Chintu started observing the minute hand of a clock at 9:20 a.m. After some time, he observed that the length swept by the tip of the minute hand of the clock was 22 cm and the area swept by the minute hand during this interval was 115.5 cm². At what time did Chintu note these observations?

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.