Let $\delta$ be the difference in minutes between hour and minute hand at $1:05$. So, the meeting times are
$1:05 + \delta$
$2:10 +2 \delta$
$3:15 + 3\delta$
$4:20 + 4\delta$
$5:25 + 5\delta$
$6:30 + 6\delta$
$7:35 +7 \delta$
$8:40 +8 \delta$
$9:45 + 9\delta$
$10:50 +10 \delta$
$11:55 +11 \delta$
We have $11\delta =5$
$\implies \delta = \dfrac{5}{11}$
minutes as the meeting time is $0:00$. So, we have $N=10.$
Alternatively,
Speed of minute hand = $360$ degrees per $60$ minutes = $6$ degrees per minute.
Speed of hour hand = $360$ degree per $12 * 60$ minutes = $0.5$ degree per minute.
For first meeting, distance traveled by minute hand = $360$ $+$ distance traveled by hour hand
Let, $x$ be the minutes after which the hands intersect.
So, $6x = 360 + 0.5 x $
$\implies 5.5x = 360 $
$\implies x = \dfrac{720}{11}$
In $12$ hours we have $12 \times 60$ minutes.
So, no. of intersections $= \dfrac {12 \times 60}{720/11} = 11$.
But the last intersection is at midnight and must be excluded as per given question. So, $N=10$.
Correct Answer: $E$