An air bubble of radius r in water is at a depth h below the water surface. If P is atmospheric pressure and d and T are the density and surface tension of water respectively. The pressure inside the bubble will be:

P + h d g - \frac{4 T}{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P + h d g + \frac{2 T}{r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P + h d g - \frac{2 T}{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P + h d g + \frac{4 T}{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $P + h d g + \frac{2 T}{r}$
Pressure just outside the bubble will be $P + h d g$
The $2 T / r$ component is due to water-air interface surface tension and is additive because water exerts pressure on the bubble
So total pressure = $P + h d g + 2 T / r$

Suggest Corrections

Similar questions

Q. An air bubble of radius

r

is formed at a depth

h

below the surface of water. The pressure inside the bubble is [

T =

surface tension,

P_{o} =

atmospheric pressure,

d =

density of water]:

Small air bubble of radius r is found to form at a depth of H from the open surface of the liquid contained in a beaker. If S is the surface tension and $ρ$ , the density of the liquid and $ρ_{0}$ , the atmospheric pressure the pressure inside the bubble is

Q. Calculate the pressure inside a small air bubble of radius