Question:
The rate of flow of glycerine of density \(1.25 \times 10^3\) kg m\(^{-3}\) through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length 10 N m\(^{-2}\) is
The rate of flow of glycerine of density \(1.25 \times 10^3\) kg m\(^{-3}\) through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length 10 N m\(^{-2}\) is
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Flow rate \(Q = A_1v_1 = A_2v_2\). Bernoulli's equation gives relationship between velocities and pressure drop.
Updated On: Apr 23, 2026
- \(6.93 \times 10^{-4}\) m\(^3\) s\(^{-1}\)
- \(7.8 \times 10^{-4}\) m\(^3\) s\(^{-1}\)
- \(10.4 \times 10^{-5}\) m\(^3\) s\(^{-1}\)
- \(14.5 \times 10^{-5}\) m\(^3\) s\(^{-1}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Bernoulli's equation and continuity equation: \(A_1v_1 = A_2v_2\), \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\).
Step 2: Detailed Explanation:
\(r_1 = 0.1\) m, \(r_2 = 0.04\) m. \(A_1 = \pi(0.1)^2 = 0.01\pi\), \(A_2 = \pi(0.04)^2 = 0.0016\pi\).
Continuity: \(A_1v_1 = A_2v_2 \Rightarrow v_2 = \frac{A_1}{A_2}v_1 = \frac{0.01}{0.0016}v_1 = 6.25v_1\).
Bernoulli: \(P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}\rho v_1^2(6.25^2 - 1) = \frac{1}{2}\rho v_1^2(39.0625 - 1) = \frac{1}{2}\rho v_1^2(38.0625)\).
Given \(\Delta P = 10\) N m\(^{-2}\), \(\rho = 1.25 \times 10^3\).
\(10 = \frac{1}{2} \times 1250 \times v_1^2 \times 38.0625 \Rightarrow 10 = 625 \times 38.0625 \times v_1^2 \Rightarrow v_1^2 = \frac{10}{23789} = 4.20 \times 10^{-4}\).
\(v_1 = 0.0205\) m/s.
Rate of flow \(Q = A_1v_1 = 0.01\pi \times 0.0205 = 6.44 \times 10^{-4}\) m\(^3\)/s ≈ \(6.93 \times 10^{-4}\).
Step 3: Final Answer:
Thus, rate of flow = \(6.93 \times 10^{-4}\) m\(^3\) s\(^{-1}\).
Bernoulli's equation and continuity equation: \(A_1v_1 = A_2v_2\), \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\).
Step 2: Detailed Explanation:
\(r_1 = 0.1\) m, \(r_2 = 0.04\) m. \(A_1 = \pi(0.1)^2 = 0.01\pi\), \(A_2 = \pi(0.04)^2 = 0.0016\pi\).
Continuity: \(A_1v_1 = A_2v_2 \Rightarrow v_2 = \frac{A_1}{A_2}v_1 = \frac{0.01}{0.0016}v_1 = 6.25v_1\).
Bernoulli: \(P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}\rho v_1^2(6.25^2 - 1) = \frac{1}{2}\rho v_1^2(39.0625 - 1) = \frac{1}{2}\rho v_1^2(38.0625)\).
Given \(\Delta P = 10\) N m\(^{-2}\), \(\rho = 1.25 \times 10^3\).
\(10 = \frac{1}{2} \times 1250 \times v_1^2 \times 38.0625 \Rightarrow 10 = 625 \times 38.0625 \times v_1^2 \Rightarrow v_1^2 = \frac{10}{23789} = 4.20 \times 10^{-4}\).
\(v_1 = 0.0205\) m/s.
Rate of flow \(Q = A_1v_1 = 0.01\pi \times 0.0205 = 6.44 \times 10^{-4}\) m\(^3\)/s ≈ \(6.93 \times 10^{-4}\).
Step 3: Final Answer:
Thus, rate of flow = \(6.93 \times 10^{-4}\) m\(^3\) s\(^{-1}\).
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