Step 1 : Understanding the Question:
This problem asks for the specific moment when the minute hand and the hour hand form a $90^\circ$ angle (a right angle). Between any two hours, the hands typically form a right angle twice. We are looking for the "first time" this happens after 4:00. To solve this, we must consider the relative speed of the clock hands. The minute hand moves faster than the hour hand, and we need to calculate how much time it takes for the minute hand to reach a position where the angular gap between them is exactly $90^\circ$.
Step 2 : Key Formulas and approach:
The hour hand moves at $0.5^\circ$ per minute.
The minute hand moves at $6^\circ$ per minute.
Relative speed = $6 - 0.5 = 5.5^\circ$ or $\frac{11}{2}^\circ$ per minute.
Angle formula: $\theta = |30H - \frac{11}{2}M|$, where $H$ is the hour and $M$ is the minutes.
The approach is to set $H=4$ and $\theta=90$, and solve for $M$.
Step 3 : Detailed Explanation:
Step 1: At 4:00, the angle of the hour hand is $4 \times 30^\circ = 120^\circ$ from the 12 o'clock position.
Step 2: For the first right angle, the minute hand must be $90^\circ$ behind the hour hand.
Step 3: The angular distance to be covered to reach this state is Initial Angle - Desired Angle = $120^\circ - 90^\circ = 30^\circ$.
Step 4: Use the relative speed formula: $\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}$.
$M = \frac{30}{11/2} = \frac{60}{11} \approx 5.45$ minutes.
Step 5: Thus, the time is approximately 4:05 and 27 seconds.
Looking at the options, 4:05 is the closest approximation for the first occurrence.
Step 4 : Final Answer:
The hands are at right angles for the first time at approximately 4:05.