The surface tension of a soap solution is 0.03 nm-1.How much work is done to produce a soap bubble of radius 0.05m?
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Answered by
36
Here T = 0.03 Nm-1
r = 0.05 m
Surface area = 4πr2
= 4×22/7×(0.05)2
No. of surface area of a soap bubble = 2 (outer and inner area)
Surface energy is given by T∆S per surface.
Total surface energy = 2 T∆S
E = 2×0.03× 4×22/7×(0.05)2 J
Answered by
6
Given:
- The surface tension of a soap bubble = 0.03 N/m
- Radius of the soap = 0.05 m
To Find:
- The work done in producing a soap bubble of 0.05 radius.
Solution:
The work done is given by,
W = 2×surface tension×surface area = 2×T×A → {equation 1}
The surface area of the soap is 4π since bubbles are circular in shape.
Substitute the values in equation 1 then we get,
W = 2×0.03×(4×3.14×0.05×0.05)
Multiplying all the values in the above formula.
W = 0.00184 J or 18.4×
∴ The work done to produce a soap bubble of 0.05 radius = 0.00184 J or 18.4× J.
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