Glycerine of density 1.25 × 10³ kg/m³ is flowing in conical shaped horizontal pipe. Crosssectional area of the pipe at its both ends is 10 cm² and 5 cm² respectively. Pressure difference at both the ends is 3 N/m². Rate of flow of liquid in the pipe is (a) 4 × 10⁻⁵ m³/s (b) 2 × 10⁻⁵ m³/s (c) 5 × 10⁻⁵ m³/s (d) 6 × 10⁻⁵ m³/s

Answer: (a). Continuity: v₂ = 10/5v₁ = 2v₁ Bernoulli: Δ P = 1/2ρ(4v₁² - v₁²) ⇒ 3 = 1/2(1250)(3v₁²) v₁² = 1.6×10⁻³ ⇒ v₁ = 0.04 m/s Q = A₁v₁ = (10×10⁻⁴)(0.04) = 4×10⁻⁵ m³/s So the answer is (a) 4×10⁻⁵ m³/s.

Glycerine of density 1.25 × 10³ kg/m³ is flowing in conical shaped… | MHT CET 2024 (2 May, Shift 1)

Q15MHT CET 2024 · 2 MAY · SHIFT 1Difficult

Bernoulli's Equation

Glycerine of density $1.25 \times 10^{3}\text{ kg/m}^3$ is flowing in conical shaped horizontal pipe. Crosssectional area of the pipe at its both ends is $10\text{ cm}^2$ and $5\text{ cm}^2$ respectively. Pressure difference at both the ends is $3\text{ N/m}^2$ . Rate of flow of liquid in the pipe is

Show answer and solution

Answer: (a)

Continuity: $v_2 = \frac{10}{5}v_1 = 2v_1$
Bernoulli: $\Delta P = \frac{1}{2}\rho(4v_1^2 - v_1^2) \Rightarrow 3 = \frac{1}{2}(1250)(3v_1^2)$
$v_1^2 = 1.6\times10^{-3} \Rightarrow v_1 = 0.04\text{ m/s}$
$Q = A_1v_1 = (10\times10^{-4})(0.04) = 4\times10^{-5}\text{ m}^3\text{/s}$
So the answer is (a) $4\times10^{-5}\text{ m}^3\text{/s}$ .

Report a mistake

Bernoulli's Equation question from MHT CET 2024 (2 May, Shift 1)

More questions on Bernoulli's Equation