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Pressure difference from the 5 cm water column equals the Bernoulli difference \(\frac12\rho(v_1^2 - v_2^2)\); use \(A_1v_1 = A_2v_2 = Q\).
Updated On: Nov 11, 2026
- 7.5 cc/s
- 8.0 cc/s
- 10.0 cc/s
- 12.5 cc/s
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The Correct Option is A
Solution and Explanation
Pressure difference: \(\Delta p = \rho g h = 1000\times10\times0.05 = 500\) Pa.
Let Q be the volume flow rate, \(A_1 = 6\times10^{-6}\ \mathrm{m^2}\), \(A_2 = 10\times10^{-6}\ \mathrm{m^2}\). Then \(v_1 = Q/A_1\), \(v_2 = Q/A_2\).
Bernoulli: \(\Delta p = \frac12\rho Q^2\left(\frac{1}{A_1^2}-\frac{1}{A_2^2}\right)\).
\(\frac{1}{A_1^2}-\frac{1}{A_2^2} = 10^{12}\left(\frac1{36}-\frac1{100}\right) = 1.778\times10^{10}\ \mathrm{m^{-4}}\).
\(500 = 500\times1.778\times10^{10}\,Q^2\), so \(Q^2 = 5.625\times10^{-11}\), \(Q = 7.5\times10^{-6}\ \mathrm{m^3/s} = 7.5\) cc/s.
Answer: (1) 7.5 cc/s
Let Q be the volume flow rate, \(A_1 = 6\times10^{-6}\ \mathrm{m^2}\), \(A_2 = 10\times10^{-6}\ \mathrm{m^2}\). Then \(v_1 = Q/A_1\), \(v_2 = Q/A_2\).
Bernoulli: \(\Delta p = \frac12\rho Q^2\left(\frac{1}{A_1^2}-\frac{1}{A_2^2}\right)\).
\(\frac{1}{A_1^2}-\frac{1}{A_2^2} = 10^{12}\left(\frac1{36}-\frac1{100}\right) = 1.778\times10^{10}\ \mathrm{m^{-4}}\).
\(500 = 500\times1.778\times10^{10}\,Q^2\), so \(Q^2 = 5.625\times10^{-11}\), \(Q = 7.5\times10^{-6}\ \mathrm{m^3/s} = 7.5\) cc/s.
Answer: (1) 7.5 cc/s
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