To solve this problem, we first need to understand the concept of harmonics in a pipe open at both ends. For a pipe open at both ends, the fundamental frequency (first harmonic) is given by:

$$ f_1 = \frac{v}{2L} $$

where $$ v $$ is the speed of sound in air, and $$ L $$ is the length of the pipe.

The second overtone in a pipe open at both ends corresponds to the third harmonic, which means:

$$ f_3 = 3 \times f_1 = \frac{3v}{2L} $$

Given that the second overtone is resonating with a tuning fork of frequency 330 Hz, we set:

$$ f_3 = 330 \, \text{Hz} $$

Substitute the given values:

$$ 330 = \frac{3 \times 330}{2L} $$

Solving for $$ L $$:

$$ \frac{3 \times 330}{2L} = 330 $$

$$ \Rightarrow \frac{990}{2L} = 330 $$

$$ \Rightarrow 2L = 990/330 $$

$$ \Rightarrow 2L = 3 $$

$$ \Rightarrow L = 1.5 \, \text{m} $$

Since the total length of the pipe is 1.5 meters, and we need to find the length of the pipe immersed in water, we consider the fact that the pipe acts as a half-open pipe when dipped in water.

In a half-open pipe (open at one end and closed at the other), the fundamental frequency is given by:

$$ f' = \frac{v}{4L'} $$

The third harmonic (second overtone) in a half-open pipe is given by:

$$ f_3' = 5 \times f' = \frac{5v}{4L'} $$

Setting this equal to the given frequency of 330 Hz:

$$ \frac{5 \times 330}{4L'} = 330 $$

$$ \Rightarrow \frac{5 \times 330}{4L'} = 330 $$

$$ \Rightarrow \frac{1650}{4L'} = 330 $$

$$ \Rightarrow 4L' = 1650/330 $$

$$ \Rightarrow 4L' = 5 $$

$$ \Rightarrow L' = \frac{5}{4} = 1.25 \, \text{m} $$

The immersed length is then given by the difference between the total length of the pipe and the length of the vibrating air column above the water:

$$ l = 1.5 \, \text{m} - 1.25 \, \text{m} = 0.25 \, \text{m} $$

So, the length of the pipe immersed in water is 0.25 meters.

Therefore, the correct answer is:

Option B: 0.25 m

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