A liquid column of height 0.04 ,cm balances e | vedclass

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

Question

(B) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{R}$, where $S$ is the surface tension and $R$ is the radius of the bubb

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{R}$, where $S$ is the surface tension and $R$ is the radius of the bubble.This pressure is balanced by the hydrostatic pressure of the liquid column: $\Delta P = ho g h$.Equating the two: $ho g h = \frac{4S}{R}$.Given: $ho = 8 	imes 10^3 \,kg \,m^{-3}$, $g = 10 \,m \,s^{-2}$, $h = 0.04 \,cm = 4 	imes 10^{-4} \,m$, and $S = 0.28 \,N \,m^{-1}$.Substituting the values: $(8 	imes 10^3) 	imes 10 	imes (4 	imes 10^{-4}) = \frac{4 	imes 0.28}{R}$.$32 = \frac{1.12}{R}$.$R = \frac{1.12}{32} \,m = 0.035 \,m = 3.5 \,cm$.The diameter $D = 2R = 2 	imes 3.5 \,cm = 7 \,cm$.

Similar Questions

If two soap bubbles of different radii are connected by a tube,

Two soap bubbles of radius $2 \ cm$ and $4 \ cm$,respectively,are in contact with each other. The radius of curvature of the common surface,in $cm$,is . . . . . . .

Pressure inside two soap bubbles are $1.01 \, atm$ and $1.02 \, atm$. The ratio between their volumes is:

The excess pressure inside a first spherical drop of water is three times that of a second spherical drop of water. Then the ratio of the mass of the first spherical drop to that of the second spherical drop is

Vedclass Test Series

Exam Paper Generator

Online Exam Module