a rectangular film of a liquid is 5 cm long and 4 cm wide if the work done in increasing is it Area 7 cm × 5 cm is 0.06 J the surface tension of the solution is
Answers
Answered by
3
a rectangular film of a liquid is 5 cm long and 4 cm wide if the work done in increasing is it Area 7 cm × 5 cm is 0.06 J the surface tension of the solution is 20 J/m^2.
- Formula to calculate work done by surface tension is,
- W = T ( 2ΔA)
- T = W / ( 2ΔA)
- Given,
- W = 0.06 J
- ΔA = (7 * 5) - (5 * 4) = 15 cm^2
- Now,
- T = 0.06 / ( 2 * 15) = 2 * 10^-3 J/cm^2
- ∴T = 20 J/m^2
Answered by
4
The tension of soap solution is T = 20 J.m^-2
Explanation:
We are given:
- Length of rectangular film = 5 m
- width of film = 4
- Work done = 0.06 J
- Area of film = 7 cm x 5 cm
Solution:
Work done by tension "W" = T (2ΔA)
T = d W / 2ΔA
T = 0.06 /2 ( 7 x 5 - 5 x 4) x 10^-4
T = 20 J.m^-2
Thus the tension of soap solution is T = 20 J.m^-2
Also learn more
The total surface area of cube is 54 CM square what is the length of its sides ?
https://brainly.in/question/2476655
Similar questions