I am stuck in this problem-

The Problem

I need to find the velocity of efflux at the hole of the container. [We can assume that the area of the hole is negligible in comparison with the base area of the container].

Here's my approach

Velocity of liquid at the upper-surface = $v_2$

Velocity of efflux (velocity of water at the hole, right?) = $v_1$

Using Bernoulli's equation for the surface and the hole -

$$ P_{atm} + \rho_2 g (h_1 + h _2) + \frac{1}{2}\rho_2 v_2^2 = P_{atm} + \rho_1 g h_1 + \rho_2 g h_2 + \frac{1}{2}\rho_1 v_1^2 \\ \implies \rho_2 g h_1 + \rho_2 g h_2 - \rho_1 g h_1 - \rho_2 g h_2 = \frac{1}{2}(\rho_1 v_1^2 - \rho_2 v_2^2) \\ \implies \frac{1}{2}(\rho_1 v_1^2 - \rho_2 v_2^2) = g h_1 (\rho_2 - \rho_1) $$

Now, let area of the base be $A_2$ and that of the hole be $A_1$

then, using equation of continuity,

$$ A_1 v_1 = A_2 v_2 \\ \implies v_2 = \frac{A_1}{A_2} v_1 \\ \implies v_2 \approx 0 (\because {A_1 << A_2}) $$

Using this value in the previous Bernoulli's relation

$$ \frac{1}{2}(\rho_1 v_1^2) = g h_1 (\rho_2 - \rho_1) \\ \implies \frac{1}{2} \rho_1 v_1^2 = g h_1 (\rho_2 - \rho_1) \\ \implies v_1 = \sqrt {\frac {2gh_1(\rho_2 - \rho_1)}{\rho_1}} $$

Which is not the correct answer.

I did get a correct answer in the chat room, but it was using a different method.

What's wrong with my method?

  • $\begingroup$ Oops, accidentally gave this a "Leave Open". VTCing now. $\endgroup$ Nov 19, 2013 at 16:51

1 Answer 1


Why do you want to use continuity at all? It is not valid in this case simply because all the fluid in the container is not going to come out of the orifice. Suppose you had a pipe whose diameter was being reduced, you'd be able to apply continuity because all fluid going through section A on the pipe would have to go through section B.

Additionally, Bernoulli's equation fully resolves the problem for you, so there is no need to use equation of continuity.

\begin{align} \frac{\rho1 * v_{efflux}^2}{2} &= P_{bottom} - P_{atm} = g * (\rho_1 * h_1 + \rho_2 * h_2) \\ v_{efflux} &= \sqrt{\frac{2g * (\rho_1 * h_1 + \rho_2 * h_2)}{\rho_1}} \end{align}