Different 'times'?

The times the hour hand and the minute hand of a clock form a right angle with each other between $0600$ and $1200$ are approximately at $0617, 0649, 0722, 0754, 0828, 0900, 0933, 1005, 1038, 1111, 1149.$

Note that this happens twice every hour, except between $0800$ and $1000$ when it happens only three times and not four times as expected.

This is because at $0900$ exactly the hands form a right angle. Thus between $0600$ and $1200$ it happens $(6 × 2) - 1= \boxed{11}$ times.

Let the angular velocity of the hour-hand and minute-hand be $\omega_h$ and $\omega_m\ ^\circ/ \text{min}$ respectively. Then $\omega_h = \frac {30}{60} = 0.5^\circ/\text{min}$ and $\omega_m = \frac {360}{60} = 6^\circ/\text{min}$ . Then the relative angular velocity, where the hour-hand is held stationary, is $\omega = \omega_m - \omega_h = 5.5^\circ/\text{min}$ . Let time $t=0\text{ min}$ at 06:00 then 12:00 is when $t=6\times 60=360\text{ min}$ and the relative angle of the two hands, where hour-hand is fixed, be $\theta(t)=5.5t^\circ$ and $\theta(0)=0^\circ$ . Then the two hands are perpendicular when $\theta(t) =$ $90^\circ, 270^\circ, 450^\circ \cdots =$ $90^\circ(2n-1)$ , where $n$ is a positive integer. Then the number of times when the two hands are perpendicular is given by the largest $n$ that satisfies the following inequality.

$\begin{aligned} 90(2n-1) & \le \theta(360) = 5.5 \times 360 = 1980 \\ n & \le \frac {\frac {1980}{90}+1}2 = 11.5 \\ \implies n & = \boxed{11} \end{aligned}$