IIT JEE , AIEEE Forum - Physics , Mechanics

nikhilwalani

when the tube of insufficent length is inserted in the liquid, the liquid does not spurts out. Is this due to change in contact angle or what

krish

In the liquid , the tube of insufficient length has the liquid inside with the same pressure and no external pressure acting on it......but when the part of the tube is kept out of the container , the pressure of the atmosphere forces the liquid to spurt out of it.....the pressure being more than that outside....so the phenomenon occurs....plz reply if u find any problem in this soln.

nikhilwalani

well this phenomenon occurs. i think u r disregarding the fact that presure difference is not the cause of capillary action. It is the force.Pressure is the result of it. if u consider that tube of insufficient length the ext pressure is not causing any net force. the surface tension balances that column of liquid by changing its angle of contact

abhaynain

the ans. in notes is h=3.041cm &wt. of water=0.024 gmf

abhaynain

but i dont understand the ans. plz help me

krishna.gopal

First thing Nikhil whenever you are asking a question please be more specific. At first go i was unable to understand that you are taking about capillary tube and the phenomenon of capilarity.

Consider a capillry tube put in a liquid (say water) whose length is more than the mximum possible capillary rise (Ho) and hence It contains water uto Ho and a concave miniscus whse radius is such that r(miniscus)=2*sigma/(rho*g*Ho)

Now as you dip the capillary into the liquid such that length of capillary above liquid surface is H (<Ho). Then water will try to come out. This will change the radius of miniscus. The new miniscus will have a radius r'(miniscus)=2*sigma/(rho*g*H). This system is still at equilibrium. So in this case there will be flatening of miniscus (Or increase in contact angle as you said) But that will never become convex to let water come out.

I hope this description clears your doubt

nikhilwalani

thanks krishna.gopal. but i still have a doubt regarding this.
it may be that the surface of contact now becomes the outer walls of the capillary(as the liquid is at the top so it can spread on outer walls) so that the tangent changes and still the angle of contact remains same.

Aatish

hey it's too simple.
have a look at it:h=2s/R*density*g
R:RADIUS OF CURVATURE
THEREFORE hR=constant therfore liqud does not flow out
as h decreases R increases and surface becomes flat

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