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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 25 revolutions in 14 s, what is the magnitude of acceleration of the stone ?

A

`90 ms^2`

B

`100 ms^2`

C

`110ms^2`

D

`120 ms^2`

Text Solution

Verified by Experts

The correct Answer is:
B
Length of the string , l = 80 cm = 0.8 m
Number of revolution = 25 , Time taken = 14 s
`:. upsilon=25/14 = 1.78 s^(-1)`
`:. omega= 2pi upsilon=2pixx1.78=11.18` rad/s
`a = omega^(2) l(11.18)^(2) xx (0.8)=99.99 m//s^2 ~~ 100 m//s^2`
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