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Q.

A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 22s, then the acceleration of the stone is

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a

16 ${ms}^{- 2}$

b

4 ${ms}^{- 2}$

c

12 ${ms}^{- 2}$

d

8 ${ms}^{- 2}$

answer is A.

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Detailed Solution

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Here, r = 100 cm = 1m

Frequency, v = $\frac{14}{22} Hz$

$ω = 2 πv = 2 \times \frac{22}{7} \times \frac{14}{22} = 4 {rads}^{- 1}$

The acceleration of the stone is

$a_{c} = ω^{2} r = {(4)}^{2} (1) = 16$ ${ms}^{- 2}$

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A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 22s, then the acceleration of the stone is