Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube.
[Take surface tension of water $$T = 7.3 \times 10^{-2}$$ N m$$^{-1}$$, angle of contact = 0, $$g = 10$$ m s$$^{-2}$$ and density of water = $$1.0 \times 10^3$$ kg m$$^{-3}$$]

We have a U-shaped tube whose two limbs are capillaries of different radii, both in contact with water. Because water wets glass (angle of contact $$\theta = 0^{\circ} \Rightarrow \cos\theta = 1$$), it rises inside each limb due to surface tension. At any horizontal level inside the continuous water column, the pressures in the two limbs must be the same. That requirement allows us to relate the capillary rises in the two limbs.

First, recall the standard capillary rise formula for a single capillary tube:

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

where

$$T = \text{surface tension}, \quad \theta = \text{angle of contact}, \quad \rho = \text{density of the liquid}, \quad g = \text{acceleration due to gravity}, \quad r = \text{radius of the tube}.$$

Because both limbs are filled with the same water, we apply this formula separately to each limb, obtaining

$$h_1 = \frac{2T}{\rho g r_1}, \qquad h_2 = \frac{2T}{\rho g r_2}$$

with $$r_1$$ and $$r_2$$ being the radii of the narrower and the wider limbs, respectively.

Inside the continuous water column, the hydrostatic pressures at the same horizontal level must balance. That means the column of water in the narrower limb must rise higher than that in the wider limb by exactly the amount that equalises the pressures. Hence the difference in heights of the two free surfaces is simply

$$\Delta h = h_1 - h_2 = \frac{2T}{\rho g}\left(\frac{1}{r_1} - \frac{1}{r_2}\right).$$

Now we substitute the numerical data. The diameters are 5.0 mm and 8.0 mm, so the radii are

$$r_1 = \frac{5.0\ \text{mm}}{2} = 2.5\ \text{mm} = 2.5 \times 10^{-3}\ \text{m},$$

$$r_2 = \frac{8.0\ \text{mm}}{2} = 4.0\ \text{mm} = 4.0 \times 10^{-3}\ \text{m}.$$

Compute the reciprocal radii:

$$\frac{1}{r_1} = \frac{1}{2.5 \times 10^{-3}} = 0.4 \times 10^{3}\ \text{m}^{-1} = 400\ \text{m}^{-1},$$

$$\frac{1}{r_2} = \frac{1}{4.0 \times 10^{-3}} = 0.25 \times 10^{3}\ \text{m}^{-1} = 250\ \text{m}^{-1}.$$

Thus

$$\frac{1}{r_1} - \frac{1}{r_2} = 400 - 250 = 150\ \text{m}^{-1}.$$

Next, evaluate the coefficient $$\dfrac{2T}{\rho g}$$:

$$2T = 2 \times 7.3 \times 10^{-2}\ \text{N\,m}^{-1} = 1.46 \times 10^{-1}\ \text{N\,m}^{-1},$$

$$\rho g = 1.0 \times 10^{3}\ \text{kg\,m}^{-3} \times 10\ \text{m\,s}^{-2} = 1.0 \times 10^{4}\ \text{N\,m}^{-3}.$$

Therefore

$$\frac{2T}{\rho g} = \frac{1.46 \times 10^{-1}}{1.0 \times 10^{4}} = 1.46 \times 10^{-5}\ \text{m}.$$

Finally, multiply to find the height difference:

$$\Delta h = 1.46 \times 10^{-5}\ \text{m} \times 150 = 2.19 \times 10^{-3}\ \text{m}.$$

Convert metres to millimetres:

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

Hence, the correct answer is Option C.