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Answer:
The wavelength of this sound wave would be longer in water than in the air.
Explanation:
Let [tex]f[/tex] denote the frequency of this sound wave (standard unit: [tex]\rm s^{-1}[/tex].)
If the speed of sound in a particular medium is [tex]v[/tex] (standard unit: [tex]{\rm m \cdot s^{-1}}[/tex],) the wavelength [tex]\lambda[/tex] of this wave in that medium would be:
[tex]\begin{aligned} \lambda = \frac{v}{f} && \genfrac{}{}{0}{}{(\text{standard unit: ${\rm m \cdot s^{-1}}$})}{(\text{standard unit: ${\rm s^{-1}}$})}\end{aligned}[/tex].
Let [tex]v_\text{water}[/tex] denote the speed of sound in water and let [tex]v_\text{air}[/tex] denote the speed of sound in the air at room temperature.
The wavelength of this sound wave in water would be:
[tex]\displaystyle \lambda_{\text{water}} = \frac{v_{\text{water}}}{f}[/tex].
The wavelength of this sound wave in the air at room temperature would be:
[tex]\displaystyle \lambda_{\text{air}} = \frac{v_{\text{air}}}{f}[/tex].
Fact: the speed of sound in water (a liquid) is greater than the speed of sound in air at room temperature. In other words:
[tex]v_{\text{water}} > v_{\text{air}}[/tex].
Given that [tex]f > 0[/tex]:
[tex]\begin{aligned} \frac{v_{\text{water}}}{f} > \frac{v_{\text{air}}}{f}\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned} \lambda_{\text{water}} > \lambda_{\text{air}}\end{aligned}[/tex].