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Question

Physics Question on Thermal Expansion

If the cold junction of a thermo-couple is kept at 0C0^{\circ} C and the hot junction is kept at TCT^{\circ} C, then the relation between neutral temperature (Tn)\left(T_{n}\right) and temperature of inversion (Ti)\left(T_{i}\right) is

A

Tn=Ti2T _{ n }=\frac{ T _{i}}{2}

B

Tn=2TiT _{ n }=2 T _{ i }

C

Tn=TiTT _{ n }= T _{ i }- T

D

Tn=Ti+TT_{n}=T_{i}+T

Answer

Tn=Ti2T _{ n }=\frac{ T _{i}}{2}

Explanation

Solution

It is found that temperature of inversion (Ti)\left(T_{i}\right) is as much above the neutral temperature (Tn)\left(T_{n}\right) as neutral temperature is above the temperature of the cold junction (T),ie(T), i e, TiTn=TnTT_{i}-T_{n} =T_{n}-T Ti=2TnTT_{i} =2 T_{n}-T But, here the cold junction is kept at 0C0^{\circ} C, hence T=0T=0. Thus, Ti=2TnT_{i}=2 T_{n} Tn=Ti2T_{n}=\frac{T_{i}}{2}