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(B) The total resistance of the primary circuit is $R_{total} = R + r_1 = 9 \, \Omega + 1 \, \Omega = 10 \, \Omega$.The current flowing through the po

Vedclass · Accepted Answer

$55.55$

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(B) The total resistance of the primary circuit is $R_{total} = R + r_1 = 9 \, \Omega + 1 \, \Omega = 10 \, \Omega$.The current flowing through the potentiometer wire $AB$ is $i_0 = \frac{E_1}{R_{total}} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$.The potential drop across the wire $AB$ is $V_{AB} = i_0 	imes R = 1 \, A 	imes 9 \, \Omega = 9 \, V$.The potential gradient along the wire is $k = \frac{V_{AB}}{l} = \frac{9 \, V}{100 \, cm} = 0.09 \, V/cm$.For the galvanometer to show no deflection,the potential drop across the length $AC$ (let it be $l_1$) must be equal to the emf $E_2$ of the secondary cell.$V_{AC} = k 	imes l_1 = E_2$$0.09 \, V/cm 	imes l_1 = 5 \, V$$l_1 = \frac{5}{0.09} = \frac{500}{9} \approx 55.55 \, cm$.

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