Question

A glass capillary of radius 0.35 mm is inclined at 60° with the vertical in water. The height of the water column in the capillary is (surface tension of water = 7 x 10$^{-2}$N/m, acceleration due to gravity, g = 10m/s², cos 0° = 1, cos 60° = 0.5)

Answer 1

8 cm

Answer 2

1. Vertical capillary rise:
   For a vertical capillary tube the rise is given by

   $$h_{\text{vert}}=\frac{2\sigma\cos\theta}{\rho g r}$$

   For water on glass, $\theta=0^\circ$ so $\cos\theta=1$. Plug in the values:

   $$h_{\text{vert}}=\frac{2\times 7\times10^{-2}}{(10^3)(10)(0.35\times10^{-3})}$$

   Calculate the denominator:

   $$(10^3)(10)(0.35\times10^{-3})= 1000\times10\times0.00035= 3.5$$

   Thus,

   $$h_{\text{vert}}=\frac{0.14}{3.5}=0.04\ \text{m}=4\ \text{cm}.$$

2. Adjustment for inclination:
   The tube is inclined at $60^\circ$ with the vertical. If $L$ is the length of the water column along the tube, its vertical component is

   $$h_{\text{vert}}= L\cos(60^\circ).$$

   So,

   $$L=\frac{h_{\text{vert}}}{\cos60^\circ}=\frac{4\ \text{cm}}{0.5}= 8\ \text{cm}.$$

Thus, the water column in the inclined capillary is 8 cm long.

---

Core Explanation:

• Compute vertical rise using $h = \frac{2\sigma}{\rho gr}$.
• Since the tube is inclined, the actual length along the tube is $L = \frac{h}{\cos60^\circ}$.