1. Capillary rise in a vertical tube

For a capillary tube dipped in a liquid, the vertical height of rise is

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

Here:

• $T = 70\,\text{dyn/cm}$
• $\theta \approx 0^\circ \Rightarrow \cos\theta = 1$
• $r = 0.1\,\text{mm}$
• $\rho = 1\,\text{g/cm}^3$ for water

Since $T$ is in cgs units, it is convenient to use cgs throughout:

• $g = 9.8\,\text{m/s}^2 = 980\,\text{cm/s}^2$
• $r = 0.1\,\text{mm} = 0.01\,\text{cm}$

So,

$h=\frac{2\times 70}{1\times 980\times 0.01}$

$h=\frac{140}{9.8}=\frac{700}{49}=\frac{100}{7}\,\text{cm}$

Thus the vertical rise is

$h=\frac{100}{7}\,\text{cm}$

2. Tube is inclined at $30^\circ$ with the vertical

If the tube makes an angle $30^\circ$ with the vertical, then the liquid rises along the tube by a length $L$ such that its vertical component equals $h$:

$L\cos 30^\circ = h$

Therefore,

$L=\frac{h}{\cos 30^\circ}$

Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$,

$L=\frac{100/7}{\sqrt{3}/2}=\frac{200}{7\sqrt{3}}\,\text{cm}$

Numerically,

$L\approx \frac{200}{7\times 1.732} \approx 16.5\,\text{cm}$

Among the options, this matches

$\frac{82}{5}=16.4\,\text{cm}$

So the intended answer is Option A.

3. Answer check

Derived answer: A

Stored correct answer: A

They agree.