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A stone tied to the end of string 100cm ...

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

A

16 m `s^(-2)`

B

4 m `s^(-2)`

C

12 m `s^(-2)`

D

8 m `s^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Here, `r=100cm=1m`
Frequency, `v=(14)/(22)rps:.omega=2piv=2xx(22)/(7)xx(14)/(22)=4rads^(-1)`
The acceleration of the stone is
`a_(c)=omega^(2)r=(4)^(2)(1)=16ms^(-1)`
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