1. Interpret the circuit

The four resistances form the four arms of quadrilateral $ABCD$:

• $AB = 40\,\Omega$
• $BC = 60\,\Omega$
• $CD = 90\,\Omega$
• $DA = 110\,\Omega$

A battery of $40\,\text{V}$ is connected across diagonal $AC$.

So between $A$ and $C$, there are two parallel branches:

• Branch 1: $A \to B \to C$ with resistance $R_1 = 40 + 60 = 100\,\Omega$
• Branch 2: $A \to D \to C$ with resistance $R_2 = 110 + 90 = 200\,\Omega$

Since both branches are directly across $AC$, each branch has the same potential difference: $V_{AC} = 40\,\text{V}$

2. Find current in each branch

For branch $ABC$: $I_1 = \frac{40}{100} = 0.4\,\text{A}$

For branch $ADC$: $I_2 = \frac{40}{200} = 0.2\,\text{A}$

3. Find potentials at points $B$ and $D$

Take $V_A = 40\,\text{V}$ and $V_C = 0\,\text{V}$.

Potential at $B$

Voltage drop across $AB$ is $V_{AB} = I_1 \cdot 40 = 0.4 \times 40 = 16\,\text{V}$ So, $V_B = V_A - 16 = 40 - 16 = 24\,\text{V}$

Potential at $D$

Voltage drop across $AD$ is $V_{AD} = I_2 \cdot 110 = 0.2 \times 110 = 22\,\text{V}$ So, $V_D = V_A - 22 = 40 - 22 = 18\,\text{V}$

4. Potential difference across $BD$

$V_{BD} = V_B - V_D = 24 - 18 = 6\,\text{V}$

Hence, the magnitude of potential difference across $BD$ is $\boxed{6}$

5. Compare with stored correct answer

Stored correct answer is $2$, but the derived answer is $6$.

So I do not agree with the stored answer. The likely correct answer is: $\boxed{6}$