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A cell of emf of 1.5V is connected in series with a potentiometer wire of 10m length and a resistance of 2980 $Ω$ in primary circuit. Resistance of potentiometer wire is 2 $Ω$ /m . A thermocouple with one junction in hot oil and the other in melting ice has a balancing point at 450cm then the thermo emf generated is

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4.5V

0.045V

0.0045V

0.45V

answer is A.

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Detailed Solution

E = emf of cell = 1.5V
E' = emf of thermocouple = ?
$R_{P} = resistance of potentiometer wire = 20 Ω$
R_T= total resistance = 2980 + 20 = 3000 $Ω$

L=length of wire=10m

$\begin{array}{l} E^{'} = \frac{E \times R_{P} \times l}{(r + R_{s} + R_{p}) L^{'}}; E^{'} = \frac{1.5 \times 20 \times 4.5}{3000 \times 10} \\ = 4.5 \times 10^{- 3}; E^{'} = 0.0045 volt \end{array}$

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