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(A) The vertical height $h$ of the water column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{ho gr}$.Given values are: $T

Vedclass · Accepted Answer

$16.49 \ cm$

Vedclass · Answer

(A) The vertical height $h$ of the water column in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{ho gr}$.Given values are: $T = 70 \ dyn/cm$,$	heta = 0^{\circ}$,$ho = 1 \ g/cm^3$,$g = 980 \ cm/s^2$,and $r = 0.1 \ mm = 0.01 \ cm$.Substituting these values into the formula:$h = \frac{2 	imes 70 	imes \cos 0^{\circ}}{1 	imes 980 	imes 0.01} = \frac{140}{9.8} = \frac{1400}{98} = \frac{100}{7} \ cm$.The tube is inclined at $30^{\circ}$ with the vertical,which means the angle $\alpha$ with the horizontal is $90^{\circ} - 30^{\circ} = 60^{\circ}$.The length $\ell$ of the water column along the tube is related to the vertical height $h$ by the relation $\ell = \frac{h}{\sin \alpha}$.Substituting $\alpha = 60^{\circ}$:$\ell = \frac{100/7}{\sin 60^{\circ}} = \frac{100/7}{\sqrt{3}/2} = \frac{200}{7\sqrt{3}} \approx 16.49 \ cm$.Thus,the correct option is $A$.

$A$ capillary tube of radius $0.1 \ mm$ is partly dipped in water (surface tension $70 \ dyn/cm$ and glass-water contact angle $\simeq 0^{\circ}$) inclined at $30^{\circ}$ with the vertical. The length of water risen in the capillary is . . . . . . $cm$. (Take $g = 980 \ cm/s^2$)

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