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A capillary tube sealed at the top has an internal radius of 0.05 cm. the tube is placed vertically in water, openend first . What should be the length of such a tube for the water column to rise in it to a length of 1 cm? Atmospheric pressure, P_(0)=1 atmosphere =1.01 xx 10^(5) Nm^(-2), and surface tension of water =70 xx 10^(-3)Nm^(-1). |
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Answer» Solution :Consider the given capillary tube of length l, RADIUS r and cross-sectional area A. Let h be the height through which the water rises in the capillary tube due to surface tension. Let R be the radius of curvature of water meniscus in the tube which is NEARLY equal to the radius of the tube because the angle of contact of water - glass surface is nearly `0^(@)`. the rising water in hte tube will compress the air inside the tube and will produce extra pressure `DeltaP(say)`. The length of air column in the tube will be =`(1-h)`. if P is the pressure of the compressed air in the tube, then using Boyal's law, we have `P(l-h) =P_(0)l` or `1P= P_(0)l(l-h)` `:. Delta P = P-P_(0)=(P_(0)l)/(l-h) P_(0)=(P_(0)h)/(l-h)` The EXCESS pressure `(=2 S//r)` produced by surface tension should be equal to the sum of pressure created by water column of height h and air compressed in the capillary i.e, `(2S)/(r) = Delta P + h rho G = (P_(0)h)/(l-h) + h rho g` On solving, we get `l = (P_(0)rh)/(2S-h rho g r)+h` `=((1.01xx10^(5))xx(0.05 xx 10^(2)) xx (1 xx 10^(-2)))/(2 xx 970 xx 10^(-3)) - (1 xx 10^(-2)) xx 10^(3) xx 9.8 +0.01` `xx(0.05 xx 10^(-2))` `=5.52m`. |
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