**Explanation:**

When a cylindrical tube is open at both ends, it can produce a fundamental note (also known as the first harmonic) with a frequency given by the formula:

f = v / (2L)

Where:
- f is the frequency of the fundamental note
- v is the speed of sound in air
- L is the length of the tube

In this case, the frequency of the fundamental note is given as 256 Hz. Let's assume the speed of sound in air is approximately 340 m/s. We can rearrange the formula to solve for the length of the tube:

L = v / (2f)

L = 340 m/s / (2 * 256 Hz) ≈ 0.664 m

This means that the length of the tube when it is open at both ends is approximately 0.664 meters.

When the tube is partially submerged in water so that half of it is in water, the effective length of the air column changes. The portion of the tube above the water level remains the same, while the portion below the water level is now considered closed. The effective length of the tube can be calculated as:

Leffective = L - ΔL

Where:
- Leffective is the effective length of the tube
- L is the original length of the tube
- ΔL is the change in length due to the water level

Since half of the tube is submerged in water, the change in length can be calculated as:

ΔL = L / 2

ΔL = 0.664 m / 2 = 0.332 m

Substituting the values into the equation, we can find the effective length of the tube:

Leffective = 0.664 m - 0.332 m = 0.332 m

Therefore, the effective length of the air column in the tube is 0.332 meters.

To determine the frequency at which the air column will be in resonance with the A tuning fork, we can use the formula for the frequency of the fundamental note:

f = v / (2Leffective)

Substituting the values, we get:

f = 340 m/s / (2 * 0.332 m) ≈ 512 Hz

Therefore, the air column in the tube will be in resonance with an A tuning fork of approximately 512 Hz when half of the tube is submerged in water.