Question

An air bubble is lying just below the surface of water.
The surface tension of water is 70-410 Nm-
and atmospheric pressure is 1.013-4 105 Nm . If
the radius of bubble is I mm, then the pressure
inside the bubble will be - [RPET 1997; AMUMED.)
2002]​

Answer 1

Pressure  inside the bubble is $=1.0144\times10^5$

Explanation:

An air bubble is lying just below the surface of water, then consider pressure outside the bubble is equivalent to atmospheric pressure but pressure inside the bubble is more than outside by,

$P_{in} - P_{out} = \frac{2T}{R}$

$P_{in} = P_{out} + \frac{2T}{R}$

$= 1.013 \times 10^{5} + \frac{140\times 10^{-3}}{0.0001}$

$P_{in}=1.0144\times10^5$

Thus, pressure  inside the bubble is $=1.0144\times10^5$

Answer 2

$\huge\tt{AnsweR}$

•GIVEN:

→Bubble is just below the water surface (${P}_{out}$)=${P}_{atm}$=1.013×10^5

→Surface tension=70×10³

→Radius of bubble=1mm

•TO FIND:

Pressure inside the bubble${P}_{in}$

→P=$\frac{2T}{r}$

→${P}_{in}$-${P}_{out}$=$\frac{2T}{r}$

→${P}_{in}$=1.013×10^5+$\frac{2T}{r}$

→${P}_{in}$=1.013×10^5+ $\frac{140×10-³}{0.0001}$

→${P}_{in}$=1.0144×10^5

•Hence the pressure inside the bubble is 1.0144×10^5