Question #60139

In a capillary tube of radius ‘R’, a straight thin metal wire of radius ‘r’ (R > r) is inserted
symmetrically and one end of the combination is dipped vertically in water such that the
lower end of the combination is at same level. The rise of water in the capillary tube is
[T = surface tension of water, ρ = density of water, g = gravitational acceleration]
1

Expert's answer

2016-05-26T10:05:03-0400

Answer on Question #60139-Physics-Mechanics-Relativity

In a capillary tube of radius ‘R’, a straight thin metal wire of radius ‘r’ (R > r) is inserted symmetrically and one end of the combination is dipped vertically in water such that the lower end of the combination is at same level. The rise of water in the capillary tube is [T = surface tension of water, p = density of water, g = gravitational acceleration]

Solution

For a vertical cylindrical tube:


LT=Aρgh.LT = A\rho gh.A=πR2πr2=π(R+r)(Rr)A = \pi R^2 - \pi r^2 = \pi(R + r)(R - r)L=2πR+2πr=2π(R+r)L = 2\pi R + 2\pi r = 2\pi(R + r)


Therefore,


2π(R+r)T=π(R+r)(Rr)ρgh.2\pi(R + r)T = \pi(R + r)(R - r)\rho gh.2T=(Rr)ρgh.2T = (R - r)\rho gh.h=2T(Rr)ρgh = \frac{2T}{(R - r)\rho g}


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