Calculate Clock Hands Perpendicular Time Post 12:00

The question asks for the first time after 12:00:00 when the hour hand and minute hand of a clock are perpendicular (form a $90^\circ$ angle).

Clock Hand Speeds and Relative Motion

• The minute hand moves $360^\circ$ in 60 minutes, which is $6^\circ$ per minute.
• The hour hand moves $360^\circ$ in 12 hours (720 minutes), which is $0.5^\circ$ per minute.
• The relative speed of the minute hand with respect to the hour hand is $6^\circ/\text{min} - 0.5^\circ/\text{min} = 5.5^\circ/\text{min}$.

Finding the Perpendicular Time

At 12:00:00, both hands are aligned at the 12. We need to find the time $t$ (in minutes) after 12:00:00 when the angle between them is $90^\circ$.

The relative angle covered by the minute hand with respect to the hour hand is given by:

For the hands to be perpendicular, the relative angle must be $90^\circ$.

Solving for $t$:

Converting Time to HH:MM:SS

Convert the fraction of minutes into minutes and seconds:

This means 16 full minutes past 12:00.

Now, convert the fractional part ($\frac{4}{11}$ minutes) into seconds:

So, the exact time is 12 hours, 16 minutes, and $21 \frac{9}{11}$ seconds ($12:16:21 \frac{9}{11}$).

Conclusion

Comparing this with the options, the first time after 12:00:00 when the hands are perpendicular is approximately 12:16:21.

Therefore, option A is the correct answer.