In order to calculate the inside diameter, we can use the following relation, since the volumetric flow is constant for any part of the tube:
[tex]\begin{gathered} Q_1=Q_2 \\ A_1V_1=A_2V_2 \end{gathered}[/tex]
Where A is the cross-section area and V is the velocity.
Using V1 = 1.5 m/s, A1 = π*1.1²/4 and V2 = 10 m/s, we have:
[tex]\begin{gathered} \frac{\pi\cdot1.1^2}{4}\cdot1.5=A_2\cdot10 \\ \frac{\pi\cdot1.1^2}{4}\cdot1.5=\frac{\pi\cdot d^2_2}{4}\cdot10 \\ 1.1^2\cdot1.5=d^2_2\cdot10 \\ 1.21\cdot1.5=d^2_2\cdot10 \\ 1.81=d^2_2\cdot10 \\ d^2_2=0.181 \\ d_2=0.42\text{ cm}=4.2\text{ mm} \end{gathered}[/tex]
Therefore the correct option is the first one.
