Capillary Rise

Question

Water rises to a height of 6 cm in a narrow glass capillary tube dipped vertically into a beaker. If a second tube of half this internal radius is used in the same water, to what height will the water rise?

RankGuru · Accepted Answer

Capillary action results from the interplay of surface tension and the adhesion between water and glass, which curves the meniscus and pulls the liquid column upward. The equilibrium rise is given by $h = \frac{2S\cos	heta}{rho g}$, showing that for fixed liquid, contact angle and tube material, the height is inversely proportional to the tube radius, $h \propto 1/r$. Therefore $h_1 r_1 = h_2 r_2$. With $h_1 = 6\,	ext{cm}$ and $r_2 = r_1/2$, we get $h_2 = h_1\frac{r_1}{r_2} = 6	imes2 = 12\,	ext{cm}$. The value 3 cm wrongly assumes height falls with radius. The value 6 cm ignores the change in radius altogether. The value 9 cm has no consistent proportionality. This is precisely the inverse-radius dependence derived in NCERT for capillarity, often called Jurin's law, and it holds because the upward surface-tension force scales with the circumference while the weight of the lifted column scales with the cross-sectional area. The contact angle for water on clean glass is nearly zero, so $\cos	heta \approx 1$ and the meniscus is almost hemispherical. As a plausibility check, a thinner tube produces a more sharply curved meniscus and a stronger lifting effect, so the narrower second tube should indeed raise the water higher, consistent with the doubled height obtained.

Capillary Rise

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