Answer:
![n=2.9* 10^9](https://img.qammunity.org/2020/formulas/physics/college/b2efc0uyx7weulpdqs9zb7b1lu4piv5s7z.png)
![A=1.88* 10^(-8)\ m^2](https://img.qammunity.org/2020/formulas/physics/college/jcuvagqlf7x4jtpyavujcs8hbv8veo01th.png)
Step-by-step explanation:
Given that
Q= 5 L/min
1 L = 10⁻³ m³/s
1 min = 60 s
Q=0.083 x 10⁻³ m³/s
d= 6 μm
v= 1 mm/s
So the discharge flow through one tube
q = A v
![A=(\pi)/(4)d^2](https://img.qammunity.org/2020/formulas/engineering/college/vro6ppnbt7qaigtxr00dwdb9t2ltf30hob.png)
![A=(\pi)/(4)* (6* 10^(-6))^2\ m^2](https://img.qammunity.org/2020/formulas/physics/college/74thqrhetsm57indt15jq7ssxnec866qkd.png)
A=2.8 x 10⁻¹¹ m²
v= 1 x 10⁻³ m/s
q= 2.8 x 10⁻¹⁴ m³/s
Lets take total number of tube is n
Q= n q
n=Q/q
![n=(0.083* 10^(-3) )/( 2.8* 10^(-14))](https://img.qammunity.org/2020/formulas/physics/college/dx6p9bet3acquyg7w1otwj8emb9y5o4w99.png)
![n=2.9* 10^9](https://img.qammunity.org/2020/formulas/physics/college/b2efc0uyx7weulpdqs9zb7b1lu4piv5s7z.png)
Surface area A
A= π d L
![A=\pi * 6* 10^(-6)* 10^(-3)\ m^2](https://img.qammunity.org/2020/formulas/physics/college/7db94l2ejeqlz40xo74gw3db3eog82odf7.png)
![A=1.88* 10^(-8)\ m^2](https://img.qammunity.org/2020/formulas/physics/college/jcuvagqlf7x4jtpyavujcs8hbv8veo01th.png)