
The excess of pressure inside a soap bubble than that of the outer pressure is
A. $\dfrac{2T}{r} \\ $
B. $\dfrac{4T}{r} \\ $
C. $\dfrac{T}{2r} \\ $
D. $\dfrac{T}{r}$
Answer
160.8k+ views
Hint:In this question, we have to derive the excess pressure inside a soap bubble. To solve this expression. First we write all the force components that are acting on the bubble and then equate them then we get the expression for the excess pressure.
Complete step by step solution:
We know in a soap bubble we have two surfaces. We consider a soap bubble of radius R and surface tension T. Due to the presence of surface tension the outer surfaces experience a net inward force. So we feel excess pressure on the inside of the soap bubble.
Let ${{p}_{i}}$ be the pressure inside the air bubble and ${{p}_{0}}$be the pressure outside the bubble. Then the excess pressure inside the bubble is
$p={{p}_{i}}-{{p}_{0}}$
Due to excess pressure inside the bubble, the free surface of the bubble will experience the net force in the outward direction due to which the bubble will expand. Let the free surface be replaced by dR under isothermal conditions. Therefore, excess pressure displaces the surface and that work will be stored in the form of potential energy.
The work done by excess pressure in displacing the surface is
$dW=\text{force}\times \text{displacement}$
$\Rightarrow dW=\rho \times 4\pi {{r}^{2}}\times dR$………………………………………………………….. (1)
Increase in potential energy is
dU= surface tension $\times $ increase in area of the free surface
$dU=T\left[ 2\left\{ 4\pi {{(R+dR)}^{2}}-4\pi {{R}^{2}} \right\} \right] \\ $
$\Rightarrow dU=T\left[ 2\left\{ 4\pi (2RdR \right\} \right] \\ $ ………………………………………………………… (2)
From (1) and (2), we get
$\rho \times 4\pi {{r}^{2}}\times dR=T\left[ 2\left\{ 4\pi (2RdR) \right\} \right] \\ $
Solving the above equation, we get
$\rho =\dfrac{4T}{R} \\ $
Where T is the surface tension and R is the radius of the soap bubble.
Thus, option B is the correct answer.
Note: Surface tension in a liquid is very strong and we can’t obtain bubbles from the liquid. When we add the soap or detergent it lowers the surface tension and the soap bubbles are formed. Due to surface tension the bubbles form a spherical shape.
Complete step by step solution:
We know in a soap bubble we have two surfaces. We consider a soap bubble of radius R and surface tension T. Due to the presence of surface tension the outer surfaces experience a net inward force. So we feel excess pressure on the inside of the soap bubble.
Let ${{p}_{i}}$ be the pressure inside the air bubble and ${{p}_{0}}$be the pressure outside the bubble. Then the excess pressure inside the bubble is
$p={{p}_{i}}-{{p}_{0}}$
Due to excess pressure inside the bubble, the free surface of the bubble will experience the net force in the outward direction due to which the bubble will expand. Let the free surface be replaced by dR under isothermal conditions. Therefore, excess pressure displaces the surface and that work will be stored in the form of potential energy.
The work done by excess pressure in displacing the surface is
$dW=\text{force}\times \text{displacement}$
$\Rightarrow dW=\rho \times 4\pi {{r}^{2}}\times dR$………………………………………………………….. (1)
Increase in potential energy is
dU= surface tension $\times $ increase in area of the free surface
$dU=T\left[ 2\left\{ 4\pi {{(R+dR)}^{2}}-4\pi {{R}^{2}} \right\} \right] \\ $
$\Rightarrow dU=T\left[ 2\left\{ 4\pi (2RdR \right\} \right] \\ $ ………………………………………………………… (2)
From (1) and (2), we get
$\rho \times 4\pi {{r}^{2}}\times dR=T\left[ 2\left\{ 4\pi (2RdR) \right\} \right] \\ $
Solving the above equation, we get
$\rho =\dfrac{4T}{R} \\ $
Where T is the surface tension and R is the radius of the soap bubble.
Thus, option B is the correct answer.
Note: Surface tension in a liquid is very strong and we can’t obtain bubbles from the liquid. When we add the soap or detergent it lowers the surface tension and the soap bubbles are formed. Due to surface tension the bubbles form a spherical shape.
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