Answer

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**Hint**The internal resistance of the potentiometer can be determined by using the internal resistance formula of the potentiometer. The internal resistance is depending on the resistance of the resistor, distance of the two balance points in the potentiometer.

**Useful formula**

The expression for the internal resistance of a cell in potentiometer is given by,

$r = R\left( {\dfrac{{{l_1}}}{{{l_2}}} - 1} \right)$

Where, $r$ is the internal resistance of the cell in the potentiometer, $R$ is the resistance of the resistor, ${l_1}$ is the distance of the initial balance point and ${l_2}$ is the distance of the final balance point.

**Complete step by step solution**

Given that,

The voltage of the potentiometer is, $V = 2.0\,V$,

The balance point of the cell in open circuit is, ${l_1} = 76.3\,cm$,

The balance point shifts to the length of the potentiometer is, ${l_2} = 64.8\,cm$.

The resistance of the resistor is, $R = 9.5\,\Omega $

Now,

The expression for the internal resistance of a cell in potentiometer is given by,

$r = R\left( {\dfrac{{{l_1}}}{{{l_2}}} - 1} \right)\,..................\left( 1 \right)$

By substituting the resistance of the resistor, the length of the balance point of the cell in open circuit and the balance point shifts to length of the potentiometer in the above equation (1), then the above equation (1) is written as,

$r = 9.5\left( {\dfrac{{76.3}}{{64.8}} - 1} \right)$

By dividing the terms in the above equation, then

$r = 9.5\left( {1.178 - 1} \right)$

By subtracting the terms in the above equation, then

$r = 9.5\left( {0.178} \right)$

By multiplying the terms in the above equation, then

$r = 1.69\,\Omega $

**Hence, the option (C) is the correct answer.**

**Note**The internal resistance of a cell is directly proportional to the resistance of the resistor and the ratio of the length of the balancing point. If the resistance of the resistor is increasing, the internal resistance of the cell is also increasing.

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