For a certain thermocouple the emf is $\begin{array}{*{20}{l}} {E = aT + \,\,b{T^2}} \end{array}$, where $t\left( {{\rm{in}}\,{\,^0}{\rm{C}}} \right)$ is the temperature of hot junction, the cold junction is at $0\,{\,^0}{\rm{C}}$. The value of contants $a\,\,{\rm{and}}\,\,b$ are $10\,\, \times \,\,{10^{ - 6}}$ and $0.02\,\, \times \,\,{10^{ - 6}}$ respectively, then the temperature of inversion $\left( {{\rm{in}}\,{\,^0}{\rm{C}}} \right)$ will be

$\begin{array}{*{20}{l}} {{\rm{500}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{150}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{250}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{750}}} \end{array}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

For a certain thermocouple the emf is

\begin{array}{llllllllllllllllllll} {E = aT + \,\,b{T^2}} \end{array}

, where

t\left( {\mathrm{in}\,{\,^0}\mathrm{C}} \right)

is the temperature of hot junction, the cold junction is at

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

. The value of contants

a\,\,\mathrm{and}\,\,b

are

10\,\, \times \,\,{10^{ - 6}}

and

0.02\,\, \times \,\,{10^{ - 6}}

respectively, then the temperature of inversion

\left( {\mathrm{in}\,{\,^0}\mathrm{C}} \right)

will be

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {\mathrm{500}} \end{array}

Option C (500) is correct because the temperature of inversion is found by setting the emf to zero and solving for the temperature.