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(A) The water rises in the annular space between the glass rod and the capillary tube. The total length of the contact line is the sum of the inner ci

Vedclass · Accepted Answer

$\frac{2T}{(r_2-r_1)\rho g}$

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(A) The water rises in the annular space between the glass rod and the capillary tube. The total length of the contact line is the sum of the inner circumference of the tube and the outer circumference of the rod,which is $L = 2\pi r_1 + 2\pi r_2 = 2\pi(r_1 + r_2)$.The upward force due to surface tension is $F = L \cdot T \cos 	heta = 2\pi(r_1 + r_2) T \cos 	heta$.The weight of the water column in the annular space is $W = 	ext{Volume} 	imes ho 	imes g = \pi(r_2^2 - r_1^2) h ho g$.Equating the upward force to the weight of the liquid column: $\pi(r_2^2 - r_1^2) h ho g = 2\pi(r_1 + r_2) T \cos 	heta$.Using the identity $r_2^2 - r_1^2 = (r_2 - r_1)(r_2 + r_1)$,we get $\pi(r_2 - r_1)(r_2 + r_1) h ho g = 2\pi(r_1 + r_2) T \cos 	heta$.For pure water,$	heta = 0^{\circ}$,so $\cos 	heta = 1$. Simplifying for $h$:$h = \frac{2T}{(r_2 - r_1)ho g}$.

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