The ratio of surface tensions of mercury and water is given to be 7.5, while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. If it is observed that mercury gets depressed by an amount $$h$$ in a capillary tube of radius $$r_1$$, while water rises by the same amount $$h$$ in a capillary tube of radius $$r_2$$, then the ratio $$\frac{r_1}{r_2}$$ is close to

For a liquid contained in a narrow capillary tube, the height of rise or depression is governed by the capillary formula

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

where $$T$$ is the surface tension of the liquid, $$\theta$$ is the contact angle with the tube material, $$\rho$$ is the density of the liquid, $$g$$ is the acceleration due to gravity, and $$r$$ is the inner radius of the capillary tube. This expression gives a positive height for rise (when $$\cos\theta>0$$) and a negative height for depression (when $$\cos\theta<0$$). However, the magnitude of the height is always

$$|h| \;=\; \frac{2T\,|\cos\theta|}{\rho g\,r}.$$

According to the statement of the problem, the magnitude of the depression of mercury equals the magnitude of the rise of water and is denoted by the same symbol $$h$$. Hence we write, for mercury (subscript $$m$$) and water (subscript $$w$$)

$$h \;=\; \frac{2T_m\,|\cos\theta_m|}{\rho_m g\,r_1} \quad\text{and}\quad h \;=\; \frac{2T_w\,|\cos\theta_w|}{\rho_w g\,r_2}.$$

Since both right-hand sides are equal to the same $$h$$, we equate them directly:

$$\frac{2T_m\,|\cos\theta_m|}{\rho_m g\,r_1} \;=\; \frac{2T_w\,|\cos\theta_w|}{\rho_w g\,r_2}.$$

The common factors $$2$$ and $$g$$ cancel from both numerators and denominators, leaving

$$\frac{T_m\,|\cos\theta_m|}{\rho_m\,r_1} \;=\; \frac{T_w\,|\cos\theta_w|}{\rho_w\,r_2}.$$

Rearranging to isolate the ratio of the radii, we obtain

$$\frac{r_1}{r_2} \;=\;\frac{T_m}{T_w}\; \frac{\rho_w}{\rho_m}\; \frac{|\cos\theta_m|}{|\cos\theta_w|}.$$

Now we substitute the numerical data supplied in the question:

Surface-tension ratio: $$\dfrac{T_m}{T_w}=7.5.$$

Density ratio: $$\dfrac{\rho_m}{\rho_w}=13.6 \;\Longrightarrow\; \dfrac{\rho_w}{\rho_m}=\dfrac{1}{13.6}.$$

Contact angles:
For mercury $$\theta_m=135^\circ\,,\quad |\cos\theta_m|=|\cos135^\circ|=\dfrac{\sqrt2}{2}\approx0.707,$$
For water $$\theta_w=0^\circ\,,\quad |\cos\theta_w|=|\cos0^\circ|=1.$$

Substituting these three pieces of information into the radius ratio gives

$$\frac{r_1}{r_2} \;=\; 7.5\; \left(\frac1{13.6}\right)\; \left(\frac{0.707}{1}\right) \;=\; \frac{7.5\times0.707}{13.6}.$$

First, multiplying $$7.5\times0.707$$ yields

$$7.5\times0.707\;\approx\;5.30.$$

Next, dividing by $$13.6$$ produces

$$\frac{5.30}{13.6}\;\approx\;0.39.$$

Converting to an exact fraction close to this decimal, we note that

$$0.39\;\text{is almost equal to}\;\frac{2}{5}=0.40.$$

Among the options offered, $$\dfrac25$$ is clearly the closest value.

Hence, the correct answer is Option D.