The given circuit diagram can be drawn as shown below



The equivalent resistance of circuit is given by
$$
\begin{aligned}
& \frac{\mathrm{l}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{l}}{\mathrm{R}_{\mathrm{AE}}}+\frac{\mathrm{l}}{\mathrm{R}_{\mathrm{DF}}} \\
& =\frac{1}{(20+30)}+\frac{1}{(30+20)} \\
& =\frac{1}{50}+\frac{1}{50}=\frac{2}{50} \\
& \Rightarrow \quad \mathrm{R}_{\mathrm{eq}}=25 \Omega \\
&
\end{aligned}
$$
As the resistance of two branches is same i.e. $50 \Omega$.
So, the current $\mathrm{I}_1=\mathrm{I}_2$
$$
\Rightarrow \quad \begin{aligned}
\mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2 \\
\frac{2}{25} & =2 \mathrm{I}_1 \Rightarrow \mathrm{I}_2=\mathrm{I}_1=\frac{1}{25} \mathrm{~A}
\end{aligned}
$$
$\therefore$ The voltage across $\mathrm{AB}$
$$
\mathrm{V}_1=\mathrm{I}_1 \mathrm{R}_1=\frac{1}{25} \times 20
$$

and voltage across $C D$
$$
\begin{aligned}
\mathrm{V}_2= & \mathrm{I}_2 \mathrm{R}_2=\frac{1}{25} \times 30 \\
\therefore \text { Voltmeter reading } & =\mathrm{V}_2-\mathrm{V}_1 \\
& =\frac{30}{25}-\frac{20}{25} \\
& =\frac{10}{25}=0.4 \mathrm{~V}
\end{aligned}
$$