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Question:
what is critical velocity derive the expression for it by the method of dimensional analysis
Answer:

 Critical velocity is the greatest velocity with which a fluid can flow through a given conduit without becoming turbulent. (or)
⇔ It is the maximum velocity of a liquid which upon exceeding leads to turbulent flow.

⇔ Velocity (V) is directly proportional to
→ density of water droplet (ρ),
→ coefficient of viscosity (η) and
→ radius of water outlet (r

 V ∞ ρ^a. η^b. r^c
⇒ Dimensional formula of density, (ρ) = [M^1 L^(-3)]
⇒  Dimensional formula of Coefficient of viscosity, (η) = [M^1 L^(-1) T^(-1)]
⇒  Dimensional formula of radius, (r) = [L^1]
⇒ Dimensional formula of critical velocity, (V) = [M^0 L^1 T^(-1)]

⇒ V = [M^1 L^(-3)]^a. [M^1 L^(-1) T^(-1)]^b. [L^1]^c
⇒ [M^0 L^1 T^(-1)] = M^a. L^(-3a). M^b. L^(-b). T^(-b). L^c

Solving a, b and c by equating L.H.S and R.H.S:

⇒ M^0 = M^(a+b)
⇒ a + b = 0
⇒ b = -a →(1)

⇒ L^(-3a-b+c) = L^1
⇒ -3a -b + c = 1
⇒ From (1), -b = a.
⇒ -3a + a + c = 1
⇒ -2a + c = 1
⇒ c = 1 + 2a →(2)

⇒ T^(-b) = T^(-1)
⇒ -b = -1
⇒ ∴b = 1.

⇒ From (1), 
⇒ a = -b
⇒ ∴a = -1

⇒ From (2),
⇒ c =  1 + 2a
⇒ ∴ c = -1

⇒ V ∞ ρ^a. η^b. r^c
⇒ V ∞ ρ^(-1). η^1. r^(-1)
⇒ V = K (ρ^(-1). η^1. r^(-1))

⇒ ∴V = Kη/ρr, where K = critical velocity constant.

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