The reading of an ideal voltmeter in the circuit shown is

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