Question

A cylindrical glass capillary of inner radius $r$ is dipped in water of density $\rho$. If surface tension of water is $T$, then

• A. Rise of water in capillary is $\frac{2T}{\rho gr}$
• B. Work done by surface tension is $\frac{4\pi T^2}{\rho g}$
• C. Work done by surface tension is $\frac{2\pi T^2}{\rho g}$
• D. Increases in gravitational potential energy is $\frac{2\pi T^2}{\rho g}$

Answer 1

Answer: (A), (B), (D)

Options (1), (2) and (4) are correct.

Answer 2

1. Capillary Rise:
   The capillary rise is given by

   $$h = \frac{2T}{\rho g r}.$$

   This confirms option (1).

2. Work Done by Surface Tension (Force Work):
   The upward force due to surface tension along the circumference is

   $$F = 2\pi rT.$$

   Thus, the work done by this force when the liquid rises by a height $h$ is

   $$W_{\text{st}} = F \times h = 2\pi rT \times \frac{2T}{\rho g r} = \frac{4\pi T^2}{\rho g}.$$

   This confirms option (2) is correct and shows that option (3) is not.

3. Increase in Gravitational Potential Energy:
   The mass of the risen water is $m=\rho\,\pi r^2h$ and its center of mass rises by $h/2$. So the increase in potential energy is:

   $$\Delta PE = m\,g\,\frac{h}{2} = \rho\,\pi r^2h \cdot g \cdot \frac{h}{2} = \frac{ \rho \pi r^2 g\,h^2}{2}.$$

   Substituting $h=\frac{2T}{\rho gr}$,

   $$\Delta PE = \frac{ \rho \pi r^2 g}{2}\left(\frac{2T}{\rho gr}\right)^2 = \frac{ \rho \pi r^2 g}{2}\left(\frac{4T^2}{\rho^2g^2r^2}\right) = \frac{4\pi T^2}{2\rho g} = \frac{2\pi T^2}{\rho g}.$$

   This confirms option (4).