Question

The surface tension of a soap solution is 0.05 Nm^(-1) How much work is done to produce a soap bubble of radius 0.03 m?

Answer 1

Solution

• Work Done in expanding the bubble is 11304 × 10^-7 J

Given

• Surface Tension (S) = 0.05 N/m

• Radius (R) = 0.03 m

To finD

Work Done in expanding the bubble

Using the Relation,

$\boxed{\boxed{\sf Work \ Done = Total \ Surface \ Area \times Surface \ Tension }}$

Here,

• A soap bubble had two faces

• Total Surface Area Of Sphere : 4πr²

Now,

$\implies \: \tt \: W \: = 2 \times 4 {\pi \: r}^{2} \times \sigma \\ \\ \implies \: \tt \: W = 2 \times 4 \times 3.14 \times (0.03) {}^{2} \times 0.05 \\ \\ \large{ \implies \: \boxed{ \boxed{ \tt W = 11304 \times {10}^{ - 7} \ J}}}$

Answer 2

Explanation:

Given,

Surface Tension of Soap Solution is 0.05 N/m¹ or N/m and radius of a bubble is 0.03 m. So we have to find out the Work done.

Surface Tension = 0.05 N/m

Radius = 0.03 m

By using this formula or equation, we get work done.

[ Work done = Total Surface Area × Surface Tension ]

As we know that the bubble of soap have 2 layer. Inner layer and outer layer.

=> Work done = 2 × 4πr² × Surface Tension

As we know that, the sphere has less surface area as compared to the other like cube, cone and cylinder or other solid faces. And surface tension is the property to remain in less surface area. So, bubble have spherical in shape.

=> Work done = 2 × 4 × 22/7 × (0.03)² × 0.5

Here, we know that 22/7 is equal to 3.1415. In round number.

[°.° π = 3.14 ]

=> Work done = 2 × 4 × 3.14 × (0.03)² × 0.5

$.°.\:Work done\:= \: 11304 × 10^{-7}$

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