1. the circuit diagram shows the use of a potentiometer to measure a small emf produced by a thermocouple connected between X and Y. The cell C, of emf 2 V, has negligible internal resistance. The potentiometer wire PQ is 1.00 m long and has resistance 5 Ω. The balance point S is found to be 400 mm from P. Calculate the value of emf V, generated by the thermo-couple.

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This answer is not currect
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I=E/R+r =2/990+5=2/995 Vr=2/995×10 (Vr=Ir×r) =20/995 v Potential gradient (k)along the wire = Vr/L K=2/995×100 K= 2×400/100×995=0.008v
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Don't send incorrect answer in internet if you are not confirmed
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THE ANSWER IS NOT CORRECT
 
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