.6\,\Omega$.The current in the circuit is $i = \frac{e}{R_{total}} = \frac{10^{-5}t}{9.6}\,	ext{A}$.The galvanometer reads the potential difference across its own resistance,which is $V_G = 8\,	ext{mV} = 8 	imes 10^{-3}\,	ext{V}$.The current through the galvanometer is also $i = \frac{V_G}{R_G} = \frac{8 	imes 10^{-3}}{8} = 10^{-3}\,	ext{A}$.Equating the two expressions for current: $\frac{10^{-5}t}{9.6} = 10^{-3}$.Solving for $t$: $t = \frac{10^{-3} 	imes 9.6}{10^{-5}} = 9.6 	imes 10^2 = 960\,^{\circ}C$.

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