I was asked this question by my boy and I want to give him the correct answer. So keep it basic please.Here is what he asked. Why execute air bubbles rise when they are released underwater?


You can start off by citing a common example: a helium filled balloon goes up. You must ask him why does it go up as soon as everything else is being pulled down? The answer is that the thickness of the balloon is far less than the density of the air.

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The very same goes for a bubble in water. The bubble includes air which is less dense than water; therefore, the bubble rises.

Why carry out less dense objects rise inside water?

The water molecules are in continuous movement and also they frequently bump right into the bubble. When they bump right into the bubble, they push the bubble. The water molecules bump right into the bubble from all sides: up, down, left, ideal, forward, earlier. Due to gravity, the press at the bottom of bubble is better than the press at the height of the bubble or in various other words, the force exerted by the water molecules which collide at the bottom of the bubble is even more than the force exerted by the water molecules at the peak of the bubble. Gravity does not exist sideways; therefore, the pressures applied from the best, left, forward and backward cancel out. You are left with a net upward pressure as a result of the water. Your bubble has some weight, but, the weight of the bubble is not considerable enough to cancel the upward pressure applied by the water. As such, the bubble rises.

Mathematical Derivation:

Consider water in a container in equilibrium.


Take a tiny block of water of elevation $h$. This block of water has actually weight but why does not it compress the water listed below and also loss down? It is because the water below applies a pressure larger than the pressure applied by the water above such that the net pressure on the block of water is zero.

$$upward force = downward force$$

$$(P + Delta P)A = PA + mg ag1$$

The mass of the water in the block have the right to be expressed in terms of its thickness.

$$m = ho V = ho Ah ag2$$

Substituting $(2)$ in $(1)$, we get:

$$(P + Delta P)A = PA + ho Ahg$$

$$P + Delta P = P + ho hg$$

$$Delta P = ho hg$$

The push below the block is $Delta P$ more than the press at the optimal of it.

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$$F_net = Delta P A - mg$$

$$F = V ho g - (Vsigma) g$$

$$F = Vg( ho - sigma)$$

The net upward pressure depends on the distinction of the thickness of the liquid and the body immersed. As the bubble is much less thick, tright here will be a net upward pressure.