Fluid Mechanics

Olympiad (NSEP) level, single correct, hard

3 of 192

Question 3 of 192

Water flows out of a small hole of area aa at the base of a large cylindrical tank of cross-sectional area AA filled to height HH. Time taken to empty the tank completely is:

Show answer and solution

Answer: (B) Aa2Hg\frac{A}{a} \sqrt{\frac{2H}{g}}

−Adhdt=a2gh  ⟹  t=Aa2g∫0Hh−1/2dh=Aa2g[2H]=Aa2Hg-A \frac{dh}{dt} = a \sqrt{2gh} \implies t = \frac{A}{a\sqrt{2g}} \int_0^H h^{-1/2} dh = \frac{A}{a\sqrt{2g}} [2\sqrt{H}] = \frac{A}{a}\sqrt{\frac{2H}{g}}.
Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Fluid mechanics simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

Olympiad (NSEP) Physics practice question on Fluid Mechanics, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.