A capillary type tube AB is connected to a ma | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The pressure inside the soap bubble at $B$ is given by $P_{in} = P_0 + \frac{4T}{r}$,where $P_0$ is the atmospheric pressure,$T$ is the surface te

Vedclass · Accepted Answer

$\frac{r \rho gh}{4}$

Vedclass · Answer

(D) The pressure inside the soap bubble at $B$ is given by $P_{in} = P_0 + \frac{4T}{r}$,where $P_0$ is the atmospheric pressure,$T$ is the surface tension,and $r$ is the radius of the bubble.The pressure at the same horizontal level in the manometer limbs is equal. Let the pressure at the liquid surface in the open limb be $P_0$. Then the pressure at the level corresponding to the lower limb $D$ is $P_0 + \rho gh$.Equating the pressures at the level of the bubble $B$ and the manometer limb $D$ (assuming $B$ is at the same level as $D$):$P_0 + \frac{4T}{r} = P_0 + \rho gh$Simplifying the equation:$\frac{4T}{r} = \rho gh$Solving for $T$:$T = \frac{r \rho gh}{4}$

Similar Questions

When a large bubble rises from the bottom of a lake to the surface,its radius doubles. If atmospheric pressure is equal to that of a column of water of height $H$,then the depth of the lake is:

The excess pressure inside a soap bubble of radius $2.5 \text{ cm}$ is $48 \text{ dyne/cm}^2$. The surface tension of the soap solution in $\text{dyne/cm}$ is:

What is the ratio of the number of moles of air in two soap bubbles of radii $R_1$ and $R_2$?

The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$) into $64$ equal drops is

In air,a charged soap bubble of radius $R$ breaks into $64$ small soap bubbles of equal radius $r$. The ratio of mechanical force per unit area of the big soap bubble to that of a small bubble is

Vedclass Test Series

Exam Paper Generator

Online Exam Module