Water rises in a capillary upto a height of 4 cm. If it is tilted to `30^(@)` from the vertical, then the length of water column in it will be-

Options

A. `(8)/(sqrt3)cm`
B. `8sqrt3cm`
C. 4 cm
D. 2 cm

Answer

To solve the problem of how the height of water in a capillary tube changes when the tube is tilted at an angle of 30 degrees, we can follow these steps:

### Step 1: Understand the initial conditions
The water rises to a height \( h_1 = 4 \, \text{cm} \) when the capillary tube is vertical. This height is determined by the balance of forces acting on the water column due to surface tension and gravity.

### Step 2: Identify the formula for capillary rise
The formula for the height of liquid rise in a capillary tube is given by:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
where:
- \( T \) = surface tension of the liquid,
- \( \theta \) = angle of contact (for vertical, \( \theta = 0 \) and \( \cos(0) = 1 \)),
- \( \rho \) = density of the liquid,
- \( g \) = acceleration due to gravity,
- \( r \) = radius of the capillary tube.

### Step 3: Analyze the situation when the capillary is tilted
When the capillary tube is tilted at an angle \( \theta = 30^\circ \), the effective gravitational force acting on the liquid changes. The vertical component of gravity acting on the liquid column is now \( g \cos(30^\circ) \).

### Step 4: Write the new height formula for the tilted capillary
The new height \( h_2 \) when the capillary is tilted can be expressed as:
\[ h_2 = \frac{2T \cos(30^\circ)}{\rho g r} \]

### Step 5: Relate the heights \( h_1 \) and \( h_2 \)
From the original height \( h_1 \) when vertical, we have:
\[ h_1 = \frac{2T}{\rho g r} \]
Now, we can find the ratio of the two heights:
\[ \frac{h_1}{h_2} = \frac{g \cos(30^\circ)}{g} \]
This simplifies to:
\[ \frac{h_1}{h_2} = \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

### Step 6: Solve for \( h_2 \)
Rearranging gives us:
\[ h_2 = \frac{2h_1}{\sqrt{3}} \]
Substituting \( h_1 = 4 \, \text{cm} \):
\[ h_2 = \frac{2 \times 4}{\sqrt{3}} = \frac{8}{\sqrt{3}} \, \text{cm} \]

### Final Answer
Thus, the length of the water column in the tilted capillary tube is:
\[ h_2 = \frac{8}{\sqrt{3}} \, \text{cm} \]

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