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(D) The total resistance of the circuit is $R_{total} = R + 10\, \Omega$.The current $i$ flowing through the potentiometer wire is $i = \frac{E}{R_{to

Vedclass · Accepted Answer

$790$

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(D) The total resistance of the circuit is $R_{total} = R + 10\, \Omega$.The current $i$ flowing through the potentiometer wire is $i = \frac{E}{R_{total}} = \frac{2}{R + 10}$.The potential drop across the entire $100\, cm$ wire is $V_{wire} = i 	imes 10 = \frac{20}{R + 10}$.The potential gradient $x$ (potential per unit length) is $x = \frac{V_{wire}}{100} = \frac{20}{100(R + 10)} = \frac{0.2}{R + 10} \, V/cm$.The emf of the source $E_s = 10\, mV = 10 	imes 10^{-3} \, V$ is balanced at length $l = 40\, cm$.Using the balancing condition $E_s = x \cdot l$,we have $10 	imes 10^{-3} = \frac{0.2}{R + 10} 	imes 40$.$10^{-2} = \frac{8}{R + 10} \Rightarrow R + 10 = \frac{8}{10^{-2}} = 800$.Therefore,$R = 800 - 10 = 790\, \Omega$.

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