One junction of thermocouple is at $0\,{\,^0}{\rm{C}}$ and the other is at $T\,{\,^0}{\rm{C}}$. The thermo emf (in volts) is given by $E = 20\,\, \times \,\,{10^{ - 6}}\,\,T - 0.02\,\, \times \,\,{10^{ - 6}}\,\,{T^2}$ The maximum value of $E$ is

$\begin{array}{*{20}{l}} {{\rm{5}}\,\,{\rm{mV}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{1}}\,\,{\rm{mV}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{10}}\,\,{\rm{mV}}} \end{array}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

One junction of thermocouple is at

0\,{\,^0}\mathrm{C}

and the other is at

T\,{\,^0}\mathrm{C}

. The thermo emf (in volts) is given by

E = 20\,\, \times \,\,{10^{ - 6}}\,\,T - 0.02\,\, \times \,\,{10^{ - 6}}\,\,{T^2}

The maximum value of

E

View answer and explanation

Correct option: A

\begin{array}{llllllllllllllllllll} {\mathrm{5}\,\,\mathrm{mV}} \end{array}

Option A (5 mV) is correct because the maximum thermo emf is found by differentiating the given equation and setting it to zero.