Check your Watch

Start at 0:00 (midnight); the hands of the clock make a $0^\circ$ angle. The angle between the two increases at a constant rate until they coincide again, a little after 1:05. After eleven of these cycles it is 12:00 (noon).

Thus in one day the angle between the hands turns through $2 \times 11 = 22$ full cycles. During each cycle, there is one moment where the minute hand is 90 degrees ahead and one moment where it is 90 degrees behind the hour hand. Thus we get $\boxed{44}$ instances of perpendicular hands.

The angular speeds of the minute and hour hands are $\omega_m = \dfrac {360^\circ}{60} = 6^\circ /\text{min}$ and $\omega_h = \dfrac {360^\circ}{12\times 60} = 0.5^\circ /\text{min}$ respectively.

Now, let time zero $t = 0 \text{ min}$ at 03:00 H, when the hands are at right angle and the time it takes the two hands are at right angle again be $t'$ , Then, the hour-hand will have moved forward $\omega_h t' = 0.5 t'^\circ$ . Within $t'$ minutes, the minute-hand will have to move forward $0.5 t'^\circ + 180^\circ$ . Therefore, $\omega_m t' = 6 t' = 0.5 t' + 180$ $\implies 5.5 t' = 180$ $\implies t' = 180/5.5 \approx 32.72727 \text{min}$ . Thereafter, the two hands are at right angle every 32.73 minutes, and there are $\lfloor 24 \times 60 \div 32.73 \rfloor = \boxed{44}$ times in a day.

Consider this: Both hands coincide at time t=0, [t]=h. The minute hand will be at the same spot at t=1 and the hour hand will have moved $\frac{1}{12}$ of the circle, which the minute hand has to catch up. In that time the hour hand will have moved another $\frac{1}{144}$ and so on. This forms a geometric series $\displaystyle \sum_{i=0}^{\infty} \frac{1}{12^i}$ which is equal to $\frac{12}{11}$ which is the period T between two coincidences of the two hands. In a day there are 24 hours, so $\frac{24}{T} =22$ such coincidences. In one coincidence there are two times when the hands are perpendicular. Therefore there are $22 × 2 = 44$ times a day when the hands are perpendicular.