A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle $$\simeq 0^\circ$$) with $$30^\circ$$ inclined with vertical. The length of water risen in the capillary is _____ cm. (Take $$g = 9.8$$ m/s$$^2$$)

The capillary rise formula for a tube kept vertical is  $$h = \dfrac{2 T \cos \phi}{\rho g r}$$  where

$$T$$ = surface tension,  $$\phi$$ = contact angle,  $$\rho$$ = density of liquid,  $$g$$ = acceleration due to gravity,  $$r$$ = radius of the tube.

Given data (in SI units):
Radius, $$r = 0.1 \text{ mm} = 0.1 \times 10^{-3}\text{ m} = 1.0 \times 10^{-4}\text{ m}$$
Surface tension, $$T = 70 \text{ dyn/cm}$$. Since $$1 \text{ dyn} = 10^{-5}\text{ N}$$ and $$1 \text{ cm} = 10^{-2}\text{ m}$$,
$$T = 70 \times 10^{-5}\text{ N} \big/ 10^{-2}\text{ m} = 70 \times 10^{-3}\text{ N/m} = 0.07\text{ N/m}$$
Contact angle with glass, $$\phi \simeq 0^\circ \;\Rightarrow\; \cos\phi = 1$$
Density of water, $$\rho = 1000\text{ kg/m}^3$$
Acceleration, $$g = 9.8\text{ m/s}^2$$

Substituting into the formula, the vertical rise is

$$h = \dfrac{2 \times 0.07 \times 1}{1000 \times 9.8 \times 1.0 \times 10^{-4}} = \dfrac{0.14}{0.98} = 0.142857\text{ m} = 14.2857\text{ cm}$$

The tube is inclined at $$30^\circ$$ to the vertical. If $$L$$ is the length of water column along the axis of the tube, its vertical component equals $$h$$, so

$$L \cos 30^\circ = h$$ $$\Rightarrow\; L = \dfrac{h}{\cos 30^\circ} = \dfrac{14.2857}{\dfrac{\sqrt3}{2}} = \dfrac{200}{7\sqrt3}\text{ cm} \approx 16.5\text{ cm}$$

Among the given options, $$\dfrac{82}{5}\text{ cm} = 16.4\text{ cm}$$ is the closest and matches the calculated value.

Hence the length of water risen in the inclined capillary is approximately $$\mathbf{16.4\;cm}$$, so the correct choice is Option A.