Let 1, 2 and 3 be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If x1, x2 and x3 are their respective angular distances in 1 minute then the factor which remains constant (k) is
- 1x1 = 2x2 = 3x3 = k
- 1 x1 = 2 x2 = 3 x3 = k
- 1 x12 = 2 x22 = 3 x32 = k
- 12 x1 = 22 x2 = 32 x3 = k
Answer
(1) 1x1 = 2x2 = 3x3 = k
Solution
Each hand rotates uniformly, so its angular distance in time t is x = t.
With the same t = 1 minute for all three hands:
1x1 = 2x2 = 3x3 = 1t = k
Answer: (1)