The pressure inside a small air bubble of radius $0.1\, mm$ situated just below the surface of water will be equal to [Take surface tension of water $70 \times {10^{ - 3}}N{m^{ - 1}}$ and atmospheric pressure = $1.013 \times {10^5}N{m^{ - 2}}$]
$2.054 \times {10^3}\,Pa$
$1.027 \times {10^3}\,Pa$
$1.027 \times {10^5}\,Pa$
$2.054 \times {10^5}\,Pa$
A capillary type tube $AB$ is connected to a manometer $M$ as shown in the figure. Stopper $S$ controls the flow of air. $AB$ is dipped into a soap solution where surface tension is $T$ . On opening the stopper for a while, a bubble is formed at $B$ end of the manometer and the level difference in manometer limbs is $h$ . If $P$ is the density of manometer soap solution and $r$ the radius of curvature of bubble, then the surface tension $T$ of the liquid is given by ...
The excess of pressure inside a soap bubble than that of the outer pressure is
In capillary pressure below the curved surface of water will be
A capillary tube of radius $r$ is dipped in a liquid of density $\rho$ and surface tension $S$. If the angle of contact is $\theta$, the pressure difference between the two surfaces in the beaker and the capillary
If a section of soap bubble (of radius $R$) through its center is considered, then force on one half due to surface tension is