Question

Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. The mass of water that will rise in a capillary tube of radius $\frac{r}{3}$ will be

• A. $3m$
• B. $\frac{m}{3}$
• C. $m$
• D. $\frac{2m}{3}$

Hint:

This principle can be memorized as a rule of thumb: for any given liquid, height is inversely proportional to radius ($h \propto 1/r$), but mass is directly proportional to radius ($m \propto r$). If the radius decreases by a factor of 3, the mass must drop by a factor of 3 automatically!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question explores capillary action in a narrow tube. We are given the mass $m$ of water that rises in a capillary tube of radius $r$. We need to find how the mass changes when the radius of the tube is reduced to a third of its original value.

Step 2: Key Formula or Approach:
1. According to Jurin's Law, the height $h$ to which a liquid rises in a capillary tube is inversely proportional to its radius $r$: $$h = \frac{2S\cos\theta}{\rho g r} \implies h \propto \frac{1}{r}$$ 2. The mass $m$ of the liquid column is equal to its density multiplied by its volume: $$m = \rho \cdot V = \rho \cdot (\pi r^2 h)$$ 3. Combine these relations to find a direct proportionality rule for mass in terms of the tube's radius.

Step 3: Detailed Explanation:
Let's substitute the dependency $h \propto \frac{1}{r}$ directly into the mass expression: $$m \propto r^2 \cdot \left(\frac{1}{r}\right) \implies m \propto r$$ This shows that the mass of the liquid lifted in a capillary tube is directly proportional to the radius of the tube. Let $m_1 = m$ be the initial mass in a tube of radius $r_1 = r$. Let $m_2$ be the new mass in a tube of radius $r_2 = \frac{r}{3}$. Setting up our direct proportionality ratio: $$\frac{m_2}{m_1} = \frac{r_2}{r_1}$$ $$\frac{m_2}{m} = \frac{\left(\frac{r}{3}\right)}{r} = \frac{1}{3}$$ Isolating the new mass $m_2$: $$m_2 = \frac{m}{3}$$

Step 4: Final Answer:
The mass of water that will rise in the new capillary tube is $\frac{m}{3}$, which corresponds to option (B).