Calculate the pressure inside a small air bubble of radius r situated at a depth h below the free surface of liquids of densities -1 and -2 and surface tension T1 and T2- The thickness of the first and second liquids are h1 and h2- respectively- Take atmosphere pressure -P0-

The correct option is B P _ 0 +ρ_ 1 gh _ 1 +ρ_ 2 g( h h _ 1 ) + 2 T _ 2 r P_c=P_B+2Tr P_B=P_A+ρ_1gh_1+ρ_2g(h h_1) where P_A=P_0 P_c ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Pascal's Law

Calculate the...

Question

Calculate the pressure inside a small air bubble of radius $r$ situated at a depth $h$ below the free surface of liquids of densities $ρ_{1}$ and $ρ_{2}$ and surface tension $T_{1}$ and $T_{2}$ . The thickness of the first and second liquids are $h_{1}$ and $h_{2}$ , respectively. Take atmosphere pressure $= P_{0}$ .

P_{0} + ρ_{1} {g h}_{1} + ρ_{2} g (h - h_{1}) - \frac{2 T_{2}}{r}

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

P_{0} + ρ_{1} {g h}_{1} + ρ_{2} g (h - h_{1}) + \frac{2 T_{2}}{r}

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

P_{0} - ρ_{1} {g h}_{1} + ρ_{2} g (h - h_{1}) + \frac{2 T_{2}}{r}

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

None of these

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

Open in App

Solution

The correct option is B $P_{0} + ρ_{1} {g h}_{1} + ρ_{2} g (h - h_{1}) + \frac{2 T_{2}}{r}$
$P_{c} = P_{B} + \frac{2 T}{r}$

$P_{B} = P_{A} + ρ_{1} g h_{1} + ρ_{2} g (h - h_{1})$

where $P_{A} = P_{0}$

$P_{c} = P_{0} + ρ_{1} g h_{1} + ρ_{2} g (h - h_{1}) + \frac{2 T_{2}}{r}$

Suggest Corrections

Similar questions

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

0.01 m m

situated at a depth of

h = 20 m

below the free surface of liquid of density

ρ_{1} = 10^{3} k g / m^{3}, ρ_{2} = 800 k m / m^{3}

and surface tension

T_{2} = 7.5 \times 10^{- 2} N / m

. The thickness of the first liquid is

h_{1} = 15 m

and

h_{2} = 25 m

Q. 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,

ρ

is the density of the liquid and

P_{0}

is the atmospheric pressure, then the pressure inside the bubble is

Q. An air bubble is formed at depth

h

below the surface of the water. What is the pressure inside the bubble?

(P_{0} =

atmospheric pressure,

r =

radius of bubble)

Q. Two immiscible liquids are poured in a

U - t u b e

having densities

ρ_{1} = 1.0 \times 10^{3} k g / m^{3}

and

ρ_{2} = 3.0 \times 10^{3} k g m^{2}

. Find the ratio of heights (of the liquids above their interface)

\frac{h_{1}}{h_{2}}

Q. A sphere of radius

R

and density

ρ_{1}

is dropped in a liquid of density

σ

. Its terminal velocity is

v_{1}

. If another sphere of radius

R

and density

ρ_{2}

is dropped in the same liquid, its terminal velocity will be:

Join BYJU'S Learning Program

Calculate the pressure inside a small air bubble of radius r situated at a depth h below the free surface of liquids of densities ρ1 and ρ2 and surface tension T1 and T2. The thickness of the first and second liquids are h1 and h2, respectively. Take atmosphere pressure =P0.

The correct option is B P0+ρ1gh1+ρ2g(h−h1)+2T2rPc=PB+2TrPB=PA+ρ1gh1+ρ2g(h−h1)where PA=P0Pc=P0+ρ1gh1+ρ2g(h−h1)+2T2r

The correct option is B $P_{0} + ρ_{1} {g h}_{1} + ρ_{2} g (h - h_{1}) + \frac{2 T_{2}}{r}$
$P_{c} = P_{B} + \frac{2 T}{r}$

$P_{B} = P_{A} + ρ_{1} g h_{1} + ρ_{2} g (h - h_{1})$

where $P_{A} = P_{0}$

$P_{c} = P_{0} + ρ_{1} g h_{1} + ρ_{2} g (h - h_{1}) + \frac{2 T_{2}}{r}$