Concept:For a horizontal flow, Bernoulli's equation shows that a decrease in fluid pressure must strictly correlate with a corresponding increase in kinetic energy (velocity).
Formula:$$ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 $$
Solution:- Set \( P_1 = P \), \( v_1 = v \), and \( P_2 = P/2 \). Let \( v_2 \) be the unknown velocity.
- \( P + \frac{1}{2}\rho v^2 = \frac{P}{2} + \frac{1}{2}\rho v_2^2 \).
- Subtract \( P/2 \) from both sides: \( \frac{P}{2} + \frac{1}{2}\rho v^2 = \frac{1}{2}\rho v_2^2 \).
- Multiply the entire equation by 2 to clear fractions: \( P + \rho v^2 = \rho v_2^2 \).
- Divide by \( \rho \): \( \frac{P}{\rho} + v^2 = v_2^2 \).
- Take the square root: \( v_2 = \sqrt{v^2 + P/\rho} \).
Why other options are incorrect:Option A suggests velocity decreases when pressure drops, violating energy conservation. Options C and D result from improperly handling the \( 1/2 \) multipliers.
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