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# rate of flow of water through a tube of radius 2mm is $8c{m^3}{s^{ - 1}}$ Under similar condition, the rate of flow of water through a tube of radius 1mm will beA)$4c{m^3}{s^{ - 1}}$ B)$2c{m^3}{s^{ - 1}}$ C) $1c{m^3}{s^{ - 1}}$ D) $0.5c{m^3}{s^{ - 1}}$

Last updated date: 17th Jun 2024
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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.

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.