ANSWER
The exponent of L must be 1
Step-by-step explanation
We have to write the dimensions of each variable in the equation. 8 and π are constants and have no units. The units of the variables, in the international system of units, are:
• Q = m³/s
,
• u = Pa·s
,
• r = m
,
• L = m
,
• P = Pa
Replace these into the equation. For this problem we can ignore the constants:
![Pa=((m^3\cdot Pa\cdot s)/(s))/(m^(4))\cdot m^{\text{?}}](https://img.qammunity.org/2023/formulas/physics/college/5cvxes3qtk6uuy446ozd4h6lzxh7y8bi8n.png)
We want to have Pascal units on both sides for this to be dimensionally consistent. Note that the seconds in the numerator's fraction get cancelled:
![Pa=(m^3\cdot Pa)/(m^4)\cdot m^{\text{?}}](https://img.qammunity.org/2023/formulas/physics/college/v6isqakhp005u8k0meg3pspadcpvss600p.png)
To have Pascal on both sides we need the same exponent in the meters. If we cancel m³ with m⁴, we have m¹ in the denominator:
![Pa=(Pa)/(m)\cdot m^{\text{?}}](https://img.qammunity.org/2023/formulas/physics/college/24tskso3a786ce92ffnt57rgts5dmwcn0v.png)
Now we can see that if L has exponent 1, then the units get cancelled correctly to obtain pressure units, as the equation shows.
![Pa=Pa](https://img.qammunity.org/2023/formulas/physics/college/z2rurslvup8z56lu192v3o4j4iv6wbjw9u.png)