There is a cylindrical capillary tube of sufficient radius R and length L with one end closed and the other end kept open. Initially the tube is placed in open atmosphere, with its open end facing down. It is then slowly dipped vertically down into a liquid, which has density , surface tension T and atmospheric pressure . Also assume that the contact angle between the capillary tube and the liquid system is zero.
Up to what depth ( y ) can this capillary tube be dipped in the liquid such that no liquid rises in the capillary tube?
The answer can be expressed as :
Find the value of .
Assumptions
Assume That
Assume That whole atmosphere is made of an Ideal Mono atomic gas.
Assume That wall of Capillary Tube is Adiabatic (No heat Loss Takes Place).
are Positive integers , and pair
and
are co-prime .
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I Will Post full Solution Later If No-one Post it , But for now the Hints and answer are :
Hints:
1)- Initially Gas is Present in Tube and when we dipped it in Liquid Then gas is Compressed by Liquid So That an Hemispherical Curved Surface ( Meniscus ) is formed at the Top most Liquid Level .
2)- For No liquid Rise The Pressure inside Capilary tube (which is exerted by gas molecules) Should be Balanced by the Lower Side of Meniscus.
3)- Now Walls are adiabatic So P V γ = c o n s t a n t Initial Pressure, Volume is Known and final Volume is can be calculate in terms of 'y' and from Here we calculate Pressure of gas finally
Note : Most Important Step of this Question is that Final Volume is equal to Vol of Tube + Vol. of Hemispherical Meniscus(Just Like an Test Tube) , Since gas is Not only contained in cylindrical Vessel !
4)- Now use P g = P o + L − y 2 T .
5)- Stablished Relation b/w all known quantity and Use ( 1 + x ) n = 1 + n x ( i f x < < 1 ) .
You will get The Answer :
y = 3 2 R + 5 P o R 6 T L .