Name: Capillary Rise
Uploaded: 2019-07-19
Duration: 10 min 29 s
Description: Have you ever seen what happens to a celery stalk dipped in red-coloured water? We notice that after a day or two, the leaves turn reddish! Let's learn the r...

Question

A circular ring has inner and outer radii equal to $r$ and $R$ respectively. Mass of the ring is $m$ It gently pulled out vertically from a water surface by a sensitive spring. When the spring is stretched $3.4 c m$ from its equilibrium position the ring is on verge of being pulled out from the water surface. If spring constant is $k$ . Find the surface tension of water.

Accepted Answer

The ring here will act as annular disc having two boundaries (inner and outer) are in contact with water surface.

Let the spring is stretched up to distance $x$ from its equilibrium position, then

Force due to spring in upward direction, $3.4 c m$

Let, $T$ be the surface tension of water, then force due to surface tension on ring will act in downward direction and is given by

$F_{S T} = 2 π R T + 2 π r T = 2 π (R + r) T$

Also, weight of the ring will act in downward direction, i.e. $W = m g$

Net force acting in downward direction is given by

$\Rightarrow F_{d o w n w a r d} = F_{S T} + m g$

$\Rightarrow F_{d o w n w a r d} = 2 π (R + r) T + m g$

Now when spring is stretched, the ring is on verge of being pulled out from the water surface. That means net downward force still balancing the spring force at that moment.

$\Rightarrow F_{d o w n w a r d} = F_{s p r i n g}$

$\Rightarrow 2 π (R + r) T + m g = k x$

$\Rightarrow$ Surface tension, $T = \frac{k x - m g}{2 π (R + r)}$

On substituting, $x = 3.4 cm = 0.034 m$ , we get

$T = \frac{0.034 k - m g}{2 π (R + r)} N m$

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