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A thermocouple of negligible resistance produces an e.m.f. of 40 $\mu$V/$^\circ C$ in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 $\mu$A/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be
A
$0.1^\circ C$
B
$0.25^\circ C$
C
$0.5^\circ C$
D
$1^\circ C$
Detailed Solution
The smallest current the galvanometer can detect is that for one division: $I_{min} = 1\ \mu A$
The thermocouple has negligible resistance, so the circuit resistance is that of the galvanometer, 10 $\Omega$.
Smallest detectable emf: $V_{min} = I_{min}\times R = 1\ \mu A\times10\ \Omega = 10\ \mu V$
The thermocouple gives 40 $\mu$V for each $^\circ C$ of temperature difference.
Smallest detectable temperature difference $= \frac{10\ \mu V}{40\ \mu V/^\circ C}$
$= 0.25^\circ C$
The thermocouple has negligible resistance, so the circuit resistance is that of the galvanometer, 10 $\Omega$.
Smallest detectable emf: $V_{min} = I_{min}\times R = 1\ \mu A\times10\ \Omega = 10\ \mu V$
The thermocouple gives 40 $\mu$V for each $^\circ C$ of temperature difference.
Smallest detectable temperature difference $= \frac{10\ \mu V}{40\ \mu V/^\circ C}$
$= 0.25^\circ C$