
Water rises to a height $h$ in capillary tube. If the length of capillary tube above the surface of water is made less than $h$ then:
(A) Water does not rise at all
(B) Water rises up to the tip of capillary tube and then starts overflowing like a fountain
(C) water rises up to the top of capillary tube and stays there without overflowing
(D) Water rises up to a point a little below the top and stays there
Answer
233.1k+ views
Hint: The rise in level of water is seen due to the adhesion of water to the walls which causes an upwards force on the liquid at the edges and results in a meniscus which turns upwards. For a given system, the product between inner radius and the rise in height water remains constant. In simple terms, we can say that height is inversely proportional to the inner radius of the capillary tube.
Complete step by step answer : The same force that pulls the water along the inside of the capillary tube also holds it there when it reaches the end. This force doesn’t just pull upward, it pulls the water along the glass. At a certain height, the weight of the water column balances this pull. In that case the pull is upward since there is water below but not above.
The rise in height of water level is given as: \[\]
$h = \dfrac{{2S}}{{\rho gR}}$
Here, $h$ is the rise in the level
$\rho $ is the density of water
$S$ is the surface tension
$g$ is the acceleration due to gravity;
$R$ is the inner radius.
Now, for a given system, we have
$hR = \dfrac{{2S}}{{\rho g}}$
As density, surface tension and acceleration due to gravity will remain constant therefore, the product of radius and height must also remain constant.
Now, as the height decreases, the radius changes and water reach the same level and stops. There is no overflow.
Therefore, option C is the correct option.
Note: The liquid rises due to various forces such as adhesion, cohesion and surface tension. The contact angle is also taken into account. The product of radius and the rise in height remains constant for a given system. When the length of the capillary tube changes, the radius changes in such a manner so that there is no overflow and the product of radius and rise in height remains constant.
Complete step by step answer : The same force that pulls the water along the inside of the capillary tube also holds it there when it reaches the end. This force doesn’t just pull upward, it pulls the water along the glass. At a certain height, the weight of the water column balances this pull. In that case the pull is upward since there is water below but not above.
The rise in height of water level is given as: \[\]
$h = \dfrac{{2S}}{{\rho gR}}$
Here, $h$ is the rise in the level
$\rho $ is the density of water
$S$ is the surface tension
$g$ is the acceleration due to gravity;
$R$ is the inner radius.
Now, for a given system, we have
$hR = \dfrac{{2S}}{{\rho g}}$
As density, surface tension and acceleration due to gravity will remain constant therefore, the product of radius and height must also remain constant.
Now, as the height decreases, the radius changes and water reach the same level and stops. There is no overflow.
Therefore, option C is the correct option.
Note: The liquid rises due to various forces such as adhesion, cohesion and surface tension. The contact angle is also taken into account. The product of radius and the rise in height remains constant for a given system. When the length of the capillary tube changes, the radius changes in such a manner so that there is no overflow and the product of radius and rise in height remains constant.
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