Two tubes of same length and diameters 4 mm | vedclass

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(A) The pressure just below the meniscus in a capillary tube of radius $r$ is given by $P = P_0 - \frac{2T}{r}$,where $P_0$ is the atmospheric pressur

Vedclass · Accepted Answer

$3.65$

Vedclass · Answer

(A) The pressure just below the meniscus in a capillary tube of radius $r$ is given by $P = P_0 - \frac{2T}{r}$,where $P_0$ is the atmospheric pressure and $T$ is the surface tension.For the two limbs with radii $R_1 = 2 \ mm = 2 	imes 10^{-3} \ m$ and $R_2 = 4 \ mm = 4 	imes 10^{-3} \ m$,the pressures at the same horizontal level $OO'$ are equal.Let $h_1$ and $h_2$ be the heights of the water columns in the two limbs above the level $OO'$. The pressure at level $OO'$ in the two limbs is:$P_{OO'} = P_0 - \frac{2T}{R_1} + ho g h_1 = P_0 - \frac{2T}{R_2} + ho g h_2$Since the total amount of water is constant,the difference in levels $x = h_1 - h_2$ is determined by the capillary rise difference:$x = h_1 - h_2 = \frac{2T}{ho g} \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Substituting the values:$x = \frac{2 	imes 7.3 	imes 10^{-2}}{1.0 	imes 10^3 	imes 10} \left( \frac{1}{2 	imes 10^{-3}} - \frac{1}{4 	imes 10^{-3}} ight)$$x = \frac{14.6 	imes 10^{-2}}{10^4} \left( \frac{2 - 1}{4 	imes 10^{-3}} ight) = 14.6 	imes 10^{-6} 	imes \frac{1}{4 	imes 10^{-3}} = 3.65 	imes 10^{-3} \ m = 3.65 \ mm$.Thus,the difference between the levels is $3.65 \ mm$.

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