The number of possible natural oscillations of air column in a pipe closed at one end of length $85 \mathrm{~cm}$ whose frequencies lie below $1250 \mathrm{~Hz}$ are (velocity of sound $=340 \mathrm{~ms}^{-1}$)

For a pipe closed at one end, the frequencies of natural oscillations are given by the formula $f_n = \dfrac{(2n-1)v}{4L}$, where $n = 1, 2, 3, \dots$ is the harmonic number, $v$ is the velocity of sound, and $L$ is the length of the pipe.

Given $v = 340 \text{ m/s}$ and $L = 85 \text{ cm} = 0.85 \text{ m}$.

Substituting the values into the formula:

$f_n = \dfrac{(2n-1) \times 340}{4 \times 0.85} = \dfrac{(2n-1) \times 340}{3.4} = (2n-1) \times 100 \text{ Hz}$.

We require $f_n < 1250 \text{ Hz}$, so:

$(2n-1) \times 100 < 1250$

$2n-1 < 12.5$

$2n < 13.5$

$n < 6.75$.

Since $n$ must be a natural number, the possible values for $n$ are $1, 2, 3, 4, 5, 6$.

The number of possible natural oscillations is 6.

Answer: 6