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Q.

In a experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a $5 Ω$ resistance and is at a length of 3 m when the cell is shunted by a $10 Ω$ resistance, the internal resistance of the cell is

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a

$1.5 Ω$

b

$10 Ω$

c

$15 Ω$

d

$1 Ω$

answer is B.

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Detailed Solution

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In case of internal resistance measurement by potentiometer,
$\frac{V_{1}}{V_{2}} = \frac{ℓ_{1}}{ℓ_{2}} = \frac{[E R_{1} / (R_{1} + r)]}{[E R_{2} / (R_{1} + r)]} = \frac{R_{1} (R_{2} + r)}{R_{2} (R_{1} + r)}$
Here $ℓ_{1} = 2 m, ℓ_{2} = 3 m, R_{1} = 5 Ω,$ and $R_{2} = 10 Ω .$ So
$\frac{2}{3} = \frac{5 (10 + r)}{10 (5 + r)} o r r = 10 Ω$

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