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(A) The density of the mixture $(ho_{	ext{mix}})$ is calculated as: $ho_{	ext{mix}} = \frac{	ext{Total Mass}}{	ext{Total Volume}} = \frac{m +

Vedclass · Accepted Answer

$217.9$

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(A) The density of the mixture $(ho_{	ext{mix}})$ is calculated as: $ho_{	ext{mix}} = \frac{	ext{Total Mass}}{	ext{Total Volume}} = \frac{m + m}{\frac{m}{0.8} + \frac{m}{1}} = \frac{2m}{1.25m + m} = \frac{2}{2.25} = \frac{200}{225} = \frac{8}{9} \, g/cm^3$.The formula for capillary rise is $h = \frac{2T \cos 	heta}{r ho g}$. Assuming the contact angle $	heta = 0^\circ$ $(\cos 0^\circ = 1)$:$T = \frac{h r ho g}{2}$.Given: $h = 5 \, cm$, $r = 1 \, mm = 0.1 \, cm$, $ho = \frac{8}{9} \, g/cm^3$, and $g = 980 \, cm/s^2$.$T = \frac{5 	imes 0.1 	imes (8/9) 	imes 980}{2}$.$T = \frac{0.5 	imes 8 	imes 980}{18} = \frac{4 	imes 980}{18} = \frac{3920}{18} \approx 217.77 \, dyne/cm$.Rounding to the nearest option, the value is $217.9 \, dyne/cm$.

In a vessel, equal masses of alcohol (specific gravity $0.8$) and water are mixed together. A capillary tube of radius $1 \, mm$ is dipped vertically in it. If the mixture rises to a height of $5 \, cm$ in the capillary tube, the surface tension of the mixture is ....... $dyne/cm$.

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