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Q.

A glass rod of diameter $d_{1} = 1 mm$ is inserted symmetrically into a glass capillary with inside diameter $d_{2} = 1.5 mm$ . Then the whole arrangement is vertically oriented and brought in contact with the surface of water. To what height (in 'cm;) will the liquid rise in the capillary. Surface tension of $water = 73 \times 10^{- 3} N / m$ .

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answer is 6.

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Detailed Solution

From $2 \frac{T}{R} = hpg, R = \frac{r_{2} - r_{1}}{cosθ}$
$\frac{2 Tcosθ}{r_{2} - r_{1}} = hρg, r_{1} = \frac{d_{1}}{2}, r_{2} = \frac{d_{2}}{2}$
$h = \frac{4 Tcosθ}{(d_{2} - d_{1}) ρg}$
$here θ ≅ 0 for water - glass interface$
$h = \frac{4 \times 73 \times 10^{- 3} \times \cos 0^{0}}{(1.5 - 1) \times 10^{- 3} \times 10^{3} \times 9.8} = 60 \times 10^{- 3} m$
$h = 6 cm$

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A glass rod of diameter d1=1mm is inserted symmetrically into a glass capillary with inside diameter d2=1.5 mm. Then the whole arrangement is vertically oriented and brought in contact with the surface of water. To what height (in 'cm;) will the liquid rise in the capillary. Surface tension of water=73×10−3N/m.