To solve this problem, we need to determine the number of natural oscillations (harmonics) of an air column in a pipe closed at one end.$\\$The fundamental frequency and its harmonics for a pipe closed at one end are given by:$\\ f_n = \frac{(2n-1) \cdot v}{4L} \\$where $n$ is the harmonic number (1, 2, 3, ...), $v$ is the velocity of sound, and $L$ is the length of the pipe.$\\$Given:$\\$• Length of the pipe $L = 85$ cm $= 0.85$ m$\\$• Velocity of sound $v = 340$ m/s$\\$• Maximum frequency $f_{max} = 1250$ Hz$\\$We need to find the maximum $n$ such that $f_n < 1250$ Hz.$\\$Substitute the values into the formula:$\\ f_n = \frac{(2n-1) \cdot 340}{4 \cdot 0.85} \\$Simplify the expression:$\\ f_n = \frac{(2n-1) \cdot 340}{3.4} \\ f_n = 100 \cdot (2n-1) \\$Set $f_n < 1250$ to find the maximum $n: \\ 100 \cdot (2n-1) < 1250 \\$Divide both sides by 100:$\\ 2n-1 < 12.5 \\$Add 1 to both sides:$\\ 2n < 13.5 \\$Divide by 2:$\\ n < 6.75 \\$Since $n$ must be a natural number, the largest possible $n$ is $6. \\$Thus, the number of possible natural oscillations is $6. \\$This corresponds to Option 4.