The excess pressure inside a spherical soap b | 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 $p = \frac{4T}{R}$.This pressure is balanced by the pressure exerted by the oil column, which

Vedclass · Accepted Answer

$0.0392$

Vedclass · Answer

(B) The excess pressure inside a soap bubble is given by $p = \frac{4T}{R}$.This pressure is balanced by the pressure exerted by the oil column, which is $p = h ho g$.Equating the two, we get $h ho g = \frac{4T}{R}$.Rearranging for surface tension $T$, we have $T = \frac{R h ho g}{4}$.Given values: $R = 1 ~cm = 10^{-2} ~m$, $h = 2 ~mm = 2 	imes 10^{-3} ~m$, $ho = 0.8 	imes 10^3 ~kg/m^3$, and $g = 9.8 ~m/s^2$.Substituting these values into the formula:$T = \frac{10^{-2} 	imes 2 	imes 10^{-3} 	imes 0.8 	imes 10^3 	imes 9.8}{4}$$T = \frac{2 	imes 0.8 	imes 9.8 	imes 10^{-2}}{4}$$T = 0.4 	imes 0.8 	imes 9.8 	imes 10^{-2} = 0.392 	imes 10^{-1} = 0.0392 ~N/m$.

The excess pressure inside a spherical soap bubble of radius $1 ~cm$ is balanced by a column of oil (specific gravity $= 0.8$), $2 ~mm$ high. The surface tension of the bubble is: (in $~N/m$)

Similar Questions

The overall goal of glycolysis,Krebs' cycle,and the electron transport system is the formation of

The particles emitted during the sequential radioactive decay of ${}^{238}U_{92}$ to ${}^{206}Pb_{82}$ are

The expression $[x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$ is a polynomial of degree :

The enthalpy of formation of ammonia is $-46.0 \, kJ \, mol^{-1}$. The enthalpy change for the reaction $2NH_{3(g)} \longrightarrow N_{2(g)} + 3H_{2(g)}$ is ..... $kJ \, mol^{-1}$.

Which of the following orders is correct for the heat of hydrogenation of these compounds?

Vedclass Test Series

Exam Paper Generator

Online Exam Module