Three liquids of densities rho_1, rho_2 and r | vedclass

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(B) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2 T \cos 	heta}{r g ho}$.Since $h, T, r,$ and $g

Vedclass · Accepted Answer

$0 \le 	heta_1 < 	heta_2 < 	heta_3 < \frac{\pi}{2}$

Vedclass · Answer

(B) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2 T \cos heta}{r g ho}$.Since $h, T, r,$ and $g$ are constant for all three liquids,we have $\cos heta \propto ho$.Given $ ho_1 > ho_2 > ho_3$,it follows that $\cos heta_1 > \cos heta_2 > \cos heta_3$.Since the cosine function is a decreasing function in the interval $[0, \frac{\pi}{2}]$,a larger value of $\cos heta$ corresponds to a smaller value of $ heta$.Therefore,$ heta_1 Since the liquids rise in the capillary,the angles of contact must be acute $(0 \le heta Thus,the correct relation is $0 \le heta_1 < heta_2 < heta_3 < \frac{\pi}{2}$.

Three liquids of densities $\rho_1, \rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$) having same value of surface tension $T$,rise to the same height in three identical capillaries. Angle of contact $\theta_1, \theta_2$ and $\theta_3$ respectively obey

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