Circular MotionHard
Question
The second’s hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be :
Options
A.2p & 0 mm/s
B.2√2 π & 4.44 mm/s
C.2√2 π & 2π mm/s
D.2π & 2√2π mm/s
Solution
ωsecond = 
= 
rad/sec.
v = ω.r =
× 0.06 m/s = 2π mm/s Ans.
= √2v = 2√2 π mm/s Ans.
= rad/sec.v = ω.r =
× 0.06 m/s = 2π mm/s Ans. = √2v = 2√2 π mm/s Ans.Create a free account to view solution
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