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The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of $10 Ω$ is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal resistance of the cell is $\frac{N}{10} Ω$ , where N is an integer then value of N is ______

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Detailed Solution

Let the emf of cell is E internal resistance is 'r' and potential gradient is x.
Only cell connected
$l_{1} = 560 c m; E \propto l_{1} \Rightarrow E = K l_{1} \Rightarrow E = 560 K l_{2} = 560 - 60 = 500 c m; W h e n r e s i s \tan c e i s c o n n e c t e d t o c e l l t h e n V = K l_{2}$

$r = (\frac{E}{V} - 1) R = (\frac{l_{1}}{l_{2}} - 1) R = (\frac{560}{500} - 1) 10 = (\frac{560 - 500}{500}) 10 = \frac{12}{10}$

$∴ n = 12$

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