ChemistryQ51–51 of 96 questions
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ChemistryMCQNEET · 2019
$A$ soap bubble,having a radius of $1 \ mm$,is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2} \ N/m$. The pressure inside the bubble is equal to the pressure at a point $Z_0$ below the free surface of water in a container. Taking $g = 10 \ m/s^2$ and the density of water $\rho = 10^3 \ kg/m^3$,the value of $Z_0$ is: (in $cm$)
A
$100$
B
$10$
C
$1$
D
$0.5$
Solution
(C) The pressure inside a soap bubble is given by $P = P_0 + \frac{4T}{R}$,where $P_0$ is the atmospheric pressure,$T$ is the surface tension,and $R$ is the radius of the bubble.
The pressure at a depth $Z_0$ below the free surface of water is given by $P = P_0 + \rho g Z_0$.
Equating the two expressions for pressure:
$P_0 + \rho g Z_0 = P_0 + \frac{4T}{R}$
This simplifies to:
$\rho g Z_0 = \frac{4T}{R}$
Substituting the given values ($T = 2.5 \times 10^{-2} \ N/m$,$R = 1 \ mm = 10^{-3} \ m$,$\rho = 10^3 \ kg/m^3$,$g = 10 \ m/s^2$):
$Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^3 \times 10 \times 10^{-3}}$
$Z_0 = \frac{10 \times 10^{-2}}{10^1} = 10^{-2} \ m$
$Z_0 = 1 \ cm$.
The pressure at a depth $Z_0$ below the free surface of water is given by $P = P_0 + \rho g Z_0$.
Equating the two expressions for pressure:
$P_0 + \rho g Z_0 = P_0 + \frac{4T}{R}$
This simplifies to:
$\rho g Z_0 = \frac{4T}{R}$
Substituting the given values ($T = 2.5 \times 10^{-2} \ N/m$,$R = 1 \ mm = 10^{-3} \ m$,$\rho = 10^3 \ kg/m^3$,$g = 10 \ m/s^2$):
$Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^3 \times 10 \times 10^{-3}}$
$Z_0 = \frac{10 \times 10^{-2}}{10^1} = 10^{-2} \ m$
$Z_0 = 1 \ cm$.
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