24ok..let me make things easy for everyone...point out mistake in soln for 2
11 Answers
24
11@ 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-hÏ*g-H/2.Ï/2.g=P0-H/2. Ï/2.g
=>h=-2T/r
plz cheak the answer.if wrong then verify my process also.
49indeed it will be 4T/r,coz, since the topmost liquid is in the upper part and so the only upward force given is by the topmost liquid's surface tension which can balance only a column of liquid whch it supported originally when contact was not made with the denser liquid
thus the answer,,,,,
49urgh..........u forgot the surface tension.........in your FBD...........
24excess pressure comes from surface tension......
i wrote that part naa....
66The 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}}