A uniform capillary tube of inner radius r is | vedclass

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(B) The pressure balance at the meniscus is given by $\frac{2 \sigma}{R} = \rho g h$,where $R$ is the radius of the meniscus.From the geometry of the

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$A, C$

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(B) The pressure balance at the meniscus is given by $\frac{2 \sigma}{R} = ho g h$,where $R$ is the radius of the meniscus.From the geometry of the meniscus,$R = \frac{r}{\cos 	heta}$,where $r$ is the radius of the capillary and $	heta$ is the angle of contact.Substituting $R$ into the height equation: $h = \frac{2 \sigma \cos 	heta}{ho g r}$.$(A)$ For a given material,$	heta$ is constant,so $h \propto \frac{1}{r}$. Thus,$h$ decreases as $r$ increases. This is true.$(B)$ From the formula,$h \propto \sigma$,so $h$ depends on $\sigma$. This is false.$(C)$ If the lift accelerates upward with acceleration $a$,the effective gravity becomes $g_{	ext{eff}} = g + a$. The new height is $h' = \frac{2 \sigma \cos 	heta}{ho (g+a) r}$. Since $g+a > g$,$h'$ decreases. This is true.$(D)$ From the formula,$h \propto \cos 	heta$,not $	heta$. This is false.Therefore,statements $(A)$ and $(C)$ are true.

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