A capillary tube stands with its lower end dipped into liquid for which the angle of contact is $90^{\circ}$. The liquid will

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r} \\ & \mathrm{~h}=\frac{2 \mathrm{~T} \cos 90}{\rho g r} \\ & \mathrm{~h}=0^{\quad \ldots\left(\cos 90^{\circ}=0\right)} \end{aligned}$
Hence, the liquid will neither rise nor fall.

A capillary tube stands with its lower end dipped into liquid for which the angle of contact is $90^{\circ}$…

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