Question

Water is being poured into a vessel at a constant rate r m3 / s . There is a small aperture of cross sectional area a at the bottom of the vessel . Find the maximum level of water in the vessel.

Answer 1

Answer: $h = \frac{r^2}{2ga^2}$

Explanation:

The maximum level of water in the vessel would be when the rate of in flow of water is equal to rate of out flow of water.

Let the maximum level of water in vessel be h

The rate of in flow of water = r m³/s

The rate of out flow of water = area of cross section × speed

                                               = a × v = a × √2gh

(∵K.E. = P.E. of water ⇒0.5 mv²=mgh⇒v=√2gh where m is the mass of water)

Rate of in flow of water = Rate of out flow of water.

⇒ r = a × √2gh

⇒ $h = \frac{r^2}{2ga^2}$

Answer 2

Answer:

Explanation:

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