Capilarity

@ eure ,when u insert capillary in the lower liquid according to ur solution there will b only denser liq. in the capillary.but in fact in the begining when u insert the capillary in the lower liquid u hav to pass it through upper one.hence the upper liquid will first enter into it .again when their will b capillary rise or fall there iz no chance of this liquid to com out of capillary. so u hav to solve as below
las we dont know wheather the liquid will rise or fall let it b rises to an height h ,then
P₀-2T/r-hρ*g-H/2.ρ/2.g=P₀-H/2. ρ/2.g
=>h=-2T/r

plz cheak the answer.if wrong then verify my process also.

ans is 4T/r

plz verify my ressult and find out the mistake

point out my mistake...

excess pressure comes from surface tension......

i wrote that part naa....

The pressure difference between the "inside" and "outside" of a curved meniscus is
Δp = 2SR
with S being the surface tension and R the radius of the curved surface.

ii) Suppose the liquid rises to a height h above the upper surface of lighter liquid. Then just inside the meniscus the pressure p is obtained by the relation
p_0-p = \dfrac{2S\cos\theta}{r}
r being the radius of the capillary tube and θ the contact angle.
Balancing pressure at the bottom of the capillary tube, we get
p+\dfrac{\rho}{2}\,g(h+H)+\rho gx = p_0+\dfrac{\rho}{2}\,gH+\rho gx
where x is the distance by which the capillary tube has been dipped in the lower liquid (having higher density).
From the above two relations we obtain,
\dfrac{\rho}{2}\,gh = p_0-p=\dfrac{2S\cos\theta}{r}
giving
\boxed{h =\dfrac{4S\cos\theta}{\rho gr}}