# Problem with Velocity of efflux [closed]

I am stuck in this 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?

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