Question

Water in a clean large cuboid aquarium forms a meniscus, as shown in the figure. The difference in height between the centre and the edge of the meniscus is h. Surface tension of water is `S = 0.073 Nm^(-1)`. Atmospheric pressure is `P_0=10^5 N//m^2` . Angle of contact between the water and aquarium wall is zero.

Answer the following 3 question above illustrated situation.

Value of height h is
A. 0.0076 m
B. 0.0019 m
C. 0.0038 m
D. 0.0152 m

Answer 1

The pressure of the water changes linearly with the increase in height. At the bottom of the meniscus it is equal to the external atmospheric pressure `p_0`, and at the top to . The average pressure exerted on the wall is `P_("average")=p_(0)-rhogh//2`. The force corresponding to this value, for an aquarium with side walls of length l, is`F_(1)=p_("average")lh`. Consider the horizontal forces acting on the volume of water enclosed by the dashed lines in the figure. The wall pushes it to the right with force `F_1`, the external air pushes it to the left with force `F_2 pi_(0)lh`, and the surface tension of the rest of the water pulls it to the right with a force `F_(3)=ls`. The resultant of these forces has to be zero, since the volume itself is at rest. This means that

`(rho_(0)-1/2 rhogh)lh-p_(0)lh+ls=0`
which we can writen as
`h=sqrt((2s)/(rhog))=sqrt((2xx0.073)/(1000xx10))=0.0038m`
Water rises by approximately 4 mm up the wall of the aquarium,
`(rho_(0)-1/2 rhogh)lh-p_(0)lh+ls=0` which we can writen as
`h=sqrt((2s)/(rhog))=sqrt((2xx0.073)/(1000xx10))=0.0038m`