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A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity (ii) 9.5 $\Omega$, the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is
A
0.25 $\Omega$
B
0.95 $\Omega$
C
0.5 $\Omega$
D
0.75 $\Omega$
Detailed Solution
Internal resistance $r = \left(\frac{E - V}{V}\right)R = \left(\frac{l_1 - l_2}{l_2}\right)R$
$r = \left(\frac{3 - 2.85}{2.85}\right)(9.5) = 0.5\ \Omega$
$r = \left(\frac{3 - 2.85}{2.85}\right)(9.5) = 0.5\ \Omega$