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Step-by-Step Solution
Step 1: Understand the phenomenon of capillary rise
When a narrow capillary tube is immersed in water, the liquid rises inside the tube due to surface tension and the adhesive forces between water and the tube walls. This is known as capillarity. The water column inside the tube has a certain mass, and its weight is balanced by the upward force due to surface tension.
Step 2: Write down the force balance equation
Let $T$ be the surface tension of water, $r$ be the radius of the capillary tube, and $\theta$ be the contact angle (which is usually very small for water-glass interaction). The net upward force due to surface tension in a circular tube is given by:
$F_{\text{surface tension}} = 2 \pi r \, T \, \cos \theta$
The downward force is due to the weight of the water in the tube, which can be written as $mg$, where $m$ is the mass of the water and $g$ is the acceleration due to gravity.
Equating the two forces, we get:
$mg = 2 \pi r \, T \, \cos \theta$
Step 3: Establish the proportional relationship
From the above equation, observe that all factors other than $m$ and $r$ are constant (i.e., $g$, $T$, and $\cos \theta$ do not change). Hence:
$m \propto r
This implies that the mass of water that rises in the tube is directly proportional to the radius of the tube.
Step 4: Compare masses for two different radii
Let $m_1$ be the mass of water in the first tube of radius $r$, and $m_2$ be the mass of water in the second tube of radius $2r$. According to the proportionality:
$\frac{m_2}{m_1} = \frac{2r}{r} = 2
We are given that $m_1 = 5 \text{ g}$. Hence:
$\frac{m_2}{5} = 2 \quad \Rightarrow \quad m_2 = 10 \text{ g}
Step 5: Conclude the answer
The mass of the water that rises in the second capillary tube (of radius $2r$) is therefore 10.0 g.
