JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the tip of the hand (in units of ms –2 ) is of the order of :
- (a) 10⁻³
- (b) 10⁻¹
- (c) 10⁻²
- (d) 10⁻⁴
Answer and solution
Answer (a) 10⁻³
Solution
Physics sir · 4 steps
The second hand turns at a steady rate, so the tip only has centripetal acceleration, ω² r.
- Angular speed. The second hand makes one full turn in T = 60 s. So ω = 2π/60 = π/30 ≈ 0.105 rad s⁻¹.
- Uniform circular motion. The hand moves continuously at constant speed. There is no tangential acceleration, and the centripetal acceleration has the same size all the time, so that size is the average acceleration asked for.
a = ω² r = (π/30)² × 0.1 m
- Put in numbers. (π/30)² = 9.87/900 ≈ 0.011 s⁻². Multiply by 0.1 m to get a ≈ 1.1 × 10⁻³ m s⁻².
- So. The acceleration of the tip is of the order of 10⁻³ m s⁻², which is option (a).
The idea to keep. For anything turning at a steady rate, find ω = 2π/T first, then a = ω² r. Carry the units through so you do not stop at ω or ω².
Answer from the NTA final answer key. Solution by PrepWiser.