Question

5. Volume of the liquid flowing per second
through a pipe of diameter d is V. At a point
where the radius is d, the velocity of flow is
2​

Answer 1

The new volume becomes $V_2= 8V_1$

Explanation:

At radius r,  volume of flowing fluid $V_1= \pi r^{2}\times velocity$

 $V_1= \pi r^{2}\times v$

Where, radius becomes double or equal to the diameter and velocity becomes 2v then volume flow rate $V_2= \pi 4r^{2}\times 2v$

$V_2=8 \pi r^{2}\times v$  

The new volume becomes $V_2= 8V_1$

Answer 2

Given that,

Diameter of pipe at entrance= d

Radius of pipe at entrance $r_{1}=\dfrac{d}{2}$

Volume = V

Radius of pipe at exit $r_{2}= d$

We need to calculate the velocity of flow

Using continuity equation

$A_{1}v_{1}=A_{2}v_{2}$

$v_{1}=\dfrac{A_{2}}{A_{1}}v_{2}$

Put the value into the formula

$v_{1}=\dfrac{\pi r_{1}^2}{\pi r_{2}^2}v_{2}$

$v_{1}=\dfrac{\pi\times(\dfrac{d}{2})^2}{\pi\times(d^2)}$

$v_{1}=\dfrac{1}{4}v_{2}$

$v_{2}=4v_{1}$

Hence, The velocity of flow is 4 v₁.