First slide

Fluid dynamics

Question

A container filled with liquid up to height h is placed on a smooth horizontal surface. The container is having a small hole at the bottom. As the liquid comes out from the hole, the container moves in a backward direction with acceleration a and finally, when all the liquid is drained out, it acquires a velocity v. Neglect mass of the container.
In this case

Difficult

A

both a and v depend on h
B

only a depends on h
C

only v depends on h
D

neither a nor v depends on h

Solution

The velocity with which the liquid comes out is $v_{0} = \sqrt{2 g h}$
Let cross-sectional area of the hole be A and cross-sectional area of the tank be A' . Then the force exerted by the ejecting fluid due to change in momentum on the container is
$F = ρ A v^{2} = m a,$ where m is the mass of the liquid inside the container
$\Rightarrow ρ A {v_{0}}^{2} = ρ A' h \times a \Rightarrow a = 2 g \frac{A}{A'}$
Acceleration is independent of h.
$v = u + a t \Rightarrow v = 2 g \frac{A}{A'} t \Rightarrow h e r e t i s t h e t i m e i n w h i c h t h e l i q u i d c o m e s o u t c o m p l e t e l y w h i c h i s d e p e n d e n t o n h .$

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