Question

A light open rigid paper frame as shown in (Fig.) floats on the surface of water. What will happen to the frame if some soap solution is dropped inside it ? What force will acts on the frame and in what direction will it act ?

Answer 1

Correct Answer - The frame will be acted upon by a force `F = (alpha_(1) - alpha_(2))l` and act out side on frame. `alpha_(1)` and `alpha_(2)` are surface tensions of water and soap solution.
If `h` is the height of water column in the capillary. The temperature of the capillary. And hence of water at this height, is
`t_(h) = (T_(UP)h)/(l)`
Water is kept in the capillary by surface tension. If `sigma_(h)` is the surface tension at the temperatrue `T_(h)`. We can write
`h = (2sigma_(h))/(rho_(w)gr)`,
Where `rho_(w)` is the density of water. Hence we obtain
`sigma_(h) = (rhogrh)/(2) = ((rhogrl)/(2))((T_(h))/(T_(up)))`.
Using the hint in the conditions of the problem, we plot the graph of the funcation `sigma(T)`. The temperature `T_(h)`b on the level of the maximum ascent of water is determined by the point of intersection of the curves describing the `(rhogl//2)T//T_(up)` and `simga(T)` dependences.

Figure shows that `T_(h) = 80^(@)C`. Consequently.
`h = (lT_(n))/(t_(up)) = 6.4 cm`.
Teh problem can also be solved analytically if we note that the `simga(T)` dependence is practically linear.