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A clock is a typical example of a circular motion where length of the track is equal to 60 km (Assume 1 minute = 1 km). Now on this track, two runners, i.e., hour hand and minute hand are running with a speed of 5 km/h and 60 km/h respectively. Since the direction of their movement is the same, so the relative speed = 55 km/h.
When will the hour hand and the minute hand of a clock be together between 1 and 2?
Hands have to be together in between 1 O’clock to 2 O’clock.
At 1 O’clock, the distance between hour and minute = 5 km
And the relative speed = 55 km/h
Time = 5/55 h = 1/11 h = 60/11 minutes = 5 5/11 minutes
= 5 300/11 s = 5:27.27s
So, the hour hand and the minute hands will be together at 1:05:27.27 s
Students can learn this value 5 minutes 27.27 s as a standard result.
So, both the hands will meet at
1:05:27.27 − Between 1 O’clock and 2 O’clock
2:10:54.54 − Between 2 O’clock and 3 O’clock
3:16:21.81 − Between 3 O’clock and 4 O’clock
and so on.
How many times in a day will the hands of a clock be together?
Using the data from the above question, hands of a clock meet at a regular interval of 5 minutes 27.27 s. So, the number of times they will the meet = Description: 3165.png = 11.3 times
So, the hands will meet for a total of 11 times.
However, it can also be observed that the hands of a clock meet once every hour except in between 11−1. They meet just once in between 11−1. So, they are meeting for 11 times.

Degree Concept of Clocks

Total angle subtended at the centre of a clock = 360°.
Angle made by an hour hand at the centre per hour = 30° per hour, or, 0.5° per minute
Angle made by the minute hand at the centre per hour = 360° per hour, or, 6° per minute
Solving 20 by this method, 
Angle between an hour hand and the minute hand at
1 O’clock = 30°
Relative speed (in terms of angle) = 5.5°/h
So, time taken = 30°/5.5° = 60/11 minutes
Mr Binod Kumar Roy goes to a market between 4 pm and 5 pm. When he comes back, he finds that the hours hand and the minutes hand have interchanged their positions. For how much time was he out of his house?
Since hands are interchanging their position, minute hand is taking the place of an hour hand and an hour hand is taking the place of minute hand. So sum of the angles formed by h hand and minute hand = 360°
Let us assume that he was out of house for ‘t’ minutes.
So, the angle formed by minutes hand = 6 × t and by hour hand = 0.5 × t
So, 0.5 × t + 6 × t = 360
Or, 6.5 × t = 360
So, t = 55.38 minutes

Important Derivations

→ The number of times hands of a clock are in a straight line (either at 0° or at 180°) in 24 h = 44
→ The number of times hands of a clock are at a right angle (at 90°) in 24 h = 44
→ Both the hands of clock are together after every 65Description: 3174.png minutes
(So if both the hands of the clock are meeting after every 65 minutes or anything less than 65Description: 3183.png minutes, then the clock is running fast and if both the hands of the clock are meeting after every 66 minutes or anything more than 65Description: 3187.png minutes, then clock is running slow)

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