As shown in the figure,a potentiometer wire o | vedclass

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(A) The potential difference across the potentiometer wire $AB$ is given by $V_{AB} = I 	imes R_{AB}$,where $I$ is the current in the primary circuit

Vedclass · Accepted Answer

$780$

Vedclass · Answer

(A) The potential difference across the potentiometer wire $AB$ is given by $V_{AB} = I 	imes R_{AB}$,where $I$ is the current in the primary circuit.The current $I$ is given by $I = \frac{E}{R + R_{AB}}$,where $E = 4\,V$ and $R_{AB} = 20\,\Omega$.So,$V_{AB} = \left( \frac{4}{R + 20} ight) 	imes 20 = \frac{80}{R + 20}$.The potential gradient along the wire is $k = \frac{V_{AB}}{L}$,where $L = 300\,cm$.The emf of the secondary cell is $E' = 20\,mV = 20 	imes 10^{-3}\,V$,and the null point length is $l = 60\,cm$.At the null point,$E' = k 	imes l = \left( \frac{V_{AB}}{L} ight) 	imes l$.Substituting the values: $20 	imes 10^{-3} = \left( \frac{80}{R + 20} ight) 	imes \left( \frac{60}{300} ight)$.$20 	imes 10^{-3} = \left( \frac{80}{R + 20} ight) 	imes \left( \frac{1}{5} ight)$.$20 	imes 10^{-3} = \frac{16}{R + 20}$.$R + 20 = \frac{16}{20 	imes 10^{-3}} = \frac{16}{0.02} = 800$.$R = 800 - 20 = 780\,\Omega$.

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