What will be the nature of flow of water form a circular tap, when its flow rate increased from $0.18 \mathrm{~L} / \mathrm{min}$ to $0.48 \mathrm{~L} / \mathrm{min}$ ? The radius of the tap and viscosity of water are 0.5 cm and $10^{-3} \mathrm{~Pa} \mathrm{~s}$, respectively. (Density of water : $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ )

(D) Steady flow to unsteady flow

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Fluid Mechanics

What will be the nature of flow of water form a circular tap, when its flow rate increased from

0.18 L / min

0.48 L / min

? The radius of the tap and viscosity of water are 0.5 cm and

1 0^{- 3} Pa s

, respectively.
(Density of water :

1 0^{3} kg / m^{3}

)

Answer: (D)

The character of the flow is fixed by the Reynolds number, which for a pipe of radius

r

carrying a volume flow rate

Q = π r^{2} v

R_{e} = \frac{ρ v D}{η} = \frac{2 ρQ}{π r η}

With

ρ = 1 0^{3} kg m^{- 3}

r = 0.5 cm = 5 \times 1 0^{- 3} m

and

η = 1 0^{- 3} Pa s

, the rate

Q_{1} = 0.18 L mi n^{- 1} = 3 \times 1 0^{- 6} m^{3} s^{- 1}

gives

R_{e 1} = \frac{2 \times 1 0 ^{3} \times 3 \times 1 0 ^{- 6}}{π \times 5 \times 1 0 ^{- 3} \times 1 0 ^{- 3}} \approx 382

and

Q_{2} = 0.48 L mi n^{- 1} = 8 \times 1 0^{- 6} m^{3} s^{- 1}

gives

R_{e 2} \approx 1019

in the same way. Flow stays steady while

R_{e}

is below

1000

and turns unsteady once

R_{e}

lies between

1000

and

2000

. Increasing the tap's delivery from

0.18

0.48 L mi n^{- 1}

therefore carries the water from steady flow to unsteady flow.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option