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(C) The height of capillary rise is given by $h = \frac{2T \cos 	heta}{r ho g}$. For a given tube and liquid,$h \propto \frac{1}{\cos 	heta}$ is n

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angle of contact increases

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(C) The height of capillary rise is given by $h = \frac{2T \cos 	heta}{r ho g}$. For a given tube and liquid,$h \propto \frac{1}{\cos 	heta}$ is not the case,rather $h \cdot r = 	ext{constant}$ for a fixed contact angle. When the tube length $L$ is less than the equilibrium height $h$,the water does not overflow. Instead,the radius of curvature $R$ of the meniscus increases to adjust to the new height $L$ such that $h \cdot r = L \cdot R$. Since $R = \frac{r}{\cos 	heta'}$,where $	heta'$ is the new contact angle,we have $h \cdot r = L \cdot \frac{r}{\cos 	heta'}$. This implies $\cos 	heta' = \frac{L}{h}$. Given $h = 5\, cm$ and $L = 3\, cm$,we have $\cos 	heta' = \frac{3}{5} = 0.6$. Since the original contact angle for water-glass is nearly $0^\circ$ $(\cos 0^\circ = 1)$,and $\cos 	heta' = 0.6 Therefore,the angle of contact increases.

In a capillary rise experiment,the water level rises to a height of $5\, cm$. If the same capillary tube is placed in water such that only $3\, cm$ of the tube projects outside the water level,then

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