A liquid of density \(750 \mathrm{kgm}^{-3}\) flows smoothly through a horizontal pipe that tapers in cross-sectional area from \(\mathrm{A}_{1}=1.2 \times 10^{-2} \mathrm{~m}^{2}\) to \(A_{2}=\frac{A_{1}}{2}\). The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ______ \(\times 10^{-3} \mathrm{~m}^{3} \mathrm{~s}^{-1}\).

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Fluid Mechanics

A liquid of density

750 kgm^{- 3}

flows smoothly through a horizontal pipe that tapers in cross-sectional area from

A_{1} = 1.2 \times 1 0^{- 2} m^{2}

A_{2} = \frac{A _{1}}{2}

. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ______

\times 1 0^{- 3} m^{3} s^{- 1}

Answer: 24

By the continuity equation,

A_{1} V_{1} = A_{2} V_{2}

. Since

A_{2} = \frac{A _{1}}{2}

, we have

V_{2} = 2 V_{1}

. Applying Bernoulli's equation for a horizontal pipe:

P_{1} - P_{2} = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}) = \frac{1}{2} ρ (4 V_{1}^{2} - V_{1}^{2}) = \frac{3}{2} ρ V_{1}^{2}

. Substituting:

4500 = \frac{3}{2} \times 750 \times V_{1}^{2}

, giving

V_{1} = 2 m/s

. The volume flow rate is

Q = A_{1} V_{1} = 1.2 \times 1 0^{- 2} \times 2 = 24 \times 1 0^{- 3} m^{3} s^{- 1}

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer