A soap film is formed on rectangular frame of length 0.05m dipping in soap solution. The frame hangs from the aim of a balance. An extra mass of 3 × 10-4kg must be placed in the other pan to balance the pull of the film. Calculate the surface tension of the soap solution.?
Calculate the energy required to split a drop of water of radius 2 × 10-3m into two thousand million droplets of the same size. Surface tension of water is 0.072 N/m.?
| We know that surface energy σ | = |  |
| but surface energy σ | = | Surface tension γ |
| Given surface tension of water γ | = | 0.072 N/m |
| Radius of the big drop R | = | 2 × 103 |
| no.of droplets after splitting n | = | 2 × 103 × 106 |
| = | 2 × 109 | |
| Let r be the radius of each small droplet | ||
| Then, volume of n droplets | = | Volume of big drop |
 |
||
| ∴ Radius of each droplet is | = | 1.6 × 10-6m |
| Surface area of each small droplet | = | 4πr2 |
| = | 4π(1.6 × 10-6)2 | |
| = | 32.2 × 10-12m2 | |
| Surface area of n droplets | = | n(4πr2) |
| = | 2 × 109(32.2 × 10- 12) | |
| = | 64.4 × 10-3m2 | |
| Surface in surface area | = | n(4πr2) - 4πR2 |
| = | 64.4 × 10-3 - [4π(2 × 10-3)2 | |
| = | 64.4 × 10-3 - 50.3 × 10-6 | |
| = | 10-3[64.4 - 50.3 × 10-3] | |
| = | 64.3 × 10-3m2 | |
| ∴ Work done | = | surface tension × increase in area |
| = | 0.072 × 64.3 × 10-3 | |
| = | 4.6 × 10-3 J |
∴ Energy required to split the drop into 2 × 109 droplets is 4.6 × 10-3 J.
Calculate the work done against the surface tension forces in blowing a soap bubble of diameter 1cm. The surface tension of soap solution is 2.5 × 10-2 N/m?
| ∴ Increase in surface area | = | 8πr2 - 0 |
| = | 8πr2 | |
| Diameter of soap bubble, given is d | = | 1cm |
| radius r | = | 0.5cm |
| = | 0.005m | |
| ∴ Work done W | = | r × 8πr2 |
| = | 2.5×10-2 × 8πr2 × (0.005)2 | |
| = | 1.57× 10-5J |