Question

Water rises to a height $h$ in a capillary tube lowered vertically into the water to a depth $l$ as shown in the figure. The lower end of the tube is now closed, the tube is then taken out of the water and opened again. The length of the water column remaining in the tube will be:

A. $2h \;if \; l\geq h\; and\; l+h\leq h$
B. $\;h \;if \; l\geq h\; and\; l+h\; if \;l\leq h$
C. $4h \;if \; l\geq h\; and\; l-h\; if \;l\leq h$
D. $\dfrac{h}{2} \;if \; l\geq h\; and\; l+h\; if \;l\leq h$

Answer 1

Hint: Surface tension is the nature of liquids to occupy as minimum surface area as possible. It is defined as the force acting per unit length of the liquid. It can also be known as the energy needed by the liquid to reduce its surface area. This is observed at the liquid- air interface of the liquid, where the force of attraction among the liquid molecules, is greater than that of the liquid-air force.

Complete answer:
We know that surface tension is an adhesive-contact force, which exerts a tangential force parallel to the liquid surface. Due to this tangential force which occurs at the liquid-air surface, there is a curvature formed at the liquid-air surface. This results in the formation of the meniscus, due to the difference in the force of attraction between the liquid-liquid interface and liquid-air interface.
Here, we have a container containing $l$ length of water. If a capillary tube of some length is immersed in it, due to surface tension, the water will rise up the tube to say length $h$.
Then the surface tension on the top of the tube is $2\;T$, when the liquid rises to $h$ height in the tube.
If the lower end of the tube is closed or sealed, and if the tube is removed from the liquid, and the seal is removed, due to the absence of the surface tension the liquid from the tube will begin to fall downwards.
Then the surface tension on the tube is $4\;T$
Now, we have two cases, $l\geq h$ and $l\leq h$
If $l\geq h$
Then the height of the liquid in the tube when the surface tension is $4\;T$ is $2\;h$, as when the surface tension was $2\;T$ the height was $h$.
If $l\leq h$
Then the height of the liquid in the tube when the surface tension is $4\;T$ is $h+l$,since, $h+l\leq 2\;h$ and $l\leq h$.

Thus the correct answer is option A. $2h \;if \; l\geq h\; and\; l+h\leq h$

Note:
Surface tension is the result of the unbalanced forces action on the liquid, namely the force of attraction between the liquid-liquid interface and liquid-air interface. Here, the maximum height till which the liquid can rise is $2\;h$ and the minimum height is $h+l$.