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

Hint:

Physics Tip : Although the water rises higher in a thinner tube ($h \times 3$), the cross-sectional area decreases significantly ($A \times 1/9$), resulting in a net decrease in the mass of the water column.

Answer 1

The Correct Option is B

Concept: Physics (Mechanical Properties of Fluids) – Capillary Action and Jurin's Law.

Step 1: Relate height to radius. According to Jurin's Law, the height $h$ to which a liquid rises in a capillary is inversely proportional to the radius $r$ ($h \propto \frac{1}{r}$).

Step 2: State the formula for the mass of the liquid column. The mass $m$ of the water in the capillary is the product of volume and density ($m = \pi r^2 h \rho$). Thus, the relationship is $m \propto r^2 h$.

Step 3: Combine the relationships. Since $h \propto \frac{1}{r}$, we substitute this into the mass proportionality: $m \propto r^2 \cdot \left(\frac{1}{r}\right) \implies m \propto r$.

Step 4: Calculate the new mass for radius $\frac{r}{3}$. Let $m_2$ be the new mass: $\frac{m_2}{m} = \frac{(r/3)}{r} \implies m_2 = \frac{m}{3}$. $$ \therefore \text{The mass of water that will rise is } \frac{m}{3}. $$