A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be _________ cm. (Take speed of sound in air as 340 m s$$^{-1}$$)

We have a tuning fork producing a frequency of $$f = 250\ \text{Hz}$$, and the speed of sound in air is given as $$v = 340\ \text{m\,s}^{-1}$$.

First, recall the relation between wave speed, frequency, and wavelength:

$$v = f\,\lambda$$

Here $$\lambda$$ is the wavelength of the sound that will resonate. Solving this equation for $$\lambda$$, we get

$$\lambda = \frac{v}{f}$$

Substituting the numerical values,

$$\lambda = \frac{340\ \text{m\,s}^{-1}}{250\ \text{Hz}}$$

$$\lambda = 1.36\ \text{m}$$

Converting metres to centimetres (since $$1\ \text{m} = 100\ \text{cm}$$), we have

$$\lambda = 1.36 \times 100\ \text{cm} = 136\ \text{cm}$$

Now, for a closed organ pipe, the fundamental (first) resonance occurs when the length of the pipe is one-fourth of the wavelength, that is

$$L = \frac{\lambda}{4}$$

Substituting $$\lambda = 136\ \text{cm}$$,

$$L = \frac{136\ \text{cm}}{4}$$

$$L = 34\ \text{cm}$$

So, the answer is $$34\ \text{cm}$$.