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(A) The thermo $e.m.f.$ produced by a thermocouple is given by $e = S 	imes 	heta$,where $S$ is the sensitivity $(30 \,\mu V/^\circ C)$ and $	heta$

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$0.5$

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(A) The thermo $e.m.f.$ produced by a thermocouple is given by $e = S 	imes 	heta$,where $S$ is the sensitivity $(30 \,\mu V/^\circ C)$ and $	heta$ is the temperature difference.Given,the minimum current $i = 3 	imes 10^{-7} \, A$ and resistance $R = 50 \,\Omega$.Using Ohm's law,$i = \frac{e}{R}$,we have $i = \frac{S 	imes 	heta}{R}$.Substituting the values: $3 	imes 10^{-7} = \frac{(30 	imes 10^{-6}) 	imes 	heta}{50}$.Solving for $	heta$: $	heta = \frac{3 	imes 10^{-7} 	imes 50}{30 	imes 10^{-6}}$.$	heta = \frac{150 	imes 10^{-7}}{30 	imes 10^{-6}} = 5 	imes 10^{-1} = 0.5^\circ C$.

The smallest temperature difference that can be measured with a combination of a thermocouple of thermo $e.m.f.$ $30\,\mu V$ per degree and a galvanometer of $50\,\Omega$ resistance,capable of measuring a minimum current of $3 \times 10^{-7} \, A$ is ............... $^\circ C$.

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