I was asked this question by my boy and also I desire to offer him the correct answer. So save it simple please.Here is what he asked. Why execute air bubbles increase once they are released underwater?


You have the right to start off by citing a widespread 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 density of the balloon is far less than the density of the air.

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

Why execute much less thick objects increase inside water?

The water molecules are in continuous activity and also they often bump right into the bubble. When they bump into the bubble, they press the bubble. The water molecules bump right into the bubble from all sides: up, dvery own, left, appropriate, forward, earlier. Due to gravity, the pressure at the bottom of bubble is better than the press at the top of the bubble or in other words, the force exerted by the water molecules which collide at the bottom of the bubble is even more than the pressure exerted by the water molecules at the height of the bubble. Gravity does not exist sideways; therefore, the forces used from the appropriate, left, forward and also backward cancel out. You are left through a net upward force because of the water. Your bubble has actually some weight, yet, the weight of the bubble is not substantial 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 however why does not it compush the water listed below and loss down? It is because the water listed below applies a pressure larger than the force used by the water above such that the net pressure on the block of water is zero.

$$upward pressure = downward force$$

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

The mass of the water in the block can 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 pressure at the top 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 force relies on the difference of the thickness of the fluid and the body immersed. As the bubble is much less thick, tbelow will be a net upward pressure.