The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densi

Q: Why do we take the absolute value of $\cos 135^\circ$?

Because the problem states that the *amount* of depression is $h$. Depression is physically a downward displacement, which corresponds to a negative $h$ in Jurin's Law (since $\cos 135^\circ$ is negative). By equating the *amounts* (magnitudes), we use the absolute value.

Q: Why does water rise while mercury gets depressed?

Water has a contact angle less than $90^\circ$ (adhesive forces dominate), leading to a concave meniscus and capillary rise. Mercury has a contact angle greater than $90^\circ$ (cohesive forces dominate), leading to a convex meniscus and capillary depression.

Q: Does the radius of the tube affect the contact angle?

No, the contact angle is a property of the liquid-solid interface and is independent of the tube's radius.

Q: What happens if the capillary tube is of insufficient length?

The liquid will rise to the top of the tube and adjust its radius of curvature to maintain equilibrium, but it will never overflow.

Q: Why did we use $1.414$ for $\sqrt{2}$?

In JEE problems, using standard approximations like $\sqrt{2} \approx 1.414$ or $\sqrt{3} \approx 1.732$ is essential for quickly evaluating numerical expressions to match the given options.

Question

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 $135^{\circ}$ and $0^{\circ}$, respectively. 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 $r_2$. The ratio $(r_1 / r_2)$, is then close to

Options:

(A) $3/5$

(B) $2/3$

(C) $2/5$

(D) $4/5$

Sigma Percentile · Accepted Answer

### Visualizing the Capillary Action

Imagine two capillary tubes dipped in different liquids. In the first one, water rises by a height $h$. In the second, mercury gets depressed by the exact same amount $h$. This beautiful symmetry in magnitudes is the key to unlocking this problem.

Water, with its strong adhesive forces to glass, forms a concave meniscus and climbs up the tube. Mercury, on the other hand, has strong cohesive forces, forming a convex meniscus and sinking below the surrounding liquid level. Despite these opposite behaviors, the problem states that the magnitude of their vertical displacement is identical.

### The Master Equation: Jurin's Law

To find the relationship between these variables, we use Jurin's Law. The magnitude of capillary rise or depression $h$ is given by:

$$h = \frac{2T |\cos	heta|}{ho g r}$$

Here, $T$ is the surface tension, $	heta$ is the contact angle, $ho$ is the density, and $r$ is the radius of the capillary tube. We take the absolute value of $\cos	heta$ because we are equating the physical magnitudes of the height, ignoring the directional sign that differentiates a rise from a depression.

### Setting Up the Balance

Since the problem states that the rise of water equals the depression of mercury, we can set their equations equal to each other:

$$\frac{2T_W \cos(0^\circ)}{ho_W g r_2} = \frac{2T_{Hg} |\cos(135^\circ)|}{ho_{Hg} g r_1}$$

Notice how the $2$ and $g$ will cancel out beautifully from both sides. Now, let's rearrange the terms to isolate the ratio of the radii, $r_1 / r_2$. We group the surface tension ratio, the density ratio, and the cosine ratio together. This makes our substitution step much cleaner:

$$\frac{r_1}{r_2} = \left( \frac{T_{Hg}}{T_W} ight) 	imes \left( \frac{ho_W}{ho_{Hg}} ight) 	imes \left( \frac{|\cos 135^\circ|}{\cos 0^\circ} ight)$$

### The Final Calculation

Let's plug in the values given in the problem. The surface tension ratio is $7.5$. The density ratio of mercury to water is $13.6$, so water to mercury is $1 / 13.6$. And the absolute value of $\cos 135^\circ$ is $1 / \sqrt{2}$.

$$\frac{r_1}{r_2} = 7.5 	imes \frac{1}{13.6} 	imes \frac{1/\sqrt{2}}{1}$$

Now for the final calculation. We know that $\sqrt{2} \approx 1.414$. Multiplying $13.6$ by $1.414$ gives us about $19.23$.

$$\frac{r_1}{r_2} = \frac{7.5}{19.23} \approx 0.39$$

Dividing $7.5$ by $19.23$ yields approximately $0.39$, which is very close to $0.4$, or $2/5$. And that's our final answer! The elegance of this problem lies in how neatly the ratios combine to give a simple fraction.

Select Answer:

Visualized Solution

Capillary Action Setup

Jurin's Law

Equating Heights

Rearranging for Radius Ratio

Substituting Given Values

Final Calculation

The Way Forward

The Sigma Insight: Surface Tension and Capillary Action

Visualizing the Capillary Action

The Master Equation

Setting Up the Balance

The Final Calculation

Similar Questions

A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $ρ$ and surface tension $T$ . The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)

Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is