1. Capillary rise in each limb

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

$h = \frac{2T\cos\theta}{\rho g r} = \frac{4T\cos\theta}{\rho g d}$

Since the tube is open at both ends and contains the same liquid, the level difference between the two limbs is the difference in capillary rises:

$\Delta h = h_1 - h_2 = \frac{4T}{\rho g}\left(\frac{1}{d_1} - \frac{1}{d_2}\right)$

Here,

• $T = 7.3 \times 10^{-2}\,\text{N m}^{-1}$
• $\theta = 0 \Rightarrow \cos\theta = 1$
• $\rho = 1.0 \times 10^3\,\text{kg m}^{-3}$
• $g = 10\,\text{m s}^{-2}$
• $d_1 = 5.0\,\text{mm} = 5.0 \times 10^{-3}\,\text{m}$
• $d_2 = 8.0\,\text{mm} = 8.0 \times 10^{-3}\,\text{m}$

2. Substitute values

First compute:

$\frac{4T}{\rho g} = \frac{4(7.3\times 10^{-2})}{(10^3)(10)} = \frac{0.292}{10^4} = 2.92\times 10^{-5}$

Now,

$\left(\frac{1}{d_1} - \frac{1}{d_2}\right) = \frac{1}{5\times 10^{-3}} - \frac{1}{8\times 10^{-3}}$

$= 200 - 125 = 75\,\text{m}^{-1}$

Therefore,

$\Delta h = 2.92\times 10^{-5} \times 75$

$= 2.19\times 10^{-3}\,\text{m}$

$= 2.19\,\text{mm}$

3. Match with options

$\boxed{\Delta h = 2.19\,\text{mm}}$

So the correct option is B.