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 ?

A light open rigid paper frame as shown in (Fig.) floats on the surfac

1.	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» 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.