A 6-volt battery with negligible internal resistance is connected...

Given $V_B = 0 \text{ V}$. The $6 \text{ V}$ battery is connected across the wire $AB$ with its positive terminal at $A$.

$V_A - V_B = 6 \text{ V} \Rightarrow V_A = 6 \text{ V}$

In the lower branch, the $4 \text{ V}$ battery has its positive terminal at $A$. Since the circuit is open at $C$, no current flows through the $1 \text{ }\Omega$ resistor, meaning there is no voltage drop across it.

$V_A - V_C = 4 \text{ V} \Rightarrow V_C = 6 - 4 = 2 \text{ V}$

Let $D$ be at a distance $x$ from $A$. The potential gradient along the wire $AB$ of length $100 \text{ cm}$ is:

$\lambda = \dfrac{V_A - V_B}{L} = \dfrac{6}{100} = 0.06 \text{ V/cm}$

The potential at $D$ is given by:

$V_D = V_A - \lambda x = 6 - 0.06x$

We require $V_D = V_C = 2 \text{ V}$:

$6 - 0.06x = 2 \Rightarrow 0.06x = 4 \Rightarrow x = \dfrac{400}{6} = 66.7 \text{ cm}$

If points $C$ and $D$ are connected, the potential difference between them is zero ($V_C = V_D = 2 \text{ V}$). Therefore, the current through the connecting wire will be zero.

If the $4 \text{ V}$ battery is replaced by a $7.5 \text{ V}$ battery, the new potential at $C$ becomes:

$V_C = V_A - 7.5 = 6 - 7.5 = -1.5 \text{ V}$

The potential on the wire $AB$ ranges from $6 \text{ V}$ at $A$ to $0 \text{ V}$ at $B$. Since $-1.5 \text{ V}$ is outside this range, no such point exists on the wire.