Water rises to a height of `2 cm` in a capillary tube. If the tube is tilted `60^@` from the vertical, water will rise in the tube to a length of

Options

A. `4.0cm`
B. `2.0cm`
C. `1.0cm`
D. water will not rise at all

Explanation

To solve the problem of how high water will rise in a capillary tube when it is tilted at an angle of \(60^\circ\) from the vertical, we can follow these steps:

### Step 1: Understand the initial height of water in the vertical tube
The initial height of water in the capillary tube when it is vertical is given as \(h = 2 \, \text{cm}\).

### Step 2: Analyze the situation when the tube is tilted
When the tube is tilted at an angle of \(60^\circ\), the height of the water column does not change; it remains \(2 \, \text{cm}\) vertically. However, the length of the water column in the tube will change due to the tilt.

### Step 3: Relate the vertical height to the length of the water column
When the tube is tilted, the effective length of the water column \(L'\) can be calculated using trigonometry. The vertical height \(h\) can be expressed in terms of the length of the water column \(L'\) as follows:

\[ h = L' \sin(60^\circ) \]

### Step 4: Solve for the new length of the water column
We know that \(h = 2 \, \text{cm}\) and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). We can rearrange the equation to find \(L'\):

\[ L' = \frac{h}{\sin(60^\circ)} = \frac{2 \, \text{cm}}{\frac{\sqrt{3}}{2}} = \frac{2 \times 2}{\sqrt{3}} = \frac{4}{\sqrt{3}} \, \text{cm} \]

### Step 5: Calculate the numerical value
Now, we can calculate the numerical value of \(L'\):

\[ L' = \frac{4}{\sqrt{3}} \approx 2.31 \, \text{cm} \]

### Step 6: Conclusion
Thus, when the capillary tube is tilted at \(60^\circ\), the water will rise to a length of approximately \(2.31 \, \text{cm}\).