Answer

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**Hint:**Apply the equation of continuity. put the area of tube and given values and we will get the final answer. In these types of questions we have to remember the area of different shapes.

**Formula used:**

${v_1}{a_1} = {v_2}{a_2}$

**Complete Step by step solution:**

${v_1}{a_1} = {v_2}{a_2}$

v is the velocity of flow.

a is the cross-section of the tube.

So, the rate of flow of liquid through the tube is va.

But, in the question there is used two tubes at the same condition having areas ${a_1}$ and ${a_2}$ with the velocity ${v_1}$ and ${v_2}$ respectively.

First we have to write given values as mentioned in the question.

${r_1} = 2mm,{v_1} = 8c{m^3}{s^{ - 1}},{r_2} = 1mm$

Now, write equation of continuity

${v_1}{a_1} = {v_2}{a_2}$ -------(1)

put the area of tube $\pi {r^2}$

let ${r_1}$ and ${r_2}$ are the radius of the two tubes respectively. so, the equation number (1) becomes.

$\pi {r_1}^2{v_1} = \pi {r_2}^2{v_2}$

Cancel $\pi $ each side of the above equation.

$ \Rightarrow {r_1}^2{v_1} = {r_2}^2{v_2}$

$ \Rightarrow \pi {r_1}^2{v_1} = \pi {r_2}^2{v_2}$

Substitute given values in the above equation.

$ \Rightarrow {v_2} = \dfrac{{{r_1}^2}}{{{r_2}^2}}{v_1}$

$ \Rightarrow {v_2} = \dfrac{{{1^2}}}{{{2^2}}} \times 8c{m^3}{s^{ - 1}}$

$ \Rightarrow {v_2} = \dfrac{1}{4} \times 8$

$\therefore {v_2} = 2c{m^3}{s^{ - 1}}$

**Hence, the correct given option is B.**

**Additional information:**

The equation of continuity expresses the law of conservation of mass.

Flow of liquid or gas through a tube says the mass of the fluid flowing per second through any cross-section of the tube remains constant.

The continuity equation for fluids first published by Euler in 1757.

Sometimes in JEE mains examination this question will be asked with different changes like they give diameter, velocity, and different fluids.

**Note:**

we know that the equation of continuity ${v_1}{a_1} = {v_2}{a_2}$, from this equation we get va is constant. So, we get $v \propto \dfrac{1}{a}$ it means the flow of any liquid through any cross-section of a tube is inversely proportional to its cross-sectional area.

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