JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Fluid Mechanics

The reading of pressure metre attached with a closed pipe is

4.5 \times 1 0^{4} N / m^{2}

. On opening the valve, water starts flowing and the reading of pressure metre falls to

2.0 \times 1 0^{4} N / m^{2}

. The velocity of water is found to be

V m / s

. The value of V is ____

Answer: 50

Solution

Bernoulli's equation between the no-flow state and the flowing state (same height, same point) says the drop in static pressure converts into kinetic energy per unit volume:

Δ P = \frac{1}{2} ρ v^{2} .

With

Δ P = (4.5 - 2.0) \times 1 0^{4} = 2.5 \times 1 0^{4} N / m^{2}

and

ρ = 1 0^{3} k g / m^{3}

v^{2} = \frac{2 \times 2.5 \times 1 0 ^{4}}{1 0 ^{3}} = 50.

V = 50

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer

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