A capillary tube with very thin walls is attached to the beam of a balance which is then equalized. The lower end of the capillry is brought in contact with the surface of water after which an additional load of `P = 0.135 gm` force is needed to regain equilibrium. If the radius of the capillary is `(lambda)/(10)cm` then find `lambda` The surface tension of water is `70 "dyne"//"cm"`. `(g = 9.8 m//s^(2))`

Explanation

To solve the problem, we need to analyze the forces acting on the water column in the capillary tube and relate them to the surface tension of the water. Here’s a step-by-step solution:

### Step 1: Understand the forces acting on the water column
When the capillary tube is immersed in water, the surface tension creates a force that pulls the water up the tube. This force is balanced by the weight of the additional load \( P \) that is needed to bring the system back to equilibrium.

### Step 2: Write the equation for equilibrium
The upward force due to surface tension can be expressed as:
\[ F_{\text{surface tension}} = 2 \pi r T \]
where \( r \) is the radius of the capillary tube and \( T \) is the surface tension of water.

The weight of the additional load \( P \) can be expressed as:
\[ P = m g \]
where \( m = 0.135 \, \text{g} = 0.135 \times 10^{-3} \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).

### Step 3: Set up the equation
At equilibrium, the force due to surface tension equals the weight of the additional load:
\[ 2 \pi r T = P \]

### Step 4: Substitute the values
We know:
- \( T = 70 \, \text{dyne/cm} = 70 \times 10^{-5} \, \text{N/m} \) (since \( 1 \, \text{dyne} = 10^{-5} \, \text{N} \))
- \( P = 0.135 \times 10^{-3} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.001323 \, \text{N} \)

Now substituting these values into the equation:
\[ 2 \pi r (70 \times 10^{-5}) = 0.001323 \]

### Step 5: Solve for \( r \)
Rearranging the equation to find \( r \):
\[ r = \frac{0.001323}{2 \pi (70 \times 10^{-5})} \]

Calculating \( r \):
\[ r = \frac{0.001323}{2 \times 3.14 \times 70 \times 10^{-5}} \approx \frac{0.001323}{0.0000314} \approx 42.14 \, \text{cm} \]

### Step 6: Convert radius to cm
Since the radius is given in terms of \( \lambda/10 \) cm, we can set:
\[ r = \frac{\lambda}{10} \, \text{cm} \]
Thus,
\[ 42.14 = \frac{\lambda}{10} \]
Multiplying both sides by 10 gives:
\[ \lambda = 421.4 \, \text{cm} \]

### Final Answer
The value of \( \lambda \) is approximately \( 421.4 \).