JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Kinematics

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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) $1 0^{- 3}$ ✓ Correct
(B) $1 0^{- 4}$
(C) $1 0^{- 2}$
(D) $1 0^{- 1}$

Answer: (A)

Solution

The tip moves in a circle of radius

ℓ = 0.1 m

with the period of the second's hand,

T = 60 s

, so

ω = \frac{2 π}{T} = \frac{π}{30} rad/s

. Its acceleration is essentially centripetal,

a = ω^{2} ℓ = (\frac{π}{30})^{2} (0.1) \approx \frac{9.87}{900} \times 0.1 \approx 1.1 \times 1 0^{- 3} m/s^{2}

, which is of the order of

1 0^{- 3}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option

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