A cylindrical capillary tube of 0.2 mm radiu | vedclass

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(B) The capillary rise is given by $h = \frac{2T \cos 	heta}{ho g R}$.For $T_1$ $(	heta = 0^{\circ})$: $h_1 = \frac{2 	imes 0.075 	imes \cos 0^{

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$1, 3, 4$

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(B) The capillary rise is given by $h = \frac{2T \cos 	heta}{ho g R}$.For $T_1$ $(	heta = 0^{\circ})$: $h_1 = \frac{2 	imes 0.075 	imes \cos 0^{\circ}}{1000 	imes 10 	imes 0.2 	imes 10^{-3}} = 0.075 \ m = 7.5 \ cm$.For $T_2$ $(	heta = 60^{\circ})$: $h_2 = \frac{2 	imes 0.075 	imes \cos 60^{\circ}}{1000 	imes 10 	imes 0.2 	imes 10^{-3}} = 0.0375 \ m = 3.75 \ cm$.$(1)$ Since the contact angles are different,the meniscus shape and the weight of the water in the meniscus will be different for both cases. Thus,the correction will be different. Statement $(1)$ is correct.$(2)$ In case $I$,$T_1$ is at the bottom. The water rises to $7.5 \ cm$. If the joint is at $5 \ cm$,the water crosses the joint. However,the pressure balance at the interface in $T_2$ would require a height $h'$ such that $ho g(5 	imes 10^{-2} + h') = \frac{2T \cos 60^{\circ}}{R}$. This leads to $h' = 3.75 - 5 = -1.25 \ cm$. Since $h' $(3)$ In case $I$,if the joint is at $8 \ cm$,the water rises in $T_1$ to its maximum height of $7.5 \ cm$. Since $7.5 \ cm $(4)$ In case $II$,$T_2$ is at the bottom. The water rises in $T_2$ to its maximum height of $3.75 \ cm$. Since $3.75 \ cm < 5 \ cm$,it does not reach the joint. Statement $(4)$ is correct.

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$A$ drop of water breaks into two droplets of equal size. In this process,which of the following statements is correct?

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