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A stone tied at the end of a string 80 c...

A stone tied at the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude of acceleration of the stone ?

A

`20 ms^(-2)`

B

`12 m//s^(-2)`

C

`9.9 ms^(-2)`

D

`8 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`omega=(14xx2pi)/25`
`:.` magnitude of acceleration
`=omega^(2)r=((14xx2pi)/25)^(2)xx80/100=9.9 m//s^(2)`
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