The fundamental frequency of an organ pipe is directly proportional to the speed of sound in the gas ($f \propto v$). The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$, which means $v \propto \sqrt{T}$, where $T$ is the absolute temperature. Therefore, $f \propto \sqrt{T}$. Given: Initial temperature, $T_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$ Initial frequency, $f_1 = 400\text{ Hz}$ Final temperature, $T_2 = 90^\circ\text{C} = 90 + 273 = 363\text{ K}$ Final frequency, $f_2 = ?$ Using the relation $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$: $f_2 = 400 \times \sqrt{\frac{363}{300}}$ $f_2 = 400 \times \sqrt{1.21}$ $f_2 = 400 \times 1.1 = 440\text{ Hz}$. The resonance frequency at $90^\circ\text{C}$ will be $440\text{ Hz}$.