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Q.

In an experiment with potentiometer to measure the internal resistance of a cell, when the cell is shunted by $5 Ω,$ the null point is obtained at 2m. When cell is shunted by $20 Ω,$ the null point is obtained at 3m. The internal resistance of cell is

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a

$2 Ω$

b

$6 Ω$

c

$4 Ω$

d

$8 Ω$

answer is C.

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Detailed Solution

$(\frac{ε R}{R + r}) = (\frac{Δ V}{L}) l$

$\Rightarrow \frac{5}{20} \times (\frac{20 + r}{5 + r}) = \frac{2}{3}$

$\Rightarrow r = 4 Ω$

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