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Question: Under isothermal condition, energy E is supplied to a soap bubble of surface tension \(\sigma \) and...

Under isothermal condition, energy E is supplied to a soap bubble of surface tension σ\sigma and radius r to double the radius of the soap bubble. The value of E is:
A. 16πr2σ\text{A}\text{. }16\pi {{r}^{2}}\sigma
B. 24πr2σ\text{B}\text{. 24}\pi {{r}^{2}}\sigma
C. 8πr2σ\text{C}\text{. 8}\pi {{r}^{2}}\sigma
D. 12πr2σ\text{D}\text{. }12\pi {{r}^{2}}\sigma

Explanation

Solution

Hint: To increase the radius of the bubble we have to provide a certain amount of energy. The value of the energy that has to be provided to increase the radius of the bubble is given as E=σ.ΔAE=\sigma .\Delta A, where ΔA\Delta A is the change in the total surface area of the bubble and σ\sigma is the surface tension.

Formula used:
E=σ.ΔAE=\sigma .\Delta A

Complete step by step answer:
Like rigid bodies, fluids also have a tension at their surfaces. However, unlike rigid bodies that produce tension inside them, fluids produce tension at their surfaces called surface tension.
Surface tension is defined as the force per unit length. It is denoted by σ\sigma .

A soap bubble also has surface tension. It is given that the surface tension of the bubble is equal to σ\sigma .

To increase the radius of the bubble we have to provide a certain amount of energy. The value of the energy that has to be provided to increase the radius of the bubble is given as
E=σ.ΔAE=\sigma .\Delta A, where ΔA\Delta A is the change in the total surface area of the bubble.
Let us calculate the energy given to the bubble.

It is given that the radius of the bubble doubles when the energy is supplied.

Let the surface area of the bubble when its radius is r be A and the surface area when its radius is 2r be A’.

Area of a sphere is 4πr24\pi {{r}^{2}}.

Note that a bubble is hollow and hence has a total surface will be the sum of the surface from inside and the surface from outside the bubble.

Therefore,

A=2×4πr2=8πr2A=2\times 4\pi {{r}^{2}}=8\pi {{r}^{2}} and A=2×4π(2r)2=32πr2A'=2\times 4\pi {{\left( 2r \right)}^{2}}=32\pi {{r}^{2}}

Hence the change in area will be ΔA=32πr28πr2=24πr2\Delta A=32\pi {{r}^{2}}-8\pi {{r}^{2}}=24\pi {{r}^{2}}.
Therefore, E=σ.ΔA=σ(24πr2)=24πr2σE=\sigma .\Delta A=\sigma \left( 24\pi {{r}^{2}} \right)=24\pi {{r}^{2}}\sigma

Hence, the correct option is B.

Note: This energy (E) depends on the surface tension of the bubble and the total surface area of the bubble. However, when some amount of energy is given to the bubble, its surface tension remains the same and only the total surface area changes.