Mercury has an angle of contact equal to 140^ | vedclass

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Question

(D) The angle of contact between mercury and soda lime glass is $	heta = 140^{\circ}$.Radius of the narrow tube is $r = 1.00 \; mm = 1.00 	imes 10^{

Vedclass · Accepted Answer

$5.34$

Vedclass · Answer

(D) The angle of contact between mercury and soda lime glass is $	heta = 140^{\circ}$.Radius of the narrow tube is $r = 1.00 \; mm = 1.00 	imes 10^{-3} \; m$.Surface tension of mercury is $s = 0.465 \; N m^{-1}$.Density of mercury is $ho = 13.6 	imes 10^{3} \; kg m^{-3}$.Acceleration due to gravity is $g = 9.8 \; m s^{-2}$.The capillary depression $h$ is given by the formula:$h = \frac{2s \cos 	heta}{r ho g}$Substituting the values:$h = \frac{2 	imes 0.465 	imes \cos(140^{\circ})}{1.00 	imes 10^{-3} 	imes 13.6 	imes 10^{3} 	imes 9.8}$Since $\cos(140^{\circ}) \approx -0.766$:$h = \frac{0.93 	imes (-0.766)}{133.28} \approx -0.00534 \; m$$h = -5.34 \; mm$.The negative sign indicates a depression in the level of mercury. Thus,the mercury dips down by $5.34 \; mm$.

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