Q. A denotes the area to the right on the cube $h$ the depth of an orifice of area of cross-section$A$, below the liquid surface. The velocity of the liquid flowing through the orifice is
A) $\sqrt{2\ gh}$
B) $\sqrt{2\ gh}\ \sqrt{\left( \frac{A^{2}}{A^{2} - a^{2}} \right)}$
C) $\sqrt{2\ gh}\ \sqrt{\left( \frac{A}{A - a} \right)}$
D) $\sqrt{2\ gh}\ \sqrt{\left( \frac{A^{2} - a^{2}}{A^{2}} \right)}$
Explanation
(b) Using Bernoulli’s theorem;
$p + \frac{1}{2}pv^{'2} + \rho gh = p + \frac{1}{2}\rho v^{2} + 0$
Where $v'$ is the velocity of the liquid at surface and $v$ is the velocity of efflux. As
$Av^{'} = av$ or $v^{'} = av/A$
$\therefore\frac{1}{2}\rho\ \left( \frac{av}{A} \right)^{2} + \rho\ gh = \frac{1}{2}\rho v^{2}$
Or $v^{2} - \frac{a^{2}v^{2}}{A^{2}} = 2gh$
Or $v = \sqrt{2gh}/\sqrt{\frac{A^{2} - a^{2}}{A^{2}}} = \sqrt{2gh}\sqrt{\frac{A^{2}}{A^{2} - a^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.