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Q.

A vertical capillary with inside diameter 0.50 mm is submerged into water so that the length of its part emerging outside the water surface is equal to 25 mm. The radius of curvature of the meniscus is ______x 10^-4m. Surface tension of water is $73 \times 10^{- 3} N / m$ .

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Detailed Solution

$From h = \frac{2 T}{ρrg} r = \frac{0.5}{2} \times 10^{- 3} = 0.25 \times 10^{- 3} m$
$h = \frac{2 \times 73 \times 10^{- 3}}{0.25 \times 10^{- 3} \times 10^{3} \times 9.8} = 59 \times 10^{3} m = 59 mm$
$here h > and radius of meniscup (R) > radius of capillary (r)$
$h^{1} = \frac{2 T}{Rρg}, h^{1} = 25 mm = 25 \times 10^{- 3} m R = \frac{2 \times 73 \times 10^{- 3}}{25 \times 10^{- 3} \times 10^{3} \times 9.8} = 0.6 \times 10^{- 3} m = 0.6 mm$

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