Glycerin of density 1.25 × 10³ kg m⁻³ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are 10 cm² and 5 cm² and pressure drop across its length is 3 Nm⁻². The rate of flow of glycerin through the pipe is x × 10⁻⁵ m³ s⁻¹. The value of x is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 4. Continuity: A₁v₁ = A₂v₂ ⇒ v₂ = 10/5v₁ = 2v₁ Bernoulli (horizontal): Δ P = 1/2ρ(v₂² - v₁²) = 1/2ρ(3v₁²) 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 x = 4.

Glycerin of density 1.25 × 10³ kg m⁻³ is flowing through the conical… | JEE Main 2023 (12 Apr, Shift 1)

Q18JEE Main 2023 · 12 APR · SHIFT 1Medium

Bernoulli's Equation

Glycerin of density $1.25 \times 10^{3}\text{ kg m}^{-3}$ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are $10\text{ cm}^2$ and $5\text{ cm}^2$ and pressure drop across its length is $3\text{ Nm}^{-2}$ . The rate of flow of glycerin through the pipe is $x \times 10^{-5}\text{ m}^3\text{ s}^{-1}$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

Continuity: $A_1v_1 = A_2v_2 \Rightarrow v_2 = \frac{10}{5}v_1 = 2v_1$
Bernoulli (horizontal): $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}\rho(3v_1^2)$
$3 = \frac{1}{2}(1250)(3v_1^2) \Rightarrow v_1^2 = 1.6\times10^{-3} \Rightarrow v_1 = 0.04\text{ m/s}$
$Q = A_1v_1 = (10\times10^{-4})(0.04) = 4\times10^{-5}\text{ m}^3\text{/s}$
So $x = 4$ .

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