A capillary tube made of glass with a radius | vedclass

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(C) Let $r$ be the radius of the capillary tube and $R$ be the radius of the meniscus.From the geometry of the meniscus,the angle of contact $	heta$

Vedclass · Accepted Answer

$0.087$

Vedclass · Answer

(C) Let $r$ be the radius of the capillary tube and $R$ be the radius of the meniscus.From the geometry of the meniscus,the angle of contact $	heta$ is related to the angle between the tangents. Since the tangents make an angle of $60^{\circ}$ with each other,the angle between the tangent and the vertical wall is $30^{\circ}$. Thus,the contact angle $	heta = 30^{\circ}$.From the geometry,$\cos 	heta = \frac{r}{R}$,so $R = \frac{r}{\cos 30^{\circ}} = \frac{r}{\sqrt{3}/2} = \frac{2r}{\sqrt{3}}$.Given $r = 0.15 	imes 10^{-3} m$,we have $R = \frac{2 	imes 0.15 	imes 10^{-3}}{\sqrt{3}} = \frac{0.3 	imes 10^{-3}}{\sqrt{3}} m$.The height $h$ of the liquid column is given by $h = \frac{2T \cos 	heta}{ho g r}$.Substituting the values: $h = \frac{2 	imes 0.05 	imes \cos 30^{\circ}}{667 	imes 10 	imes 0.15 	imes 10^{-3}}$.Alternatively,using $h = \frac{2T}{ho g R}$:$h = \frac{2 	imes 0.05}{667 	imes 10 	imes (\frac{0.3 	imes 10^{-3}}{\sqrt{3}})} = \frac{0.1 	imes \sqrt{3}}{6670 	imes 0.3 	imes 10^{-3}} = \frac{0.1732}{2.001} \approx 0.0865\, m$.Rounding to the nearest value,$h \approx 0.087\, m$.

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