Question

A liquid of density $\rho $ and surface tension S rises in a capillary tube up to its equilibrium position. Let $\theta $ is the angle of contact between the liquid and capacity tube. Choose the correct option(s) regarding capillary rise.

A) The magnitude of work done by the surface tension force during complete rise is $\dfrac{{4\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$.
B) The work done by gravity force during complete rise is $ - \dfrac{{2\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$.
C) The heat liberated during complete rise is $\dfrac{{2\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$.
D) The heat liberated during complete rise is $\dfrac{{6\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$.

Answer 1

Hint: Here due to the complete rise of the fluid in the capillary tube there is substantial amount of heat released. To calculate the amount of heat released we need to calculate the work done by two forces one is tensional force which is working against the gravity and other one is work done by the gravity on the liquid.

Complete step by step solution:
Step 1: Consider the height that liquid rises in the capillary tube and let it be h.
Now, we know the formula for surface tension as:
$S = \dfrac{{\rho grh}}{{2\cos \theta }}$ ;
Where:
S = Surface tension.
g = Acceleration due to gravity.
r = Radius of the tube.
h = height of the tube.
$\rho $= Density.
Write the above equation in terms of h.
$ \Rightarrow \dfrac{{2S\cos \theta }}{{\rho gr}} = h$;
Now, the force which is acting upward due to tensional force is: ${F_{up}} = \left( {S\cos \theta } \right)\left( {2\pi r} \right)$.
Now, we know that the work done by a body is equal to the force applied times the distance.
$W = F.d$ ;
Similarly, the work done by the surface tension is given by:
${W_{st}} = {F_u}.h$. …(Here, d = h)
Put the value of force and height in the above equation.
${W_{st}} = \left( {S\cos \theta } \right)\left( {2\pi r} \right) \times \dfrac{{2S\cos \theta }}{{\rho gr}}$;
$ \Rightarrow {W_{st}} = \dfrac{{4\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$;

Step 2: Calculate the risen mass of liquid in the capillary tube.
$m = \rho \left( {\pi {r^2}h} \right)$;
The center of mass of the risen liquid is $\dfrac{h}{2}$
So, the work done by the gravitational force is:
${W_g} = - mg\dfrac{h}{2}$;
Put the value of $m = \rho \left( {\pi {r^2}h} \right)$in the above equation and solve:
${W_g} = - \rho \left( {\pi {r^2}gh} \right) \times \dfrac{h}{2}$;
$ \Rightarrow {W_g} = \left( { - \rho \pi {r^2}g \times \dfrac{{2S\cos \theta }}{{\rho gr}}} \right) \times \dfrac{{2S\cos \theta }}{{2\rho gr}}$;
Cancel out the common factors:
$ \Rightarrow {W_g} = \left( { - \pi \times 2S\cos \theta } \right) \times \dfrac{{S\cos \theta }}{{\rho g}}$;
$ \Rightarrow {W_g} = - \dfrac{{2\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$;

Step 3:
Now, the heat liberated will be equal to the work done by the tension force in addition with the work done by gravity.
$H = {W_{st}} + {W_g}$; ….(H = heat liberated)
Put in the given values of ${W_{st}}$ and ${W_g}$ in the above equation and solve.
$ \Rightarrow H = \dfrac{{4\pi {S^2}{{\cos }^2}\theta }}{{\rho g}} - \dfrac{{2\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$;
$ \Rightarrow H = \dfrac{{2\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$;

Final Answer: Option” C” is correct. The heat liberated during complete rise is $\dfrac{{2\pi {S^2}{{\cos }^2}\theta }}{{\rho g}}$.

Note: Here the process is very lengthy so, go step by step. First solve for the work done by the surface tension then similarly solve for the work done by the gravity and after that calculate the heat liberated by adding the work done by gravity and tensional force.