A capillary tube of radius 1 mm is dipped in water of surface tension 0.073 N `m^(-1)`. Find the weight of water that rises in the tube. Take angle of contact `theta = 0^(@)`
Hint : `h rho g = (2 S cos theta)/(2)`
Required wt = `pi r^(2) h rho g`

Explanation

To solve the problem of finding the weight of water that rises in a capillary tube, we will follow these steps:

### Step 1: Understand the given data
- Radius of the capillary tube, \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( S = 0.073 \, \text{N/m} \)
- Angle of contact, \( \theta = 0^\circ \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \) (approximately)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
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