Let r and θ be the respective radius and the angle swept by the minute hand of the clock. At 9:20 a.m., the minute hand will be at the 4 mark.

It is given that the length swept by the tip of the minute hand of the clock is 22 cm.

$\therefore l=\frac{\theta }{360°}\times 2\mathrm{\pi r}=22 \mathrm{cm}$

It is also given that the area swept by the minute hand of the clock is 115.5 cm².

$\therefore A=\frac{\theta }{360°}\times {\mathrm{\pi r}}^2=115.5 {\mathrm{cm}}^2$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{\pi r}}^2}{2\mathrm{\pi r}}=\frac{115.5}{22} \\ \Rightarrow \frac{r}{2}=5.25 \\ \Rightarrow r=10.5 \end{gathered}$$

Substituting r = 10.5 cm, we obtain

$$\begin{gathered} \frac{\theta }{360°}\times 2\times \frac{22}{7}\times 10.5=22 \\ \Rightarrow \theta =120° \end{gathered}$$

Time taken by the minute hand to rotate by 360° = 60 minutes

Time taken by the minute hand to rotate by1° $=\frac{60}{360} \min$

Time taken by the minute hand to rotate by120° $=120°\times \frac{60}{360} \min$ = 20 minutes

Thus, Chintu noted the observation 20 minutes after 9:20 a.m.

Therefore, the time would be 9:40 a.m.