Physics - Fluid Question with Solution | TestHub

The height of water in a capillary tube is given by the formula:

$h=\frac{2T\cos \theta }{\rho r{g}_{eff}}$

where $g_{eff}$ is the effective acceleration due to gravity.

For $h$ to increase, the value of $g_{eff}$ must decrease.

In a lift going upward with acceleration $a$, $g_{eff}=g+a>g$, so $h$ decreases.

In an accelerating train with acceleration $a$, $g_{eff}=\sqrt{g^2+a^2}>g$, so $h$ decreases.

In a satellite rotating close to earth, $g_{eff}=0$. The water will rise to the full length of the tube, and the concept of a finite equilibrium height $h$ does not apply.

In a lift going down with acceleration $a<g$, $g_{eff}=g-a$. Since $g_{eff}<g$, the height $h$ will increase to a new finite equilibrium value.

Therefore, the value of $h$ increases when the setup is kept in a lift going down with acceleration less than $g$.

Answer: a lift going down with acceleration less than '$g$'