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**Hint:**Assume the liquid is incompressible and therefore its density remains constant in both the sections. The equation of continuity states that the volume rate of flow remains the same throughout the flow. The volume rate of flow is given as, \[Av\], where, A is the area of cross-section and v is the velocity of the flow.

**Formula used:**

Equation of continuity, \[{A_1}{v_1} = {A_2}{v_2}\]

Here, \[{A_1}\] is the area of the wider section, \[{v_1}\] is the velocity in the wider section, \[{A_2}\] is the area of narrower section and \[{v_2}\] is the velocity of the fluid in the narrower section.

**Complete step by step answer:**

Let the liquid enter a wider section of area \[{A_1}\] and leave the narrow section of area \[{A_2}\]. We assume the liquid is incompressible and therefore its density remains constant in both the sections. Therefore, the liquid will move faster in the narrow section to accommodate liquid particles behind it. Let the velocity of the flow at section \[{A_1}\] is \[{v_1}\] and velocity of flow at section \[{A_2}\] is \[{v_2}\].

We have the equation of continuity which states that the volume rate of flow remains the same throughout the flow. We can express the continuity equation as,

\[{A_1}{v_1} = {A_2}{v_2}\]

Let \[{r_1}\] be the radius of the wider section and \[{r_2}\] be the radius of the narrow section.Rewriting the above equation by expressing the area of respective ends as follows,

\[\left( {\pi r_1^2} \right){v_1} = \left( {\pi {{\left( {\dfrac{{{r_1}}}{2}} \right)}^2}} \right){v_2}\]

\[ \Rightarrow {v_1} = \dfrac{{{v_2}}}{4}\]

\[ \Rightarrow {v_2} = 4{v_1}\]

Substituting 3 m/s for \[{v_1}\] in the above equation, we get,

\[{v_2} = 4\left( 3 \right)\]

\[ \therefore {v_2} = 12\,{\text{m/s}}\]

**So, the correct answer is option B.**

**Note:**The continuity equation that we have used is valid only for the incompressible fluid. For the compressible fluid, students have to consider the density of the fluid at both the sections. While solving these types of questions, make sure that you have given whether the area of the narrower section is half or the radius is the half of the wider section.

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