Answer

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**Hint:**Here, by using the velocity formula, we have to calculate the radius of the pipes and separate the diameter, then proceed to find the flow of water in a particular diameter and then the vector defines the solution and compares the diameter of the other pipe.

**Formula used:**

According to that velocity of pipes

${A_x}{v_x} = {A_y}{v_y}$

Where,

${v_x}$ and ${v_y}$ are the velocity of water in pipes\[X\]and $Y$respectively.

The radius of $X,\,{r_x} = \dfrac{{dx}}{2}$

The radius of $Y,{r_y} = \dfrac{{dy}}{2}$

$dx,dy$ is differentiation with respect to a particular variable.

**Complete step by step solution:**

Given by ,

The diameter of the vector pipes

Let, $X = 2cm$ and $Y = 4cm$ respectively

We know that ,

Radius formula

The radius of $X,\,{r_x} = \dfrac{{dx}}{2}$

The radius of $Y,{r_y} = \dfrac{{dy}}{2}$

The value of $dx = 2$ and \[dy = 4\]

Substituting the given value in above equation,

We get,

The radius of $X$ and \[Y\]

${r_x} = 0.01\,m$

${r_y} = 0.02\,m$

According to the equation of continuity:

$\Rightarrow {A_x}{v_x} = {A_y}{v_y}$

$\Rightarrow {A_x} = \pi \left( {{r_x}} \right)$ and ${A_y} = \pi \left( {{r_y}} \right)$

Substituting the given value in above equation,

we get,

$\Rightarrow \pi {\left( {0.01\,m} \right)^2} \times {v_x} = \pi {\left( {0.02\,m} \right)^2}{v_y}$

Rearranging the given equation,

We have,

$\Rightarrow {v_x} = \dfrac{{{{\left( {0.02\,m} \right)}^2}}}{{{{\left( {0.01\,m} \right)}^2}}}{v_y}$

On simplifying,

We get, ${v_x} = 4{v_y}$

Therefore the velocity of flow of water in the pipe of $2cm$ diameter $X$ in four times that of in pipe $Y$

**Hence, option A is the correct answer.**

**Note:**If we have to calculate the pipe radius and compare the diameter of the other pipe. A flow rate increase or decrease will result in a corresponding velocity increase or decrease. Velocity is also influenced by pipe size. Reducing pipe size increases the velocity, which increases friction, given a constant flow rate.

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