Glycerine of density \(1.25 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\) is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is \(10 \mathrm{~cm}^{2}\) and \(5 \mathrm{~cm}^{2}\) and pressure drop across its length is \(3 \mathrm{Nm}^{-2}\). The rate of flow of glycerine through the pipe is \(\mathrm{x} \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}\). The value of x is \_\_\_\_ .

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Fluid Mechanics

Glycerine of density

1.25 \times 1 0^{3} kg m^{- 3}

is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is

10 cm^{2}

and

5 cm^{2}

and pressure drop across its length is

3 Nm^{- 2}

. The rate of flow of glycerine through the pipe is

x \times 1 0^{- 5} m^{3} s^{- 1}

. The value of x is ____ .

Answer: 4

Continuity gives

A_{1} v_{1} = A_{2} v_{2}

, so

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

. For a horizontal pipe Bernoulli's equation gives

Δ P = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} ρ v_{2}^{2} (1 - \frac{1}{4}) = \frac{3}{8} ρ v_{2}^{2}

. With

Δ P = 3 N m^{- 2}

and

ρ = 1.25 \times 1 0^{3} kg m^{- 3}

v_{2}^{2} = \frac{8 Δ P}{3 ρ} = \frac{8 \times 3}{3 \times 1.25 \times 1 0 ^{3}} = 6.4 \times 1 0^{- 3}

, so

v_{2} = 8 \times 1 0^{- 2} m s^{- 1}

. The flow rate is

Q = A_{2} v_{2} = 5 \times 1 0^{- 4} \times 8 \times 1 0^{- 2} = 4 \times 1 0^{- 5} m^{3} s^{- 1}

, so

x = 4

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer