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A thermocouple of negligible resistance ...

A thermocouple of negligible resistance produces an e.m.f. of `40mu (V)//""^(0)C` in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 `muA//"division"`, is employed with the thermocouple. The smallest value of temperature difference that can be detected by system will be

A

`16^@C`

B

`12^@C`

C

`8^@C`

D

`20^@C`

Text Solution

Verified by Experts

Thermo- emf of the thermocouple =`25muV//^@C`
Let `theta` be the smallest temperature difference. Therefore, after connecting the thermocouple with the galvanometer, thermo-emf `E=25muV//^@C) times theta (^@C)=250 times 10^-6V`
Potential drop developed across the galvanometer `=IR=10^-5 times 40=4 times 10^-4V`
`therefore 4 times 10^-4=250 times 10^-6 or theta=4/25 times 10^2=16^@C`
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