Explanation:
D Given,
\(\mathrm{r}_{\mathrm{P}}=4 \mathrm{~m}, \mathrm{r}_{\mathrm{Q}}=9 \mathrm{~m}\)

Now, intensity of propagation of sound of power $P$ is
$\mathrm{I}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}$
Since, $\mathrm{P}$ is constant.
So,
$\mathrm{I} \propto \frac{1}{\mathrm{r}^{2}}$
Then,
$\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}=\left(\frac{\mathrm{r}_{\mathrm{Q}}}{\mathrm{r}_{\mathrm{p}}}\right)^{2}=\left(\frac{9}{4}\right)^{2}$
We know that,
$\mathrm{I} \propto \mathrm{A}^{2}$
Where, $\mathrm{A}$ is amplitude.
Hence, $\frac{A_{P}}{A_{Q}}=\sqrt{\frac{I_{P}}{I_{Q}}}$
$\frac{\mathrm{A}_{\mathrm{P}}}{\mathrm{A}_{\mathrm{Q}}}=\frac{9}{4}$