B

Step 1: Understand the potentiometer arrangement. The potentiometer wire is 100 cm long, with the balance point lengths given as follows:
- Without resistor: 70 cm
- With 9.5 Ω resistor: 60 cm

Step 2: Calculate the potential drop across the potentiometer for both scenarios. The total voltage (V) across the potentiometer wire is 2.0 V, giving a uniform potential drop along the wire. The potential drop can be calculated using the formula \( V = k \cdot L \) where k is the potential gradient.

Step 3: Determine k:
- Without the resistor: \( k = \frac{2.0 \text{ V}}{100 \text{ cm}} = 0.02 \text{ V/cm} \)
- With the 9.5 Ω resistor: \( k = \frac{V}{60} \text{ cm} \)
Hence, total current flowing through the circuit is given as \( I = \frac{1.5 \text{ V}}{R_{internal} + 9.5} \)

Step 4: Set up equations for potential drops:
The potential difference across the internal resistance will also be \( 1.5 \text{ V} = k \cdot 70 ext{ cm} \) for the first case and \( I (R_{internal}) = k \cdot 60 ext{ cm} \) for the second case.

Step 5: Equate and solve using calculations:
\( 1.5 = 0.02 \cdot 70 \Rightarrow 1.5 = 1.4 + I(R_{internal}) \Rightarrow I(R_{internal}) = 1.5 - 1.4 \Rightarrow I(R_{internal}) = 0.1\) .

Step 6: Substitute the current: \( I = \frac{1.5}{R_{internal} + 9.5} \Rightarrow 0.1 = \frac{1.5}{R_{internal} + 9.5} \Rightarrow 0.1(R_{internal} + 9.5) = 1.5\Rightarrow R_{internal} + 9.5 = 15 \Rightarrow R_{internal} = 5.5\).
Therefore, the internal resistance of the cell is 5.5 Ω which corresponds to option B.