Q. A liquid of density 800 kg $m^{3}$ is filled in a tank open at the top. The pressure of the liquid at the bottom of the tank is 6.4 atm. The velocity of efflux through a hole at the bottom is(1 atm = $10^{5}$ ${Nm}^{- 2}$)
A) 10 ${ms}^{- 1}$
B) 20${\ ms}^{- 1}$
C) 30${\ ms}^{- 1}$
D) 40${\ ms}^{- 1}$
Explanation
(d) Velocity of efflux $v = \sqrt{2gh}$
But $h\rho g = p$
$\therefore\ \ hg = \frac{p}{\rho}$
$\therefore\ v = \sqrt{\frac{2p}{\rho}}$
$\ \ \ = \sqrt{\frac{2 \times 6.4 \times 10^{5}}{800}{ms}^{- 1}}$
$\ \ \ = 40{\ ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.