Question

As shown in the figure, a pump is designed as a horizontal cylinder with a piston having area A and an outer orifice having an area 'a'. The piston moves with a constant velocity under the action of force F. If the density of the liquid is ρ, then the speed of liquid emerging from the orifice is(assume A>>a)

• A. \(\sqrt{\frac{F}{\rho A}}\)
• B. \(\frac{a}{A}\sqrt{\frac{F}{\rho A}}\)
• C. \(\sqrt{\frac{2F}{\rho A}}\)
• D. \(\frac{A}{a}\sqrt{\frac{2F}{\rho A}}\)

Answer 1

The Correct Option is C

We start from Bernoulli’s equation: 

\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]

Assume a fluid is at rest initially, so \( v_1 = 0 \), and pressure at point 1 is \( P_1 = P_0 + AF \), where \( A \) is area and \( F \) is force applied.

At point 2 (after acceleration), the pressure is \( P_2 = P_0 \), and the velocity is \( v_2 = v \).

Substituting into Bernoulli’s equation:

\[ P_0 + AF + 0 = P_0 + \frac{1}{2} \rho v^2 \]

Cancel \( P_0 \) from both sides:

\[ AF = \frac{1}{2} \rho v^2 \]

Solving for \( v \):

\[ v = \sqrt{\frac{2F}{\rho A}} \]

Hence, the correct answer is Option (C): \( \sqrt{\frac{2F}{\rho A}} \)