1. Given data

• Frequency of tuning fork: $f = 340\,\text{Hz}$
• Length of air column initially resonating in fundamental mode: $L_1 = 125\,\text{cm} = 1.25\,\text{m}$
• Speed of sound: $v = 340\,\text{m s}^{-1}$

The tube is closed at one end, so resonance lengths are:

$L = \frac{(2n-1)\lambda}{4}, \quad n=1,2,3,\dots$

2. Find the wavelength

Using

$v = f\lambda$

we get

$\lambda = \frac{v}{f} = \frac{340}{340} = 1\,\text{m} = 100\,\text{cm}$

3. Check the initial resonating length

For a closed pipe, allowed lengths are:

$\frac{\lambda}{4},\; \frac{3\lambda}{4},\; \frac{5\lambda}{4},\dots$

Since $\lambda = 100\,\text{cm}$,

$\frac{5\lambda}{4} = \frac{5\times 100}{4} = 125\,\text{cm}$

So the given $125\,\text{cm}$ corresponds to the resonance length

$L_1 = \frac{5\lambda}{4}$

4. Find the next resonance when water is poured

When water is poured in, the air column length decreases. The next smaller resonance length is:

$L_2 = \frac{3\lambda}{4} = \frac{3\times 100}{4} = 75\,\text{cm}$

5. Height of water required

Initial air column length = $125\,\text{cm}$

New air column length for next resonance = $75\,\text{cm}$

Hence water must rise by

$h = 125 - 75 = 50\,\text{cm}$

6. Final answer

$\boxed{50}$

The minimum height of water required is $50\,\text{cm}$.