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Question
A u-tube is made up of capillaries of bore 1 mm and 2 mm respectively. The tube is held vertically and partially filled with a liquid of surface tension 49 dyne/cm and zero angles of contact. Calculate the density of the liquid, if the difference in the levels of the meniscus is 1.25 cm. take g = 980 cm/s2
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Solution
Let ‘r1’ be the radius of one bore and ‘r2’ be the radius of the second bore of the U-tube. Then, if ‘h1’ and ‘h2’ are the height of water on two sides, then
h1 = `(2T cosθ)/(r_1ρg)` ..........(i)
and h2 = `(2T cosθ)/(r_2ρg)` .................(ii)
Subtracting equation (ii) from equation (i),
h1 - h2 = `(2T cosθ)/(r_1ρg) - (2T cosθ)/(r_2ρg) = (2T cosθ)/(ρg)[1/r_1 - 1/r_2]`
∴ ρ = `(2Tcosθ)/((h_1 - h_2)g)[1/r_1 - 1/r_2]`
Given: T = 49 dyne/cm, θ = 0°,
h1 − h2 = 1.25 cm, g = 980 cm/s2
r1 = `1/2` mm = 0.5 mm = 5 × 10-2 cm
r2 = `2/2` mm = 1 mm = 10-1 cm
∴ ρ = `(2 xx 49 xx cos0^circ)/(1.25 xx 980) xx [1/(5 xx 10^-2) - 1/10^-1]`
= `1/12.5[20 - 10]`
= `10/12.5 = 1/1.25`
Using reciprocal table,
ρ = 0.8 g/cm3
The density of liquid is ρ = 0.8 g/cm3.
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